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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_4 | B | 32 | The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$ | [
"Let $a$ be the bigger number and $b$ be the smaller.\n$a + b = 5(a - b)$\nMultiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives\n$\\frac{a}{b} = \\frac32$ , so the answer is $\\boxed{32}$",
"Without loss of generality, let the two numbers be $3$ and $2$ , as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\\boxed{32}$"
] |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_4 | E | 166 | The sum of two prime numbers is $85$ . What is the product of these two prime numbers?
$\textbf{(A) }85\qquad\textbf{(B) }91\qquad\textbf{(C) }115\qquad\textbf{(D) }133\qquad \textbf{(E) }166$ | [
"Since the two prime numbers sum to an odd number, one of them must be even. The only even prime number is $2$ . The other prime number is $85-2=83$ , and the product of these two numbers is $83\\cdot2=\\boxed{166}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_8 | D | 7 | The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | [
"Let the three numbers be equal to $a$ $b$ , and $c$ . We can now write three equations:\n$a+b=12$\n$b+c=17$\n$a+c=19$\nAdding these equations together, we get that\n$2(a+b+c)=48$ and\n$a+b+c=24$\nSubstituting the original equations into this one, we find\n$c+12=24$\n$a+17=24$\n$b+19=24$\nTherefore, our numbers are 12, 7, and 5. The middle number is $\\boxed{7}$",
"Let the three numbers be $a$ $b$ and $c$ and $a<b<c$ . We get the three equations:\n$a+b=12$\n$a+c=17$\n$b+c=19$\nTo isolate $b$ , We add the first and last equations and then subtract the second one.\n$(a+b)+(b+c)-(a+c) = 12+19-17 \\Rightarrow 2b=14 \\Rightarrow b = 7$\nBecause $b$ is the middle number, the middle number is $\\boxed{7}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_6 | D | 7 | The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | [
"Let the three numbers be equal to $a$ $b$ , and $c$ . We can now write three equations:\n$a+b=12$\n$b+c=17$\n$a+c=19$\nAdding these equations together, we get that\n$2(a+b+c)=48$ and\n$a+b+c=24$\nSubstituting the original equations into this one, we find\n$c+12=24$\n$a+17=24$\n$b+19=24$\nTherefore, our numbers are 12, 7, and 5. The middle number is $\\boxed{7}$",
"Let the three numbers be $a$ $b$ and $c$ and $a<b<c$ . We get the three equations:\n$a+b=12$\n$a+c=17$\n$b+c=19$\nTo isolate $b$ , We add the first and last equations and then subtract the second one.\n$(a+b)+(b+c)-(a+c) = 12+19-17 \\Rightarrow 2b=14 \\Rightarrow b = 7$\nBecause $b$ is the middle number, the middle number is $\\boxed{7}$"
] |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_7 | B | 38 | The symbolism $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$ . For example, $\lfloor 3 \rfloor = 3,$ and $\lfloor 9/2 \rfloor = 4$ . Compute \[\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor.\]
$\textbf{(A) } 35 \qquad\textbf{(B) } 38 \qquad\textbf{(C) } 40 \qquad\textbf{(D) } 42 \qquad\textbf{(E) } 136$ | [
"The first three values in the sum are equal to $1,$ the next five equal to $2,$ the next seven equal to $3,$ and the last one equal to $4.$ For example, since $2^2=4$ any square root of a number less than $4$ must be less than $2.$ Sum them all together to get\n\\[3\\cdot1 + 5\\cdot2 + 7\\cdot3 + 1\\cdot4 = 3+10+21+4 = \\boxed{38}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_9 | null | 25 | The system of equations \begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\ \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ \end{eqnarray*}
has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ . Find $y_{1} + y_{2}$ | [
"Since $\\log ab = \\log a + \\log b$ , we can reduce the equations to a more recognizable form:\n\\begin{eqnarray*} -\\log x \\log y + \\log x + \\log y - 1 &=& 3 - \\log 2000\\\\ -\\log y \\log z + \\log y + \\log z - 1 &=& - \\log 2\\\\ -\\log x \\log z + \\log x + \\log z - 1 &=& -1\\\\ \\end{eqnarray*}\nLet $a,b,c$ be $\\log x, \\log y, \\log z$ respectively. Using SFFT , the above equations become (*)\n\\begin{eqnarray*}(a - 1)(b - 1) &=& \\log 2 \\\\ (b-1)(c-1) &=& \\log 2 \\\\ (a-1)(c-1) &=& 1 \\end{eqnarray*}\nSmall note from different author: $-(3 - \\log 2000) = \\log 2000 - 3 = \\log 2000 - \\log 1000 = \\log 2.$\nFrom here, multiplying the three equations gives\n\\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &=& (\\log 2)^2\\\\ (a-1)(b-1)(c-1) &=& \\pm\\log 2\\end{eqnarray*}\nDividing the third equation of (*) from this equation, $b-1 = \\log y - 1 = \\pm\\log 2 \\Longrightarrow \\log y = \\pm \\log 2 + 1$ . (Note from different author if you are confused on this step: if $\\pm$ is positive then $\\log y = \\log 2 + 1 = \\log 2 + \\log 10 = \\log 20,$ so $y=20.$ if $\\pm$ is negative then $\\log y = 1 - \\log 2 = \\log 10 - \\log 2 = \\log 5,$ so $y=5.$ ) This gives $y_1 = 20, y_2 = 5$ , and the answer is $y_1 + y_2 = \\boxed{025}$",
"Subtracting the second equation from the first equation yields \\begin{align*} \\log 2000xy-\\log 2yz-((\\log x)(\\log y)-(\\log y)(\\log z)) &= 3 \\\\ \\log\\frac{2000xy}{2yz}-\\log y(\\log x-\\log z) &= 3 \\\\ \\log1000+\\log\\frac{x}{z}-\\log y(\\log\\frac{x}{z}) &= 3 \\\\ 3+\\log\\frac{x}{z}-\\log y(\\log\\frac{x}{z}) &= 3 \\\\ \\log\\frac{x}{z}(1-\\log y) &= 0 \\\\ \\end{align*} If $1-\\log y=0$ then $y=10$ . Substituting into the first equation yields $\\log20000=4$ which is not possible.\nIf $\\log\\frac{x}{z}=0$ then $\\frac{x}{z}=1\\Longrightarrow x=z$ . Substituting into the third equation gets \\begin{align*} \\log x^2-(\\log x)(\\log x) &= 0 \\\\ \\log x^2-\\log x^x &= 0 \\\\ \\log x^{2-x} &= 0 \\\\ x^{2-x} &= 1 \\\\ \\end{align*} Thus either $x=1$ or $2-x=0\\Longrightarrow x=2$ . (Note that here $x\\neq-1$ since logarithm isn't defined for negative number.)\nSubstituting $x=1$ and $x=2$ into the first equation will obtain $y=5$ and $y=20$ , respectively. Thus $y_1+y_2=\\boxed{025}$",
"Let $a = \\log x$ $b = \\log y$ and $c = \\log z$ . Then the given equations become:\n\\begin{align*} \\log 2 + a + b - ab = 1 \\\\ \\log 2 + b + c - bc = 1 \\\\ a+c = ac \\\\ \\end{align*}\nEquating the first and second equations, solving, and factoring, we get $a(1-b) = c(1-b) \\implies{a = c}$ . Plugging this result into the third equation, we get $c = 0$ or $2$ . Substituting each of these values of $c$ into the second equation, we get $b = 1 - \\log 2$ and $b = 1 + \\log 2$ . Substituting backwards from our original substitution, we get $y = 5$ and $y = 20$ , respectively, so our answer is $\\boxed{025}$"
] |
https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_3 | null | 943 | The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n\,$ fish for various values of $n\,$
In the newspaper story covering the event, it was reported that
What was the total number of fish caught during the festival? | [
"Suppose that the number of fish is $x$ and the number of contestants is $y$ . The $y-(9+5+7)=y-21$ fishers that caught $3$ or more fish caught a total of $x - \\left(0\\cdot(9) + 1\\cdot(5) + 2\\cdot(7)\\right) = x - 19$ fish. Since they averaged $6$ fish,\nSimilarily, those who caught $12$ or fewer fish averaged $5$ fish per person, so\nSolving the two equation system, we find that $y = 175$ and $x = \\boxed{943}$ , the answer.",
"Let $f$ be the total number of fish caught by the contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish and let $a$ be the number of contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish. From $\\text{(b)}$ , we know that $\\frac{69+f+65+28+15}{a+31}=6\\implies f=6a+9$ . From $\\text{(c)}$ we have $\\frac{f+69+14+5}{a+44}=5\\implies f=5a+132$ . Using these two equations gets us $a=123$ . Plug this back into the equation to get $f=747$ . Thus, the total number of fish caught is $5+14+69+f+65+28+15=\\boxed{943}$ - Heavytoothpaste"
] |
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_17 | D | 15 | The table below gives the percent of students in each grade at Annville and Cleona elementary schools:
\[\begin{tabular}{rccccccc}&\textbf{\underline{K}}&\textbf{\underline{1}}&\textbf{\underline{2}}&\textbf{\underline{3}}&\textbf{\underline{4}}&\textbf{\underline{5}}&\textbf{\underline{6}}\\ \textbf{Annville:}& 16\% & 15\% & 15\% & 14\% & 13\% & 16\% & 11\%\\ \textbf{Cleona:}& 12\% & 15\% & 14\% & 13\% & 15\% & 14\% & 17\%\end{tabular}\]
Annville has 100 students and Cleona has 200 students. In the two schools combined, what percent of the students are in grade 6?
$\text{(A)}\ 12\% \qquad \text{(B)}\ 13\% \qquad \text{(C)}\ 14\% \qquad \text{(D)}\ 15\% \qquad \text{(E)}\ 28\%$ | [
"By the tables, Annville has $11$ 6th graders and Cleona has $34$ . Together they have $45$ 6th graders and $300$ total students, so the percent is $\\frac{45}{300}=\\frac{15}{100}= \\boxed{15}$"
] |
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_15 | B | 62.5 | The table below shows the distance $s$ in feet a ball rolls down an inclined plane in $t$ seconds.
\[\begin{tabular}{|c|c|c|c|c|c|c|}\hline t & 0 & 1 & 2 & 3 & 4 & 5\\ \hline s & 0 & 10 & 40 & 90 & 160 & 250\\ \hline\end{tabular}\] The distance $s$ for $t = 2.5$ is:
$\textbf{(A)}\ 45\qquad \textbf{(B)}\ 62.5\qquad \textbf{(C)}\ 70\qquad \textbf{(D)}\ 75\qquad \textbf{(E)}\ 82.5$ | [
"Looking at the pattern, we can determine that $t=10s^2$ . Applying the relationship, we can see that $s = \\boxed{62.5}$ when $t=2.5$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_8 | E | 75 | The table shows some of the results of a survey by radiostation KACL. What percentage of the males surveyed listen to the station?
$\begin{tabular}{|c|c|c|c|}\hline & Listen & Don't Listen & Total\\ \hline Males & ? & 26 & ?\\ \hline Females & 58 & ? & 96\\ \hline Total & 136 & 64 & 200\\ \hline\end{tabular}$
$\textbf{(A)}\ 39\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 55\qquad\textbf{(E)}\ 75$ | [
"Filling out the chart, it becomes\n$\\begin{tabular}{|c|c|c|c|}\\hline & Listen & Don't Listen & Total\\\\ \\hline Males & 78 & 26 & 104\\\\ \\hline Females & 58 & 38 & 96\\\\ \\hline Total & 136 & 64 & 200\\\\ \\hline\\end{tabular}$\nThus, the percentage of males surveyed that listen to the station is $100 \\cdot \\frac{78}{104} \\%= \\boxed{75}$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_10 | C | 3.3 | The taxi fare in Gotham City is $2.40 for the first $\frac12$ mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for $10?
$\textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75$ | [
"Let $x$ be the number of miles you ride. The number of miles you ride after the first half mile is $x-0.5.$ We can write this equation:\n\\begin{align*} 10 &= 2.4 + 0.2 \\times \\frac{x-0.5}{0.1} + 2\\\\ 5.6 &= 2(x-0.5)\\\\ 2.8 &= x-0.5\\\\ x &= \\boxed{3.3}",
"\\begin{array}{|c|c|c|}\nMiles & Money & Remark\\\\\n\\hline\n0 & 8 & \\text{2 for tip} \\\\\n\\hline\n0.5 & 5.6 & \\text{-2.4 for first 0.5 mi} \\\\\n\\hline\n3.3 & 0 & \\text{-5.6 for 2.8 mi}\n\\end{array}\nMiles travelled = $\\boxed{3.3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_2 | null | 781 | The teams $T_1$ $T_2$ $T_3$ , and $T_4$ are in the playoffs. In the semifinal matches, $T_1$ plays $T_4$ , and $T_2$ plays $T_3$ . The winners of those two matches will play each other in the final match to determine the champion. When $T_i$ plays $T_j$ , the probability that $T_i$ wins is $\frac{i}{i+j}$ , and the outcomes of all the matches are independent. The probability that $T_4$ will be the champion is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | [
"There are two scenarios in which $T_4$ wins. The first scenario is where $T_4$ beats $T_1$ $T_3$ beats $T_2$ , and $T_4$ beats $T_3$ , and the second scenario is where $T_4$ beats $T_1$ $T_2$ beats $T_3$ , and $T_4$ beats $T_2$ . Consider the first scenario. The probability $T_4$ beats $T_1$ is $\\frac{4}{4+1}$ , the probability $T_3$ beats $T_2$ is $\\frac{3}{3+2}$ , and the probability $T_4$ beats $T_3$ is $\\frac{4}{4+3}$ . Therefore the first scenario happens with probability $\\frac{4}{4+1}\\cdot\\frac{3}{3+2}\\cdot\\frac{4}{4+3}$ . Consider the second scenario. The probability $T_4$ beats $T_1$ is $\\frac{4}{1+4}$ , the probability $T_2$ beats $T_3$ is $\\frac{2}{2+3}$ , and the probability $T_4$ beats $T_2$ is $\\frac{4}{4+2}$ . Therefore the second scenario happens with probability $\\frac{4}{1+4}\\cdot\\frac{2}{2+3}\\cdot\\frac{4}{4+2}$ . By summing these two probabilities, the probability that $T_4$ wins is $\\frac{4}{4+1}\\cdot\\frac{3}{3+2}\\cdot\\frac{4}{4+3}+\\frac{4}{1+4}\\cdot\\frac{2}{2+3}\\cdot\\frac{4}{4+2}$ . Because this expression is equal to $\\frac{256}{525}$ , the answer is $256+525=\\boxed{781}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_2 | A | 8,671 | The ten-letter code $\text{BEST OF LUCK}$ represents the ten digits $0-9$ , in order. What 4-digit number is represented by the code word $\text{CLUE}$
$\textbf{(A)}\ 8671 \qquad \textbf{(B)}\ 8672 \qquad \textbf{(C)}\ 9781 \qquad \textbf{(D)}\ 9782 \qquad \textbf{(E)}\ 9872$ | [
"We can derive that $C=8$ $L=6$ $U=7$ , and $E=1$ . Therefore, the answer is $\\boxed{8671}$ ~edited by Owencheng"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_2 | null | 195 | The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of the first, last, and middle terms of the original sequence. | [
"If the sum of the original sequence is $\\sum_{i=1}^{n} a_i$ then the sum of the new sequence can be expressed as $\\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \\sum_{i=1}^{n} a_i.$ Therefore, $836 = n^2 + 715 \\rightarrow n=11.$ Now the middle term of the original sequence is simply the average of all the terms, or $\\frac{715}{11} = 65,$ and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or $\\boxed{195}.$",
"After the adding of the odd numbers, the total of the sequence increases by $836 - 715 = 121 = 11^2$ . Since the sum of the first $n$ positive odd numbers is $n^2$ , there must be $11$ terms in the sequence, so the mean of the sequence is $\\dfrac{715}{11} = 65$ . Since the first, last, and middle terms are centered around the mean, our final answer is $65 \\cdot 3 = \\boxed{195}$",
"Proceed as in Solution 2 until it is noted that there are 11 terms in the sequence. Since the sum of the terms of the original arithmetic sequence is 715, we note that $\\frac{2a_1 + 10d}{2} \\cdot 11 = 715$ or $2a_1 + 10d = 130$ for all sets of first terms and common differences that fit the conditions given in the problem. Assume WLOG that $a_1 = 60$ and $d = 1$ . Then the first term of the corresponding arithmetic sequence will be $60$ , the sixth (middle) term will be $65$ , and the eleventh (largest) term will be $70$ . Thus, our final answer is $60 + 65 + 70 = \\boxed{195}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_13 | null | 90 | The terms of the sequence $(a_i)$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$ | [
"This question is guessable but let's prove our answer\n\\[a_{n + 2} = \\frac {a_n + 2009} {1 + a_{n + 1}}\\]\n\\[a_{n + 2}(1 + a_{n + 1})= a_n + 2009\\]\n\\[a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009\\]\nlets put $n+1$ into $n$ now\n\\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009\\]\nand set them equal now\n\\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n\\]\n\\[a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n\\]\nlet's rewrite it\n\\[(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n\\]\nLet's make it look nice and let $b_n=a_{n + 2}-a_n$\n\\[(b_{n+1})(a_{n + 2}+1)= b_n\\]\nSince $b_n$ and $b_{n+1}$ are integers, we can see $b_n$ is divisible by $b_{n+1}$\nBut we can't have an infinite sequence of proper factors, unless $b_n=0$\nThus, $a_{n + 2}-a_n=0$\n\\[a_{n + 2}=a_n\\]\nSo now, we know $a_3=a_1$\n\\[a_{3} = \\frac {a_1 + 2009} {1 + a_{2}}\\]\n\\[a_{1} = \\frac {a_1 + 2009} {1 + a_{2}}\\]\n\\[a_{1}+a_{1}a_{2} = a_1 + 2009\\]\n\\[a_{1}a_{2} = 2009\\]\nTo minimize $a_{1}+a_{2}$ , we need $41$ and $49$\nThus, our answer $= 41+49=\\boxed{090}$",
"If $a_{n} \\ne \\frac {2009}{a_{n+1}}$ , then either \\[a_{n} = \\frac {a_{n}}{1} < \\frac {a_{n} + 2009}{1 + a_{n+1}} < \\frac {2009}{a_{n+1}}\\]\nor\n\\[\\frac {2009}{a_{n+1}} < \\frac {2009 + a_{n}}{a_{n+1} + 1} < \\frac {a_{n}}{1} = a_{n}\\]\nAll the integers between $a_{n}$ and $\\frac {2009}{a_{n+1}}$ would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term.\nSo $a_{n} = \\frac {2009}{a_{n+1}}$ , which $a_{n} \\cdot a_{n+1} = 2009$ . When $n = 1$ $a_{1} \\cdot a_{2} = 2009$ . The smallest sum of two factors which have a product of $2009$ is $41 + 49=\\boxed{090}$",
"Essentially you see that it must be an integer for infinite numbers, which doesn't quite seem probable. The most logical explanation is that the sequence repeats, and the numbers in the sequence that repeat are integers. We list out some terms. \\begin{align*} a_{1} &= a \\\\ a_{2} &= b \\\\ a_{3} &=\\frac{a+2009}{1+b} \\\\ a_{4} &=\\frac{(b+1)(b+2009)}{a+b+2010} \\\\ \\end{align*} The terms get more and more wacky, so we just solve for $a,b$ such that $a_{1}=a_{3}$ and $a_{2}=a_{4}.$\nSolving we find both equations end up to the equation $ab=2009$ in which we see to minimize we see that $a = 49$ and $b=41$ or vice versa for an answer of $\\boxed{90}.$ This solution is VERY non rigorous and not recommended."
] |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_7 | E | 130 | The third exit on a highway is located at milepost 40 and the tenth exit is at milepost 160. There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost would you expect to find this service center?
$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 110 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 130$ | [
"There are $160-40=120$ miles between the third and tenth exits, so the service center is at milepost $40+(3/4)(120) = 40+90=\\boxed{130}$"
] |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_16 | D | 4 | The three row sums and the three column sums of the array
\[\left[\begin{matrix}4 & 9 & 2\\ 8 & 1 & 6\\ 3 & 5 & 7\end{matrix}\right]\]
are the same. What is the least number of entries that must be altered to make all six sums different from one another?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | [
"If you change $3$ numbers, then you either change one number in each column and row (ie sudoku-style):\n\\[\\left[\\begin{matrix}* & 9 & 2\\\\ 8 & * & 6\\\\ 3 & 5 & *\\end{matrix}\\right]\\]\nOr you leave at least one row and one column unchanged:\n\\[\\left[\\begin{matrix}* & 9 & 2\\\\ * & * & 6\\\\ 3 & 5 & 7\\end{matrix}\\right]\\]\nIn the first case, you are changing just one common number in two sums, so you wind up with three pairs of sums. (In the example given, the sum in row $x$ is the same as in column $x$ .)\nIn the second case, since two of the sums are unchanged, and the sums started out equal, they must remain equal. (In the second example given, row $3$ and column $3$ are untouched.)\nEither way, $3$ changes is not enough. However, building on the second example, if you change either the untouched column or the untouched row, you will get a possible answer:\n\\[\\left[\\begin{matrix}* & 9 & 2\\\\ * & * & 6\\\\ 3 & * & 7\\end{matrix}\\right]\\]\nLetting the $*$ be a zero does indeed give $6$ different sums, so the answer is $4$ , which is option $\\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_1 | C | 6 | The three-digit number $2a3$ is added to the number $326$ to give the three-digit number $5b9$ . If $5b9$ is divisible by 9, then $a+b$ equals
$\text{(A)}\ 2\qquad\text{(B)}\ 4\qquad\text{(C)}\ 6\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ | [
"If $5b9$ is divisible by $9$ , this must mean that $5 + b + 9$ is a multiple of $9$ . So, \\[5 + b + 9 = 9, 18, 27, 36...\\]\nBecause $5 + 9 = 14$ and $b$ is in between 0 and 9,\n\\[5 + b + 9 = 18\\] \\[b = 4\\]\nThe question states that \\[2a3 + 326 = 549\\] so \\[2a3 = 549 - 326\\] \\[2a3 = 223\\] \\[a = 2\\]\n\\[a + b = 6\\] which is answer choice $\\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_11 | B | 64 | The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$ . In feet, how tall is the taller tree?
$\textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112$ | [
"Let the height of the taller tree be $h$ and let the height of the smaller tree be $h-16$ . Since the ratio of the smaller tree to the larger tree is $\\frac{3}{4}$ , we have $\\frac{h-16}{h}=\\frac{3}{4}$ . Solving for $h$ gives us $h=64 \\Rightarrow \\boxed{64}$",
"To answer this problem, you have to make it so that we have the same proportion as 3:4, but the difference between them is 16. Since the two numbers are consecutive, if we multiply both of them by 16, we would get a difference of 16 between them. So, it would be 48:64 and since we need to find the height of the taller tree, we get $h=64 \\Rightarrow \\boxed{64}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_24 | E | 2,013 | The tower function of twos is defined recursively as follows: $T(1) = 2$ and $T(n + 1) = 2^{T(n)}$ for $n\ge1$ . Let $A = (T(2009))^{T(2009)}$ and $B = (T(2009))^A$ . What is the largest integer $k$ for which \[\underbrace{\log_2\log_2\log_2\ldots\log_2B}_{k\text{ times}}\] is defined?
$\textbf{(A)}\ 2009\qquad \textbf{(B)}\ 2010\qquad \textbf{(C)}\ 2011\qquad \textbf{(D)}\ 2012\qquad \textbf{(E)}\ 2013$ | [
"(Note: This for the people who need clear, concise reasoning in the form of mostly words, instead of a compact proof written in set theory symbols and the like. I figured if I had trouble understanding the above solutions, others would too.)\nWe begin by contemplating what $B$ actually is. Calculating the first few values of $T(n)$ , we see that it quickly becomes absolutely, massively, well, MASSIVE. So actually calculating $T(2009)$ , is sadly, not an option (this is a final five problem, after all). Next we consider if it is possible to express $B$ differently. We have:\n$B = T(2009)^{T(2009)^{T(2009)}} \\Rightarrow \\text{calculating.... ERROR}$\n(I didn't actually try plug it into a computer program; the point is, it's not that easy)\nSo, the only thing left to do is jump right into the problem. We start by applying the $\\log$ s one at a time, and seeing where it goes. We have:\n$\\log_2{\\left(T(2009)^{\\left(T(2009)^{T(2009)}\\right)}\\right)}$\n(Uhhhhhh....)\nBut wait! We can use the rule $\\log_b{x^y} = y*\\log_b{x}$ . Then the expression becomes\n$(T(2009)^{T(2009)}*\\log_2{T(2009)} \\Rightarrow (T(2009)^{T(2009)}*T(2008)$\n(because $\\log_2{T(n)} = {T(n-1)}$\nNow we try $\\log$ ing it again:\n$\\log_2{(T(2009)^{T(2009)}*T(2008)}$\nThis time, we can make use of the $\\log$ addition rule $\\log_b{x*y} = \\log_b{x}+\\log_b{y}$ . The expression becomes\n$\\log_2{(T(2009)^{T(2009)})} + \\log_2{T(2008)}$\nPulling out the exponent again and further simplifying, we have\n$\\log_2{(\\log_2{(B)})} = T(2009)*T(2008)+T(2007)$\nNow we $\\log$ it a third time... and we're stuck 😐. There is no pretty formula for taking the $\\log$ of two things added together. So we rack our brains, sigh in frustration... and give up.\nJUST KIDDING! (Don't give up)\nWhenever you get stuck, you want to \"zoom out\" and re-gain perspective of what it is you're actually trying to find. The problem asks us to find the number of times we can keep taking the $\\log$ of this expression, until we no longer can (once you reach a negative value, you can't take the log of it anymore; i.e. $\\log$ s are only defined for positive inputs). So, for the context of this problem, what we want to know is:\n\"How many times can I take the $\\log$ of $B$ and get a positive value?\"\nLet's think about what $\\log$ actually means. We want a number $x$ , such that when $2$ is raised to the power of $x$ , we get the expression in the $\\log$ . So, when we think about\n$\\log_2{T(2009)*T(2008)+T(2007)}$\nWhat actually matters? Well, even if you have a big number, say, $124986068$ , what happens when you raise $2$ to the power of that number? We would end up getting another one of these:\nCalculating... ERROR\n(lol 🤣)\nBut in all seriousness, whenever we consider tower functions, the next tower number is always much, much bigger. So, back to the problem:\n$\\log_2{(T(2009)*T(2008)+T(2007))}$\n$T(2007)$ is huge... but when you put it next to $T(2008)$ , it looks like an ant. Even more so when it's next to $T(2009)*T(2008)$ . So... what does that mean? Well, if we could get rid of the $T(2007)$ , we could at least go another step. So... let's get rid of it! (At this point I can almost see the audience: *surprised pikachu face)\n\\[\\text{WARNING: THIS SECTION IS FOR CALCULUS USERS ONLY}\\]\n(Just kidding, but yes, the following part of the explanation does incorporate some calculus-related ideas)\nIf we were to consider just\n$\\log_2{(T(2009)*T(2008))}$\nwe would get\n$\\log_2{T(2009)} + \\log_2{T(2008)} \\Rightarrow T(2008) + T(2007)$\nBut what if we wanted to get\n$T(2008) + T(2007)+1$\nas our answer? What would the argument have to be? (In this context, \"argument\" refers to the value in the parentheses). Well, we can work backwards:\n$T(2008) + T(2007)+1$\n$\\Rightarrow \\log_2{T(2009)} + \\log_2{T(2008)} + \\log_2{2}$\n$\\Rightarrow \\log_2{(T(2009)*T(2008)*2)}$\nSo the argument would have to be at least\n$2*T(2009)*T(2008) = T(2009)*T(2008)+T(2009)*T(2008)$\nBut when we compare\n$T(2009)*T(2008)+T(2009)*T(2008)$\nto\n$T(2009)*T(2008)+T(2007)$\nClearly\n$T(2009)*T(2008) > T(2007)$\nSo,\n$\\log_2{(T(2009)*T(2008)+T(2007))}$\nequals some value between\n$T(2008) + T(2007)$\nand\n$T(2008) + T(2007) + 1$\nLet's call that small excess value $\\varepsilon$ (lowercase epsilon, literally \"infinitely small but not 0\"). We want to keep taking the $\\log$ of the expression, and $\\varepsilon$ is going to be pitifully small compared to $T(2008)$ or $T(2007)$ . So, within the context of this problem,\n$T(2008) + T(2007) + \\varepsilon \\approx T(2008) + T(2007)$\nNow we take the $\\log$ again:\n$\\log_2{(T(2008)+T(2007))}$\nLook familiar? It should, because this is almost exactly the same scenario as the one I just made a big fuss about (Take that, you no-know-calculus noobs! On an off note, it's completely fine if you don't know calculus; AMC problems are supposed to solvable by students who don't know calculus.) So, applying the same idea as earlier we have\n$\\log_2{(T(2008)+T(2007))} \\approx \\log_2{T(2008)}$\nFrom here on out, it's calm waters. To recap,\n$\\log_2{(\\log_2{(\\log_2{(B)})})} \\approx T(2008)$\nTaking the $\\log$ $2007$ more times(for a total of $2010$ ), we get\n$\\log_2\\log_2... T(2008) = T(1) = 2$\nTaking the log again, we get\n$\\log_2{2} = 1$\n(Now we're at $2011$ times)\nOnce more and we have\n$\\log_2{1} = 0$\n$2012$ times)\nBut wait, there's more! (IYKYK lol) Remember how we ignored the $\\varepsilon$ ? Well, It's time to bring it back. Every time we took the log, we would get $T(\\text{something})$ $\\varepsilon$ because we defined $\\varepsilon$ to be some very, very small (but finite) number. Let's go back a few steps:\n$\\log_2{2} = 1$\nshould really be\n$\\log_2{(2+\\varepsilon)} = 1+\\epsilon$\nand\n$\\log_2{1} = 0$\nshould really be\n$\\log_2{(1+\\varepsilon)} = \\varepsilon$\nSince $\\varepsilon$ is still a positive number, we can take the $\\log$ of it one last time before it becomes negative. So, after a long and very tiring journey (for both you, the reader, and me, the writer) we have $k$ (the number of times we took the $\\log$ $= 2013 \\Rightarrow \\boxed{2013}$ .\n~TheAsian"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_15 | B | 47 | The town of Hamlet has $3$ people for each horse, $4$ sheep for each cow, and $3$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
$\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66$ | [
"Let the amount of people be $p$ , horses be $h$ , sheep be $s$ , cows be $c$ , and ducks be $d$ . \nWe know \\[3h=p\\] \\[4c=s\\] \\[3p=d\\] Then the total amount of people, horses, sheep, cows, and ducks may be written as $p+h+s+c+d = 3h+h+4c+c+(3\\times3h)$ . This is equivalent to $13h+5c$ . Looking through the options, we see $47$ is impossible to make for integer values of $h$ and $c$ . So the answer is $\\boxed{47}$",
"As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as $13d+5s$ . However, instead of going through each of the solutions and testing the options, you can use the Chicken McNugget Theorem to find the greatest number of people, horses, sheep, cows, and ducks that cannot be written in the form $13d+5s$ \\[13\\cdot 5-13-5=47,\\] so our answer is $\\boxed{47}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_7 | C | 4.5 | The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD?
[asy] size(250); defaultpen(linewidth(0.55)); pair A=(-6,0), B=origin, C=(0,6), D=(0,12); pair ac=C+2.828*dir(45), ca=A+2.828*dir(225), ad=D+2.828*dir(A--D), da=A+2.828*dir(D--A), ab=(2.828,0), ba=(-6-2.828, 0); fill(A--C--D--cycle, gray); draw(ba--ab); draw(ac--ca); draw(ad--da); draw((0,-1)--(0,15)); draw((1/3, -1)--(1/3, 15)); int i; for(i=1; i<15; i=i+1) { draw((-1/10, i)--(13/30, i)); } label("$A$", A, SE); label("$B$", B, SE); label("$C$", C, SE); label("$D$", D, SE); label("$3$", (1/3,3), E); label("$3$", (1/3,9), E); label("$3$", (-3,0), S); label("Main", (-3,0), N); label(rotate(45)*"Aspen", A--C, SE); label(rotate(63.43494882)*"Brown", A--D, NW); [/asy]
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$ | [
"The area of a triangle is $\\frac12 bh$ . If we let $CD$ be the base of the triangle, then the height is $AB$ , and the area is $\\frac12 \\cdot 3 \\cdot 3 = \\boxed{4.5}$",
"We can see that there is a big triangle encasing $ACD$ . The area of that triangle is $\\frac12 \\cdot 3 \\cdot 6 = 9$ . We can easily see that triangle $ABC$ is $\\frac12 \\cdot 3 \\cdot 3$ , which is $4.5$ $9-4.5$ is $4.5$ , so the answer is $\\boxed{4.5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_4 | C | 13 | The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$
[asy] unitsize(8mm); for (int i=0; i<7; ++i) { draw((i,0)--(i,7),gray); draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]
$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$ | [
"We count $3 \\cdot 3=9$ unit squares in the middle, and $8$ small triangles, which gives 4 rectangles each with an area of $1$ . Thus, the answer is $9+4=\\boxed{13}$",
"We can see here that there are $9$ total squares in the middle. We also see that the triangles that make the corners of the shape have an area half the squares' area. Then, we can easily find that each corner has an area of one square and there are $4$ corners so we add that to the original 9 squares to get $9+4=\\boxed{13}$ . That is how I did it.",
"We can apply Pick's Theorem here. There are $8$ lattice points, and $12$ lattice points on the boundary. Then,\n\\[8 + 12 \\div 2 - 1 = \\boxed{13}.\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_19 | C | 64 | The two circles pictured have the same center $C$ . Chord $\overline{AD}$ is tangent to the inner circle at $B$ $AC$ is $10$ , and chord $\overline{AD}$ has length $16$ . What is the area between the two circles?
[asy] unitsize(45); import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); draw((2,1)--(4,0)); dot((0,0),ds); label("$A$", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label("$B$", (1.97,-0.31),NE*lsf); dot((2,1),ds); label("$C$", (1.96,1.09),NE*lsf); dot((4,0),ds); label("$D$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); [/asy]
$\textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi$ | [
"Since $\\triangle ACD$ is isosceles, $CB$ bisects $AD$ . Thus $AB=BD=8$ . From the Pythagorean Theorem, $CB=6$ . Thus the area between the two circles is $100\\pi - 36\\pi=64\\pi$ $\\boxed{64}$"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_17 | B | 18 | The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?
$\mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 45$ | [
"If Jack's current age is $\\overline{ab}=10a+b$ , then Bill's current age is $\\overline{ba}=10b+a$\nIn five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$\nWe are given that $10a+b+5=2(10b+a+5)$\nThus $8a=19b+5 \\Rightarrow a=\\dfrac{19b+5}{8}$\nFor $b=1$ we get $a=3$ . For $b=2$ and $b=3$ the value $\\frac{19b+5}8$ is not an integer, and for $b\\geq 4$ $a$ is more than $9$ . Thus the only solution is $(a,b)=(3,1)$ , and the difference in ages is $31-13=\\boxed{18}$",
"Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.\nThe age difference is $(10a+b)-(10b+a)=9(a-b)$ , hence it is a multiple of $9$ . Thus Bill's current age modulo $9$ must be $4$\nThus Bill's age is in the set $\\{13,22,31,40,49,58,67,76,85,94\\}$\nAs Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options $\\{13,49,58,67\\}$\nChecking each of them, we see that only $13$ works, and gives the solution $31-13=\\boxed{18}$"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_15 | B | 18 | The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?
$\mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 45$ | [
"If Jack's current age is $\\overline{ab}=10a+b$ , then Bill's current age is $\\overline{ba}=10b+a$\nIn five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$\nWe are given that $10a+b+5=2(10b+a+5)$\nThus $8a=19b+5 \\Rightarrow a=\\dfrac{19b+5}{8}$\nFor $b=1$ we get $a=3$ . For $b=2$ and $b=3$ the value $\\frac{19b+5}8$ is not an integer, and for $b\\geq 4$ $a$ is more than $9$ . Thus the only solution is $(a,b)=(3,1)$ , and the difference in ages is $31-13=\\boxed{18}$",
"Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.\nThe age difference is $(10a+b)-(10b+a)=9(a-b)$ , hence it is a multiple of $9$ . Thus Bill's current age modulo $9$ must be $4$\nThus Bill's age is in the set $\\{13,22,31,40,49,58,67,76,85,94\\}$\nAs Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options $\\{13,49,58,67\\}$\nChecking each of them, we see that only $13$ works, and gives the solution $31-13=\\boxed{18}$"
] |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_9 | C | 3 | The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$ . How long is the third altitude of the triangle?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | [
"We find that the area of the triangle is $\\frac{6\\times 2\\sqrt{3}}{2} =6\\sqrt{3}$ . By the Pythagorean Theorem , we have that the length of the hypotenuse is $\\sqrt{(2\\sqrt{3})^2+6^2}=4\\sqrt{3}$ . Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.\nLet $h$ be the third height of the triangle. We have $\\frac{4\\sqrt{3}h}{2}=6\\sqrt{3}\\implies h = \\frac{6 \\cdot 2}{4} \\implies h=\\boxed{3}$",
"By the Pythagorean Theorem, we have that the length of the hypotenuse is $\\sqrt{(2\\sqrt{3})^2+6^2}=4\\sqrt{3}$ . Notice that we now have a 30-60-90 triangle, with the angle between sides $2\\sqrt{3}$ and $4\\sqrt{3}$ equal to $60^{\\circ}$ . Dropping an altitude from the right angle to the hypotenuse, we see that our desired height is $\\boxed{3}$ (We can also check from the other side)."
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_12 | D | 79 | The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?
[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$1$",(0,.5)); label("$3$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$5$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy] [asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$2$",(0,.5)); label("$4$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$6$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]
$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{2}{3}\qquad\textbf{(C)}\ \frac{3}{4}\qquad\textbf{(D)}\ \frac{7}{9}\qquad\textbf{(E)}\ \frac{5}{6}$ | [
"The possible sums are \\[\\begin{tabular}{c|ccc} & 1 & 3 & 5 \\\\ \\hline 2 & 3 & 5 & 7 \\\\ 4 & 5 & 7 & 9 \\\\ 6 & 7 & 9 & 11 \\end{tabular}\\]\nOnly $9$ is not prime, so there are $7$ prime numbers and $9$ total numbers for a probability of $\\boxed{79}$"
] |
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_4 | null | 185 | The two squares shown share the same center $O_{}$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n.$
[asy] //code taken from thread for problem real alpha = 25; pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin; pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z; draw(W--X--Y--Z--cycle^^w--x--y--z--cycle); pair A=intersectionpoint(Y--Z, y--z), C=intersectionpoint(Y--X, y--x), E=intersectionpoint(W--X, w--x), G=intersectionpoint(W--Z, w--z), B=intersectionpoint(Y--Z, y--x), D=intersectionpoint(Y--X, w--x), F=intersectionpoint(W--X, w--z), H=intersectionpoint(W--Z, y--z); dot(O); label("$O$", O, SE); label("$A$", A, dir(O--A)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$D$", D, dir(O--D)); label("$E$", E, dir(O--E)); label("$F$", F, dir(O--F)); label("$G$", G, dir(O--G)); label("$H$", H, dir(O--H));[/asy] | [
"Triangles $AOB$ $BOC$ $COD$ , etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length $1$ in the circumcircle of the squares pass through $B$ , etc.), and each area is $\\frac{\\frac{43}{99}\\cdot\\frac{1}{2}}{2}$ . Since the area of a triangle is $bh/2$ , the area of all $8$ of them is $\\frac{86}{99}$ and the answer is $\\boxed{185}$",
"Define the two possible distances from one of the labeled points and the corners of the square upon which the point lies as $x$ and $y$ . The area of the octagon (by subtraction of areas) is $1 - 4\\left(\\frac{1}{2}xy\\right) = 1 - 2xy$\nBy the Pythagorean theorem \\[x^2 + y^2 = \\left(\\frac{43}{99}\\right)^2\\]\nAlso, \\begin{align*}x + y + \\frac{43}{99} &= 1\\\\ x^2 + 2xy + y^2 &= \\left(\\frac{56}{99}\\right)^2\\end{align*}\nSubstituting, \\begin{align*}\\left(\\frac{43}{99}\\right)^2 + 2xy &= \\left(\\frac{56}{99}\\right)^2 \\\\ 2xy = \\frac{(56 + 43)(56 - 43)}{99^2} &= \\frac{13}{99} \\end{align*}\nThus, the area of the octagon is $1 - \\frac{13}{99} = \\frac{86}{99}$ , so $m + n = \\boxed{185}$"
] |
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_6 | A | 0 | The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | [
"Within six consecutive integers, there must be a number with a factor of $5$ and an even integer with a factor of $2$ . Multiplied together, these would produce a number that is a multiple of $10$ and has a units digit of $\\boxed{0}$",
"We can easily compute the product of the first 6 positive integers: $(1*2*3*4*5*6)=6!=720$ Therefore the units digit must be $\\boxed{0}$"
] |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_14 | E | 9 | The units digit of $3^{1001} 7^{1002} 13^{1003}$ is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$ | [
"First, we notice that $3^0$ is congruent to $1 \\ \\text{(mod 10)}$ $3^1$ is $3 \\ \\text{(mod 10)}$ $3^2$ is $9 \\ \\text{(mod 10)}$ $3^3$ is $7 \\ \\text{(mod 10)}$ $3^4$ is $1 \\ \\text{(mod 10)}$ , and so on. This turns out to be a cycle repeating every $4$ terms, so $3^{1001}$ is congruent to $3 \\ \\text{(mod 10)}$\nThe number $7$ has a similar cycle, going $1, 7, 9, 3, 1, ...$ Hence we have that $7^{1002}$ is congruent to $9 \\ \\text{(mod 10)}$ . Finally, $13^{1003}$ is congruent to $3^{1003} \\equiv 7 \\ \\text{(mod 10)}$ . Thus the required units digit is $3\\cdot 9\\cdot 7 \\equiv 9 \\ \\text{(mod 10)}$ , so the answer is $\\boxed{9}$"
] |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_7 | D | 10,000,000,000 | The value of $\frac{(487,000)(12,027,300)+(9,621,001)(487,000)}{(19,367)(.05)}$ is closest to
$\text{(A)}\ 10,000,000 \qquad \text{(B)}\ 100,000,000 \qquad \text{(C)}\ 1,000,000,000 \qquad \text{(D)}\ 10,000,000,000 \qquad \text{(E)}\ 100,000,000,000$ | [
"We can make the approximations \\begin{align*} 487,000 &\\approx 500,000 \\\\ 12,027,300 &\\approx 12,000,000 \\\\ 9,621,001 &\\approx 10,000,000 \\\\ 19,367 &\\approx 20,000. \\end{align*}\nUsing these instead of the original numbers for an estimate, we have \\begin{align*} \\frac{(500,000)(12,000,000)+(10,000,000)(500,000)}{(20000)(.05)} &= \\frac{500,000\\times 22,000,000}{1000} \\\\ &= 500,000 \\times 22,000 \\\\ &= 1.1\\times 10^{10} \\\\ &\\approx 10,000,000,000 \\rightarrow \\boxed{10,000,000,000}"
] |
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_5 | A | 4 | The value of $\left(256\right)^{.16}\left(256\right)^{.09}$ is:
$\textbf{(A)}\ 4 \qquad \\ \textbf{(B)}\ 16\qquad \\ \textbf{(C)}\ 64\qquad \\ \textbf{(D)}\ 256.25\qquad \\ \textbf{(E)}\ -16$ | [
"When we multiply numbers with exponents, we add the exponents together and leave the bases unchanged. We can apply this concept to computate $256^{0.16} \\cdot 256^{0.09}$ \\[256^{0.16} \\cdot 256^{0.09} = 256^{0.16+0.09}=256^{0.25}.\\] Now we can convert the decimal exponent to a fraction: \\[256^{0.25} = 256^{\\frac{1}{4}}.\\] Now, let us convert the expression into radical form. Since $4$ is the denominator of the fractional exponent, it will be the index exponent: \\[256^{\\frac{1}{4}}=\\sqrt[4]{256}.\\] Since $256 =16^2=(4^2)^2=4^4$ , we can solve for the fourth root of $256$ \\[\\sqrt[4]{256}=\\sqrt[4]{4^4}=4.\\] Therefore, $(256)^{.16} \\cdot (256)^{.09}=\\boxed{4}.$"
] |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_25 | D | 5 | The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$ | [
"$\\log_{5}\\frac{(125)(625)}{25}$ can be simplified to $\\log_{5}\\ (125)(25)$ since $25^2 = 625$ $125 = 5^3$ and $5^2 = 25$ so $\\log_{5}\\ 5^5$ would be the simplest form. In $\\log_{5}\\ 5^5$ $5^x = 5^5$ . Therefore, $x = 5$ and the answer is $\\boxed{5}$",
"$\\log_{5}\\frac{(125)(625)}{25}$ can be also represented as $\\log_{5}\\frac{(5^3)(5^4)}{5^2}= \\log_{5}\\frac{(5^7)}{5^2}= \\log_{5} 5^5$ which can be solved to get $\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_9 | B | 14 | The value of $x - y^{x - y}$ when $x = 2$ and $y = -2$ is:
$\textbf{(A)}\ -18 \qquad \textbf{(B)}\ -14\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 256$ | [
"Just plug in the numbers and follow the order of operations: \\[2-(-2)^{2-(-2)}\\] \\[2-(-2)^4\\] \\[2-16\\] \\[\\boxed{14}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_3 | null | 103 | The value of $x$ that satisfies $\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [
"Let $\\log _{2^x}3^{20}=\\log _{2^{x+3}}3^{2020}=n$ . Based on the equation, we get $(2^x)^n=3^{20}$ and $(2^{x+3})^n=3^{2020}$ . Expanding the second equation, we get $8^n\\cdot2^{xn}=3^{2020}$ . Substituting the first equation in, we get $8^n\\cdot3^{20}=3^{2020}$ , so $8^n=3^{2000}$ . Taking the 100th root, we get $8^{\\frac{n}{100}}=3^{20}$ . Therefore, $(2^{\\frac{3}{100}})^n=3^{20}$ , and using the our first equation( $2^{xn}=3^{20}$ ), we get $x=\\frac{3}{100}$ and the answer is $\\boxed{103}$ .\n~rayfish",
"Recall the identity $\\log_{a^n} b^{m} = \\frac{m}{n}\\log_{a} b$ (which is easily proven using exponents or change of base).\nThen this problem turns into \\[\\frac{20}{x}\\log_{2} 3 = \\frac{2020}{x+3}\\log_{2} 3\\] Divide $\\log_{2} 3$ from both sides. And we are left with $\\frac{20}{x}=\\frac{2020}{x+3}$ .Solving this simple equation we get \\[x = \\tfrac{3}{100} \\Rightarrow \\boxed{103}\\] ~mlgjeffdoge21",
"$\\log_{2^x} 3^{20} = 2^{xy} = 3^{20}$\n$\\log_{2^{x+3}} 3^{2020} = (2^{x+3})^y = 3^{2020}$\n$3^{2020} = (3^{20})^{101}$\n$(2^{xy})^{101} = (2^{x+3})^y = 3^{2020}$\n$(2^{xy})^{101} = (2^{x+3})^y \\Rightarrow 2^{101xy} = 2^{xy+3y} \\Rightarrow 101xy = xy + 3y \\Rightarrow 101xy = y(x+3)$\n$101x = x + 3$\n$100x = 3$\n$x = \\frac{3}{100}$\n$100 + 3 = \\boxed{103}$ ~Airplanes2007"
] |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_20 | D | 3 | The value of the expression $\frac{(304)^5}{(29.7)(399)^4}$ is closest to
$\text{(A)}\ .003 \qquad \text{(B)}\ .03 \qquad \text{(C)}\ .3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 30$ | [
"\\[\\frac{(304)^5}{(29.7)(399)^4} \\approx \\frac{300^5}{30\\cdot400^4} = \\frac{3^5 \\cdot 10^{10}}{3\\cdot 4^4 \\cdot 10^9} = \\frac{3^4\\cdot 10}{4^4} = \\frac{810}{256}\\] Which is closest to $3\\rightarrow\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_9 | E | 16 | The vertex of the parabola $y = x^2 - 8x + c$ will be a point on the $x$ -axis if the value of $c$ is:
$\textbf{(A)}\ - 16 \qquad \textbf{(B) }\ - 4 \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 8 \qquad \textbf{(E) }\ 16$ | [
"Notice that if the vertex of a parabola is on the x-axis, then the x-coordinate of the vertex must be a solution to the quadratic. Since the quadratic is strictly increasing on either side of the vertex, the solution must have double multiplicity, or the quadratic is a perfect square trinomial. This means that for the vertex of $y = x^2 - 8x + c$ to be on the x-axis, the trinomial must be a perfect square, and have discriminant of zero. So,\n\\begin{align*} 0 &= b^2-4ac\\\\ 0 &= (-8)^2-4c\\\\ c &= 64\\\\ c &= 16\\\\ \\end{align*}\nTherefore $c=16$ , and our answer is $\\boxed{16}$"
] |
https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_12 | null | 344 | The vertices of $\triangle ABC$ are $A = (0,0)\,$ $B = (0,420)\,$ , and $C = (560,0)\,$ . The six faces of a die are labeled with two $A\,$ 's, two $B\,$ 's, and two $C\,$ 's. Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$ , and points $P_2\,$ $P_3\,$ $P_4, \dots$ are generated by rolling the die repeatedly and applying the rule: If the die shows label $L\,$ , where $L \in \{A, B, C\}$ , and $P_n\,$ is the most recently obtained point, then $P_{n + 1}^{}$ is the midpoint of $\overline{P_n L}$ . Given that $P_7 = (14,92)\,$ , what is $k + m\,$ | [
"If we have points $(p,q)$ and $(r,s)$ and we want to find $(u,v)$ so $(r,s)$ is the midpoint of $(u,v)$ and $(p,q)$ , then $u=2r-p$ and $v=2s-q$ . So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have \\[P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\\bmod{560},\\ 2y_n\\bmod{420})\\] Then $P_7=(14,92)$ , so $x_7=14$ and $y_7=92$ , and we get \\[\\begin{array}{c||ccccccc} n & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\\\ \\hline\\hline x_n & 14 & 28 & 56 & 112 & 224 & 448 & 336 \\\\ \\hline y_n & 92 & 184 & 368 & 316 & 212 & 4 & 8 \\end{array}\\]\nSo the answer is $\\boxed{344}$",
"Let $L_1$ be the $n^{th}$ roll that directly influences $P_{n + 1}$\nNote that $P_7 = \\cfrac{\\cfrac{\\cfrac{P_1 + L_1}2 + L_2}2 + \\cdots}{2\\ldots} = \\frac {(k,m)}{64} + \\frac {L_1}{64} + \\frac {L_2}{32} + \\frac {L_3}{16} + \\frac {L_4}8 + \\frac {L_5}4 + \\frac {L_6}2 = (14,92)$\nThen quickly checking each addend from the right to the left, we have the following information (remembering that if a point must be $(0,0)$ , we can just ignore it!):\nfor $\\frac {L_6}2,\\frac {L_5}4$ , since all addends are nonnegative, a non- $(0,0)$ value will result in a $x$ or $y$ value greater than $14$ or $92$ , respectively, and we can ignore them,\nfor $\\frac {L_4}8,\\frac {L_3}{16},\\frac {L_2}{32}$ in a similar way, $(0,0)$ and $(0,420)$ are the only possibilities,\nand for $\\frac {L_1}{64}$ , all three work.\nAlso, to be in the triangle, $0\\le k\\le560$ and $0\\le m\\le420$\nSince $L_1$ is the only point that can possibly influence the $x$ coordinate other than $P_1$ , we look at that first.\nIf $L_1 = (0,0)$ , then $k = 2^6\\cdot14 = 64\\cdot14 > 40\\cdot14 = 560$\nso it can only be that $L_1 = (560,0)$ , and $k + 560 = 2^6\\cdot14$\n$\\implies k = 64\\cdot14 - 40\\cdot14 = 24\\cdot14 = 6\\cdot56 = 336$\nNow, considering the $y$ coordinate, note that if any of $L_2,L_3,L_4$ are $(0,0)$ $L_2$ would influence the least, so we test that),\nthen $\\frac {L_2}{32} + \\frac {L_3}{16} + \\frac {L_4}8 < \\frac {420}{16} + \\frac {420}8 = 79\\pm\\epsilon < 80$\nwhich would mean that $P_1 > 2^6\\cdot(92 - 80) = 64\\cdot12 > 42\\cdot10 = 420\\ge m$ , so $L_2,L_3,L_4 = (0,420)$\nand now $\\frac {P_1}{64} + \\frac {420}{2^5} + \\frac {420}{2^4} + \\frac {420}{2^3} = 92$\n$\\implies P_1$\n$= 64\\cdot92 - 420(2 + 4 + 8)$\n$= 64\\cdot92 - 420\\cdot14= 64(100 - 8) - 14^2\\cdot30$\n$= 6400 - 512 - (200 - 4)\\cdot30$\n$= 6400 - 512 - 6000 + 120$\n$= - 112 + 120$\n$= 8$\nand finally, $k + m = 336 + 8 = \\boxed{344}$"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_17 | null | 12 | The vertices of a quadrilateral lie on the graph of $y=\ln{x}$ , and the $x$ -coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$ . What is the $x$ -coordinate of the leftmost vertex?
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13$ | [
"Let the coordinates of the quadrilateral be $(n,\\ln(n)),(n+1,\\ln(n+1)),(n+2,\\ln(n+2)),(n+3,\\ln(n+3))$ . We have by shoelace's theorem, that the area is \\begin{align*} &\\frac{\\ln(n)(n+1) + \\ln(n+1)(n+2) + \\ln(n+2)(n+3)+n\\ln(n+3)}{2} - \\frac{\\ln(n+1)(n) + \\ln(n+2)(n+1) + \\ln(n+3)(n+2)+\\ln(n)(n+3)}{2} \\\\ &=\\frac{\\ln \\left( \\frac{n^{n+1}(n+1)^{n+2}(n+2)^{n+3}(n+3)^n}{(n+1)^n(n+2)^{n+1}(n+3)^{n+2}n^{n+3}}\\right)}{2} \\\\ &= \\ln \\left( \\sqrt{\\frac{(n+1)^2(n+2)^2}{n^2(n+3)^2}} \\right) \\\\ &= \\ln \\left(\\frac{(n+1)(n+2)}{n(n+3)}\\right) \\\\ &= \\ln \\left( \\frac{91}{90} \\right). \\end{align*} We know that the numerator must have a factor of $13$ , so given the answer choices, $n$ is either $12$ or $11$ . If $n=11$ , the expression $\\frac{(n+1)(n+2)}{n(n+3)}$ does not evaluate to $\\frac{91}{90}$ , but if $n=12$ , the expression evaluates to $\\frac{91}{90}$ . Hence, our answer is $\\boxed{12}$",
"Like above, use the shoelace formula to find that the area of the quadrilateral is equal to $\\ln\\frac{(n+1)(n+2)}{n(n+3)}$ . Because the final area we are looking for is $\\ln\\frac{91}{90}$ , the numerator factors into $13$ and $7$ , which one of $n+1$ and $n+2$ has to be a multiple of $13$ and the other has to be a multiple of $7$ . Clearly, the only choice for that is $\\boxed{12}$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_5 | null | 144 | The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3. Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane. Find the number of distinguishable acceptable arrangements. | [
"First, we determine which possible combinations of digits $1$ through $9$ will yield sums that are multiples of $3$ . It is simplest to do this by looking at each of the digits $\\bmod{3}$\nWe see that the numbers $1, 4,$ and $7$ are congruent to $1 \\pmod{3}$ , that the numbers $2, 5,$ and $8$ are congruent to $2 \\pmod{3}$ , and that the numbers $3, 6,$ and $9$ are congruent to $0 \\pmod{3}$ . In order for a sum of three of these numbers to be a multiple of three, the mod $3$ sum must be congruent to $0$ . Quick inspection reveals that the only possible combinations are $0+0+0, 1+1+1, 2+2+2,$ and $0+1+2$ . However, every set of three consecutive vertices must sum to a multiple of three, so using any of $0+0+0, 1+1+1$ , or $2+2+2$ would cause an adjacent sum to include exactly $2$ digits with the same $\\bmod{3}$ value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different $\\bmod{3}$ values.\nWe see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to $1 \\pmod{3}$ can be located counterclockwise of a digit congruent to $0$ and clockwise of a digit congruent to $2 \\pmod{3}$ , or the reverse can be true.\nWe set the first digit as $3$ avoid overcounting rotations, so we have one option as a choice for the first digit. The other two $0 \\pmod{3}$ numbers can be arranged in $2!=2$ ways. The three $1 \\pmod{3}$ and three $2 \\pmod{3}$ can both be arranged in $3!=6$ ways. Therefore, the desired result is $2(2 \\times 6 \\times 6)=\\boxed{144}$",
"Notice that there are three triplets of congruent integers $\\mod 3$ $(1,4,7),$ $(2,5,8),$ and $(3,6,9).$ There are $3!$ ways to order each of the triplets individually and $3!$ ways to order the triplets as a group (see solution 1). Rotations are indistinguishable, so there are $(3!)^4/9=\\boxed{144}$ total arrangements."
] |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_24 | C | 108 | The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$ | [
"WLOG, let the centroid of $\\triangle ABC$ be $I = (-1,-1)$ . The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, $A = (1,1)$ , so $AI = BI = CI = 2\\sqrt{2}$ , so since $\\triangle AIB$ is isosceles and $\\angle AIB = 120^{\\circ}$ , then by the Law of Cosines $AB = 2\\sqrt{6}$ . Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to $\\frac {s}{\\sqrt{3}}$ . Therefore, the area of the triangle is $\\frac{(2\\sqrt{6})^2\\sqrt{3}}4 = 6\\sqrt{3}$ , so the square of the area of the triangle is $\\boxed{108}$",
"Without loss of generality, let the centroid of $\\triangle ABC$ be $G = (-1,-1)$ . Then, one of the vertices must be the other curve of the hyperbola. Without loss of generality, let $A = (1,1)$ . Then, point $B$ must be the reflection of $C$ across the line $y=x$ , so let $B = \\left(a,\\frac{1}{a}\\right)$ and $C=\\left(\\frac{1}{a},a\\right)$ , where $a <-1$ . Because $G$ is the centroid, the average of the $x$ -coordinates of the vertices of the triangle is $-1$ . So we know that $a + 1/a+ 1 = -3$ . Multiplying by $a$ and solving gives us $a=-2-\\sqrt{3}$ . So $B=(-2-\\sqrt{3},-2+\\sqrt{3})$ and $C=(-2+\\sqrt{3},-2-\\sqrt{3})$ . So $BC=2\\sqrt{6}$ , and finding the square of the area gives us $\\boxed{108}$",
"Without loss of generality, let the centroid of $\\triangle ABC$ be $G = (1, 1)$ and let point $A$ be $(-1, -1)$ . It is known that the centroid is equidistant from the three vertices of $\\triangle ABC$ . Because we have the coordinates of both $A$ and $G$ , we know that the distance from $G$ to any vertice of $\\triangle ABC$ is $\\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\\sqrt{2}$ . Therefore, $AG=BG=CG=2\\sqrt{2}$ . It follows that from $\\triangle ABG$ , where $AG=BG=2\\sqrt{2}$ and $\\angle AGB = \\dfrac{360^{\\circ}}{3} = 120^{\\circ}$ $[\\triangle ABG]= \\dfrac{(2\\sqrt{2})^2 \\cdot \\sin(120)}{2} = 4 \\cdot \\dfrac{\\sqrt{3}}{2} = 2\\sqrt{3}$ using the formula for the area of a triangle with sine $\\left([\\triangle ABC]= \\dfrac{1}{2} AB \\cdot BC \\sin(\\angle ABC)\\right)$ . Because $\\triangle ACG$ and $\\triangle BCG$ are congruent to $\\triangle ABG$ , they also have an area of $2\\sqrt{3}$ . Therefore, $[\\triangle ABC] = 3(2\\sqrt{3}) = 6\\sqrt{3}$ . Squaring that gives us the answer of $\\boxed{108}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_24 | null | 108 | The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$ | [
"Without loss of generality, let the centroid of $\\triangle ABC$ be $G = (1, 1)$ . Assuming we don't know one vertex is $(-1, -1)$ we let the vertices be $A\\left(x_1, \\frac{1}{x_1}\\right), B\\left(x_2, \\frac{1}{x_2}\\right), C\\left(x_3, \\frac{1}{x_3}\\right).$\nSince the centroid coordinates are the average of the vertex coordinates, we have that $\\frac{x_1+x_2+x_3}{3}=1$ and $\\frac{\\frac{1}{x_1}+\\frac{1}{x_2}+\\frac{1}{x_3}}{3}=1.$\nWe also know that the centroid is the orthocenter in an equilateral triangle, so $CG \\perp AB.$ Examining slopes, we simplify the equation to $x_1x_2x_3 = -1$ . From the equation $\\frac{\\frac{1}{x_1}+\\frac{1}{x_2}+\\frac{1}{x_3}}{3}=1,$ we get that $x_1x_2+x_1x_3+x_2x_3 = -3$ . These equations are starting to resemble Vieta's:\n$x_1+x_2+x_3=3$\n$x_1x_2+x_1x_3+x_2x_3 = -3$\n$x_1x_2x_3=-1$\n$x_1,x_2,x_3$ are the roots of the equation $x^3 - 3x^2 - 3x + 1 = 0$ . This factors as $(x+1)(x^2-4x+1)=0 \\implies x = -1, 2 \\pm \\sqrt3,$ for the points $(-1, -1), (2+\\sqrt3, 2-\\sqrt3), (2-\\sqrt3, 2+\\sqrt3)$ . The side length is clearly $\\sqrt{24}$ , so the square of the area is $\\boxed{108}.$"
] |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_25 | B | 32 | The volume of a certain rectangular solid is $8$ cm , its total surface area is $32$ cm , and its three dimensions are in geometric progression. The sums of the lengths in cm of all the edges of this solid is
$\mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 40 \qquad \mathrm{(E) \ }44$ | [
"As the dimensions are in geometric progression, let them be $\\frac{b}{r}$ $b$ , and $br$ cm, so the volume is $\\left(\\frac{b}{r}\\right)(b)(br) = b^3$ , giving $b^3 = 8$ and thus $b = 2$ . The surface area condition now yields \\begin{align*}2\\left(\\frac{2}{r}\\right)(2)+2(2)(2r)+2(2r)\\left(\\frac{2}{r}\\right) = 32 &\\iff \\frac{8}{r}+8+8r = 32 \\\\ &\\iff r+\\frac{1}{r} = 3 \\\\ &\\iff r^2-3r+1 = 0 \\\\ &\\iff r = \\frac{3 \\pm \\sqrt{5}}{2}.\\end{align*}\nSince \\[\\frac{3-\\sqrt{5}}{2} = \\frac{1}{\\left(\\frac{3+\\sqrt{5}}{2}\\right)},\\] the two possible values of $r$ do not actually give different dimensions, but merely determine whether they are in increasing or decreasing order. Therefore, without loss of generality, we take $r = \\frac{3+\\sqrt{5}}{2}$ , giving the dimensions (in cm) as \\begin{align*}&\\frac{2}{\\left(\\frac{3+\\sqrt{5}}{2}\\right)}, 2, \\text{ and } 2\\left(\\frac{3+\\sqrt{5}}{2}\\right) \\\\ &= \\frac{4}{3+\\sqrt{5}}, 2, \\text{ and } 3+\\sqrt{5} \\\\ &= 3-\\sqrt{5}, 2, \\text{ and } 3+\\sqrt{5}.\\end{align*}\nAs there are $4$ edges with each of these distinct lengths, it follows that the sum of all the edge lengths (in cm) is \\begin{align*}4\\left(3-\\sqrt{5}+2+3+\\sqrt{5}\\right) &= 4(8) \\\\ &= \\boxed{32}",
"Similarly to in Solution 1, we let the dimensions (in cm) be $b$ $br$ , and $br^2$ , so that the volume condition gives $8 = b^3r^3 = (br)^3$ and thus $br = 2$ . The surface area condition now becomes \\begin{align*}2(b)(br)+2(br)\\left(br^2\\right)+2\\left(br^2\\right)(b) = 32 &\\iff b^2r+b^2r^2+b^2r^3 = 16 \\\\&\\iff br\\left(b+br+br^2\\right) = 16,\\end{align*} so substituting $br = 2$ from above immediately gives \\[b+br+br^2 = \\frac{16}{2} = 8,\\] and hence, without needing to actually compute the dimensions, we deduce that the sum of the edge lengths (in cm) is \\[4b + 4br + 4br^2 = 4(8) = \\boxed{32}.\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_32 | A | 9 | The volume of a pyramid whose base is an equilateral triangle of side length 6 and whose other edges are each of length $\sqrt{15}$ is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 9/2 \qquad \textbf{(C)}\ 27/2 \qquad \textbf{(D)}\ \frac{9\sqrt3}{2} \qquad \textbf{(E)}\ \text{none of these}$ | [
"\nDraw an altitude towards the equilateral triangle base. By symmetry (this can also be proved by HL), the base of the altitude is equidistant from the three points of the equilateral triangle. This means that the distance from the base of the altitude to one of the points of the equilateral triangle is $2\\sqrt{3}$\n\nUsing the Pythagorean Theorem , the length of the altitude is $\\sqrt{3}$ , so the volume of the triangular pyramid is $\\tfrac13 \\cdot \\tfrac{6^2 \\cdot \\sqrt{3}}{4} \\cdot \\sqrt{3} = \\boxed{9}$"
] |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_21 | B | 24 | The volume of a rectangular solid each of whose side, front, and bottom faces are $12\text{ in}^{2}$ $8\text{ in}^{2}$ , and $6\text{ in}^{2}$ respectively is:
$\textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ \text{None of these}$ | [
"If the sidelengths of the cubes are expressed as $a, b,$ and $c,$ then we can write three equations:\n\\[ab=12, bc=8, ac=6.\\]\nThe volume is $abc.$ Notice symmetry in the equations. We can find $abc$ my multiplying all the equations and taking the positive square root.\n\\begin{align*} (ab)(bc)(ac) &= (12)(8)(6)\\\\ a^2b^2c^2 &= 576\\\\ abc &= \\boxed{24}"
] |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_10 | null | 4 | The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$ . At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path $AJABCHCHIJA$ , which has $10$ steps. Let $n$ be the number of paths with $15$ steps that begin and end at point $A$ . Find the remainder when $n$ is divided by $1000.$ | [
"We divide this up into casework. The \"directions\" the bug can go are $\\text{Clockwise}$ $\\text{Counter-Clockwise}$ , and $\\text{Switching}$ . Let an $I$ signal going clockwise (because it has to be in the inner circle), an $O$ signal going counter-clockwise, and an $S$ switching between inner and outer circles. An example string of length fifteen that gets the bug back to $A$ would be $ISSIIISOOSISSII$ .\nFor the bug to end up back at $A$ , the difference between the number of $I$ 's and $O$ 's must be a multiple of $5$\nSo, the total number of ways is $1+2002+1001=3004$ which gives $\\boxed{004}$ as the answer.",
"Define $A_n$ to be the number of sequences of length $n$ that ends at $A$ and similarly for the other spokes. Also let \\[S_n=A_n+B_n+C_n+D_n+E_n+F_n+G_n+H_n+I_n+J_n\\] Apparently everytime the bug has $2$ choices for its next move, thus we have $S_n=2^n$ . Now we attempt to find a recursive formula for $A_n$ \\begin{align*} A_n&=J_{n-1}+E_{n-1} \\\\ &=(I_{n-2}+A_{n-2})+(D_{n-2}+F_{n-2}) \\\\ &=(H_{n-3}+B_{n-3})+(J_{n-3}+E_{n-3})+(G_{n-3}+C_{n-3})+(J_{n-3}+E_{n-3}) \\\\ &=(B_{n-3}+C_{n-3}+E_{n-3}+G_{n-3}+H_{n-3}+J_{n-3})+(J_{n-3}+E_{n-3}) \\\\ &=(S_{n-3}-(I_{n-3}+A_{n-3}+D_{n-3}+F_{n-3}))+A_{n-2} \\\\ &=2^{n-3}-(J_{n-2}+E_{n-2})+A_{n-2} \\\\ &=2^{n-3}-A_{n-1}+A_{n-2} \\\\ \\end{align*} Computing a few easy terms we have $A_0=0$ $A_1=0$ $A_2=1$ $A_3=0$ $A_4=3$ . Continuing the process yields $A_{15}=3\\boxed{004}$"
] |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_13 | D | 36 | The word " HELP " in block letters is painted in black with strokes $1$ unit wide on a $5$ by $15$ rectangular white sign with dimensions as shown. The area of the white portion of the sign, in square units, is
[asy] unitsize(12); fill((0,0)--(0,5)--(1,5)--(1,3)--(2,3)--(2,5)--(3,5)--(3,0)--(2,0)--(2,2)--(1,2)--(1,0)--cycle,black); fill((4,0)--(4,5)--(7,5)--(7,4)--(5,4)--(5,3)--(7,3)--(7,2)--(5,2)--(5,1)--(7,1)--(7,0)--cycle,black); fill((8,0)--(8,5)--(9,5)--(9,1)--(11,1)--(11,0)--cycle,black); fill((12,0)--(12,5)--(15,5)--(15,2)--(13,2)--(13,0)--cycle,black); fill((13,3)--(14,3)--(14,4)--(13,4)--cycle,white); draw((0,0)--(15,0)--(15,5)--(0,5)--cycle); [/asy]
$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 34 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 38$ | [
"Count the number of black squares in each letter. H has 11, E has 11, L has 7, and P has 10, giving the number of black squares to be $11+11+7+10=39$ . The total number of squares is $(15)(5)=75$ and the number of white squares is $75-39=\\boxed{36}$"
] |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_4 | null | 450 | The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $100$ workers can produce $300$ widgets and $200$ whoosits. In two hours, $60$ workers can produce $240$ widgets and $300$ whoosits. In three hours, $50$ workers can produce $150$ widgets and $m$ whoosits. Find $m$ | [
"Suppose that it takes $x$ hours for one worker to create one widget, and $y$ hours for one worker to create one whoosit.\nTherefore, we can write that (note that two hours is similar to having twice the number of workers, and so on):\nSolve the system of equations with the first two equations to find that $(x,y) = \\left(\\frac{1}{7}, \\frac{2}{7}\\right)$ . Substitute this into the third equation to find that $1050 = 150 + 2m$ , so $m = \\boxed{450}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_4 | B | 4 | The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?
$\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25$ | [
"The palindrome right after 2002 is 2112. The product of the digits of 2112 is $\\boxed{4}$",
"The palindrome formula is to add 110 to the number in order to get the next palindrome, a palindrome needs to be in the form as ABBA . We can use this in this case to get 2112. 2*1*1*2=4. Therefore, the answer is $\\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_18 | A | 12 | The yearly changes in the population census of a town for four consecutive years are,
respectively, 25% increase, 25% increase, 25% decrease, 25% decrease.
The net change over the four years, to the nearest percent, is:
$\textbf{(A)}\ -12 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 12$ | [
"A 25% increase means the new population is $\\frac{5}{4}$ of the original population. A 25% decrease means the new population is $\\frac{3}{4}$ of the original population.\nThus, after four years, the population is $1 \\cdot \\frac{5}{4} \\cdot \\frac{5}{4} \\cdot \\frac{3}{4} \\cdot \\frac{3}{4} = \\frac{225}{256}$ times the original population.\nThus, the net change is -12%, so the answer is $\\boxed{12}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_23 | C | 16 | The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a?$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$ | [
"By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.\nBecause the zeros are integral, the discriminant of the function, $a^2 - 8a$ , is a perfect square, say $k^2$ . Then adding 16 to both sides and completing the square yields \\[(a - 4)^2 = k^2 + 16.\\] Therefore $(a-4)^2 - k^2 = 16$ and \\[((a-4) - k)((a-4) + k) = 16.\\] Let $(a-4) - k = u$ and $(a-4) + k = v$ ; then, $a-4 = \\dfrac{u+v}{2}$ and so $a = \\dfrac{u+v}{2} + 4$ . Listing all possible $(u, v)$ pairs (not counting transpositions because this does not affect ( $u + v$ ), $(2, 8), (4, 4), (-2, -8), (-4, -4)$ , yields $a = 9, 8, -1, 0$ . These $a$ sum to $16$ , so our answer is $\\boxed{16}$",
"Let $r_1$ and $r_2$ be the integer zeroes of the quadratic. Since the coefficient of the $x^2$ term is $1$ , the quadratic can be written as \\[(x - r_1)(x - r_2)=x^2 - (r_1 + r_2)x + r_1r_2\\]\nBy comparing this with $x^2 - ax + 2a$ \\[r_1 + r_2 = a\\text{ and }r_1r_2 = 2a.\\]\nPlugging the first equation in the second, \\[r_1r_2 = 2 (r_1 + r_2).\\] Rearranging gives \\[r_1r_2 - 2r_1 - 2r_2 = 0\\implies (r_1 - 2)(r_2 - 2) = 4.\\] These factors can be $(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2),$ or $(-2, -2).$\nWe want the number of distinct $a = r_1 + r_2$ , and these factors gives $a = -1, 0, 8, 9$ . So the answer is $-1 + 0 + 8 + 9 = \\boxed{16}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | C | 16 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$ | [
"The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.\nThe quadratic formula gives the roots of the quadratic equation: $x=\\frac{a\\pm\\sqrt{a^2-8a}}{2}$\nAs long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant $a^2 - 8a$ equals $k^2$ , for some nonnegative integer $k$\n$a^2-8a=k^2$\n$a(a-8)=k^2$\n$((a-4)+4)((a-4)-4)=k^2$\n$(a-4)^2-4^2=k^2$\n$(a-4)^2=k^2+4^2$\nFrom this last equation, we are given a hint of the Pythagorean theorem. Thus, $(k,4,|a-4|)$ must be a Pythagorean triple unless $k = 0$\nIn the case $k=0$ , the equation simplifies to $|a-4|=4$ . From this equation, we have $a=0,8$ . For both $a=0$ and $a=8$ $\\frac{a\\pm\\sqrt{a^2-8a}}{2}$ yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as \"two integers.\")\nIf $k$ is a positive integer, then only one Pythagorean triple could match the triple $(k,4,|a - 4|)$ because the only Pythagorean triple with a $4$ as one of the values is the classic $(3,4,5)$ triple. Here, $k=3$ and $|a-4|=5$ . Hence, $a=-1,9$ . Again, $\\frac{a\\pm\\sqrt{a^2-8a}}{2}$ yields two integers for both $a=-1$ and $a=9$ , so these two values also satisfy the original constraints.\nThere are a total of four possible values for $a$ $-1,0,8,$ and $9$ . Hence, the sum of all of the possible values of $a$ is $\\boxed{16}$",
"By the quadratic formula, the roots $r$ can be represented by \\[r=\\frac{a\\pm\\sqrt{a^2-8a}}{2}\\] For $r\\in\\mathbb{Z}$ $a\\in\\mathbb{Z}$ , since $\\frac{\\sqrt{a^2-8a}}{2}$ and $\\frac{a}{2}$ will have different mantissas (mantissae?).\nNow observe the discriminant $\\sqrt{a^2-8a}=\\sqrt{a(a-8)}$ and have two cases.\nPositive $a$\n$a\\geq8$ and $a\\leq0$ , since $1\\geq a \\geq7$ gives imaginary roots. Testing positive $a$ values, quickly see that $a\\leq9$ . After $16$ and $36$ , the difference between the closest nonzero factor pairs of perfect squares exceeds $8$ . For $8\\geq a \\geq9$ $a=8,9$ . Checking both yields an integer.\nNegative $a$\nWe can instead test with $\\sqrt{-a(8-a)}$ . If $b=8-a$ , we have our original expression. Thus, for the same reasons, $b=8,9\\implies 8,9=8-a$ $a=-1$ (0 does not affect the answer).\n$-1+8+9=16\\implies\\boxed{16}$",
"Let $m$ and $n$ be the roots of $x^2-ax+2a$\nBy Vieta's Formulas, $n+m=a$ and $mn=2a$\nSubstituting gets us $n+m=\\frac{mn}{2}$\n$2n-mn+2m=0$\nUsing Simon's Favorite Factoring Trick:\n$n(2-m)+2m=0$\n$-n(2-m)-2m=0$\n$-n(2-m)-2m+4=4$\n$(2-n)(2-m)=4$\nThis means that the values for $(m,n)$ are $(0,0),(4,4),(3,6),(1,-2)$ giving us $a$ values of $-1,0,8,$ and $9$ . Adding these up gets $\\boxed{16}$",
"The quadratic formula gives \\[x = \\frac{a \\pm \\sqrt{a(a-8)}}{2}\\] . For $x$ to be an integer, it is necessary (and sufficient!) that $a(a-8)$ to be a perfect square. So we have $a(a-8) = b^2$ ; this is a quadratic in itself and the quadratic formula gives \\[a = 4 \\pm \\sqrt{16 + b^2}\\]\nWe want $16 + b^2$ to be a perfect square. From smartly trying small values of $b$ , we find $b = 0, b = 3$ as solutions, which correspond to $a = -1, 0, 8, 9$ . These are the only ones; if we want to make sure then we must hand check up to $b=8$ . Indeed, for $b \\geq 9$ we have that the differences between consecutive squares are greater than $16$ so we can't have $b^2 + 16$ be a perfect square. So summing our values for $a$ we find $\\boxed{16}$ . as the answer.",
"First of all, we know that $a$ is the sum of the quadratic's two roots, by Vieta's formulas. Thus, $a$ must be an integer. Then, we notice that the discriminant $a^2-8a$ must be equal to a perfect square so that the roots are integers. Thus, $a(a-8)=b^2$ where $b$ is an integer.\nWe can complete the square and rearrange to get $(a-4)^2-b^2=16$ . Let's define $m=a-4$ , just to make things a little easier to write, so now we have $(m+b)(m-b)=16$ . We can now list out the integer factor pairs of 16 and the resulting values of $m$ and $b$ . (Note that $m$ and $b$ must both be integers)\n$(m+b)(m-b)=16$\n$16*1=16$ $\\Rightarrow$ Doesn't work\n$8*2=16$ $\\Rightarrow$ $m=5, b=3$\n$4*4=16$ $\\Rightarrow$ $m=4, b=0$\n$2*8=16$ $\\Rightarrow$ $m=5, b=-3$\n$1*16=16$ $\\Rightarrow$ Doesn't work\n$-16*-1=16$ $\\Rightarrow$ Doesn't work\n$-8*-2=16$ $\\Rightarrow$ $m=-5, b=-3$\n$-4*-4=16$ $\\Rightarrow$ $m=-4, b=0$\n$-2*-8=16$ $\\Rightarrow$ $m=-5, b=-3$\n$-1*-16=16$ $\\Rightarrow$ Doesn't work\nWe want the possible values of $m$ , which are $-5,-4,4,$ and $5$ . As $m+4=a$ $a$ can equal $-1,0,8,$ or $9.$ Adding all of that up gets us our answer, $\\boxed{16}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_16 | B | 3 | There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520$ . Let $T > 0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$
$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6$ | [
"If we have horses, $a_1, a_2, \\ldots, a_n$ , then any number that is a multiple of all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that $\\text{LCM}(1,2,3,2\\cdot2,2\\cdot3) = 12$ . Finally, $1+2 = \\boxed{3}$",
"We are trying to find the smallest number that has $5$ one-digit divisors. Therefore we try to find the LCM for smaller digits, such as $1$ $2$ $3$ , or $4$ . We quickly consider $12$ since it is the smallest number that is the LCM of $1$ $2$ $3$ and $4$ . Since $12$ has $5$ single-digit divisors, namely $1$ $2$ $3$ $4$ , and $6$ , our answer is $1+2 = \\boxed{3}$",
"First, for 5 horses to simultaneously pass the starting line after $T$ seconds, $T$ must be divisible by the amount of seconds it takes each of the 5 horses to pass the starting line, meaning all of the horses must be divisors of $T$ , and therefore meaning $T$ must have at least $5$ $1$ -digit divisors. Since we want to minimize $T$ , we will start by guessing the lowest natural number, $1$ $1$ has only $1$ factor, so it does not work, we now repeat the process for the numbers between $2$ and $12$ (This should not take more than a minute) to get that $12$ is the first number to have $5$ or more single-digit divisors ( $1, 2, 3, 4, 6$ ). The sum of the digits of $12$ is $1+2 = \\boxed{3}$ , which is our answer.",
"By inspection, $(1, 2, 3, 4, 6)$ yields the lowest answer of $12$ and the sum of the digits is $1+2 \\Longrightarrow \\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_12 | B | 3 | There are $10$ horses, named Horse 1, Horse 2, $\ldots$ , Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S = 2520$ . Let $T>0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$ | [
"We know that Horse $k$ will be at the starting point after $n$ minutes if $k|n$ . Thus, we are looking for the smallest $n$ such that at least $5$ of the numbers $\\{1,2,\\cdots,10\\}$ divide $n$ . Thus, $n$ has at least $5$ positive integer divisors.\nWe quickly see that $12$ is the smallest number with at least $5$ positive integer divisors and that $1,2,3,4,6$ are each numbers of horses. Thus, our answer is $1+2=\\boxed{3}$",
"In order for at least $5$ horses to finish simultaneously, the current time needs to have at least $5$ divisors. Thus the number must have a form of either $p^4$ or $p^2*q$ , which have $5$ and $6$ factors respectively. The smallest number of the first form is $16$ , and the smallest number of the second form is $12.$ Thus, our answer is $1+2=\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_17 | C | 13 | There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know each other?
$\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15$ | [
"Consider the $10$ people to be standing in a circle, where two people opposite each other form a diameter of the circle.\nLet us use casework on the number of pairs that form a diameter of the circle.\nCase 1: $0$ diameters\nThere are $2$ ways: either $1$ pairs with $2$ $3$ pairs with $4$ , and so on or $10$ pairs with $1$ $2$ pairs with $3$ , etc.\nCase 2: $1$ diameter\nThere are $5$ possible diameters to draw (everyone else pairs with the person next to them).\nNote that there cannot be $2$ diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.\nCase 3: $3$ diameters\nThere are $5$ possible sets of $3$ diameters to draw.\nNotice we are technically choosing the number of ways to choose a pair of two diameters that are neighbors to each other. This means we can choose the first diameter in the pair, and have only two diameters to choose from for the second in the pair. This means we have $5*2=10$ possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the $10$ possibilities by $2$ to get $5$\nNote that there cannot be a case with $4$ diameters because then there would have to be $5$ diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.\nCase 4: $5$ diameters\nThere is only $1$ way to do this.\nThus, in total there are $2+5+5+1=\\boxed{13}$ possible ways.\n- Minor edits by Pearl2008",
"If each person knows exactly $3$ people, that means we form \" $4$ -person groups\". That is, all the people in the group knows every other person. We can rotate until the circle is filled. We want pairs of people, so the first pair has $\\dbinom{4}{2}=6$ . The $2$ nd pair is just $\\dbinom{2}{2} =1$ . We need to multiply these together since these are $1$ group. The $3$ rd pair would be $\\dbinom{4}{2}=6$ . The $4$ th pair is $\\dbinom{2}{2}=1$ . We multiply these $2$ together and get $6$ . The final group would be $\\dbinom{2}{2}=1$ . So we add these up and we have $6 + 6 + 1 = \\boxed{13}$ possible ways."
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_15 | C | 13 | There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know each other?
$\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15$ | [
"Consider the $10$ people to be standing in a circle, where two people opposite each other form a diameter of the circle.\nLet us use casework on the number of pairs that form a diameter of the circle.\nCase 1: $0$ diameters\nThere are $2$ ways: either $1$ pairs with $2$ $3$ pairs with $4$ , and so on or $10$ pairs with $1$ $2$ pairs with $3$ , etc.\nCase 2: $1$ diameter\nThere are $5$ possible diameters to draw (everyone else pairs with the person next to them).\nNote that there cannot be $2$ diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.\nCase 3: $3$ diameters\nThere are $5$ possible sets of $3$ diameters to draw.\nNotice we are technically choosing the number of ways to choose a pair of two diameters that are neighbors to each other. This means we can choose the first diameter in the pair, and have only two diameters to choose from for the second in the pair. This means we have $5*2=10$ possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the $10$ possibilities by $2$ to get $5$\nNote that there cannot be a case with $4$ diameters because then there would have to be $5$ diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.\nCase 4: $5$ diameters\nThere is only $1$ way to do this.\nThus, in total there are $2+5+5+1=\\boxed{13}$ possible ways.\n- Minor edits by Pearl2008",
"If each person knows exactly $3$ people, that means we form \" $4$ -person groups\". That is, all the people in the group knows every other person. We can rotate until the circle is filled. We want pairs of people, so the first pair has $\\dbinom{4}{2}=6$ . The $2$ nd pair is just $\\dbinom{2}{2} =1$ . We need to multiply these together since these are $1$ group. The $3$ rd pair would be $\\dbinom{4}{2}=6$ . The $4$ th pair is $\\dbinom{2}{2}=1$ . We multiply these $2$ together and get $6$ . The final group would be $\\dbinom{2}{2}=1$ . So we add these up and we have $6 + 6 + 1 = \\boxed{13}$ possible ways."
] |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_19 | B | 40 | There are $120$ seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?
$\text{(A)}\ 30 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 41 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 119$ | [
"Let $p$ be a person seated and $o$ is an empty seat\nThe pattern of seating that results in the fewest occupied seats is $\\text{opoopoopoo...po}$ .\nWe can group the seats in 3s like this: $\\text{opo opo opo ... opo}.$\nThere are a total of $40=\\boxed{40}$ groups"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_14 | D | 95,000 | There are $20$ cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all $20$ cities?
[asy] // made by SirCalcsALot size(300); pen shortdashed=linetype(new real[] {6,6}); for (int i = 2000; i < 9000; i = i + 2000) { draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey); label(string(i), (0,i), W); } for (int i = 500; i < 9300; i=i+500) { draw((0,i)--(150,i),linewidth(1.25)); if (i % 2000 == 0) { draw((0,i)--(250,i),linewidth(1.25)); } } int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750}; int data_length = 20; int r = 550; for (int i = 0; i < data_length; ++i) { fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey); draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)); } draw((0,4750)--(11450,4750),shortdashed); label("Cities", (11450*0.5,0), S); label(rotate(90)*"Population", (0,9000*0.5), 10*W); // axis draw((0,0)--(0,9300), linewidth(1.25)); draw((0,0)--(11550,0), linewidth(1.25)); [/asy]
$\textbf{(A) }65000 \qquad \textbf{(B) }75000 \qquad \textbf{(C) }85000 \qquad \textbf{(D) }95000 \qquad \textbf{(E) }105000$ | [
"We can see that the dotted line is exactly halfway between $4500$ and $5000$ , so it is at $4750$ . As this is the average population of all $20$ cities, the total population is simply $4750 \\cdot 20 = \\boxed{95000}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_13 | C | 3 | There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | [
"By PIE (Property of Inclusion/Exclusion), we have\n$|A_1 \\cup A_2 \\cup A_3| = \\sum |A_i| - \\sum |A_i \\cap A_j| + |A_1 \\cap A_2 \\cap A_3|.$ Number of people in at least two sets is $\\sum |A_i \\cap A_j| - 2|A_1 \\cap A_2 \\cap A_3| = 9.$ So, $20 = (10 + 13 + 9) - (9 + 2x) + x,$ which gives $x = \\boxed{3}.$",
"The total number of classes taken among the 20 students is $10 + 13 + 9 = 32$ . Each student is taking at least one class so let's subtract the $20$ classes ( $1$ per each of the $20$ students) from $32$ classes to get $12$ $12$ classes is the total number of extra classes taken by the students who take $2$ or $3$ classes. Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \\boxed{3}$ students that are taking all $3$ classes.",
"Total class count is 32. Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$ . Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$ . There are $9$ students taking two or three classes, so $b+c = 9$ . Solving this system of equations gives us $c=\\boxed{3}$",
"Let us assign the following variables and put them in our Venn Diagram [1] $a$ which designates the number of people taking exactly Bridge and Yoga. $b$ which designates the number of people taking exactly Bridge and Painting. $c$ which designates the number of people that took all $3$ classes or what we want to find. $d$ which designates the number of people taking exactly Yoga and Painting.\nLet's now recall what information we have given: There are exactly $9$ people that are taking at least $2$ classes meaning in other words, $9$ people total are taking strictly $2$ classes or strictly all the available classes meaning that $a+b+c+d=9$\nLet's now start filling out the Venn Diagram:\nStrictly taking Bridge, no other classes: We know in total, the number is $13$ , however this includes the people taking other classes too meaning we'd need to do some subtraction. From our Venn Diagram we see that we'd need to subtract the following variables to get our wanted outcome here, $a, b, c$ . Giving our answer as $13-(a+b+c)$\nHowever, this equation seems complicated as it has $3$ different variables, so to make this look a lot less complicated we can use our earlier equation: $a+b+c+d=9$ to see that $a+b+c=9-d$ . This means that this can also be written as $13-(9-d)=4+d$\nStrictly taking Yoga Only: The total number of people is $10$ , but this would also count people taking other classes too along with it, so we need to subtract this overcount which is visible in the Venn Diagram giving us: $10-(a+c+d)$\nAgain, we can use substitution to see that $a+c+d=9-b$ . This simplifies our equation to $10-(9-b)=1+b$\nStrictly taking Painting Only: We know again, in total this number is $9$ , which also accounts for the people taking other classes too. From our Venn Diagram it is visible that we need to subtract: $b, c, d$ giving $9-(b+c+d)$\nAgain, through substitution of our first equation we see that $b+c+d=9-a$ meaning we can simplify this equation to $9-(9-a)=a$\nIf we add these newly made equations of strictly taking one class, we get the total number of people taking exactly one class as these equations each were a subcase for it. We can also find the exact number for this because we are given that there are exactly $20$ students in total, and $9$ students are taking exactly $2$ or $3$ classes, meaning that if we do $20-9$ we get our answer for the number of students taking exactly $1$ class because those taking exactly one class have no overlap with those taking exactly $2$ or exactly $3$ classes as shown in our Venn Diagram and because $1$ $2$ , and $3$ classes are subcases for finding the total number of students as we know that each student is in exactly $1, 2$ , or $3$ classes. This means that exactly $11$ students took strictly $1$ class.\nWe can add up our equations we found to equal $11$ because those equations were for subcases of having exactly $1$ class giving: $(4+d)+(1+b)+a=11$ $=5+a+b+d=11$\nFrom our equation $a+b+c+d=9$ , we can substitute $9-c$ for $a+b+d$ giving us: $5+9-c=11$\nThis gives $c=3$ . We assigned $c$ for the number of people taking exactly $3$ classes meaning that when we find $c$ , we find the answer. This means our answer is $\\boxed{3}$",
"We are told that there are $20$ students in all, and $10$ take yoga, $13$ take bridge, and $9$ take painting.\nRepresenting each student as a number from $1$ to $20$ , we can then make a list of which classes they are taking. Students 1-10 take yoga, and students $11$ to $20$ take bridge. However, this means that only $10$ students are taking bridge. To make up for it, we go back to the top of the list and start over from student $1$ . Students $1$ $2$ , and $3$ will also take bridge giving the desired count of $13$ total bridge students.\nNow, all that is left are the students who take painting. There are $9$ students who take painting, so students $4$ through students $9$ take painting. Note that because $9$ people take at least two classes, students $10$ through $20$ are unable to take more than one class. This means that we must once more start over from the top. Already $6$ painting slots have been filled, so students $1$ $2$ , and $3$ will also take painting. This gives a total of $3$ students (students $1$ $2$ , and $3$ ) who take all three classes. Therefore, our answer is $\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_17 | D | 12 | There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$ | [
"Note that these $z$ such that $z^{24}=1$ are $e^{\\frac{ni\\pi}{12}}$ for integer $0\\leq n<24$ . So\n$z^6=e^{\\frac{ni\\pi}{2}}$\nThis is real if $\\frac{n}{2}\\in \\mathbb{Z} \\Leftrightarrow (n$ is even $)$ . Thus, the answer is the number of even $0\\leq n<24$ which is $\\boxed{12}$",
"$z = \\sqrt[24]{1} = 1^{\\frac{1}{24}}$\nBy Euler's identity $1 = e^{0 \\times i} = \\cos (0+2k\\pi) + i \\sin(0+2k\\pi)$ , where $k$ is an integer.\nUsing De Moivre's Theorem , we have $z = 1^{\\frac{1}{24}} = {\\cos (\\frac{k\\pi}{12}) + i \\sin (\\frac{k\\pi}{12})}$ , where $0 \\leq k<24$ that produce $24$ unique results.\nUsing De Moivre's Theorem again, we have $z^6 = {\\cos (\\frac{k\\pi}{2}) + i \\sin (\\frac{k\\pi}{2})}$\nFor $z^6$ to be real, $\\sin(\\frac{k\\pi}{2})$ has to equal $0$ to negate the imaginary component. This occurs whenever $\\frac{k\\pi}{2}$ is an integer multiple of $\\pi$ , requiring that $k$ is even. There are exactly $\\boxed{12}$",
"From the start, recall from the Fundamental Theorem of Algebra that $z^{24} = 1$ must have $24$ solutions (and these must be distinct since the equation factors into $0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)$ ), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be $24$ . Notice that $1 = z^{24} = (z^6)^4$ , so for any solution $z$ $z^6$ will be one of the 4th roots of unity ( $1$ $i$ $-1$ , or $-i$ ). Then $6$ solutions $z$ will satisfy $z^6 = 1$ $6$ will satisfy $z^6 = -1$ (and this is further justified by knowledge of the 6th roots of unity), so there must be $\\boxed{12}$ such $z$",
"Let $a\\in\\mathbb{R}$ and $a = z^6.$ We have \\[a^4 = 1 \\implies a = 1,-1.\\] $z^6 = \\pm 1$ has 6 solutions for $1$ and $-1$ respectively, so $6+6=\\boxed{12}.$ \\[\\] -svyn"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | null | 548 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | [
"Let $a_{n}$ be the number of ways to form $n$ -letter strings made up of As and Bs such that no more than $3$ adjacent letters are identical.\nNote that, at the end of each $n$ -letter string, there are $3$ possibilities for the last letter chain: it must be either $1$ $2$ , or $3$ letters long. Removing this last chain will make a new string that is $n-1$ $n-2$ , or $n-3$ letters long, respectively.\nTherefore we can deduce that $a_{n}=a_{n-1}+a_{n-2}+a_{n-3}$\nWe can see that \\[a_{1}=2\\] \\[a_{2}=2^{2}=4\\] \\[a_{3}=2^{3}=8\\] so using our recursive relation we find \\[a_{4}=14\\] \\[a_{5}=26\\] \\[a_{6}=48\\] \\[a_{7}=88\\] \\[a_{8}=162\\] \\[a_{9}=298\\] \\[a_{10}=\\boxed{548}\\]",
"The solution is a simple recursion:\nWe have three cases for the ending of a string: three in a row, two in a row, and a single:\n...AAA $(1)$ ...BAA $(2)$ ...BBA or ...ABA $(3)$\n(Here, WLOG each string ends with A. This won't be the case when we actually solve for values in recursion.)\nFor case $(1)$ , we could only add a B to the end, making it a case $(3)$ . \nFor case $(2)$ , we could add an A or a B to the end, making it a case $(1)$ if you add an A, or a case $(3)$ if you add a B.\nFor case $(3)$ , we could add an A or a B to the end, making it a case $(2)$ or a case $(3)$\nLet us create three series to represent the number of permutations for each case: $\\{a\\}$ $\\{b\\}$ , and $\\{c\\}$ representing case $(1)$ $(2)$ , and $(3)$ respectively.\nThe series have the following relationship:\n$a_n=b_{n-1}$ $b_n=c_{n-1}$ $c_n=c_{n-1}+a_{n-1}+b_{n-1}$\nFor $n=3$ $a_3$ and $b_3$ both equal $2$ $c_3=4$ . With some simple math, we have: $a_{10}=88$ $b_{10}=162$ , and $c_{10}=298$ .\nSumming the three up we have our solution: $88+162+298=\\boxed{548}$",
"This is a recursion problem. Let $a_n$ be the number of valid strings of $n$ letters, where the first letter is $A$ . Similarly, let $b_n$ be the number of valid strings of $n$ letters, where the first letter is $B$\nNote that $a_n=b_{n-1}+b_{n-2}+b_{n-3}$ for all $n\\ge4$\nSimilarly, we have $b_n=a_{n-1}+a_{n-2}+a_{n-3}$ for all $n\\ge4$\nHere is why: every valid strings of $n$ letters $(n\\ge4)$ where the first letter is $A$ must begin with one of the following:\n$AAAB$ - and the number of valid ways is $b_{n-3}$\n$AAB$ - and the number of valid ways is $b_{n-2}$\n$AB$ - and there are $b_{n-1}$ ways.\nWe know that $a_1=1$ $a_2=2$ , and $a_3=4$ . Similarly, we have $b_1=1$ $b_2=2$ , and $b_3=4$ . We can quickly check our recursion to see if our recursive formula works. By the formula, $a_4=b_3+b_2+b_1=7$ , and listing out all $a_4$ , we can quickly verify our formula.\nTherefore, we have the following:\n$\\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\\\\hline a & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\\\\b & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\\end{tabular}$\nThe total number of valid $10$ letter strings is equal to $a_{10}+b_{10}=274+274=\\boxed{548}$",
"Playing around with strings gives this approach: We have a certain number of As, then Bs, and so on. Therefore, what if we denoted each solution with numbers like $(3,3,3,1)$ to denote AAABBBAAAB or vice versa (starting with Bs)? Every string can be represented like this!\nWe can have from 4 to 10 numbers in our parentheses. For each case, we will start with the largest number possible, usually a bunch of 3s, then go down systematically. Realize also that if we are left with just 2s and 1s, there is only one number of 2s and 1s that adds up to the leftover amount. Our final answer is the sum of all of these parenthetical sets [each set multiplied by its permutations, as order matters] multiplied by two [starting with either A or B, and alternating as we go along].\n$4 \\rightarrow (3,3,3,1) = 4, (3,3,2,2) = 6$\n$5 \\rightarrow (3, 3, 2, 1, 1) = 30, (3, 2, 2, 2, 1) = 20, (2,2,2,2,2)=1$\n$6 \\rightarrow (3,3,1,1,1,1) = 15, (3,2,2,1,1,1) = 60, (2,2,2,2,1,1) = 15$\n$7 \\rightarrow (3,2,1,1,1,1,1) = 42, (2,2,2,1,1,1,1) = 35$\n$8 \\rightarrow (3,1...1) = 8, (2,2,1...1) = 28$\n$9 \\rightarrow (2,1...1) = 9$\n$10 \\rightarrow (1,1....1) =1$\nAdding them all up gives you 274; multiplying by 2 gives $\\boxed{548}$",
"We are going to build the string, 1 character at a time. And, we are going to only care about the streak of letters at the end of the string.\nLet $a_n$ be the number of strings of length n that satisfy the problem statement and also has a \"streak\" of length 1 at the end. ABABBA has a streak of length 1.\nLet $b_n$ be the number of strings of length n that satisfy the problem statement and also has of length 2 at the end. ABABBAA has a streak of length 2. There are 2 \"A\" s at the end of the string.\nLet $c_n$ be the number of string of length n that that satisfy the problem statement and also has a \"streak\" of length 3 at the end. ABABBAAA has a streak of length 3. There are 3 \"A\" s at the end of the string.\nLet's establish a recursive relationship. $a_{n+1} = a_n+b_n+c_n$ , since you can simply break the streak. $b_{n+1} = a_n$ , and $c_{n+1} = b_n$ Since you can just add to the streak.\nWe can log everything using a table.\n$\\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\\\\hline a & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 & 298\\\\b & 0 & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 \\\\c & 0 & 0 & 2 & 2 &4 & 8 & 14 & 26 & 48 & 88\\end{tabular}$\nAdding $a_{10}$ $b_{10}$ $c_{10}$ gets the total number of numbers that doesn't have more than 3 concecutive letters.\nThat gets a total of $298+162+88 = \\boxed{548}$",
"Let $S_n$ be the number of n-letter strings satisfying the problem criteria. Then we can easily see $S_1 = 2, S_2 = 4, S_3 = 8, S_4 = 16-2 =14$ . For $n \\ge 4$ consider the last three elements of the list. For example to get $S_5$ we could try to take $S_4$ and multiply by 2 since for each $S_4$ string we have two choices for the fifth character. However we have to be careful if the last three characters of $S_4$ are all the same, as in that case we would be overcounting, (i.e if we have a string $\\dots aaa$ , we can only add $b$ to the end so that the problem criteria is satisfied. Additionally strings such as $\\dots aaa$ or $\\dots bbb$ should only be counted once as we only have one choice for the $n$ th character to add ( $b$ and $a$ respectively).\nThus to compute $S_n$ we start with $S_{n-1}$ then take out the strings ending in $\\dots bbb$ or $\\dots aaa$ . There are $S_{n-4}$ remaining valid strings (i.e if we pick any of the $S_{n-4}$ valid strings, examine the last character, if it is a $b$ then we append $aaa$ to it, and if if it an $a$ we append $bbb$ to it, hence the number of $S_{n-1}$ strings are in one-to-one correspondence with the number of strings in $S_{n-4}$ . Thus we have the recursion $S_n = 2(S_{n-1} - S_{n-4}) + S_{n-4}$ (where we are first taking away $S_{n-4}$ , doubling the result, then adding $S_{n-4}$ back in signifying that we only count it once, as described above).\nThe recursion simplifies to $S_n = 2S_{n-1} - S_{n-4}$ and we can now quickly compute the remaining values: \\[S_5 = 2S_4 - S_1 = 26, S_6 = 2(26)-4 = 48, S_7 = 2(48) - 8 = 88, S_8 = 2(88) - 14 = 162, S_9 = 2(162) - 26 = 298, S_{10} = 2S_9 - 48 = \\boxed{548}\\]",
"Let us define a \"run\" as a set of consecutive letters that are all the same and let a \"maximal run\" be a run that is not the proper subset of any other run, in other words, it cannot be expanded.\nFrom now on, we consider all runs to be maximal.\nNote that the minimum number of runs in the $10$ -letter string is $4$ (or else the Pigeonhole Principle tells us that at least $1$ run has $4$ or more letters in it, contradiction), and the maximum number of runs is clearly $10.$\nSince an arbitrary run can have anywhere from $1$ to $3$ letters in it, it follows that the number of $10$ -letter strings with $n$ runs (where $4 \\leq n \\leq 10$ ) is \\[2[x^{10}] (x+x^2+x^3)^n,\\] i.e. twice the coefficient of $x^{10}$ in the expansion of $(x+x^2+x^3)^n.$ (Note that we multiplied by $2$ because there are two choices for which letter the first run has and then the rest are fixed).\nHence, we wish to find \\[2\\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n.\\] First, we can rewrite this as\n\\begin{align*}\n\\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n &= \\sum_{n=4}^{10} [x^{10}] \\ x^n \\cdot (1+x+x^2)^n \\\\\n&= \\sum_{n=4}^{10} [x^{10-n}] \\left(\\frac{1-x^3}{1-x}\\right)^n \\\\\n&= \\sum_{n=4}^{10} [x^{10-n}] \\frac{(1-x^3)^n}{(1-x)^n}. \\\\\n\\end{align*}\nNow, we proceed case by case, utilizing the Binomial Theorem for the numerator in all of the cases:\nFor $n=4,$ we have\n\\begin{align*}\n[x^6] \\frac{(1-x^3)^4}{(1-x)^4} &= \\binom{4}{0} [x^6] (1-x)^{-4} - \\binom{4}{1} [x^3] (1-x)^{-4} + \\binom{4}{2} [x^0] (1-x)^{-4} \\\\\n&= 1\\cdot\\binom{9}{3} - 4\\cdot\\binom{6}{3} + 6\\cdot\\binom{3}{3} \\\\\n&= 1\\cdot84-4\\cdot20+6\\cdot1 \\\\\n&= 84-80+6 \\\\\n&= \\underline{10}.\n\\end{align*}\nFor $n=5,$ we have\n\\begin{align*}\n[x^5] \\frac{(1-x^3)^5}{(1-x)^5} &= \\binom{5}{0} [x^5] (1-x)^{-5} - \\binom{5}{1}[x^2] (1-x)^{-5} \\\\\n&= 1\\cdot\\binom{9}{4}-5\\cdot\\binom{6}{4} \\\\\n&= 1\\cdot126-5\\cdot15 \\\\\n&= 126-75 \\\\\n&= \\underline{51}.\n\\end{align*}For $n=6,$ we have\n\\begin{align*}\n[x^4] \\frac{(1-x^3)^6}{(1-x)^6} &= \\binom{6}{0} [x^4](1-x)^{-6} - \\binom{6}{1}[x^1](1-x)^{-6} \\\\\n&= 1 \\cdot \\binom{9}{5} - 6 \\cdot \\binom{6}{5} \\\\\n&= 1 \\cdot 126 - 6 \\cdot 6 \\\\\n&= 126-36 \\\\\n&= \\underline{90}.\n\\end{align*}For $n=7,$ we have\n\\begin{align*}\n[x^3] \\frac{(1-x^3)^7}{(1-x)^7} &= \\binom{7}{0}[x^3](1-x)^{-7} - \\binom{7}{1}[x^0](1-x)^{-7} \\\\\n&= 1\\cdot \\binom{9}{6} - 7\\cdot\\binom{6}{6} \\\\\n&= 1\\cdot84-7\\cdot1\\\\\n&= 84-7 \\\\\n&= \\underline{77}.\n\\end{align*}For $n=8,$ we have\n\\begin{align*}\n[x^2] \\frac{(1-x^3)^8}{(1-x)^8} &=\\binom{8}{0} [x^2](1-x)^{-8} \\\\\n&= 1\\cdot\\binom{9}{7} \\\\\n&= 1\\cdot36 \\\\\n&= \\underline{36}.\n\\end{align*}For $n=9,$ we have\n\\begin{align*}\n[x^1] \\frac{(1-x^3)^9}{(1-x)^9} &= \\binom{9}{0} [x^1] (1-x)^{-9}\\\\\n&= 1\\cdot\\binom{9}{8} \\\\\n&= 1\\cdot9 \\\\\n&= \\underline{9}.\n\\end{align*}For $n=10,$ we have\n\\begin{align*}\n[x^0] \\frac{(1-x^3)^{10}}{(1-x)^{10}} &= [x^0] (1-x)^{-10} \\\\\n&= 1\\cdot\\binom{9}{9} \\\\\n&= 1\\cdot1 \\\\\n&= \\underline{1}.\n\\end{align*}Hence, the answer is\n\\begin{align*}\n2\\sum_{n=4}^{10} [x^{10-n}] \\frac{(1-x^3)^n}{(1-x)^n} &= 2(10+51+90+77+36+9+1) \\\\\n&= 2\\cdot274 \\\\\n&= \\boxed{548}.\n\\end{align*}"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_14 | null | 840 | There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $\mid z \mid = 1$ . These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$ , where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ | [
"$z$ can be written in the form $\\text{cis\\,}\\theta$ . Rearranging, we find that $\\text{cis\\,}{28}\\theta = \\text{cis\\,}{8}\\theta+1$\nSince the real part of $\\text{cis\\,}{28}\\theta$ is one more than the real part of $\\text{cis\\,} {8}\\theta$ and their imaginary parts are equal, it is clear that either $\\text{cis\\,}{28}\\theta = \\frac{1}{2}+\\frac {\\sqrt{3}}{2}i$ and $\\text{cis\\,} {8}\\theta = -\\frac{1}{2}+\\frac {\\sqrt{3}}{2}i$ , or $\\text{cis\\,}{28}\\theta = \\frac{1}{2} - \\frac{\\sqrt{3}}{2}i$ and $\\text{cis\\,} {8}\\theta = -\\frac{1}{2}- \\frac{\\sqrt{3}}{2}i$\nSetting up and solving equations, $Z^{28}= \\text{cis\\,}{60^\\circ}$ and $Z^8= \\text{cis\\,}{120^\\circ}$ , we see that the solutions common to both equations have arguments $15^\\circ , 105^\\circ, 195^\\circ,$ and $\\ 285^\\circ$ . We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values.\nAgain setting up equations ( $Z^{28}= \\text{cis\\,}{300^\\circ}$ and $Z^{8} = \\text{cis\\,}{240^\\circ}$ ) we see that the common solutions have arguments of $75^\\circ, 165^\\circ, 255^\\circ,$ and $345^\\circ$\nListing all of these values, we find that $\\theta_{2} + \\theta_{4} + \\ldots + \\theta_{2n}$ is equal to $(75 + 165 + 255 + 345) ^\\circ$ which is equal to $\\boxed{840}$ degrees. We only want the sum of a certain number of theta, not all of it."
] |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_11 | D | 5 | There are $52$ people in a room. what is the largest value of $n$ such that the statement "At least $n$ people in this room have birthdays falling in the same month" is always true?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 12$ | [
"Pretend you have $52$ people you want to place in $12$ boxes, because there are $12$ months in a year. By the Pigeonhole Principle , one box must have at least $\\left\\lceil \\frac{52}{12} \\right\\rceil$ people $\\longrightarrow \\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_14 | null | 440 | There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$ $m$ divides $a_{n+m} - a_{n}$ for all integers $n$ with $1 \leq n < n+m \leq 30$ . Find the remainder when $N$ is divided by $1000$ | [
"Be wary of \"position\" versus \"number\" in the solution!\nEach POSITION in the 30-position permutation is uniquely defined by an ordered triple $(i, j, k)$ . The $n$ th position is defined by this ordered triple where $i$ is $n \\mod 2$ $j$ is $n \\mod 3$ , and $k$ is $n \\mod 5$ . There are 2 choices for $i$ , 3 for $j$ , and 5 for $k$ , yielding $2 \\cdot 3 \\cdot 5=30$ possible triples. Because the least common multiple of 2, 3, and 5 is 30, none of these triples are repeated and all are used. By the conditions of the problem, if $i$ is the same in two different triples, then the two numbers in these positions must be equivalent mod 2. If $j$ is the same, then the two numbers must be equivalent $\\mod 3$ , and if $k$ is the same, the two numbers must be equivalent mod 5. Take care to note that that doesn't mean that the number 1 has to have $1 \\mod 2$ ! It's that the POSITION which NUMBER 1 occupies has $1 \\mod 2$\nThe ordered triple (or position) in which 1 can be placed has 2 options for i, 3 for j, and 5 for k, resulting in 30 different positions of placement.\nThe ordered triple where 2 can be placed in is somewhat constrained by the placement of 1. Because 1 is not equivalent to 2 in terms of mod 2, 3, or 5, the i, j, and k in their ordered triples must be different. Thus, for the number 2, there are (2-1) choices for i, (3-1) choices for j, and (5-1) choices for k. Thus, there are 1*2*4=8 possible placements for the number two once the number one is placed.\nBecause 3 is equivalent to 1 mod 2, it must have the same i as the ordered triple of 1. Because 3 is not equivalent to 1 or 2 in terms of mod 3 or 5, it must have different j and k values. Thus, there is 1 choice for i, (2-1) choices for j, and (4-1) choices for k, for a total of $1\\cdot 1 \\cdot 3=3$ choices for the placement of 3.\nAs above, 4 is even, so it must have the same value of i as 2. It is also 1 mod 3, so it must have the same j value of 1. 4 is not equivalent to 1, 2, or 3 mod 5, so it must have a different k value than that of 1, 2, and 3. Thus, there is 1 choice for i, 1 choice for j, and (3-1) choices for k, yielding a total of $1 \\cdot 1 \\cdot 2=2$ possible placements for 4.\n5 is odd and is equivalent to 2 mod 3, so it must have the same i value as 1 and the same j value of 2. 5 is not equivalent to 1, 2, 3, or 4 mod 5, so it must have a different k value from 1, 2, 3, and 4. However, 4 different values of k are held by these four numbers, so 5 must hold the one remaining value. Thus, only one possible triple is found for the placement of 5.\nAll numbers from 6 to 30 are also fixed in this manner. All values of i, j, and k have been used, so every one of these numbers will have a unique triple for placement, as above with the number five. Thus, after 1, 2, 3, and 4 have been placed, the rest of the permutation is fixed.\nThus, $N = 30 \\cdot 8 \\cdot 3 \\cdot 2=30 \\cdot 48=1440$ . Thus, the remainder when $N$ is divided by $1000$ is $\\boxed{440}.$",
"Note that $30=2\\cdot 3\\cdot 5$ . Since $\\gcd(2, 3, 5)=1$ , by CRT, for each value $k=0\\ldots 29$ modulo $30$ there exists a unique ordered triple of values $(a, b, c)$ such that $k\\equiv a\\pmod{2}$ $k\\equiv b\\pmod{3}$ , and $k\\equiv c\\pmod{5}$ . Therefore, we can independently assign the residues modulo $2, 3, 5$ , so $N=2!\\cdot 3!\\cdot 5!=1440$ , and the answer is $\\boxed{440}$",
"First let's look at the situation when $m$ is equal to 2. It isn't too difficult to see the given conditions are satisfied iff the sequences $S_1 = a_1,a_3,a_5,a_7...$ and $S_2 =a_2,a_4,a_6...$ each are assigned either $\\equiv 0 \\pmod 2$ or $\\equiv 1 \\pmod 2$ . Another way to say this is each element in the sequence $a_1,a_3,a_5,a_7...$ would be the same mod 2, and similarly for the other sequence. There are 2! = 2 ways to assign the mods to the sequences.\nNow when $m$ is equal to 3, the sequences are $T_1 = a_1, a_4,a_7..$ $T_2 = a_2,a_5,a_8,a_11...$ , and $T_3 = a_3,a_6,a_9...$ . Again, for each sequence, all of its elements are congruent either $0$ $1$ , or $2$ mod $3$ . There are $3! = 6$ ways to assign the mods to the sequences.\nFinally do the same thing for $m = 5$ . There are $5!$ ways. In total there are $2 * 6*120 = 1440$ and the answer is $\\boxed{440}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_15 | null | 332 | There are $n$ mathematicians seated around a circular table with $n$ seats numbered $1,$ $2,$ $3,$ $...,$ $n$ in clockwise order. After a break they again sit around the table. The mathematicians note that there is a positive integer $a$ such that
Find the number of possible values of $n$ with $1 < n < 1000.$ | [
"It is a well-known fact that the set $0, a, 2a, ... (n-1)a$ forms a complete set of residues if and only if $a$ is relatively prime to $n$\nThus, we have $a$ is relatively prime to $n$ . In addition, for any seats $p$ and $q$ , we must have $ap - aq$ not be equivalent to either $p - q$ or $q - p$ modulo $n$ to satisfy our conditions. These simplify to $(a-1)p \\not\\equiv (a-1)q$ and $(a+1)p \\not\\equiv (a+1)q$ modulo $n$ , so multiplication by both $a-1$ and $a+1$ must form a complete set of residues mod $n$ as well.\nThus, we have $a-1$ $a$ , and $a+1$ are relatively prime to $n$ . We must find all $n$ for which such an $a$ exists. $n$ obviously cannot be a multiple of $2$ or $3$ , but for any other $n$ , we can set $a = n-2$ , and then $a-1 = n-3$ and $a+1 = n-1$ . All three of these will be relatively prime to $n$ , since two numbers $x$ and $y$ are relatively prime if and only if $x-y$ is relatively prime to $x$ . In this case, $1$ $2$ , and $3$ are all relatively prime to $n$ , so $a = n-2$ works.\nNow we simply count all $n$ that are not multiples of $2$ or $3$ , which is easy using inclusion-exclusion. We get a final answer of $998 - (499 + 333 - 166) = \\boxed{332}$"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_25 | D | 7,425 | There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
$\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$ | [
"We begin by narrowing down the possibilities. If the larger number were twice the smaller number, then the smallest possibility for the larger number is $2457\\times2=4914$ , since $2457$ is the smallest number in the set. The largest possibility would have to be twice the largest number in the set such that when it is multiplied by $2$ , it is less than or equal to $7542$ , the largest number in the set. This happens to be $2754\\times2=5508$ . Therefore, the number would have to be between $4914$ and $5508$ , and also even. The only even numbers in the set and in this range are $5472$ and $5274$ . A quick check reveals that neither of these numbers is twice a number in the set. The number can't be quadruple or more another number in the set since $2457\\times4=9828$ , well past the range of the set. Therefore, the number must be triple another number in the set. The least possibility is $2457\\times3=7371$ and the greatest is $2475\\times3=7425$ , since any higher number in the set multiplied by $3$ would be out of the range of the set. Reviewing, we find that the upper bound does in fact work, so the multiple is $2475\\times3=7425, \\boxed{7425}$",
"There are only 5 options for the problem so we can just try them. It is easy since that we only need try to use $2$ $3$ to divide them. Even $5$ will leads to a solution start with $1$ which we don't need.\n$5724/2=2862$ $5724/3=1908$ $7245/3=2415$ $7254/2=3612$ $7254/3=2418$ $7425/3=2475$ . The answer is $\\boxed{7425}$ . You can obtain the answer in only 6 calculations."
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_9 | B | 3 | There are 3 numbers A, B, and C, such that $1001C - 2002A = 4004$ , and $1001B + 3003A = 5005$ . What is the average of A, B, and C?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E) }\text{Not uniquely determined}$ | [
"Notice that we don't need to find what $A, B,$ and $C$ actually are, just their average. In other words, if we can find $A+B+C$ , we will be done.\nAdding up the equations gives $1001(A+B+C)=9009=1001(9)$ so $A+B+C=9$ and the average is $\\frac{9}{3}=3$ . Our answer is $\\boxed{3}$",
"Start by isolating $B$ and $C$ in both of the equations, in order to represent the variables $C$ and $B$ in terms of A. Ending up with the two equations $C = 2A +4$ and $B = -3A + 5$ , we have to calculate the value of the expression $\\frac{A+B+C}{3}$ . Plugging in $2A + 4$ for $C$ and $-3A + 5$ for $B$ , we add them up and end up with a value of 9. Dividing 9 by 3 to compute the average, we get our answer of $\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_24 | D | 27,720 | There are exactly $77,000$ ordered quadruplets $(a, b, c, d)$ such that $\gcd(a, b, c, d) = 77$ and $\operatorname{lcm}(a, b, c, d) = n$ . What is the smallest possible value for $n$
$\textbf{(A)}\ 13,860\qquad\textbf{(B)}\ 20,790\qquad\textbf{(C)}\ 21,560 \qquad\textbf{(D)}\ 27,720 \qquad\textbf{(E)}\ 41,580$ | [
"Let $A=\\frac{a}{77},\\ B=\\frac{b}{77}$ , etc., so that $\\gcd(A,B,C,D)=1$ . Then for each prime power $p^k$ in the prime factorization of $N=\\frac{n}{77}$ , at least one of the prime factorizations of $(A,B,C,D)$ has $p^k$ , at least one has $p^0$ , and all must have $p^m$ with $0\\le m\\le k$\nLet $f(k)$ be the number of ordered quadruplets of integers $(m_1,m_2,m_3,m_4)$ such that $0\\le m_i\\le k$ for all $i$ , the largest is $k$ , and the smallest is $0$ . Then for the prime factorization $N=2^{k_2}3^{k_3}5^{k_5}\\ldots$ we must have $77000=f(k_2)f(k_3)f(k_5)\\ldots$ So let's take a look at the function $f(k)$ by counting the quadruplets we just mentioned.\nThere are $14$ quadruplets which consist only of $0$ and $k$ . Then there are $36(k-1)$ quadruplets which include three different values, and $12(k-1)(k-2)$ with four. Thus $f(k)=14+12(k-1)(3+k-2)=14+12(k^2-1)$ and the first few values from $k=1$ onwards are \\[14,50,110,194,302,434,590,770,\\ldots\\] Straight away we notice that $14\\cdot 50\\cdot 110=77000$ , so the prime factorization of $N$ can use the exponents $1,2,3$ . To make it as small as possible, assign the larger exponents to smaller primes. The result is $N=2^33^25^1=360$ , so $n=360\\cdot 77=27720$ which is answer $\\boxed{27,720}$"
] |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_19 | E | 78 | There are exactly $N$ distinct rational numbers $k$ such that $|k|<200$ and \[5x^2+kx+12=0\] has at least one integer solution for $x$ . What is $N$
$\textbf{(A) }6\qquad \textbf{(B) }12\qquad \textbf{(C) }24\qquad \textbf{(D) }48\qquad \textbf{(E) }78\qquad$ | [
"Factor the quadratic into \\[\\left(5x + \\frac{12}{n}\\right)\\left(x + n\\right) = 0\\] where $-n$ is our integer solution. Then, \\[k = \\frac{12}{n} + 5n,\\] which takes rational values between $-200$ and $200$ when $|n| \\leq 39$ , excluding $n = 0$ . This leads to an answer of $2 \\cdot 39 = \\boxed{78}$",
"Solve for $k$ so \\[k=-\\frac{12}{x}-5x.\\] Note that $x$ can be any integer in the range $[-39,0)\\cup(0,39]$ so $k$ is rational with $\\lvert k\\rvert<200$ . Hence, there are $39+39=\\boxed{78}.$",
"Plug in $k=200$ to find the upper limit. You will find the limit to be a number from $0<x<-1$ and one that is just below $-39.$ All the integer values from $-1$ to $-39$ can be attainable through some value of $k$ . Since the question asks for the absolute value of $k$ , we see that the answer is $39\\cdot2 = \\boxed{78.}$"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_13 | B | 3 | There are integers $a, b,$ and $c,$ each greater than $1,$ such that
\[\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]
for all $N \neq 1$ . What is $b$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$ | [
"$\\sqrt[a]{N\\sqrt[b]{N\\sqrt[c]{N}}}$ can be simplified to $N^{\\frac{1}{a}+\\frac{1}{ab}+\\frac{1}{abc}}.$\nThe equation is then $N^{\\frac{1}{a}+\\frac{1}{ab}+\\frac{1}{abc}}=N^{\\frac{25}{36}}$ which implies that $\\frac{1}{a}+\\frac{1}{ab}+\\frac{1}{abc}=\\frac{25}{36}.$\n$a$ has to be $2$ since $\\frac{25}{36}>\\frac{7}{12}$ $\\frac{7}{12}$ is the result when $a, b,$ and $c$ are $3, 2,$ and $2$\n$b$ being $3$ will make the fraction $\\frac{2}{3}$ which is close to $\\frac{25}{36}$\nFinally, with $c$ being $6$ , the fraction becomes $\\frac{25}{36}$ . In this case $a, b,$ and $c$ work, which means that $b$ must equal $\\boxed{3.}$ ~lopkiloinm",
"As above, notice that you get $\\frac{1}{a}+\\frac{1}{ab}+\\frac{1}{abc}=\\frac{25}{36}.$\nNow, combine the fractions to get $\\frac{bc+c+1}{abc}=\\frac{25}{36}$\nLet us assume $bc+c+1=25$ and $abc=36$ as this is the most convenient. (EDIT: This used to say WLOG but that is inaccurate)\nFrom the first equation, we get $c(b+1)=24$ . Note also that from the second equation, $b$ and $c$ must be factors of 36.\nAfter listing out the factors of 36 and utilising trial and error, we find that $c=6$ and $b=3$ works, with $a=2$ . So our answer is $\\boxed{3.}$",
"Collapsed, $\\sqrt[a]{N\\sqrt[b]{N\\sqrt[c]{N}}} = \\sqrt[abc]{N^{bc+c+1}}$ . Comparing this to $\\sqrt[36]{N^{25}}$ , observe that $bc+c+1=25$ and $abc=36$ . The first can be rewritten as $c(b+1)=24$ . Then, $b+1$ has to factor into 24 while 1 less than that also must factor into 36. The prime factorizations are as follows $36=2^2 3^2$ and $24=2^33$ . Then, $b=\\boxed{3}$ , as only 4 and 3 factor into 36 and 24 while being 1 apart."
] |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_5 | A | 119 | There are many two-digit multiples of 7, but only two of the multiples have a digit sum of 10. The sum of these two multiples of 7 is
$\text{(A)}\ 119 \qquad \text{(B)}\ 126 \qquad \text{(C)}\ 140 \qquad \text{(D)}\ 175 \qquad \text{(E)}\ 189$ | [
"Writing out all two digit numbers that have a digital sum of $10$ , you get $19, 28, 37, 46, 55, 64, 73, 82,$ and $91$ . The two numbers on that list that are divisible by $7$ are $28$ and $91$ . Their sum is $28+91=119$ , choice $\\boxed{119}$",
"Writing out all the two digit multiples of $7$ , you get $14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84,$ and $91$ . Again you find $28$ and $91$ have a digital sum of $10$ , giving answer $\\boxed{119}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_10 | null | 80 | There are nonzero integers $a$ $b$ $r$ , and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$ . For each possible combination of $a$ and $b$ , let ${p}_{a,b}$ be the sum of the zeros of $P(x)$ . Find the sum of the ${p}_{a,b}$ 's for all possible combinations of $a$ and $b$ | [
"Since $r+si$ is a root, by the Complex Conjugate Root Theorem, $r-si$ must be the other imaginary root. Using $q$ to represent the real root, we have\n$(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65$\nApplying difference of squares, and regrouping, we have\n$(x-q)(x^2 - 2rx + (r^2 + s^2)) = x^3 -ax^2 + bx -65$\nSo matching coefficients, we obtain\n$q(r^2 + s^2) = 65$\n$b = r^2 + s^2 + 2rq$\n$a = q + 2r$\nBy Vieta's each ${p}_{a,b} = a$ so we just need to find the values of $a$ in each pair.\nWe proceed by determining possible values for $q$ $r$ , and $s$ and using these to determine $a$ and $b$\nIf $q = 1$ $r^2 + s^2 = 65$ so (r, s) = $(\\pm1, \\pm 8), (\\pm8, \\pm 1), (\\pm4, \\pm 7), (\\pm7, \\pm 4)$\nSimilarly, for $q = 5$ $r^2 + s^2 = 13$ so the pairs $(r,s)$ are $(\\pm2, \\pm 3), (\\pm3, \\pm 2)$\nFor $q = 13$ $r^2 + s^2 = 5$ so the pairs $(r,s)$ are $(\\pm2, \\pm 1), (\\pm1, \\pm 2)$\nNow we can disregard the plus minus signs for s because those cases are included as complex conjugates of the counted cases.\nThe positive and negative values of r will cancel, so the sum of the ${p}_{a,b} = a$ for $q = 1$ is $q$ times the number of distinct $r$ values (as each value of $r$ generates a pair $(a,b)$ ).\nOur answer is then $(1)(8) + (5)(4) + (13)(4) = \\boxed{080}$"
] |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | null | 880 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$ | [
"Add the two equations to get that $\\log x+\\log y+2(\\log(\\gcd(x,y))+2(\\log(\\text{lcm}(x,y)))=630$ .\nThen, we use the theorem $\\log a+\\log b=\\log ab$ to get the equation, $\\log (xy)+2(\\log(\\gcd(x,y))+\\log(\\text{lcm}(x,y)))=630$ .\nUsing the theorem that $\\gcd(x,y) \\cdot \\text{lcm}(x,y)=x\\cdot y$ , along with the previously mentioned theorem, we can get the equation $3\\log(xy)=630$ .\nThis can easily be simplified to $\\log(xy)=210$ , or $xy = 10^{210}$\n$10^{210}$ can be factored into $2^{210} \\cdot 5^{210}$ , and $m+n$ equals to the sum of the exponents of $2$ and $5$ , which is $210+210 = 420$ .\nMultiply by two to get $2m +2n$ , which is $840$ .\nThen, use the first equation ( $\\log x + 2\\log(\\gcd(x,y)) = 60$ ) to show that $x$ has to have lower degrees of $2$ and $5$ than $y$ (you can also test when $x>y$ , which is a contradiction to the restrains we set before). Therefore, $\\gcd(x,y)=x$ . Then, turn the equation into $3\\log x = 60$ , which yields $\\log x = 20$ , or $x = 10^{20}$ .\nFactor this into $2^{20} \\cdot 5^{20}$ , and add the two 20's, resulting in $m$ , which is $40$ .\nAdd $m$ to $2m + 2n$ (which is $840$ ) to get $40+840 = \\boxed{880}$",
"First simplifying the first and second equations, we get that\n\\[\\log_{10}(x\\cdot\\text{gcd}(x,y)^2)=60\\] \\[\\log_{10}(y\\cdot\\text{lcm}(x,y)^2)=570\\]\nThus, when the two equations are added, we have that \\[\\log_{10}(x\\cdot y\\cdot\\text{gcd}^2\\cdot\\text{lcm}^2)=630\\] When simplified, this equals \\[\\log_{10}(x^3y^3)=630\\] so this means that \\[x^3y^3=10^{630}\\] so \\[xy=10^{210}.\\]\nNow, the following cannot be done on a proof contest but let's (intuitively) assume that $x<y$ and $x$ and $y$ are both powers of $10$ . This means the first equation would simplify to \\[x^3=10^{60}\\] and \\[y^3=10^{570}.\\] Therefore, $x=10^{20}$ and $y=10^{190}$ and if we plug these values back, it works! $10^{20}$ has $20\\cdot2=40$ total factors and $10^{190}$ has $190\\cdot2=380$ so \\[3\\cdot 40 + 2\\cdot 380 = \\boxed{880}.\\]",
"Let $x=10^a$ and $y=10^b$ and $a<b$ . Then the given equations become $3a=60$ and $3b=570$ . Therefore, $x=10^{20}=2^{20}\\cdot5^{20}$ and $y=10^{190}=2^{190}\\cdot5^{190}$ . Our answer is $3(20+20)+2(190+190)=\\boxed{880}$",
"We will use the notation $(a, b)$ for $\\gcd(a, b)$ and $[a, b]$ as $\\text{lcm}(a, b)$ . \nWe can start with a similar way to Solution 1. We have, by logarithm properties, $\\log_{10}{x}+\\log_{10}{(x, y)^2}=60$ or $x(x, y)^2=10^{60}$ . We can do something similar to the second equation and our two equations become \\[x(x, y)^2=10^{60}\\] \\[y[x, y]^2=10^{570}\\] Adding the two equations gives us $xy(x, y)^2[x, y]^2=10^{630}$ . Since we know that $(a, b)\\cdot[a, b]=ab$ $x^3y^3=10^{630}$ , or $xy=10^{210}$ . We can express $x$ as $2^a5^b$ and $y$ as $2^c5^d$ . Another way to express $(x, y)$ is now $2^{min(a, c)}5^{min(b, d)}$ , and $[x, y]$ is now $2^{max(a, c)}5^{max(b, d)}$ . We know that $x<y$ , and thus, $a<c$ , and $b<d$ . Our equations for $lcm$ and $gcd$ now become \\[2^a5^b(2^a5^a)^2=10^{60}\\] or $a=b=20$ . Doing the same for the $lcm$ equation, we have $c=d=190$ , and $190+20=210$ , which satisfies $xy=210$ . Thus, $3m+2n=3(20+20)+2(190+190)=\\boxed{880}$ .\n~awsomek",
"Let $x=d\\alpha, y=d\\beta, (\\alpha, \\beta)=1$ . Simplifying, $d^3\\alpha=10^{60}, d^3\\alpha^2\\beta^3=10^{510} \\implies \\alpha\\beta^3 = 10^{510}=2^{510} \\cdot 5^{510}$ . Notice that since $\\alpha, \\beta$ are coprime, and $\\alpha < 5^{90}$ (Prove it yourself !) , $\\alpha=1, \\beta = 10^{170}$ . Hence, $x=10^{20}, y=10^{190}$ giving the answer $\\boxed{880}$",
"Add the two equations and use the fact that $\\gcd\\left(x,y\\right)\\cdot\\mathrm{lcm}\\left(x,y\\right)=xy$ to find that $xy=10^{210}$ . So let $x=2^a5^b$ and $y=2^{210-a}5^{210-b}$ for $0\\leq a,b\\leq210$ . If $a\\geq105$ then the exponent of $2$ in $x\\cdot\\gcd\\left(x,y\\right)^2=10^{60}$ is $a+2\\left(210-a\\right)=420-a$ , so $a=360$ , contradiction. So $a<105$ . Then the exponent of $2$ in $x\\cdot\\gcd\\left(x,y\\right)^2$ is $a+2a=3a$ , so $a=20$ . Similarly, $b=20$ . Then $3m+2n=3\\left(a+b\\right)+2\\left(420-a-b\\right)=\\boxed{880}$ as desired."
] |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_23 | C | 36 | There are positive integers that have these properties:
The product of the digits of the largest integer with both properties is
$\text{(A)}\ 7 \qquad \text{(B)}\ 25 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 60$ | [
"Five-digit numbers will have a minimum of $1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$ as the sum of their squares if the five digits are distinct and non-zero. If there is a zero, it will be forced to the left by rule #2.\nNo digit will be greater than $7$ , as $8^2 = 64$\nTrying four digit numbers $WXYZ$ , we have $w^2 + x^2 + y^2 + z^2 = 50$ with $0 < w < x < y < z < 8$\n$z=7$ will not work, since the other digits must be at least $1^2 + 2^2 + 3^2 = 14$ , and the sum of the squares would be over $50$\n$z=6$ will give $w^2 + x^2 + y^2 = 14$ $(w,x,y) = (1,2,3)$ will work, giving the number $1236$ . No other number with $z=6$ will work, as $w, x,$ and $y$ would have to be greater.\n$z=5$ will give $w^2 + x^2 + y^2 = 25$ $y=4$ forces $x=3$ and $w=0$ , which has a leading zero, and then we have 345 which is a 3-digit number.\n$z=4$ can only give the number $1234$ , which does not satisfy the condition of the problem.\nThus, the number in question is $1236$ , and the product of the digits is $36$ , giving $\\boxed{36}$ as the answer."
] |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | null | 330 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | [
"By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are complex conjugates. Let $z=m+\\sqrt{n}\\cdot i$ and $\\overline{z}=m-\\sqrt{n}\\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\\overline{z}.$\nWe know that \\begin{align*} z+\\overline{z}&=2m, & (1) \\\\ z\\overline{z}&=m^2+n. & (2) \\end{align*} Applying Vieta's Formulas to $x^3+ax+b,$ we have $-20+z+\\overline{z}=0.$ Substituting $(1)$ into this equation, we get $m=10.$\nApplying Vieta's Formulas to $x^3+cx^2+d,$ we have $-21z-21\\overline{z}+z\\overline{z}=0,$ or $-21(z+\\overline{z})+z\\overline{z}=0.$ Substituting $(1)$ and $(2)$ into this equation, we get $n=320.$\nFinally, the answer is $m+n=\\boxed{330}.$",
"$(-20)^{3} + (-20)a + b = 0$ , hence $-20a + b = 8000$\nAlso, $(-21)^{3} + c(-21)^{2} + d = 0$ , hence $441c + d = 9261$\n$m + i \\sqrt{n}$ satisfies both $\\Rightarrow$ we can put it in both equations and equate to 0.\nIn the first equation, we get $(m + i \\sqrt{n})^{3} + a(m + i \\sqrt{n}) + b = 0$ Simplifying this further, we get $(m^{3} - 3mn + am + b) + i(3m^{2} \\sqrt{n} - n\\sqrt{n} + a\\sqrt{n}) = 0$\nHence, $m^{3} - 3mn + am + b = 0$ and $3m^{2} \\sqrt{n} - n\\sqrt{n} + a\\sqrt{n} = 0 \\Rightarrow 3m^{2} - n + a = 0 \\rightarrow (1)$\nIn the second equation, we get $(m + i \\sqrt{n})^{3} + c(m + i \\sqrt{n})^{2} + d = 0$ Simplifying this further, we get $(m^{3} + m^{2}c - nc - 3mn + d) + i(3m^{2} \\sqrt{n} - n\\sqrt{n} + 2mc\\sqrt{n}) = 0$\nHence, $m^{3} + m^{2}c - nc - 3mn + d = 0$ and $3m^{2} \\sqrt{n} - n\\sqrt{n} + 2mc\\sqrt{n} = 0 \\Rightarrow 3m^{2} - n + 2mc = 0 \\rightarrow (2)$\nComparing (1) and (2),\n$a = 2mc$ and $am + b = m^{2}c - nc + d \\rightarrow (3)$\n$b = 8000 + 20a \\Rightarrow b = 40mc + 8000$ $d = 9261 - 441c$\nSubstituting these in $(3)$ gives, $2m^{2}c + 8000 + 40mc = m^{2}c - nc + 9261 - 441c$\nThis simplifies to $m^{2}c + nc + 40mc + 441c = 1261 \\Rightarrow c(m^{2} + n + 40m + 441) = 1261$\nHence, $c|1261 \\Rightarrow c \\in {1,13,97,1261}$\nConsider case of $c = 1$\n$c = 1 \\Rightarrow d = 8820$ Also, $a = 2m, b = 8000 + 40m$\n$am + b = m^{2} - n + 8820$ (because c = 1)\nAlso, $m^{2} + n + 40m = 820 \\rightarrow (4)$ Also, Equation (2) gives $3m^{2} - n + 2m = 0 \\rightarrow (5)$\nSolving (4) and (5) simultaneously gives $m = 10, n = 320$\n[AIME can not have more than one answer, so we can stop here ... Not suitable for Subjective exam]\nHence, $m + n = 10 + 320 = \\boxed{330}$",
"start off by applying vieta's and you will find that $a=m^2+n-40m$ $b=20m^2+20n$ $c=21-2m$ and $d=21m^2+21n$ . After that, we have to use the fact that $-20$ and $-21$ are roots of $x^3+ax+b$ and $x^3+cx^2+d$ , respectively. Since we know that if you substitute the root of a function back into the function, the output is zero, therefore $(-20)^3-20(a)+b=0$ and $(-21)^3+c*(-21)^2+d=0$ and you can set these two equations equal to each other while also substituting the values of $a$ $b$ $c$ , and $d$ above to give you $21m^2+21n-1682m+8000=0$ , then you can rearrange the equation into $21n = -21m^2+1682m-8000$ . With this property, we know that $-21m^2+1682m-8000$ is divisible by $21$ therefore that means $1682m-8000=0(mod 21)$ which results in $2m-20=0(mod 21)$ which finally gives us m=10 mod 21. We can test the first obvious value of $m$ which is $10$ and we see that this works as we get $m=10$ and $n=320$ . That means your answer will be $m + n = 10 + 320 = \\boxed{330}$",
"We note that $x^3 + ax + b = (x+20)P(x)$ and $x^3 + cx^2 + d = (x+21)Q(x)$ for some polynomials $P(x)$ and $Q(x)$\nThrough synthetic division (ignoring the remainder as we can set $b$ and $d$ to constant values such that the remainder is zero), $P(x) = x^2 - 20x + (400+a)$ , and $Q(x) = x^2 + (c-21)x + (441 - 21c)$\nBy the complex conjugate root theorem, we know that $P(x)$ and $Q(x)$ share the same roots, and they share the same leading coefficient, so $P(x) = Q(x)$\nTherefore, $c-21 = -20$ and $441-21c = 400 + a$ . Solving the system of equations, we get $a = 20$ and $c = 1$ , so $P(x) = Q(x) = x^2 - 20x + 420$\nFinally, by the quadratic formula, we have roots of $\\frac{20 \\pm \\sqrt{400 - 1680}}{2} = 10 \\pm \\sqrt{320}i$ , so our final answer is $10 + 320 = \\boxed{330}$",
"We plug -20 into the equation obtaining $(-20)^3-20a+b$ , likewise, plugging -21 into the second equation gets $(-21)^3+441c+d$\nBoth equations must have 3 solutions exactly, so the other two solutions must be $m + \\sqrt{n} \\cdot i$ and $m - \\sqrt{n} \\cdot i$\nBy Vieta's, the sum of the roots in the first equation is $0$ , so $m$ must be $10$\nNext, using Vieta's theorem on the second equation, you get $x1x2+x2x3+x1x3 = 0$ . However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so $-21*(20)+(m^2+n)=0$\nGiven that $m$ is $10$ , then $n$ is equal to $320$\nTherefore, the answer to the equation is $\\boxed{330}$",
"Since $m+i\\sqrt{n}$ is a common root and all the coefficients are real, $m-i\\sqrt{n}$ must be a common root, too.\nNow that we know all three roots of both polynomials, we can match coefficients (or more specifically, the zero coefficients).\nFirst, however, the product of the two common roots is: \\begin{align*} &&&(x-m-i\\sqrt{n})(x-m+i\\sqrt{n})\\\\ &=&&x^2-x(m+i\\sqrt{n}+m-i\\sqrt{n})+(m+i\\sqrt{n})(m-i\\sqrt{n})\\\\ &=&&x^2-2xm+(m^2-i^2n)\\\\ &=&&x^2-2xm+m^2+n \\end{align*}\nNow, let's equate the two forms of both the polynomials: \\[x^3+ax+b=(x^2-2xm+m^2+n)(x+20)\\] \\[x^3+cx^2+d=(x^2-2xm+m^2+n)(x+21)\\] Now we can match the zero coefficients. \\[-2m+20=0\\to m=10\\text{ and}\\] \\[-42m+m^2+n=0\\to-420+100+n=0\\to n=320\\text{.}\\] Thus, $m+n=10+320=\\boxed{330}$"
] |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_11 | B | 4 | There are several sets of three different numbers whose sum is $15$ which can be chosen from $\{ 1,2,3,4,5,6,7,8,9 \}$ . How many of these sets contain a $5$
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$ | [
"Let the three-element set be $\\{ a,b,c \\}$ and suppose that $a=5$\nWe need $b+c=10$ and $b\\neq c$ . This gives us four solutions, so there are $4$ sets with a $5$ also with the desired properties $\\rightarrow \\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_12 | B | 5,724 | There are twenty-four $4$ -digit numbers that use each of the four digits $2$ $4$ $5$ , and $7$ exactly once. Listed in numerical order from smallest to largest, the number in the $17\text{th}$ position in the list is
$\text{(A)}\ 4527 \qquad \text{(B)}\ 5724 \qquad \text{(C)}\ 5742 \qquad \text{(D)}\ 7245 \qquad \text{(E)}\ 7524$ | [
"For each choice of the thousands digit, there are $6$ numbers with that as the thousands digit. Thus, the six smallest are in the two thousands, the next six are in the four thousands, and then we need $5$ more numbers.\nWe can just list from here: $5247,5274,5427,5472,5724 \\rightarrow \\boxed{5724}$"
] |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_23 | D | 23 | There are two cards; one is red on both sides and the other is red on one side and blue on the other. The cards have the same probability (1/2) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is red, then the probability that the under-side is also red is
$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac12 \qquad \textbf{(D)}\ \frac23 \qquad \textbf{(E)}\ \frac34$ | [
"There are three red faces, and two are on the card that is completely red, so our answer is $\\frac{2}{3}$ , which is $\\boxed{23}$"
] |
https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_21 | B | 392 | There are two natural ways to inscribe a square in a given isosceles right triangle.
If it is done as in Figure 1 below, then one finds that the area of the square is $441 \text{cm}^2$ .
What is the area (in $\text{cm}^2$ ) of the square inscribed in the same $\triangle ABC$ as shown in Figure 2 below?
[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((-25,0)--(-15,0)--(-25,10)--cycle); draw((-20,0)--(-20,5)--(-25,5)); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label("A", (-25,10), W); label("B", (-25,0), W); label("C", (-15,0), E); label("Figure 1", (-20, -5)); label("Figure 2", (5, -5)); label("A", (0,10), W); label("B", (0,0), W); label("C", (10,0), E); [/asy]
$\textbf{(A)}\ 378 \qquad \textbf{(B)}\ 392 \qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 441 \qquad \textbf{(E)}\ 484$ | [
"We are given that the area of the inscribed square is $441$ , so the side length of that square is $21$ . Since the square divides the $45-45-90$ larger triangle into 2 smaller congruent $45-45-90$ , then the legs of the larger isosceles right triangle ( $BC$ and $AB$ ) are equal to $42$ \nWe now have that $3S=42\\sqrt{2}$ , so $S=14\\sqrt{2}$ . But we want the area of the square which is $S^2=(14\\sqrt{2})^2= \\boxed{392}$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_10 | A | 16 | There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$ . What is the sum of those values of $a$
$\textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20$ | [
"quadratic equation has exactly one root if and only if it is a perfect square . So set\n$4x^2 + ax + 8x + 9 = (mx + n)^2$\n$4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2$\nTwo polynomials are equal only if their coefficients are equal, so we must have\n$m^2 = 4, n^2 = 9$\n$m = \\pm 2, n = \\pm 3$\n$a + 8= 2mn = \\pm 2\\cdot 2\\cdot 3 = \\pm 12$\n$a = 4$ or $a = -20$\nSo the desired sum is $(4)+(-20)=\\boxed{16}$",
"Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have \\[(a+8)^2 - 4(4)(9) = 0 \\implies a^2 + 16a - 80.\\] We can use the quadratic formula to solve for its roots (we can ignore the expression in the radical sign as it will cancel out due to the $\\pm$ sign when added). So we must have \\[\\frac{-16 + \\sqrt{\\text{something}}}{2} + \\frac{-16 - \\sqrt{\\text{something}}}{2}.\\] $\\frac{-32}{2} = \\boxed{16}$",
"There is only one positive value for $k$ such that the quadratic equation would have only one solution. $k-8$ and $-k-8$ are the values of $a$ $-8-8=-16$ , so the answer is $\\boxed{16}$"
] |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_25 | B | 9 | There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ . Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$ | [
"Multiply out the $7!$ to get\n\\[5 \\cdot 6! = (3 \\cdot 4 \\cdots 7)a_2 + (4 \\cdots 7)a_3 + (5 \\cdot 6 \\cdot 7)a_4 + 42a_5 + 7a_6 + a_7 .\\]\nBy Wilson's Theorem (or by straightforward division), $a_7 + 7(a_6 + 6a_5 + \\cdots) \\equiv 5 \\cdot 6! \\equiv -5 \\equiv 2 \\pmod{7}$ , so $a_7 = 2$ . Then we move $a_7$ to the left and divide through by $7$ to obtain\n\\[\\frac{5 \\cdot 6!-2}{7} = 514 = 360a_2 + 120a_3 + 30a_4 + 6a_5 + a_6.\\]\nWe then repeat this procedure $\\pmod{6}$ , from which it follows that $a_6 \\equiv 514 \\equiv 4 \\pmod{6}$ , and so forth. Continuing, we find the unique solution to be $(a_2, a_3, a_4, a_5, a_6, a_7) = (1,1,1,0,4,2)$ (uniqueness is assured by the Division Theorem ). The answer is $9 \\Longrightarrow \\boxed{9}$",
"We start by multiplying both sides by $7!$ , and we get: \\[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\\] After doing some guess and check, we find that the answer is $\\boxed{9}$",
"Let's clear up the fractions: \\[\\frac{5}{7}=\\frac{2520a_2+840a_3+210a_4+42a_5+7a_6+a_7}{7!}\\] \\[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\\] Notice that if we divide everything by $7$ then we would have: \\[514+\\frac{2}{7}=360a_2+120a_3+30a_4+6a_5+a_6+\\frac{1}{7}a_7\\] Since $0 \\le a_7<7$ and $a_7$ must be an integer, then we have $\\frac{2}{7}=\\frac{1}{7}a_7$ , so $a_7=2$\nSimilarly, if we divide everything by $6$ , then we would have: \\[85+\\frac{4}{6}=60a_2+20a_3+5a_4+a_5+\\frac{1}{6}a_6\\] Again, since $0 \\le a_6<6$ and $a_6$ must be an integer, we have $\\frac{4}{6}=\\frac{1}{6}a_6$ , so $a_6=4$\nThe pattern repeats itself, so in the end we have $a_2=1$ $a_3=1$ $a_4=1$ $a_5=0$ $a_6=4$ $a_7=2$ . So $a_2+a_3+a_4+a_5+a_6+a_7=\\boxed{9}$",
"By multiplying both sides by $7$ we get\n\\[5 = \\frac72 a_2 + \\frac76 a_3+ \\frac{7}{24} a_4 + \\frac{7}{120} a_5 + \\frac{7}{720} a_6 + \\frac{a_7}{720}\\]\nsince $0 \\le a_2 < 2$ , if $a_2 = 0$ the rest of the right hand side will not add up to be $5$ , so $a_2 = 1$\n\\[\\frac32 = \\frac76 a_3+ \\frac{7}{24} a_4 + \\frac{7}{120} a_5 + \\frac{7}{720} a_6 + \\frac{a_7}{720}\\]\nIf $a_3 = 2$ $\\frac76 a_3 = \\frac73 > \\frac32$ , so $a_3 = 1$\n\\[\\frac13 = \\frac{7}{24} a_4 + \\frac{7}{120} a_5 + \\frac{7}{720} a_6 + \\frac{a_7}{720}\\]\nIf $a_4 = 2$ $\\frac{7}{24} a_4 = \\frac{7}{12} > \\frac13$ , so $a_4 = 1$\n\\[\\frac{1}{24} = \\frac{7}{120} a_5 + \\frac{7}{720} a_6 + \\frac{a_7}{720}\\]\nIf $a_5 = 1$ $\\frac{7}{120} a_5 = \\frac{7}{120} > \\frac{1}{24}$ , so $a_5 = 0$\n\\[\\frac{1}{24} = \\frac{7}{720} a_6 + \\frac{a_7}{720}\\]\n$30 = 7 a_6 + a_7$ . Since $a_7 < 7$ $a_7 = 2$ and $a_6=4$\nTherefore, $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} = 1 + 1 + 1 + 0 + 4 + 2= \\boxed{9}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_4 | null | 21 | There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ unique integers $a_k$ $1\le k\le r$ ) with each $a_k$ either $1$ or $- 1$ such that \[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\] Find $n_1 + n_2 + \cdots + n_r$ | [
"In base $3$ , we find that $\\overline{2008}_{10} = \\overline{2202101}_{3}$ . In other words,\nIn order to rewrite as a sum of perfect powers of $3$ , we can use the fact that $2 \\cdot 3^k = 3^{k+1} - 3^k$\nThe answer is $7+5+4+3+2+0 = \\boxed{021}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_4 | null | 80 | There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$ | [
"Completing the square $y^2 = x^2 + 84x + 2008 = (x+42)^2 + 244$ . Thus $244 = y^2 - (x+42)^2 = (y - x - 42)(y + x + 42)$ by difference of squares\nSince $244$ is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since $244 = 2^2 \\cdot 61$ , the factors must be $2$ and $122$ . Since $x,y > 0$ , we have $y - x - 42 = 2$ and $y + x + 42 = 122$ ; the latter equation implies that $x + y = \\boxed{080}$",
"We complete the square like in the first solution: $y^2 = (x+42)^2 + 244$ . Since consecutive squares differ by the consecutive odd numbers, we note that $y$ and $x+42$ must differ by an even number. We can use casework with the even numbers, starting with $y-(x+42)=2$ \\begin{align*}2(x+42)+1+2(x+42)+3&=244\\\\ \\Leftrightarrow x&=18\\end{align*}\nThus, $y=62$ and the answer is $\\boxed{080}$",
"We see that $y^2 \\equiv x^2 + 4 \\pmod{6}$ . By quadratic residues , we find that either $x \\equiv 0, 3 \\pmod{6}$ . Also, $y^2 \\equiv (x+42)^2 + 244 \\equiv (x+2)^2 \\pmod{4}$ , so $x \\equiv 0, 2 \\mod{4}$ . Combining, we see that $x \\equiv 0 \\mod{6}$\nTesting $x = 6$ and other multiples of $6$ , we quickly find that $x = 18, y = 62$ is the solution. $18+62=\\boxed{080}$",
"We solve for x: $x^2 + 84x + 2008-y^2 = 0$\n$x=\\dfrac{-84+\\sqrt{84^2-4\\cdot 2008+4y^2}}{2}=-42+\\sqrt{y^2-244}$\nSo $y^2-244$ is a perfect square. Since 244 is even, the difference $\\sqrt{y^2-244} -y^2$ is even, so we try $y^2-244=(y-2)^2$ $-244=-4y+4$ $y=62$\nPlugging into our equation, we find that $x=18$ , and $(x,y)=(18,62)$ indeed satisfies the original equation. $x+y=\\boxed{080}$",
"Let $y=x+d$ for some $d>0$ , substitute into the original equation to get $84x + 2008 = 2xd + d^2$\nAll terms except for the last one are even, hence $d^2$ must be even, hence let $d=2e$ . We obtain $21x + 502 = xe + e^2$ . Rearrange to $502-e^2 = x(e-21)$\nObviously for $0<e<21$ the right hand side is negative and the left hand side is positive. Hence $e\\geq 21$ . Let $e=21+f$ , then $f\\geq 0$\nWe have $502 - (21+f)^2 = xf$ . Left hand side simplifies to $61 - 42f + f^2$ . As $x$ must be an integer, $f$ must divide the left hand side. But $61$ is a prime, which only leaves two options: $f=1$ and $f=61$\nOption $f=61$ gives us a negative $x$ . Option $f=1$ gives us $x=61/f - 42 + f = 18$ , and $y = x + d= x + 2e = x + 2(21+f) = 18 + 44 = 62$ , hence $x+y=\\boxed{080}$",
"First complete the square to get $y^2 = (x+42)^2 + 244$ . Remember that squares are the sums of consecutive odd integers, so when the difference between the two squares is 244, the two squares must be an even number of odd integers apart. However, there is only one distinct solution, as the problem states, and very quickly you will realize that only two odd integers work. When there are four, then the numbers are not odd, and when it is any other even integer it does not divide. So we need two consecutive odd integers that sum to 244. Easily we find 121 and 123. 121 is the 61st odd integer and 123 is the 62nd odd integer, so $(x+42)^2$ is the sum of the first 60 odd integers, or $(60)^2$ , while $y^2$ is $62^2$ for the same reasons. That way we get $x=12$ $y=62$ , hence $x+y=\\boxed{080}$"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10 | null | 944 | There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .) | [
"Define $\\left\\{ x \\right\\} = x - \\left\\lfloor x \\right\\rfloor$\nFirst, we bound $U$\nWe establish an upper bound of $U$ . We have \\begin{align*} U & \\leq \\sum_{n=1}^{2023} \\frac{n^2 - na}{5} \\\\ & = \\frac{1}{5} \\sum_{n=1}^{2023} n^2 - \\frac{a}{5} \\sum_{n=1}^{2023} n \\\\ & = \\frac{1012 \\cdot 2023}{5} \\left( 1349 - a \\right) \\\\ & \\triangleq UB . \\end{align*}\nWe establish a lower bound of $U$ . We have \\begin{align*} U & = \\sum_{n=1}^{2023} \\left( \\frac{n^2 - na}{5} - \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\right) \\\\ & = \\sum_{n=1}^{2023} \\frac{n^2 - na}{5} - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & = UB - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & \\geq UB - \\sum_{n=1}^{2023} \\mathbf 1 \\left\\{ \\frac{n^2 - na}{5} \\notin \\Bbb Z \\right\\} . \\end{align*}\nWe notice that if $5 | n$ , then $\\frac{n^2 - na}{5} \\in \\Bbb Z$ .\nThus, \\begin{align*} U & \\geq UB - \\sum_{n=1}^{2023} \\mathbf 1 \\left\\{ \\frac{n^2 - na}{5} \\notin \\Bbb Z \\right\\} \\\\ & \\geq UB - \\sum_{n=1}^{2023} \\mathbf 1 \\left\\{ 5 \\nmid n \\right\\} \\\\ & = UB - \\left( 2023 - \\left\\lfloor \\frac{2023}{5} \\right\\rfloor \\right) \\\\ & = UB - 1619 \\\\ & \\triangleq LB . \\end{align*}\nBecause $U \\in \\left[ - 1000, 1000 \\right]$ and $UB - LB = 1619 < \\left( 1000 - \\left( - 1000 \\right) \\right)$ , we must have either $UB \\in \\left[ - 1000, 1000 \\right]$ or $LB \\in \\left[ - 1000, 1000 \\right]$\nFor $UB \\in \\left[ - 1000, 1000 \\right]$ , we get a unique $a = 1349$ .\nFor $LB \\in \\left[ - 1000, 1000 \\right]$ , there is no feasible $a$\nTherefore, $a = 1349$ . Thus $UB = 0$\nNext, we compute $U$\nLet $n = 5 q + r$ , where $r = {\\rm Rem} \\ \\left( n, 5 \\right)$\nWe have \\begin{align*} \\left\\{ \\frac{n^2 - na}{5} \\right\\} & = \\left\\{ \\frac{\\left( 5 q + r \\right)^2 - \\left( 5 q + r \\right)\\left( 1350 - 1 \\right)}{5} \\right\\} \\\\ & = \\left\\{ 5 q^2 + 2 q r - \\left( 5 q + r \\right) 270 + q + \\frac{r^2 + r}{5} \\right\\} \\\\ & = \\left\\{\\frac{r^2 + r}{5} \\right\\} \\\\ & = \\left\\{ \\begin{array}{ll} 0 & \\mbox{ if } r = 0, 4 \\\\ \\frac{2}{5} & \\mbox{ if } r = 1, 3 \\\\ \\frac{1}{5} & \\mbox{ if } r = 2 \\end{array} \\right. . \\end{align*}\nTherefore, \\begin{align*} U & = \\sum_{n=1}^{2023} \\left( \\frac{n^2 - na}{5} - \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\right) \\\\ & = UB - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & = - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & = - \\sum_{q=0}^{404} \\sum_{r=0}^4 \\left\\{\\frac{r^2 + r}{5} \\right\\} + \\left\\{ \\frac{0^2 - 0 \\cdot a}{5} \\right\\} + \\left\\{ \\frac{2024^2 - 2024a}{5} \\right\\} \\\\ & = - \\sum_{q=0}^{404} \\left( 0 + 0 + \\frac{2}{5} + \\frac{2}{5} + \\frac{1}{5} \\right) + 0 + 0 \\\\ & = - 405 . \\end{align*}\nTherefore, $a + U = 1349 - 405 = \\boxed{944}$",
"We define $U' = \\sum^{2023}_{n=1} {\\frac{n^2-na}{5}}$ . Since for any real number $x$ $\\lfloor x \\rfloor \\le x \\le \\lfloor x \\rfloor + 1$ , we have $U \\le U' \\le U + 2023$ . Now, since $-1000 \\le U \\le 1000$ , we have $-1000 \\le U' \\le 3023$\nNow, we can solve for $U'$ in terms of $a$ . We have: \\begin{align*} U' &= \\sum^{2023}_{n=1} {\\frac{n^2-na}{5}} \\\\ &= \\sum^{2023}_{n=1} {\\frac{n^2}{5} - \\frac{na}{5}} \\\\ &= \\sum^{2023}_{n=1} {\\frac{n^2}{5}} - \\sum^{2023}_{n=1} {\\frac{na}{5}} \\\\ &= \\frac{\\sum^{2023}_{n=1} {{n^2}} - \\sum^{2023}_{n=1} {na}}{5} \\\\ &= \\frac{\\frac{2023(2023+1)(2023 \\cdot 2 + 1)}{6} - \\frac{a \\cdot 2023(2023+1)}{2} }{5} \\\\ &= \\frac{2023(2024)(4047-3a)}{30} \\\\ \\end{align*} So, we have $U' = \\frac{2023(2024)(4047-3a)}{30}$ , and $-1000 \\le U' \\le 3023$ , so we have $-1000 \\le \\frac{2023(2024)(4047-3a)}{30} \\le 3023$ , or $-30000 \\le 2023(2024)(4047-3a) \\le 90690$ . Now, $2023 \\cdot 2024$ is much bigger than $90690$ or $30000$ , and since $4047-3a$ is an integer, to satsify the inequalities, we must have $4047 - 3a = 0$ , or $a = 1349$ , and $U' = 0$\nNow, we can find $U - U'$ . We have: \\begin{align*} U - U' &= \\sum^{2023}_{n=1} {\\lfloor \\frac{n^2-1349n}{5} \\rfloor} - \\sum^{2023}_{n=1} {\\frac{n^2-1349n}{5}} \\\\ &= \\sum^{2023}_{n=1} {\\lfloor \\frac{n^2-1349n}{5} \\rfloor - \\frac{n^2-1349n}{5}} \\end{align*}. Now, if $n^2-1349n \\equiv 0 \\text{ (mod 5)}$ , then $\\lfloor \\frac{n^2-1349n}{5} \\rfloor - \\frac{n^2-1349n}{5} = 0$ , and if $n^2-1349n \\equiv 1 \\text{ (mod 5)}$ , then $\\lfloor \\frac{n^2-1349n}{5} \\rfloor - \\frac{n^2-1349n}{5} = -\\frac{1}{5}$ , and so on. Testing with $n \\equiv 0,1,2,3,4, \\text{ (mod 5)}$ , we get $n^2-1349n \\equiv 0,2,1,2,0 \\text{ (mod 5)}$ respectively. From 1 to 2023, there are 405 numbers congruent to 1 mod 5, 405 numbers congruent to 2 mod 5, 405 numbers congruent to 3 mod 5, 404 numbers congruent to 4 mod 5, and 404 numbers congruent to 0 mod 5. So, solving for $U - U'$ , we get: \\begin{align*} U - U' &= \\sum^{2023}_{n=1} {\\lfloor \\frac{n^2-1349n}{5} \\rfloor - \\frac{n^2-1349n}{5}} \\\\ &= 404 \\cdot 0 - 405 \\cdot \\frac{2}{5} - 405 \\cdot \\frac{1}{5} - 405 \\cdot \\frac{2}{5} - 404 \\cdot t0 \\\\ &= -405(\\frac{2}{5}+\\frac{1}{5}+\\frac{2}{5}) \\\\ &= -405 \\end{align*} Since $U' = 0$ , this gives $U = -405$ , and we have $a + U = 1349-405 = \\boxed{944}$",
"We can view the floor function in this problem as simply subtracting the remainder of $n^2 - na$ (mod $5$ ) from the numerator of $\\frac{n^2-na}{5}$ . For example, $\\left\\lfloor \\frac{7}{5} \\right\\rfloor = \\frac{7-2}{5} = 1$\nNote that the congruence of $n^2 - na$ (mod $5$ ) loops every time $n$ increases by 5. Also, note that the congruence of $a$ (mod $5$ ) determines the set of congruences of $n^2 - na$ for each congruence of $n$ (mod $5$ ).\nFor example, if $a \\equiv 1$ (mod $5$ ), the set of remainders is $(0, 2, 1, 2, 0)$ for $n \\equiv 1,2,3,4,0$ (mod $5$ ). Let the sum of these elements be $s$ . Note that for each “loop” of the numerator (mod $5$ ), each element of the set will be subtracted exactly once, meaning $s$ is subtracted once for each loop. The value of the numerator will loop $404$ times (mod $5$ ) throughout the sum, as $5 \\cdot 404=2020$ . Then\n$U \\approx \\frac {\\left( \\frac {n(n+1)(2n+1)}{6} - \\frac{(a)(n)(n+1)}{2} -404s \\right)}{5}$\nWhere $n=2023$ . Note that since $5 \\cdot 404=2020$ , this is an approximation for $U$ because the equation disregards the remainder (mod $5$ ) when $n=2021, 2022$ , and $2023$ so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of $U$ (*). However, we will find that this is a negligible error when it comes to the inequality $-1000<U<1000$ , so we can proceed with this approximation to solve for $a$\nFactoring our approximation gives $U \\approx \\frac {\\frac{(n)(n+1)(2n+1 - 3a)}{6}-404s}{5}$\nWe set $a= \\frac{(2n+1)}{3} = 1349$ to make $\\frac{(n)(n+1)(2n+1 - 3a)}{6}=0$ , accordingly minimizing $|U|$ , yielding $U \\approx \\frac{-404s}{5}$\nIf $a$ increases or decreases by $1$ , then $U$ changes by $\\frac {(n)(n+1)}{2 \\cdot 5} = \\frac {2023 \\cdot 2024}{10}$ which clearly breaks the inequality on $U$ . Therefore $a=1349 \\equiv 4$ (mod $5$ ) giving the set of remainders $(2,1,2,0,0)$ , so $s=5$ and our approximation yields $U \\approx -404$ . However, we must subtract 2, 1, and 2 (*) giving us $U = - 404 - \\frac{(2+1+2)}{5} = - 405$ , giving an answer of $1349-405= \\boxed{944}$",
"Consider the integral \\[\\int_{0}^{2023} \\dfrac{n^2-na}{5} \\, dn.\\] We hope this will give a good enough appoximation of $U$ to find $a.$ However, this integral can be easily evaluated(if you know calculus) to be \\[\\dfrac{1}{15}2023^3-\\dfrac{a}{10}2023^2=2023^2\\left(\\dfrac{2023}{15}-\\dfrac{a}{10}\\right).\\] Because we want this to be as close to $0$ as possible, we find that $a$ should equal $1349.$ Then, evaluating the sum becomes trivial. Set \\[U'=\\sum_{n=1}^{2023}\\dfrac{n^2-1349n}{5}\\] and \\[U''=\\sum_{n=1}^{2023}\\{\\dfrac{n^2-1349n}{5}\\}.\\] Then $U=U'-U''.$ We can evaluate $U'$ to be $0$ and $U''$ to be $-405$ (using some basic number theory). Thus, $U=-405$ and the answer is \\[1349+(-405)=\\boxed{944}.\\] BS2012"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_21 | C | 137 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | [
"First, substitute $2^{17}$ with $x$ . \nThen, the given equation becomes $\\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+x^{14}...-x^1+x^0$ by sum of powers factorization.\nNow consider only $x^{16}-x^{15}$ . This equals $x^{15}(x-1)=x^{15} \\cdot (2^{17}-1)$ .\nNote that $2^{17}-1$ equals $2^{16}+2^{15}+...+1$ , by difference of powers factorization (or by considering the expansion of $2^{17}=2^{16}+2^{15}+...+2+2$ ).\nThus, we can see that $x^{16}-x^{15}$ forms the sum of 17 different powers of 2. \nApplying the same method to each of $x^{14}-x^{13}$ $x^{12}-x^{11}$ , ... , $x^{2}-x^{1}$ , we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us $17 \\cdot 8=136$ .\nBut we must count also the $x^0$ term. \nThus, Our answer is $136+1=\\boxed{137}$",
"Multiply both sides by $2^{17}+1$ to get \\[2^{289}+1=2^{a_1} + 2^{a_2} + … + 2^{a_k} + 2^{a_1+17} + 2^{a_2+17} + … + 2^{a_k+17}.\\]\nNotice that $a_1 = 0$ , since there is a $1$ on the LHS. However, now we have an extra term of $2^{18}$ on the right from $2^{a_1+17}$ . To cancel it, we let $a_2 = 18$ . The two $2^{18}$ 's now combine into a term of $2^{19}$ , so we let $a_3 = 19$ . And so on, until we get to $a_{18} = 34$ . Now everything we don't want telescopes into $2^{35}$ . We already have that term since we let $a_2 = 18 \\implies a_2+17 = 35$ . Everything from now on will automatically telescope to $2^{52}$ . So we let $a_{19}$ be $52$\nAs you can see, we will have to add $17$ $a_n$ 's at a time, then \"wait\" for the sum to automatically telescope for the next $17$ numbers, etc, until we get to $2^{289}$ . We only need to add $a_n$ 's between odd multiples of $17$ and even multiples. The largest even multiple of $17$ below $289$ is $17\\cdot16$ , so we will have to add a total of $17\\cdot 8$ $a_n$ 's. However, we must not forget we let $a_1=0$ at the beginning, so our answer is $17\\cdot8+1 = \\boxed{137}$",
"In order to shorten expressions, $\\#$ will represent $16$ consecutive $0$ s when expressing numbers. Think of the problem in binary. We have $\\frac{1\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#1_2}{1\\#1_2}$ Note that $(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \\cdots + 2^{272}) = 2^0(2^{17} + 1) + 2^{34}(2^{17} + 1) + 2^{68}(2^{17} + 1) + \\cdots 2^{272}(2^{17} + 1)$ $= 1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1_2$ and $(2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \\cdots + 2^{255}) = 2^{17}(2^{17} + 1) + 2^{51}(2^{17} + 1) + 2^{85}(2^{17} + 1) + \\cdots 2^{255}(2^{17} + 1)$ $= 1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#0_2$ Since $\\phantom{=\\ } 1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1_2$ $-\\ \\phantom{1\\#} 1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#0_2$ $= 1\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#1_2$ this means that $(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \\cdots + 2^{272}) - (2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \\cdots + 2^{255}) = 2^{289}$ so $\\frac{2^{289}+1}{2^{17}+1} = (2^0 + 2^{34} + 2^{68} + \\cdots + 2^{272}) - (2^{17} + 2^{51} + 2^{85} + \\cdots + 2^{255})$ $= 2^0 + (2^{34} - 2^{17}) + (2^{68} - 2^{51}) + \\cdots + (2^{272} - 2^{255})$ Expressing each of the pairs of the form $2^{n + 17} - 2^n$ in binary, we have $\\phantom{=\\ } 1000000000000000000 \\cdots 0_2$ $-\\ \\phantom{10000000000000000} 10 \\cdots 0_2$ $= \\phantom{1} 111111111111111110 \\cdots 0_2$ or $2^{n + 17} - 2^n = 2^{n + 16} + 2^{n + 15} + 2^{n + 14} + \\cdots + 2^{n}$ This means that each pair has $17$ terms of the form $2^n$ Since there are $8$ of these pairs, there are a total of $8 \\cdot 17 = 136$ terms. Accounting for the $2^0$ term, which was not in the pair, we have a total of $136 + 1 = \\boxed{137}$ terms. ~ emerald_block"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_21 | null | 137 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | [
"Notice that the only answer choices that are spaced one apart are $136$ and $137$ . It's likely that people will forget to include the final term so the answer is $\\boxed{137}$"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_19 | C | 137 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | [
"First, substitute $2^{17}$ with $x$ . \nThen, the given equation becomes $\\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+x^{14}...-x^1+x^0$ by sum of powers factorization.\nNow consider only $x^{16}-x^{15}$ . This equals $x^{15}(x-1)=x^{15} \\cdot (2^{17}-1)$ .\nNote that $2^{17}-1$ equals $2^{16}+2^{15}+...+1$ , by difference of powers factorization (or by considering the expansion of $2^{17}=2^{16}+2^{15}+...+2+2$ ).\nThus, we can see that $x^{16}-x^{15}$ forms the sum of 17 different powers of 2. \nApplying the same method to each of $x^{14}-x^{13}$ $x^{12}-x^{11}$ , ... , $x^{2}-x^{1}$ , we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us $17 \\cdot 8=136$ .\nBut we must count also the $x^0$ term. \nThus, Our answer is $136+1=\\boxed{137}$",
"Multiply both sides by $2^{17}+1$ to get \\[2^{289}+1=2^{a_1} + 2^{a_2} + … + 2^{a_k} + 2^{a_1+17} + 2^{a_2+17} + … + 2^{a_k+17}.\\]\nNotice that $a_1 = 0$ , since there is a $1$ on the LHS. However, now we have an extra term of $2^{18}$ on the right from $2^{a_1+17}$ . To cancel it, we let $a_2 = 18$ . The two $2^{18}$ 's now combine into a term of $2^{19}$ , so we let $a_3 = 19$ . And so on, until we get to $a_{18} = 34$ . Now everything we don't want telescopes into $2^{35}$ . We already have that term since we let $a_2 = 18 \\implies a_2+17 = 35$ . Everything from now on will automatically telescope to $2^{52}$ . So we let $a_{19}$ be $52$\nAs you can see, we will have to add $17$ $a_n$ 's at a time, then \"wait\" for the sum to automatically telescope for the next $17$ numbers, etc, until we get to $2^{289}$ . We only need to add $a_n$ 's between odd multiples of $17$ and even multiples. The largest even multiple of $17$ below $289$ is $17\\cdot16$ , so we will have to add a total of $17\\cdot 8$ $a_n$ 's. However, we must not forget we let $a_1=0$ at the beginning, so our answer is $17\\cdot8+1 = \\boxed{137}$",
"In order to shorten expressions, $\\#$ will represent $16$ consecutive $0$ s when expressing numbers. Think of the problem in binary. We have $\\frac{1\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#1_2}{1\\#1_2}$ Note that $(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \\cdots + 2^{272}) = 2^0(2^{17} + 1) + 2^{34}(2^{17} + 1) + 2^{68}(2^{17} + 1) + \\cdots 2^{272}(2^{17} + 1)$ $= 1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1_2$ and $(2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \\cdots + 2^{255}) = 2^{17}(2^{17} + 1) + 2^{51}(2^{17} + 1) + 2^{85}(2^{17} + 1) + \\cdots 2^{255}(2^{17} + 1)$ $= 1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#0_2$ Since $\\phantom{=\\ } 1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1_2$ $-\\ \\phantom{1\\#} 1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#1\\#0_2$ $= 1\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#0\\#1_2$ this means that $(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \\cdots + 2^{272}) - (2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \\cdots + 2^{255}) = 2^{289}$ so $\\frac{2^{289}+1}{2^{17}+1} = (2^0 + 2^{34} + 2^{68} + \\cdots + 2^{272}) - (2^{17} + 2^{51} + 2^{85} + \\cdots + 2^{255})$ $= 2^0 + (2^{34} - 2^{17}) + (2^{68} - 2^{51}) + \\cdots + (2^{272} - 2^{255})$ Expressing each of the pairs of the form $2^{n + 17} - 2^n$ in binary, we have $\\phantom{=\\ } 1000000000000000000 \\cdots 0_2$ $-\\ \\phantom{10000000000000000} 10 \\cdots 0_2$ $= \\phantom{1} 111111111111111110 \\cdots 0_2$ or $2^{n + 17} - 2^n = 2^{n + 16} + 2^{n + 15} + 2^{n + 14} + \\cdots + 2^{n}$ This means that each pair has $17$ terms of the form $2^n$ Since there are $8$ of these pairs, there are a total of $8 \\cdot 17 = 136$ terms. Accounting for the $2^0$ term, which was not in the pair, we have a total of $136 + 1 = \\boxed{137}$ terms. ~ emerald_block"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_19 | null | 137 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | [
"Notice that the only answer choices that are spaced one apart are $136$ and $137$ . It's likely that people will forget to include the final term so the answer is $\\boxed{137}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_2 | null | 107 | There is a $40\%$ chance of rain on Saturday and a $30\%$ chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Find $a+b$ | [
"Let $x$ be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have $\\dfrac{4}{5}x+\\dfrac{2}{5}2x = \\dfrac{3}{10} \\implies \\dfrac{7}{5}x=\\dfrac{3}{10}$ $\\implies x=\\dfrac{3}{14}$ . Therefore, the probability that it doesn't rain on either day is $\\left(1-\\dfrac{3}{14}\\right)\\left(\\dfrac{3}{5}\\right)=\\dfrac{33}{70}$ . Therefore, the probability that rains on at least one of the days is $1-\\dfrac{33}{70}=\\dfrac{37}{70}$ , so adding up the $2$ numbers, we have $37+70=\\boxed{107}$"
] |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_9 | null | 902 | There is a collection of $25$ indistinguishable white chips and $25$ indistinguishable black chips. Find the number of ways to place some of these chips in the $25$ unit cells of a $5\times5$ grid such that: | [
"The problem says \"some\", so not all cells must be occupied.\nWe start by doing casework on the column on the left. There can be 5,4,3,2, or 1 black chip. The same goes for white chips, so we will multiply by 2 at the end. There is $1$ way to select $5$ cells with black chips. Because of the 2nd condition, there can be no white, and the grid must be all black- $1$ way . There are $5$ ways to select 4 cells with black chips. We now consider the row that does not contain a black chip. The first cell must be blank, and the remaining $4$ cells have $2^4-1$ different ways( $-1$ comes from all blank). This gives us $75$ ways. Notice that for 3,2 or 1 black chips on the left there is a pattern. Once the first blank row is chosen, the rest of the blank rows must be ordered similarly. For example, with 2 black chips on the left, there will be 3 blank rows. There are 15 ways for the first row to be chosen, and the following 2 rows must have the same order. Thus, The number of ways for 3,2,and 1 black chips is $10*15$ $10*15$ $5*15$ . Adding these up, we have $1+75+150+150+75 = 451$ . Multiplying this by 2, we get $\\boxed{902}$ .\n~westwoodmonster",
"Note that the answer is equivalent to the number of ways to choose rows and columns that the white chips occupy, as once those are chosen, there is only one way to place the chips, and every way to place the chips corresponds to a set of rows and columns occupied by the white pieces.\nIf the white pieces occupy none of the rows, then because they don't appear on the board, they will not occupy any of the columns. Similar logic can be applied to show that if white pieces occupy all of the rows, they will also occupy all of the columns.\nThe number of sets of rows and columns that white can occupy are $2^{5} - 2 = 30$ each, accounting for the empty and full set.\nSo, including the board with 25 white pieces and the board with 25 black pieces, the answer is $30^{2}+2 = \\boxed{902}$",
"Case 1: All chips on the grid have the same color.\nIn this case, all cells are occupied with chips with the same color.\nTherefore, the number of configurations in this case is 2.\nCase 2: Both black and white chips are on the grid.\nObservation 1: Each colored chips must occupy at least one column and one row.\nThis is because, for each given color, there must be at least one chip. Therefore, all chips placed in the cells that are in the same row or the same column with this given chip must have the same color with this chip.\nObservation 2: Each colored chips occupy at most 4 rows and 4 columns.\nThis directly follows from Observation 1.\nObservation 3: For each color, if all chips in this color occupy columns with $x$ -coordinates $\\left\\{ x_1, \\cdots, x_m \\right\\}$ and rows with $y$ -coordinates $\\left\\{ y_1, \\cdots, y_n \\right\\}$ , then every cell $\\left( x, y \\right)$ with $x \\in \\left\\{ x_1, \\cdots , x_m \\right\\}$ and $y \\in \\left\\{ y_1, \\cdots , y_n \\right\\}$ is occupied by a chip with the same color.\nThis is because, if there is any cell in this region occupied by a chip with a different color, it violates Condition 2.\nIf there is any cell in this region that is empty, then it violates Condition 3.\nObservation 4: For each color, if all chips in this color occupy columns with $x$ -coordinates $\\left\\{ x_1, \\cdots, x_m \\right\\}$ and rows with $y$ -coordinates $\\left\\{ y_1, \\cdots, y_n \\right\\}$ , then every cell $\\left( x, y \\right)$ with $x \\notin \\left\\{ x_1, \\cdots , x_m \\right\\}$ and $y \\in \\left\\{ y_1, \\cdots , y_n \\right\\}$ , or $x \\in \\left\\{ x_1, \\cdots , x_m \\right\\}$ and $y \\notin \\left\\{ y_1, \\cdots , y_n \\right\\}$ is empty.\nThis is because, if there is any cell in this region occupied by a chip with a different color, it violates Condition 2.\nObservation 5: For each color, if all chips in this color occupy columns with $x$ -coordinates $\\left\\{ x_1, \\cdots, x_m \\right\\}$ and rows with $y$ -coordinates $\\left\\{ y_1, \\cdots, y_n \\right\\}$ , then every cell $\\left( x, y \\right)$ with $x \\notin \\left\\{ x_1, \\cdots , x_m \\right\\}$ and $y \\notin \\left\\{ y_1, \\cdots , y_n \\right\\}$ is occupied by chips with the different color.\nThis follows from Condition 3.\nBy using the above observations, the number of feasible configurations in this case is given by\n\\begin{align*}\n\\sum_{n=1}^4 \\sum_{m=1}^4 \\binom{5}{n} \\binom{5}{m}\n& = \\left( \\sum_{n=1}^4 \\binom{5}{n} \\right)\n\\left( \\sum_{m=1}^4 \\binom{5}{m} \\right) \\\\\n& = \\left( 2^5 - 2 \\right)^2 \\\\\n& = 900 .\n\\end{align*}\nPutting all cases together, the total number of feasible configurations is $2 + 900 = \\boxed{902}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_2 | null | 697 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$ | [
"Let $z = a + 164i$\nThen \\[\\frac {a + 164i}{a + 164i + n} = 4i\\] and \\[a + 164i = \\left (4i \\right ) \\left (a + n + 164i \\right ) = 4i \\left (a + n \\right ) - 656.\\]\nBy comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,\nwe conclude that \\[a = -656.\\]\nBy equating the imaginary terms on each side of the equation,\nwe conclude that \\[164i = 4i \\left (a + n \\right ) = 4i \\left (-656 + n \\right ).\\]\nWe now have an equation for $n$ \\[4i \\left (-656 + n \\right ) = 164i,\\]\nand this equation shows that $n = \\boxed{697}.$",
"\\[\\frac {z}{z+n}=4i\\]\n\\[1-\\frac {n}{z+n}=4i\\]\n\\[1-4i=\\frac {n}{z+n}\\]\n\\[\\frac {1}{1-4i}=\\frac {z+n}{n}\\]\n\\[\\frac {1+4i}{17}=\\frac {z}{n}+1\\]\nSince their imaginary part has to be equal,\n\\[\\frac {4i}{17}=\\frac {164i}{n}\\]\n\\[n=\\frac {(164)(17)}{4}=697\\]\n\\[n = \\boxed{697}.\\]",
"Below is an image of the complex plane. Let $\\operatorname{Im}(z)$ denote the imaginary part of a complex number $z$ $z$ must lie on the line $\\operatorname{Im}(z)=164$ $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$\nConsider $z$ and $z+n$ in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have $z_1z_2 = r_1\\angle\\theta_1 \\cdot r_2\\angle\\theta_2 = r_1r_2\\angle(\\theta_1+\\theta_2)$ and $\\frac{z_1}{z_2} = \\frac{r_1\\angle\\theta_1}{r_2\\angle\\theta_2} = \\frac{r_1}{r_2}\\angle(\\theta_1-\\theta_2)$ , where $r$ is the magnitude and $\\theta$ is the phase, and $z_n=r_n\\angle\\theta_n$\nSince $4i$ has magnitude $4$ and phase $90^\\circ$ (since the positive imaginary axis points in a direction $90^\\circ$ counterclockwise from the positive real axis), $z$ must have a magnitude $4$ times that of $z+n$ . We denote the length from the origin to $z+n$ with the value $x$ and the length from the origin to $z$ with the value $4x$ . Additionally, $z$ , the origin, and $z+n$ must form a right angle, with $z$ counterclockwise from $z+n$\nThis means that $z$ , the origin, and $z+n$ form a right triangle. The hypotenuse is the length from $z$ to $z+n$ and has length $n$ , since $n$ is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as $\\frac{x \\cdot 4x}{2}$ , or using the hypotenuse and its corresponding altitude, as $\\frac{164n}{2}$ , so $\\frac{x \\cdot 4x}{2} = \\frac{164n}{2} \\implies x^2 = 41n$ . By Pythagorean Theorem, $x^2+(4x)^2 = n^2 \\implies 17x^2 = n^2$ . Substituting out $x^2$ using the earlier equation, we get $17\\cdot41n = n^2 \\implies n = \\boxed{697}$ . ~ emerald_block",
"Taking the reciprocal of our equation gives us $1 + \\frac{n}{z} = \\frac{1}{4i}.$ Therefore, \\[\\frac{n}{z} = \\frac{1-4i}{4i} = \\frac{17}{-16+4i}.\\] Since $z$ has an imaginary part of $164$ , we must multiply both sides of our RHS fraction by $\\frac{164}{4} = 41$ so that its denominator's imaginary part matches the LHS fraction's denominator's imaginary part. That looks like this: \\[\\frac{n}{z} = \\frac{697}{-656 + 164i}.\\] Therefore, we can conclude the the real part of $z$ is $-656$ and $n = \\boxed{697}.$ (it wasn't necessary to find the real part)"
] |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_23 | B | 6 | There is a list of seven numbers. The average of the first four numbers is $5$ , and the average of the last four numbers is $8$ . If the average of all seven numbers is $6\frac{4}{7}$ , then the number common to both sets of four numbers is
$\text{(A)}\ 5\frac{3}{7}\qquad\text{(B)}\ 6\qquad\text{(C)}\ 6\frac{4}{7}\qquad\text{(D)}\ 7\qquad\text{(E)}\ 7\frac{3}{7}$ | [
"Remember that if a list of $n$ numbers has an average of $k$ , then the sum $S$ of all the numbers on the list is $S = nk$\nSo if the average of the first $4$ numbers is $5$ , then the first four numbers total $4 \\cdot 5 = 20$\nIf the average of the last $4$ numbers is $8$ , then the last four numbers total $4 \\cdot 8 = 32$\nIf the average of all $7$ numbers is $6\\frac{4}{7}$ , then the total of all seven numbers is $7 \\cdot 6\\frac{4}{7} = 7\\cdot 6 + 4 = 46$\nIf the first four numbers are $20$ , and the last four numbers are $32$ , then all \"eight\" numbers are $20 + 32 = 52$ . But that's counting one number twice. Since the sum of all seven numbers is $46$ , then the number that was counted twice is $52 - 46 = 6$ , and the answer is $\\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_13 | null | 220 | There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$ | [
"Because $0 < x < 1$ , we have \\begin{align*} P \\left( x \\right) & = \\sum_{a=0}^6 \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\binom{6}{a} x^{2310a} \\left( - 1 \\right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\\\ & = \\sum_{a=0}^6 \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\left( - 1 \\right)^{6-a} x^{2310 a + 105 b + 70 c + 42 d + 30 e} . \\end{align*}\nDenote by $c_{2022}$ the coefficient of $P \\left( x \\right)$ .\nThus, \\begin{align*} c_{2022} & = \\sum_{a=0}^6 \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\left( - 1 \\right)^{6-a} \\Bbb I \\left\\{ 2310 a + 105 b + 70 c + 42 d + 30 e = 2022 \\right\\} \\\\ & = \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\left( - 1 \\right)^{6-0} \\Bbb I \\left\\{ 2310 \\cdot 0 + 105 b + 70 c + 42 d + 30 e = 2022 \\right\\} \\\\ & = \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\Bbb I \\left\\{ 105 b + 70 c + 42 d + 30 e = 2022 \\right\\} . \\end{align*}\nNow, we need to find the number of nonnegative integer tuples $\\left( b , c , d , e \\right)$ that satisfy \\[ 105 b + 70 c + 42 d + 30 e = 2022 . \\hspace{1cm} (1) \\]\nModulo 2 on Equation (1), we have $b \\equiv 0 \\pmod{2}$ .\nHence, we can write $b = 2 b'$ . Plugging this into (1), the problem reduces to finding the number of \nnonnegative integer tuples $\\left( b' , c , d , e \\right)$ that satisfy \\[ 105 b' + 35 c + 21 d + 15 e = 1011 . \\hspace{1cm} (2) \\]\nModulo 3 on Equation (2), we have $2 c \\equiv 0 \\pmod{3}$ .\nHence, we can write $c = 3 c'$ . Plugging this into (2), the problem reduces to finding the number of\nnonnegative integer tuples $\\left( b' , c' , d , e \\right)$ that satisfy \\[ 35 b' + 35 c' + 7 d + 5 e = 337 . \\hspace{1cm} (3) \\]\nModulo 5 on Equation (3), we have $2 d \\equiv 2 \\pmod{5}$ .\nHence, we can write $d = 5 d' + 1$ . Plugging this into (3), the problem reduces to finding the number of\nnonnegative integer tuples $\\left( b' , c' , d' , e \\right)$ that satisfy \\[ 7 b' + 7 c' + 7 d' + e = 66 . \\hspace{1cm} (4) \\]\nModulo 7 on Equation (4), we have $e \\equiv 3 \\pmod{7}$ .\nHence, we can write $e = 7 e' + 3$ . Plugging this into (4), the problem reduces to finding the number of\nnonnegative integer tuples $\\left( b' , c' , d' , e' \\right)$ that satisfy \\[ b' + c' + d' + e' = 9 . \\hspace{1cm} (5) \\]\nThe number of nonnegative integer solutions to Equation (5) is $\\binom{9 + 4 - 1}{4 - 1} = \\binom{12}{3} = \\boxed{220}$",
"Note that $2022 = 210\\cdot 9 +132$ . Since the only way to express $132$ in terms of $105$ $70$ $42$ , or $30$ is $132 = 30+30+30+42$ , we are essentially just counting the number of ways to express $210*9$ in terms of these numbers. Since $210 = 2*105=3*70=5*42=7*30$ , it can only be expressed as a sum in terms of only one of the numbers ( $105$ $70$ $42$ , or $30$ ). Thus, the answer is (by sticks and stones) \\[\\binom{12}{3} = \\boxed{220}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_6 | C | 10 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | [
"\\[\\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \\cdot n! \\\\ \\Rightarrow \\ &n![n+1 + (n+2)(n+1)] = 440 \\cdot n! \\\\ \\Rightarrow \\ &n + 1 + n^2 + 3n + 2 = 440 \\\\ \\Rightarrow \\ &n^2 + 4n - 437 = 0\\end{split}\\]\nSolving by the quadratic formula, $n = \\frac{-4\\pm \\sqrt{16+437\\cdot4}}{2} = \\frac{-4\\pm 42}{2} = \\frac{38}{2} = 19$ (since clearly $n \\geq 0$ ). The answer is therefore $1 + 9 = \\boxed{10}$",
"Dividing both sides by $n!$ gives \\[(n+1)+(n+2)(n+1)=440 \\Rightarrow n^2+4n-437=0 \\Rightarrow (n-19)(n+23)=0.\\] Since $n$ is non-negative, $n=19$ . The answer is $1 + 9 = \\boxed{10}$",
"Dividing both sides by $n!$ as before gives $(n+1)+(n+1)(n+2)=440$ . Now factor out $(n+1)$ , giving $(n+1)(n+3)=440$ . By considering the prime factorization of $440$ , a bit of experimentation gives us $n+1=20$ and $n+3=22$ , so $n=19$ , so the answer is $1 + 9 = \\boxed{10}$",
"Since $(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \\cdot n!$ , the result can be factored into $(n+1)(n+3)n!=440 \\cdot n!$ and divided by $n!$ on both sides to get $(n+1)(n+3)=440$ . From there, it is easier to complete the square with the quadratic $(n+1)(n+3) = n^2 + 4n + 3$ , so $n^2+4n+4=441 \\Rightarrow (n+2)^2=441$ . Solving for $n$ results in $n=19,-23$ , and since $n>0$ $n=19$ and the answer is $1 + 9 = \\boxed{10}$",
"Rewrite $(n+1)! + (n+2)! = 440 \\cdot n!$ as $(n+1)n! + (n+2)(n+1)n! = 440 \\cdot n!.$ Factoring out the $n!$ we get $n!(n + 1 + (n+1)(n+2)) = 440 \\cdot n!.$ Expand this to get $n!(n^2 + 4n + 3) = 440 \\cdot n!.$ Factor this and divide by $n!$ to get $(n + 1)(n + 3) = 440.$ If we take the prime factorization of $440$ we see that it is $2^3 * 5 * 11.$ Intuitively, we can find that $n + 1 = 20$ and $n + 3 = 22.$ Therefore, $n = 19.$ Since the problem asks for the sum of the didgits of $n$ , we finally calculate $1 + 9 = 10$ and get answer choice $\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | C | 10 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | [
"\\[\\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \\cdot n! \\\\ \\Rightarrow \\ &n![n+1 + (n+2)(n+1)] = 440 \\cdot n! \\\\ \\Rightarrow \\ &n + 1 + n^2 + 3n + 2 = 440 \\\\ \\Rightarrow \\ &n^2 + 4n - 437 = 0\\end{split}\\]\nSolving by the quadratic formula, $n = \\frac{-4\\pm \\sqrt{16+437\\cdot4}}{2} = \\frac{-4\\pm 42}{2} = \\frac{38}{2} = 19$ (since clearly $n \\geq 0$ ). The answer is therefore $1 + 9 = \\boxed{10}$",
"Dividing both sides by $n!$ gives \\[(n+1)+(n+2)(n+1)=440 \\Rightarrow n^2+4n-437=0 \\Rightarrow (n-19)(n+23)=0.\\] Since $n$ is non-negative, $n=19$ . The answer is $1 + 9 = \\boxed{10}$",
"Dividing both sides by $n!$ as before gives $(n+1)+(n+1)(n+2)=440$ . Now factor out $(n+1)$ , giving $(n+1)(n+3)=440$ . By considering the prime factorization of $440$ , a bit of experimentation gives us $n+1=20$ and $n+3=22$ , so $n=19$ , so the answer is $1 + 9 = \\boxed{10}$",
"Since $(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \\cdot n!$ , the result can be factored into $(n+1)(n+3)n!=440 \\cdot n!$ and divided by $n!$ on both sides to get $(n+1)(n+3)=440$ . From there, it is easier to complete the square with the quadratic $(n+1)(n+3) = n^2 + 4n + 3$ , so $n^2+4n+4=441 \\Rightarrow (n+2)^2=441$ . Solving for $n$ results in $n=19,-23$ , and since $n>0$ $n=19$ and the answer is $1 + 9 = \\boxed{10}$",
"Rewrite $(n+1)! + (n+2)! = 440 \\cdot n!$ as $(n+1)n! + (n+2)(n+1)n! = 440 \\cdot n!.$ Factoring out the $n!$ we get $n!(n + 1 + (n+1)(n+2)) = 440 \\cdot n!.$ Expand this to get $n!(n^2 + 4n + 3) = 440 \\cdot n!.$ Factor this and divide by $n!$ to get $(n + 1)(n + 3) = 440.$ If we take the prime factorization of $440$ we see that it is $2^3 * 5 * 11.$ Intuitively, we can find that $n + 1 = 20$ and $n + 3 = 22.$ Therefore, $n = 19.$ Since the problem asks for the sum of the didgits of $n$ , we finally calculate $1 + 9 = 10$ and get answer choice $\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | null | 112 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [
"Define $a$ to be $\\log_{20x} (22x) = \\log_{2x} (202x)$ , what we are looking for. Then, by the definition of the logarithm, \\[\\begin{cases} (20x)^{a} &= 22x \\\\ (2x)^{a} &= 202x. \\end{cases}\\] Dividing the first equation by the second equation gives us $10^a = \\frac{11}{101}$ , so by the definition of logs, $a = \\log_{10} \\frac{11}{101}$ . This is what the problem asked for, so the fraction $\\frac{11}{101}$ gives us $m+n = \\boxed{112}$",
"We could assume a variable $v$ which equals to both $\\log_{20x} (22x)$ and $\\log_{2x} (202x)$\nSo that $(20x)^v=22x \\textcircled{1}$ and $(2x)^v=202x \\textcircled{2}$\nExpress $\\textcircled{1}$ as: $(20x)^v=(2x \\cdot 10)^v=(2x)^v \\cdot \\left(10^v\\right)=22x \\textcircled{3}$\nSubstitute $\\textcircled{{2}}$ to $\\textcircled{3}$ $202x \\cdot (10^v)=22x$\nThus, $v=\\log_{10} \\left(\\frac{22x}{202x}\\right)= \\log_{10} \\left(\\frac{11}{101}\\right)$ , where $m=11$ and $n=101$\nTherefore, $m+n = \\boxed{112}$",
"We have \\begin{align*} \\log_{20x} (22x) & = \\frac{\\log_k 22x}{\\log_k 20x} \\\\ & = \\frac{\\log_k x + \\log_k 22}{\\log_k x + \\log_k 20} . \\end{align*}\nWe have \\begin{align*} \\log_{2x} (202x) & = \\frac{\\log_k 202x}{\\log_k 2x} \\\\ & = \\frac{\\log_k x + \\log_k 202 }{\\log_k x + \\log_k 2} . \\end{align*}\nBecause $\\log_{20x} (22x)=\\log_{2x} (202x)$ , we get \\[ \\frac{\\log_k x + \\log_k 22}{\\log_k x + \\log_k 20} = \\frac{\\log_k x + \\log_k 202 }{\\log_k x + \\log_k 2} . \\]\nWe denote this common value as $\\lambda$\nBy solving the equality $\\frac{\\log_k x + \\log_k 22}{\\log_k x + \\log_k 20} = \\lambda$ , we get $\\log_k x = \\frac{\\log_k 22 - \\lambda \\log_k 20}{\\lambda - 1}$\nBy solving the equality $\\frac{\\log_k x + \\log_k 202 }{\\log_k x + \\log_k 2} = \\lambda$ , we get $\\log_k x = \\frac{\\log_k 202 - \\lambda \\log_k 2}{\\lambda - 1}$\nBy equating these two equations, we get \\[ \\frac{\\log_k 22 - \\lambda \\log_k 20}{\\lambda - 1} = \\frac{\\log_k 202 - \\lambda \\log_k 2}{\\lambda - 1} . \\]\nTherefore, \\begin{align*} \\log_{20x} (22x) & = \\lambda \\\\ & = \\frac{\\log_k 22 - \\log_k 202}{\\log_k 20 - \\log_k 2} \\\\ & = \\frac{\\log_k \\frac{11}{101}}{\\log_k 10} \\\\ & = \\log_{10} \\frac{11}{101} . \\end{align*}\nTherefore, the answer is $11 + 101 = \\boxed{112}$",
"Let $a$ be the exponent such that $(20x)^a = 22x$ and $(2x)^a = 202x$ . Dividing, we get \\begin{align*} \\dfrac{(20x)^a}{(2x)^a} &= \\dfrac{22x}{202x}. \\\\ \\left(\\dfrac{20x}{2x}\\right)^a &= \\dfrac{22x}{202x}. \\\\ 10^a &= \\dfrac{11}{101}. \\\\ \\end{align*} Thus, we see that $\\log_{10} \\left(\\dfrac{11}{101}\\right) = a = \\log_{20x} 22x$ , so the answer is $11 + 101 = \\boxed{112}$",
"By the change of base rule, we have $\\frac{\\log 22x}{\\log 20x}=\\frac{\\log 202x}{\\log 2x}$ , or $\\frac{\\log 22 +\\log x}{\\log 20 +\\log x}=\\frac{\\log 202 +\\log x}{\\log 2 +\\log x}=k$ . We also know that if $a/b=c/d$ , then this also equals $\\frac{a-c}{b-d}$ . We use this identity and find that $k=\\frac{\\log 202 -\\log 22}{\\log 2 -\\log 20}=-\\log\\frac{202}{22}=\\log\\frac{11}{101}$ . The requested sum is $11+101=\\boxed{112}.$",
"By change of base formula, \\[\\frac{\\log_{2x} 22x}{\\log_{2x} 20x} = \\frac{{\\log_{2x} 11} + 1}{{\\log_{2x} 10} + 1} = {\\log_{2x} 101} + 1\\] \\[\\log_{2x} 11 + 1 = (\\log_{2x} 10)(\\log_{2x} 101) + \\log{2x} 1010 + 1\\] \\[\\frac{\\log_{2x} \\frac{11}{1010}}{\\log_{2x} 10} = \\log_{2x} 101\\] \\[\\log_{10} {\\frac{11}{1010}} = \\log_{2x} 101\\] \\[\\log_{10} {\\frac{11}{1010}} + 1 = \\log_{2x} 101 + 1 = \\log_{2x} 202x = \\log_{20x} {22x}\\] Thus, \\[\\log_{20x} 22x = \\log_{10} \\left( \\frac{11}{1010} \\times 10 \\right) = \\log_{10} \\frac{11}{101}\\] The requested answer is $11 + 101 = \\boxed{112}$"
] |
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_7 | null | 650 | There is a set of 1000 switches, each of which has four positions, called $A, B, C$ , and $D$ . When the position of any switch changes, it is only from $A$ to $B$ , from $B$ to $C$ , from $C$ to $D$ , or from $D$ to $A$ . Initially each switch is in position $A$ . The switches are labeled with the 1000 different integers $(2^{x})(3^{y})(5^{z})$ , where $x, y$ , and $z$ take on the values $0, 1, \ldots, 9$ . At step i of a 1000-step process, the $i$ -th switch is advanced one step, and so are all the other switches whose labels divide the label on the $i$ -th switch. After step 1000 has been completed, how many switches will be in position $A$ | [
"For each $i$ th switch (designated by $x_{i},y_{i},z_{i}$ ), it advances itself only one time at the $i$ th step; thereafter, only a switch with larger $x_{j},y_{j},z_{j}$ values will advance the $i$ th switch by one step provided $d_{i}= 2^{x_{i}}3^{y_{i}}5^{z_{i}}$ divides $d_{j}= 2^{x_{j}}3^{y_{j}}5^{z_{j}}$ . Let $N = 2^{9}3^{9}5^{9}$ be the max switch label. To find the divisor multiples in the range of $d_{i}$ to $N$ , we consider the exponents of the number $\\frac{N}{d_{i}}= 2^{9-x_{i}}3^{9-y_{i}}5^{9-z_{i}}$ . In general, the divisor-count of $\\frac{N}{d}$ must be a multiple of 4 to ensure that a switch is in position A:\nWe consider the cases where the 3 factors above do not contribute multiples of 4.\nThe number of switches in position A is $1000-125-225 = \\boxed{650}$"
] |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_14 | D | 7 | There is a set of five positive integers whose average (mean) is 5, whose median is 5, and whose only mode is 8. What is the difference between the largest and smallest integers in the set?
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$ | [
"When these numbers are ordered in ascending order, 5, the median, falls right in the middle, which is the third integer from the left. Since there is a unique mode of 8, both integers to the right of 5 must be 8s. Since the mean is 5, the sum of the integers is 25, which means the 2 leftmost integers have to sum to 4. 2 and 2 does not work because that would result in two modes. However, 1 and 3 does, and so our answer is 8-1=7. $\\boxed{7}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24 | B | 9 | There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$ | [
"The acceleration must be zero at the $x$ -intercept; this intercept must be an inflection point for the minimum $a$ value.\nDerive $f(x)$ so that the acceleration $f''(x)=0$ . Using the power rule, \\begin{align*} f(x) &= x^3-ax^2+bx-a \\\\ f’(x) &= 3x^2-2ax+b \\\\ f’’(x) &= 6x-2a \\end{align*} So $x=\\frac{a}{3}$ for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at $x=a/3$ (if the slope is greater than zero, there will be two complex roots and we do not want that).\nThe function with the minimum $a$\n\\[f(x)=\\left(x-\\frac{a}{3}\\right)^3\\] \\[x^3-ax^2+\\left(\\frac{a^2}{3}\\right)x-\\frac{a^3}{27}\\] Since this is equal to the original equation $x^3-ax^2+bx-a$ , equating the coefficients, we get that\n\\[\\frac{a^3}{27}=a\\rightarrow a^2=27\\rightarrow a=3\\sqrt{3}\\] \\[b=\\frac{a^2}{3}=\\frac{27}{3}=\\boxed{9}\\]",
"Let the roots of the polynomial be $r, s, t$ . By Vieta's formulas we have $r+s+t=a$ $rs+st+rt=b$ , and $rst=a$ . Since both $a$ and $b$ are positive, it follows that all 3 roots $r, s, t$ are positive as well, and so we can apply AM-GM to get \\[\\tfrac 13 (r+s+t) \\ge \\sqrt[3]{rst} \\quad \\Rightarrow \\quad a \\ge 3\\sqrt[3]{a}.\\] Cubing both sides and then dividing by $a$ (since $a$ is positive we can divide by $a$ and not change the sign of the inequality) yields \\[a^2 \\ge 27 \\quad \\Rightarrow \\quad a \\ge 3\\sqrt{3}.\\] Thus, the smallest possible value of $a$ is $3\\sqrt{3}$ which is achieved when all the roots are equal to $\\sqrt{3}$ . For this value of $a$ , we can use Vieta's to get $b=\\boxed{9}$",
"We see that with cubics, the number $3$ comes up a lot, and as $9=3\\cdot3$ has the most relation to $3$ , we can assume $b=\\boxed{9}$"
] |
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