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[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 3.5 | A domino is a 1-by-2 or 2-by-1 rectangle. A domino tiling of a region of the plane is a way of covering it (and only it) completely by nonoverlapping dominoes. For instance, there is one domino tiling of a 2-by-1 rectangle and there are 2 tilings of a 2-by-2 rectangle (one consisting of two horizontal dominoes and one consisting of two vertical dominoes). How many domino tilings are there of a 2-by-10 rectangle? | The number of tilings of a 2-by-$n$, rectangle is the $n$th Fibonacci number $F_{n}$, where $F_{0}=F_{1}=1$ and $F_{n}=F_{n-1}+F_{n-1}$ for $n \geq 2$. (This is not hard to show by induction.) The answer is 89. | 89 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Curves -> Other",
"Mathematics -> Algebra -> Equations and Inequalities -> Other"
] | 5 | Let $\mathcal{C}$ be the hyperbola $y^{2}-x^{2}=1$. Given a point $P_{0}$ on the $x$-axis, we construct a sequence of points $\left(P_{n}\right)$ on the $x$-axis in the following manner: let $\ell_{n}$ be the line with slope 1 passing through $P_{n}$, then $P_{n+1}$ is the orthogonal projection of the point of intersection of $\ell_{n}$ and $\mathcal{C}$ onto the $x$-axis. (If $P_{n}=0$, then the sequence simply terminates.) Let $N$ be the number of starting positions $P_{0}$ on the $x$-axis such that $P_{0}=P_{2008}$. Determine the remainder of $N$ when divided by 2008. | Let $P_{n}=\left(x_{n}, 0\right)$. Then the $\ell_{n}$ meet $\mathcal{C}$ at $\left(x_{n+1}, x_{n+1}-x_{n}\right)$. Since this point lies on the hyperbola, we have $\left(x_{n+1}-x_{n}\right)^{2}-x_{n+1}^{2}=1$. Rearranging this equation gives $$x_{n+1}=\frac{x_{n}^{2}-1}{2x_{n}}$$ Choose a $\theta_{0} \in(0, \pi)$ with $\cot \theta_{0}=x_{0}$, and define $\theta_{n}=2^{n} \theta_{0}$. Using the double-angle formula, we have $$\cot \theta_{n+1}=\cot \left(2 \theta_{n}\right)=\frac{\cot^{2} \theta_{n}-1}{2 \cot \theta_{n}}$$ It follows by induction that $x_{n}=\cot \theta_{n}$. Then, $P_{0}=P_{2008}$ corresponds to $\cot \theta_{0}=\cot \left(2^{2008} \theta_{0}\right)$ (assuming that $P_{0}$ is never at the origin, or equivalently, $2^{n} \theta$ is never an integer multiple of $\pi$ ). So, we need to find the number of $\theta_{0} \in(0, \pi)$ with the property that $2^{2008} \theta_{0}-\theta_{0}=k \pi$ for some integer $k$. We have $\theta_{0}=\frac{k \pi}{2^{2008}-1}$, so $k$ can be any integer between 1 and $2^{2008}-2$ inclusive (and note that since the denominator is odd, the sequence never terminates). It follows that the number of starting positions is $N=2^{2008}-2$. Finally, we need to compute the remainder when $N$ is divided by 2008. We have $2008=2^{3} \times 251$. Using Fermat's Little Theorem with 251, we get $2^{2008} \equiv\left(2^{250}\right)^{4} \cdot 256 \equiv 1^{4} \cdot 5=5(\bmod 251)$. So we have $N \equiv 3(\bmod 251)$ and $N \equiv-2(\bmod 8)$. Using Chinese Remainder Theorem, we get $N \equiv 254$ $(\bmod 2008)$. | 254 | HMMT_2 |
[
"Mathematics -> Number Theory -> Factorization"
] | 5 | For how many integers $a(1 \leq a \leq 200)$ is the number $a^{a}$ a square? | 107 If $a$ is even, we have $a^{a}=\left(a^{a / 2}\right)^{2}$. If $a$ is odd, $a^{a}=\left(a^{(a-1) / 2}\right)^{2} \cdot a$, which is a square precisely when $a$ is. Thus we have 100 even values of $a$ and 7 odd square values $\left(1^{2}, 3^{2}, \ldots, 13^{2}\right)$ for a total of 107. | 107 | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Other"
] | 5 | Find all values of $x$ that satisfy $x=1-x+x^{2}-x^{3}+x^{4}-x^{5}+\cdots$ (be careful; this is tricky). | Multiplying both sides by $1+x$ gives $(1+x) x=1$, or $x=\frac{-1 \pm \sqrt{5}}{2}$. However, the series only converges for $|x|<1$, so only the answer $x=\frac{-1+\sqrt{5}}{2}$ makes sense. | x=\frac{-1+\sqrt{5}}{2} | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 4.5 | An omino is a 1-by-1 square or a 1-by-2 horizontal rectangle. An omino tiling of a region of the plane is a way of covering it (and only it) by ominoes. How many omino tilings are there of a 2-by-10 horizontal rectangle? | There are exactly as many omino tilings of a 1-by-$n$ rectangle as there are domino tilings of a 2-by-$n$ rectangle. Since the rows don't interact at all, the number of omino tilings of an $m$-by-$n$ rectangle is the number of omino tilings of a 1-by-$n$ rectangle raised to the $m$ th power, $F_{n}^{m}$. The answer is $89^{2}=7921$. | 7921 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 4 | Let $a, b$, and $c$ be real numbers such that $a+b+c=100$, $ab+bc+ca=20$, and $(a+b)(a+c)=24$. Compute all possible values of $bc$. | We first expand the left-hand-side of the third equation to get $(a+b)(a+c)=a^{2}+ac+ab+bc=24$. From this, we subtract the second equation to obtain $a^{2}=4$, so $a=\pm 2$. If $a=2$, plugging into the first equation gives us $b+c=98$ and plugging into the second equation gives us $2(b+c)+bc=20 \Rightarrow 2(98)+bc=20 \Rightarrow bc=-176$. Then, if $a=-2$, plugging into the first equation gives us $b+c=102$, and plugging into the second equation gives us $-2(b+c)+bc=20 \Rightarrow -2(102)+bc=20 \Rightarrow bc=224$. Therefore, the possible values of $bc$ are $224,-176$. | 224, -176 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Algebra -> Intermediate Algebra -> Exponential Functions"
] | 3.5 | Draw a square of side length 1. Connect its sides' midpoints to form a second square. Connect the midpoints of the sides of the second square to form a third square. Connect the midpoints of the sides of the third square to form a fourth square. And so forth. What is the sum of the areas of all the squares in this infinite series? | The area of the first square is 1, the area of the second is $\frac{1}{2}$, the area of the third is $\frac{1}{4}$, etc., so the answer is $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots=2$. | 2 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics",
"Mathematics -> Geometry -> Plane Geometry -> Other (Chessboard problems) -> Other"
] | 5.25 | Let $n$ be the maximum number of bishops that can be placed on the squares of a $6 \times 6$ chessboard such that no two bishops are attacking each other. Let $k$ be the number of ways to put $n$ bishops on an $6 \times 6$ chessboard such that no two bishops are attacking each other. Find $n+k$. (Two bishops are considered to be attacking each other if they lie on the same diagonal. Equivalently, if we label the squares with coordinates $(x, y)$, with $1 \leq x, y \leq 6$, then the bishops on $(a, b)$ and $(c, d)$ are attacking each other if and only if $|a-c|=|b-d|$.) | Color the square with coordinates $(i, j)$ black if $i+j$ is odd and white otherwise, for all $1 \leq i, j \leq 6$. Looking at the black squares only, we note that there are six distinct diagonals which run upward and to the right, but that two of them consist only of a corner square; we cannot simultaneously place bishops on both of these corner squares. Consequently, we can place at most five bishops on black squares. (This can be achieved by placing bishops on $(1,2),(1,4),(6,1),(6,3),(6,5)$.) If there are five bishops on black squares, there must be exactly one bishop on one of the two black corner squares, $(6,1)$ and $(1,6)$ : suppose without loss of generality that we place a bishop on $(1,6)$. Then, exactly one of $(3,6)$ and $(1,4)$ must also contain a bishop, and there are 2 ways to place two bishops on the four remaining black squares that are not yet under attack. Thus, we have a total of $2 \cdot 2 \cdot 2$ possible placements on black squares. Similarly, there are at most 5 bishops which can be placed on white squares and $2^{3}$ ways to place them, so that $n=10$ and $k=2^{6}$. Finally, $n+k=10+2^{6}=74$. | 74 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 5.25 | Amy and Ben need to eat 1000 total carrots and 1000 total muffins. The muffins can not be eaten until all the carrots are eaten. Furthermore, Amy can not eat a muffin within 5 minutes of eating a carrot and neither can Ben. If Amy eats 40 carrots per minute and 70 muffins per minute and Ben eats 60 carrots per minute and 30 muffins per minute, what is the minimum number of minutes it will take them to finish the food? | Amy and Ben will continuously eat carrots, then stop (not necessarily at the same time), and continuously eat muffins until no food is left. Suppose that Amy and Ben finish eating the carrots in $T_{1}$ minutes and the muffins $T_{2}$ minutes later; we wish to find the minimum value of $T_{1}+T_{2}$. Furthermore, suppose Amy finishes eating the carrots at time $a_{1}$, and Ben does so at time $b_{1}$, so that $T_{1}=\max \left(a_{1}, b_{1}\right)$. First, suppose that $a_{1} \leq b_{1}$, and let $b_{1}-a_{1}=c$. We have $40\left(T_{1}-c\right)+60 T_{1}=1000$, so $T_{1}$ is minimized when $c=0$. Also, $30\left(T_{2}-5\right)+70\left(T_{2}-\max (5-c, 0)\right)=1000$. We see that $T_{1}+T_{2}$ is minimized when $c=5$, and $T_{1}+T_{2}=23.5$. In a similar way, we see that when $b_{1} \leq a_{1}, T_{1}+T_{2}>23.5$, so our answer is 23.5 . | 23.5 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Probability -> Other"
] | 4 | Another professor enters the same room and says, 'Each of you has to write down an integer between 0 and 200. I will then compute $X$, the number that is 3 greater than half the average of all the numbers that you will have written down. Each student who writes down the number closest to $X$ (either above or below $X$) will receive a prize.' One student, who misunderstood the question, announces to the class that he will write the number 107. If among the other 99 students it is common knowledge that all 99 of them will write down the best response, and there is no further communication between students, what single integer should each of the 99 students write down? | Use the same logic to get 7. Note 6 and 8 do not work. | 7 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other"
] | 4.5 | Let $p$ denote the proportion of teams, out of all participating teams, who submitted a negative response to problem 5 of the Team round (e.g. "there are no such integers"). Estimate $P=\lfloor 10000p\rfloor$. An estimate of $E$ earns $\max (0,\lfloor 20-|P-E|/20\rfloor)$ points. If you have forgotten, problem 5 of the Team round was the following: "Determine, with proof, whether there exist positive integers $x$ and $y$ such that $x+y, x^{2}+y^{2}$, and $x^{3}+y^{3}$ are all perfect squares." | Of the 88 teams competing in this year's Team round, 49 of them answered negatively, 9 (correctly) provided a construction, 16 answered ambiguously or did not provide a construction, and the remaining 14 teams did not submit to problem 5. Thus $p=\frac{49}{88} \approx 0.5568$. | 5568 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics",
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 5 | A lattice point in the plane is a point of the form $(n, m)$, where $n$ and $m$ are integers. Consider a set $S$ of lattice points. We construct the transform of $S$, denoted by $S^{\prime}$, by the following rule: the pair $(n, m)$ is in $S^{\prime}$ if and only if any of $(n, m-1),(n, m+1),(n-1, m)$, $(n+1, m)$, and $(n, m)$ is in $S$. How many elements are in the set obtained by successively transforming $\{(0,0)\} 14$ times? | Transforming it $k \geq 1$ times yields the 'diamond' of points $(n, m)$ such that $|n|+|m| \leq k$. The diamond contains $(k+1)^{2}+k^{2}$ lattice points (this can be seen by rotating the plane 45 degrees and noticing the lattice points in the transforms form two squares, one of which is contained in the other), so the answer is 421. | 421 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations",
"Mathematics -> Calculus -> Differential Calculus -> Applications of Derivatives",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Let $\ldots, a_{-1}, a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of positive integers satisfying the following relations: $a_{n}=0$ for $n<0, a_{0}=1$, and for $n \geq 1$, $a_{n}=a_{n-1}+2(n-1) a_{n-2}+9(n-1)(n-2) a_{n-3}+8(n-1)(n-2)(n-3) a_{n-4}$. Compute $\sum_{n \geq 0} \frac{10^{n} a_{n}}{n!}$ | Let $y=\sum_{n \geq 0} \frac{x^{n} a_{n}}{n!}$. Then $y^{\prime}=\left(1+2x+9x^{2}+8x^{3}\right) y$ by definition. So $y=C \exp \left(x+x^{2}+3x^{3}+2x^{4}\right)$. Take $x=0$ to get $C=1$. Take $x=10$ to get the answer. | e^{23110} | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 4 | How many ways are there of using diagonals to divide a regular 6-sided polygon into triangles such that at least one side of each triangle is a side of the original polygon and that each vertex of each triangle is a vertex of the original polygon? | The number of ways of triangulating a convex $(n+2)$-sided polygon is $\binom{2 n}{n} \frac{1}{n+1}$, which is 14 in this case. However, there are two triangulations of a hexagon which produce one triangle sharing no sides with the original polygon, so the answer is $14-2=12$. | 12 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5 | Three points, $A, B$, and $C$, are selected independently and uniformly at random from the interior of a unit square. Compute the expected value of $\angle A B C$. | Since $\angle A B C+\angle B C A+\angle C A B=180^{\circ}$ for all choices of $A, B$, and $C$, the expected value is $60^{\circ}$. | 60^{\circ} | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Other"
] | 4.5 | The Antarctican language has an alphabet of just 16 letters. Interestingly, every word in the language has exactly 3 letters, and it is known that no word's first letter equals any word's last letter (for instance, if the alphabet were $\{a, b\}$ then $a a b$ and aaa could not both be words in the language because $a$ is the first letter of a word and the last letter of a word; in fact, just aaa alone couldn't be in the language). Given this, determine the maximum possible number of words in the language. | 1024 Every letter can be the first letter of a word, or the last letter of a word, or possibly neither, but not both. If there are $a$ different first letters and $b$ different last letters, then we can form $a \cdot 16 \cdot b$ different words (and the desired conditions will be met). Given the constraints $0 \leq a, b ; a+b \leq 16$, this product is maximized when $a=b=8$, giving the answer. | 1024 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 4.5 | A deck of 100 cards is labeled $1,2, \ldots, 100$ from top to bottom. The top two cards are drawn; one of them is discarded at random, and the other is inserted back at the bottom of the deck. This process is repeated until only one card remains in the deck. Compute the expected value of the label of the remaining card. | Note that we can just take averages: every time you draw one of two cards, the EV of the resulting card is the average of the EVs of the two cards. This average must be of the form $$2^{\bullet} \cdot 1+2^{\bullet} \cdot 2+2^{\bullet} \cdot 3+\cdots+2^{\bullet} \cdot 100$$ where the $2^{\bullet}$ add up to 1. Clearly, the cards further down in the deck get involved in one less layer of averaging, and therefore 1 through 72 are weighted $2^{-7}$ while the rest are weighted $2^{-6}$. To compute the average now, we just add it up to get $\frac{467}{8}$. | \frac{467}{8} | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5 | Find the number of 20-tuples of integers $x_{1}, \ldots, x_{10}, y_{1}, \ldots, y_{10}$ with the following properties: - $1 \leq x_{i} \leq 10$ and $1 \leq y_{i} \leq 10$ for each $i$; - $x_{i} \leq x_{i+1}$ for $i=1, \ldots, 9$; - if $x_{i}=x_{i+1}$, then $y_{i} \leq y_{i+1}$. | By setting $z_{i}=10 x_{i}+y_{i}$, we see that the problem is equivalent to choosing a nondecreasing sequence of numbers $z_{1}, z_{2}, \ldots, z_{10}$ from the values $11,12, \ldots, 110$. Making a further substitution by setting $w_{i}=z_{i}-11+i$, we see that the problem is equivalent to choosing a strictly increasing sequence of numbers $w_{1}, \ldots, w_{10}$ from among the values $1,2, \ldots, 109$. There are $\binom{109}{10}$ ways to do this. | \binom{109}{10} | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5 | Two $4 \times 4$ squares are randomly placed on an $8 \times 8$ chessboard so that their sides lie along the grid lines of the board. What is the probability that the two squares overlap? | $529 / 625$. Each square has 5 horizontal $\cdot 5$ vertical $=25$ possible positions, so there are 625 possible placements of the squares. If they do not overlap, then either one square lies in the top four rows and the other square lies in the bottom four rows, or one square lies in the left four columns and the other lies in the right four columns. The first possibility can happen in $2 \cdot 5 \cdot 5=50$ ways (two choices of which square goes on top, and five horizontal positions for each square); likewise, so can the second. However, this double-counts the 4 cases in which the two squares are in opposite corners, so we have $50+50-4=96$ possible non-overlapping arrangements $\Rightarrow 25^{2}-96=529$ overlapping arrangements. | 529/625 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Define the sequence $b_{0}, b_{1}, \ldots, b_{59}$ by $$ b_{i}= \begin{cases}1 & \text { if } \mathrm{i} \text { is a multiple of } 3 \\ 0 & \text { otherwise }\end{cases} $$ Let \left\{a_{i}\right\} be a sequence of elements of \{0,1\} such that $$ b_{n} \equiv a_{n-1}+a_{n}+a_{n+1} \quad(\bmod 2) $$ for $0 \leq n \leq 59\left(a_{0}=a_{60}\right.$ and $\left.a_{-1}=a_{59}\right)$. Find all possible values of $4 a_{0}+2 a_{1}+a_{2}$. | Try the four possible combinations of values for $a_{0}$ and $a_{1}$. Since we can write $a_{n} \equiv$ $b_{n-1}-a_{n-2}-a_{n-1}$, these two numbers completely determine the solution $\left\{a_{i}\right\}$ beginning with them (if there is one). For $a_{0}=a_{1}=0$, we can check that the sequence beginning $0,0,0,0,1,1$ and repeating every 6 indices is a possible solution for $\left\{a_{i}\right\}$, so one possible value for $4 a_{0}+2 a_{1}+a_{2}$ is 0 . The other three combinations for $a_{0}$ and $a_{1}$ similarly lead to valid sequences (produced by repeating the sextuples $0,1,1,1,0,1 ; 1,0,1,1,1,0$; $1,1,0,1,0,1$, respectively); we thus obtain the values 3,5 , and 6. | 0, 3, 5, 6 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5 | Let triangle $A B C$ have $A B=5, B C=6$, and $A C=7$, with circumcenter $O$. Extend ray $A B$ to point $D$ such that $B D=5$, and extend ray $B C$ to point $E$ such that $O D=O E$. Find $C E$. | Because $O D=O E, D$ and $E$ have equal power with respect to the circle, so $(E C)(E B)=(D B)(D A)=50$. Letting $E C=x$, we have $x(x+6)=50$, and taking the positive root gives $x=\sqrt{59}-3$. | \sqrt{59}-3 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 4.5 | Let $N$ denote the sum of the decimal digits of $\binom{1000}{100}$. Estimate the value of $N$. | http://www.wolframalpha.com/input/?i=sum+of+digits+of $+\mathrm{nCr}(1000,100)$. To see this, one can estimate there are about 150 digits, and we expect the digits to be roughly random, for $150 \cdot 4.5 \approx 675$, which is already very close to the actual answer. The actual number of digits is 140, and here $140 \cdot 4.5=630$ is within 9 of the actual answer. | 621 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 4 | Let $P(x)$ be the monic polynomial with rational coefficients of minimal degree such that $\frac{1}{\sqrt{2}}$, $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{4}}, \ldots, \frac{1}{\sqrt{1000}}$ are roots of $P$. What is the sum of the coefficients of $P$? | For irrational $\frac{1}{\sqrt{r}},-\frac{1}{\sqrt{r}}$ must also be a root of $P$. Therefore $P(x)=\frac{\left(x^{2}-\frac{1}{2}\right)\left(x^{2}-\frac{1}{3}\right) \cdots\left(x^{2}-\frac{1}{1000}\right)}{\left(x+\frac{1}{2}\right)\left(x+\frac{1}{3}\right) \cdots\left(x+\frac{1}{31}\right)}$. We get the sum of the coefficients of $P$ by setting $x=1$, so we use telescoping to get $P(1)=\frac{\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{999}{1000}}{\frac{3}{2} \cdot \frac{4}{3} \cdots \frac{32}{31}}=\frac{1}{16000}$. | \frac{1}{16000} | HMMT_2 |
[
"Mathematics -> Algebra -> Prealgebra -> Absolute Values -> Other"
] | 5 | Suppose $a, b, c, d$ are real numbers such that $$|a-b|+|c-d|=99 ; \quad|a-c|+|b-d|=1$$ Determine all possible values of $|a-d|+|b-c|$. | 99 If $w \geq x \geq y \geq z$ are four arbitrary real numbers, then $|w-z|+|x-y|=$ $|w-y|+|x-z|=w+x-y-z \geq w-x+y-z=|w-x|+|y-z|$. Thus, in our case, two of the three numbers $|a-b|+|c-d|,|a-c|+|b-d|,|a-d|+|b-c|$ are equal, and the third one is less than or equal to these two. Since we have a 99 and a 1, the third number must be 99. | 99 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Logic",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | This question forms a three question multiple choice test. After each question, there are 4 choices, each preceded by a letter. Please write down your answer as the ordered triple (letter of the answer of Question \#1, letter of the answer of Question \#2, letter of the answer of Question \#3). If you find that all such ordered triples are logically impossible, then write 'no answer' as your answer. If you find more than one possible set of answers, then provide all ordered triples as your answer. When we refer to 'the correct answer to Question $X$ ' it is the actual answer, not the letter, to which we refer. When we refer to 'the letter of the correct answer to question $X$ ' it is the letter contained in parentheses that precedes the answer to which we refer. You are given the following condition: No two correct answers to questions on the test may have the same letter. Question 1. If a fourth question were added to this test, and if the letter of its correct answer were $(\mathrm{C})$, then: (A) This test would have no logically possible set of answers. (B) This test would have one logically possible set of answers. (C) This test would have more than one logically possible set of answers. (D) This test would have more than one logically possible set of answers. Question 2. If the answer to Question 2 were 'Letter (D)' and if Question 1 were not on this multiple-choice test (still keeping Questions 2 and 3 on the test), then the letter of the answer to Question 3 would be: (A) Letter (B) (B) Letter (C) (C) Letter $(\mathrm{D})$ (D) Letter $(\mathrm{A})$ Question 3. Let $P_{1}=1$. Let $P_{2}=3$. For all $i>2$, define $P_{i}=P_{i-1} P_{i-2}-P_{i-2}$. Which is a factor of $P_{2002}$ ? (A) 3 (B) 4 (C) 7 (D) 9 | (A, C, D). Question 2: In order for the answer to be consistent with the condition, 'If the answer to Question 2 were Letter (D),' the answer to this question actually must be 'Letter (D).' The letter of this answer is (C). Question 1: If a fourth question had an answer with letter (C), then at least two answers would have letter (C) (the answers to Questions 2 and 4). This is impossible. So, (A) must be the letter of the answer to Question 1. Question 3: If we inspect the sequence $P_{i}$ modulo 3, 4, 7, and 9 (the sequences quickly become periodic), we find that 3,7 , and 9 are each factors of $P_{2002}$. We know that letters (A) and (C) cannot be repeated, so the letter of this answer must be (D). | (A, C, D) | HMMT_2 |
[
"Mathematics -> Number Theory -> Divisibility -> Other",
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 6 | Find the least positive integer $N>1$ satisfying the following two properties: There exists a positive integer $a$ such that $N=a(2 a-1)$. The sum $1+2+\cdots+(N-1)$ is divisible by $k$ for every integer $1 \leq k \leq 10$. | The second condition implies that 16 divides $a(2 a-1)\left(2 a^{2}-a-1\right)$, which shows that $a \equiv 0$ or 1 modulo 16. The case $a=1$ would contradict the triviality-avoiding condition $N>1$. $a$ cannot be 16, because 7 does not divide $a(2 a-1)\left(2 a^{2}-a-1\right)$. a cannot be 17, because 9 does not divide $a(2 a-1)\left(2 a^{2}-a-1\right)$. It can be directly verified that $a=32$ is the smallest positive integer for which $1+2+\cdots+(N-1)=2^{4} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 13 \cdot 31$ which is divisible by $1,2, \ldots, 10$. For this $a$, we compute $N=32(2 \cdot 32-1)=2016$. | 2016 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 4 | How many sequences of 0s and 1s are there of length 10 such that there are no three 0s or 1s consecutively anywhere in the sequence? | We can have blocks of either 1 or 20s and 1s, and these blocks must be alternating between 0s and 1s. The number of ways of arranging blocks to form a sequence of length $n$ is the same as the number of omino tilings of a $1-b y-n$ rectangle, and we may start each sequence with a 0 or a 1, making $2 F_{n}$ or, in this case, 178 sequences. | 178 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 4 | Points $P$ and $Q$ are 3 units apart. A circle centered at $P$ with a radius of $\sqrt{3}$ units intersects a circle centered at $Q$ with a radius of 3 units at points $A$ and $B$. Find the area of quadrilateral APBQ. | The area is twice the area of triangle $A P Q$, which is isosceles with side lengths $3,3, \sqrt{3}$. By Pythagoras, the altitude to the base has length $\sqrt{3^{2}-(\sqrt{3} / 2)^{2}}=\sqrt{33} / 2$, so the triangle has area $\frac{\sqrt{99}}{4}$. Double this to get $\frac{3 \sqrt{11}}{2}$. | \frac{3 \sqrt{11}}{2} | HMMT_2 |
[
"Mathematics -> Number Theory -> Prime Numbers"
] | 5 | Let $N$ be a three-digit integer such that the difference between any two positive integer factors of $N$ is divisible by 3 . Let $d(N)$ denote the number of positive integers which divide $N$. Find the maximum possible value of $N \cdot d(N)$. | We first note that all the prime factors of $n$ must be 1 modulo 3 (and thus 1 modulo 6 ). The smallest primes with this property are $7,13,19, \ldots$ Since $7^{4}=2401>1000$, the number can have at most 3 prime factors (including repeats). Since $7 \cdot 13 \cdot 19=1729>1000$, the most factors $N$ can have is 6 . Consider the number $7^{2} \cdot 19=931$, which has 6 factors. For this choice of $N, N \cdot d(N)=5586$. For another $N$ to do better, it must have at least 6 factors, for otherwise, $N \cdot d(N)<1000 \cdot 5=5000$. It is easy to verify that $7^{2} \cdot 19$ is the greatest number with 6 prime factors satisfying our conditions, so the answer must be 5586 . | 5586 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5 | Consider the two hands of an analog clock, each of which moves with constant angular velocity. Certain positions of these hands are possible (e.g. the hour hand halfway between the 5 and 6 and the minute hand exactly at the 6), while others are impossible (e.g. the hour hand exactly at the 5 and the minute hand exactly at the 6). How many different positions are there that would remain possible if the hour and minute hands were switched? | 143 We can look at the twelve-hour cycle beginning at midnight and ending just before noon, since during this time, the clock goes through each possible position exactly once. The minute hand has twelve times the angular velocity of the hour hand, so if the hour hand has made $t$ revolutions from its initial position $(0 \leq t<1)$, the minute hand has made $12 t$ revolutions. If the hour hand were to have made $12 t$ revolutions, the minute hand would have made $144 t$. So we get a valid configuration by reversing the hands precisely when $144 t$ revolutions land the hour hand in the same place as $t$ revolutions - i.e. when $143 t=144 t-t$ is an integer, which clearly occurs for exactly 143 values of $t$ corresponding to distinct positions on the clock $(144-1=143)$. | 143 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 4 | Points $X$ and $Y$ are inside a unit square. The score of a vertex of the square is the minimum distance from that vertex to $X$ or $Y$. What is the minimum possible sum of the scores of the vertices of the square? | Let the square be $A B C D$. First, suppose that all four vertices are closer to $X$ than $Y$. Then, by the triangle inequality, the sum of the scores is $A X+B X+C X+D X \geq A B+C D=2$. Similarly, suppose exactly two vertices are closer to $X$ than $Y$. Here, we have two distinct cases: the vertices closer to $X$ are either adjacent or opposite. Again, by the Triangle Inequality, it follows that the sum of the scores of the vertices is at least 2 . On the other hand, suppose that $A$ is closer to $X$ and $B, C, D$ are closer to $Y$. We wish to compute the minimum value of $A X+B Y+C Y+D Y$, but note that we can make $X=A$ to simply minimize $B Y+C Y+D Y$. We now want $Y$ to be the Fermat point of triangle $B C D$, so that \measuredangle B Y C=$ \measuredangle C Y D=\measuredangle D Y B=120^{\circ}$. Note that by symmetry, we must have \measuredangle B C Y=\measuredangle D C Y=45^{\circ}$, so \measuredangle C B Y=\measuredangle C D Y=15^{\circ}$ And now we use the law of sines: $B Y=D Y=\frac{\sin 45^{\circ}}{\sin 120^{\circ}}$ and $C Y=\frac{\sin 15^{\circ}}{\sin 120^{\circ}}$. Now, we have $B Y+C Y+$ $D Y=\frac{\sqrt{2}+\sqrt{6}}{2}$, which is less than 2 , so this is our answer. | \frac{\sqrt{6}+\sqrt{2}}{2} | HMMT_2 |
[
"Mathematics -> Calculus -> Integral Calculus -> Techniques of Integration -> Single-variable",
"Mathematics -> Algebra -> Intermediate Algebra -> Other"
] | 5 | Compute $$\sum_{n=1}^{\infty} \frac{2 n+5}{2^{n} \cdot\left(n^{3}+7 n^{2}+14 n+8\right)}$$ | First, we manipulate using partial fractions and telescoping: $$\begin{aligned} \sum_{n=1}^{\infty} \frac{2 n+5}{2^{n} \cdot\left(n^{3}+7 n^{2}+14 n+8\right)} & =\frac{1}{2} \cdot \sum_{n=1}^{\infty} \frac{1}{2^{n}}\left(\frac{2}{n+1}-\frac{1}{n+2}-\frac{1}{n+4}\right) \\ & =\frac{1}{4}-\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{2^{n} \cdot(n+4)} \end{aligned}$$ Now, consider the function $f(r, k):=\sum_{n=1}^{\infty} \frac{r^{n}}{n^{k}}$. We have $$\begin{aligned} \frac{\partial f(r, k)}{\partial r} & =\frac{\partial}{\partial r} \sum_{n=1}^{\infty} \frac{r^{n}}{n^{k}}=\sum_{n=1}^{\infty} \frac{\partial}{\partial r}\left[\frac{r^{n}}{n^{k}}\right]=\sum_{n=1}^{\infty} \frac{r^{n-1}}{n^{k-1}}=\frac{1}{r} f(r, k-1) \\ \frac{d f(r, 1)}{d r} & =\frac{1}{r} \sum_{n=1}^{\infty} \frac{r^{n}}{n^{0}}=\frac{1}{r} \cdot \frac{r}{1-r}=\frac{1}{1-r} \\ f(r, 1) & =\int \frac{d r}{1-r}=-\ln (1-r)+f(0,1) \end{aligned}$$ By inspection, $f(0,1)=0$, so $f\left(\frac{1}{2}, 1\right)=\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^{n}}=\ln (2)$. It is easy to compute the desired sum in terms of $f\left(\frac{1}{2}, 1\right)$, and we find $\sum_{n=1}^{\infty} \frac{1}{2^{n}(n+4)}=16 \ln (2)-\frac{131}{12}$. Hence, our final answer is $\frac{137}{24}-8 \ln (2)$. | \frac{137}{24}-8 \ln 2 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Call the pentominoes found in the last problem square pentominoes. Just like dominos and ominos can be used to tile regions of the plane, so can square pentominoes. In particular, a square pentomino tiling of a region of the plane is a way of covering it (and only it) completely by nonoverlapping square pentominoes. How many square pentomino tilings are there of a 12-by-12 rectangle? | Since 5 does not divide 144, there are 0. | 0 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Let $S_{7}$ denote all the permutations of $1,2, \ldots, 7$. For any \pi \in S_{7}$, let $f(\pi)$ be the smallest positive integer $i$ such that \pi(1), \pi(2), \ldots, \pi(i)$ is a permutation of $1,2, \ldots, i$. Compute \sum_{\pi \in S_{7}} f(\pi)$. | Extend the definition of $f$ to apply for any permutation of $1,2, \ldots, n$, for any positive integer $n$. For positive integer $n$, let $g(n)$ denote the number of permutations \pi$ of $1,2, \ldots, n$ such that $f(\pi)=n$. We have $g(1)=1$. For fixed $n, k$ (with $k \leq n$ ), the number of permutations \pi$ of $1,2, \ldots, n$ such that $f(\pi)=k$ is $g(k)(n-k)$ !. This gives us the recursive formula $g(n)=$ $n!-\sum_{k=1}^{n-1} g(k)(n-k)$ !. Using this formula, we find that the first 7 values of $g$ are $1,1,3,13,71,461,3447$. Our sum is then equal to \sum_{k=1}^{7} k \cdot g(k)(7-k)$ !. Using our computed values of $g$, we get that the sum evaluates to 29093 . | 29093 | HMMT_2 |
[
"Mathematics -> Number Theory -> Prime Numbers"
] | 5 | Compute the prime factorization of 1007021035035021007001. | The number in question is $$\sum_{i=0}^{7}\binom{7}{i} 1000^{i}=(1000+1)^{7}=1001^{7}=7^{7} \cdot 11^{7} \cdot 13^{7}$$ | 7^{7} \cdot 11^{7} \cdot 13^{7} | HMMT_2 |
[
"Mathematics -> Algebra -> Abstract Algebra -> Other"
] | 5 | A sequence $s_{0}, s_{1}, s_{2}, s_{3}, \ldots$ is defined by $s_{0}=s_{1}=1$ and, for every positive integer $n, s_{2 n}=s_{n}, s_{4 n+1}=s_{2 n+1}, s_{4 n-1}=s_{2 n-1}+s_{2 n-1}^{2} / s_{n-1}$. What is the value of $s_{1000}$? | 720 Some experimentation with small values may suggest that $s_{n}=k$!, where $k$ is the number of ones in the binary representation of $n$, and this formula is in fact provable by a straightforward induction. Since $1000_{10}=1111101000_{2}$, with six ones, $s_{1000}=6!=720$. | 720 | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Exponential Functions",
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 4 | Find the set consisting of all real values of $x$ such that the three numbers $2^{x}, 2^{x^{2}}, 2^{x^{3}}$ form a non-constant arithmetic progression (in that order). | The empty set, $\varnothing$. Trivially, $x=0,1$ yield constant arithmetic progressions; we show that there are no other possibilities. If these numbers do form a progression, then, by the AM-GM (arithmetic mean-geometric mean) inequality, $$2 \cdot 2^{x^{2}}=2^{x}+2^{x^{3}} \geq 2 \sqrt{2^{x} \cdot 2^{x^{3}}} \Rightarrow 2^{x^{2}} \geq 2^{\left(x+x^{3}\right) / 2} \Rightarrow x^{2} \geq\left(x+x^{3}\right) / 2 \Rightarrow x(x-1)^{2}=x^{3}-2 x^{2}+x \leq 0$$ Assuming $x \neq 0,1$, we can divide by $(x-1)^{2}>0$ and obtain $x<0$. However, then $2^{x}, 2^{x^{3}}$ are less than 1, while $2^{x^{2}}$ is more than 1, so the given sequence cannot possibly be an arithmetic progression. | \varnothing | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Probability -> Other"
] | 5.25 | Ash and Gary independently come up with their own lineups of 15 fire, grass, and water monsters. Then, the first monster of both lineups will fight, with fire beating grass, grass beating water, and water beating fire. The defeated monster is then substituted with the next one from their team's lineup; if there is a draw, both monsters get defeated. Gary completes his lineup randomly, with each monster being equally likely to be any of the three types. Without seeing Gary's lineup, Ash chooses a lineup that maximizes the probability $p$ that his monsters are the last ones standing. Compute $p$. | First, we show Ash cannot do better. Notice there is a $\frac{2^{15}}{3^{15}}$ chance that Gary's $i$-th monster ties or defeats Ash's $i$-th monster for each $i$. If this is the case, Ash cannot win, as Ash's $i$-th monster will always be defeated by Gary's $i$-th monster, if not sooner. Thus, Ash wins with probability at most $1-\frac{2^{15}}{3^{15}}$. It remains to show this is achievable. Ash uses the lineup fire-grass-water repeated 5 times. Then, none of Gary's monsters can defeat more than one monster in Ash's lineup, so Ash will win unless Gary manages to take down exactly one monster with each of his. In particular, this means the $i$-th monster Gary has must tie or defeat Ash's $i$-th monster, which occurs with $\frac{2}{3}$ chance with each $i$. Thus this construction achieves the answer of $1-\frac{2^{15}}{3^{15}}$. | 1-\frac{2^{15}}{3^{15}} | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 3.5 | Divide an $m$-by-$n$ rectangle into $m n$ nonoverlapping 1-by-1 squares. A polyomino of this rectangle is a subset of these unit squares such that for any two unit squares $S, T$ in the polyomino, either (1) $S$ and $T$ share an edge or (2) there exists a positive integer $n$ such that the polyomino contains unit squares $S_{1}, S_{2}, S_{3}, \ldots, S_{n}$ such that $S$ and $S_{1}$ share an edge, $S_{n}$ and $T$ share an edge, and for all positive integers $k<n, S_{k}$ and $S_{k+1}$ share an edge. We say a polyomino of a given rectangle spans the rectangle if for each of the four edges of the rectangle the polyomino contains a square whose edge lies on it. What is the minimum number of unit squares a polyomino can have if it spans a 128-by343 rectangle? | To span an $a \times b$ rectangle, we need at least $a+b-1$ squares. Indeed, consider a square of the polyomino bordering the left edge of the rectangle and one bordering the right edge. There exists a path connecting these squares; suppose it runs through $c$ different rows. Then the path requires at least $b-1$ horizontal and $c-1$ vertical steps, so it uses at least $b+c-1$ different squares. However, since the polyomino also hits the top and bottom edges of the rectangle, it must run into the remaining $a-c$ rows as well, so altogether we need at least $a+b-1$ squares. On the other hand, this many squares suffice - just consider all the squares bordering the lower or right edges of the rectangle. So, in our case, the answer is $128+343-1=470$. | 470 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 3.5 | The Dyslexian alphabet consists of consonants and vowels. It so happens that a finite sequence of letters is a word in Dyslexian precisely if it alternates between consonants and vowels (it may begin with either). There are 4800 five-letter words in Dyslexian. How many letters are in the alphabet? | 12 Suppose there are $c$ consonants, $v$ vowels. Then there are $c \cdot v \cdot c \cdot v \cdot c+$ $v \cdot c \cdot v \cdot c \cdot v=(c v)^{2}(c+v)$ five-letter words. Thus, $c+v=4800 /(c v)^{2}=3 \cdot(40 / c v)^{2}$, so $c v$ is a divisor of 40. If $c v \leq 10$, we have $c+v \geq 48$, impossible for $c, v$ integers; if $c v=40$, then $c+v=3$ which is again impossible. So $c v=20$, giving $c+v=12$, the answer. As a check, this does have integer solutions: $(c, v)=(2,10)$ or $(10,2)$. | 12 | HMMT_2 |
[
"Mathematics -> Precalculus -> Trigonometric Functions"
] | 3.5 | Find all values of $x$ with $0 \leq x<2 \pi$ that satisfy $\sin x+\cos x=\sqrt{2}$. | Squaring both sides gives $\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=1+\sin 2 x=2$, so $x=\frac{\pi}{4}, \frac{5 \pi}{4}$. | x=\frac{\pi}{4}, \frac{5 \pi}{4} | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 6 | Let $P(x)=x^{3}+a x^{2}+b x+2015$ be a polynomial all of whose roots are integers. Given that $P(x) \geq 0$ for all $x \geq 0$, find the sum of all possible values of $P(-1)$. | Since all the roots of $P(x)$ are integers, we can factor it as $P(x)=(x-r)(x-s)(x-t)$ for integers $r, s, t$. By Viete's formula, the product of the roots is $r s t=-2015$, so we need three integers to multiply to -2015. $P(x)$ cannot have two distinct positive roots $u, v$ since otherwise, $P(x)$ would be negative at least in some infinitesimal region $x<u$ or $x>v$, or $P(x)<0$ for $u<x<v$. Thus, in order to have two positive roots, we must have a double root. Since $2015=5 \times 13 \times 31$, the only positive double root is a perfect square factor of 2015, which is at $x=1$, giving us a possibility of $P(x)=(x-1)^{2}(x+2015)$. Now we can consider when $P(x)$ only has negative roots. The possible unordered triplets are $(-1,-1,-2015),(-1,-5,-(-1,-31,-65),(-5,-13,-31)$ which yield the polynomials $(x+1)^{2}(x+2015),(x+1)(x+5)(x+403),(x+1)(x+13)(x+155),(x+1)(x+31)(x+65),(x+5)(x+13)(x+31)$, respectively. Noticing that $P(-1)=0$ for four of these polynomials, we see that the nonzero values are $P(-1)=(-1-1)^{2}(2014),(5-1)(13-1)(31-1)$, which sum to $8056+1440=9496$. | 9496 | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers",
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5.25 | Over all pairs of complex numbers $(x, y)$ satisfying the equations $$x+2y^{2}=x^{4} \quad \text{and} \quad y+2x^{2}=y^{4}$$ compute the minimum possible real part of $x$. | Note the following observations: (a) if $(x, y)$ is a solution then $(\omega x, \omega^{2} y)$ is also a solution if $\omega^{3}=1$ and $\omega \neq 1$. (b) we have some solutions $(x, x)$ where $x$ is a solution of $x^{4}-2x^{2}-x=0$. These are really the only necessary observations and the first does not need to be noticed immediately. Indeed, we can try to solve this directly as follows: first, from the first equation, we get $y^{2}=\frac{1}{2}(x^{4}-x)$, so inserting this into the second equation gives $$\begin{aligned} \frac{1}{4}(x^{4}-x)^{2}-2x^{2} & =y \\ \left((x^{4}-x)^{2}-8x^{2}\right)^{2}-8x^{4}+8x & =0 \\ x^{16}+\cdots+41x^{4}+8x & =0 \end{aligned}$$ By the second observation, we have that $x(x^{3}-2x-1)$ should be a factor of $P$. The first observation gives that $(x^{3}-2\omega x-1)(x^{3}-2\omega^{2} x-1)$ should therefore also be a factor. Now $(x^{3}-2\omega x-1)(x^{3}-2\omega^{2} x-1)=x^{6}+2x^{4}-2x^{3}+4x^{2}-2x+1$ since $\omega$ and $\omega^{2}$ are roots of $x^{2}+x+1$. So now we see that the last two terms of the product of all of these is $-5x^{4}-x$. Hence the last two terms of the polynomial we get after dividing out should be $-x^{3}-8$, and given what we know about the degree and the fact that everything is monic, the quotient must be exactly $x^{6}-x^{3}-8$ which has roots being the cube roots of the roots to $x^{2}-x-8$, which are $\sqrt[3]{\frac{1 \pm \sqrt{33}}{2}}$. Now $x^{3}-2x-1$ is further factorable as $(x-1)(x^{2}-x-1)$ with roots $1, \frac{1 \pm \sqrt{5}}{2}$ so it is not difficult to compare the real parts of all roots of $P$, especially since 5 are real and non-zero, and we have that $\operatorname{Re}(\omega x)=-\frac{1}{2} x$ if $x \in \mathbb{R}$. We conclude that the smallest is $\sqrt[3]{\frac{1-\sqrt{33}}{2}}$. | \sqrt[3]{\frac{1-\sqrt{33}}{2}} | HMMT_2 |
[
"Mathematics -> Number Theory -> Prime Numbers"
] | 4 | Define $\varphi^{k}(n)$ as the number of positive integers that are less than or equal to $n / k$ and relatively prime to $n$. Find $\phi^{2001}\left(2002^{2}-1\right)$. (Hint: $\phi(2003)=2002$.) | $\varphi^{2001}\left(2002^{2}-1\right)=\varphi^{2001}(2001 \cdot 2003)=$ the number of $m$ that are relatively prime to both 2001 and 2003, where $m \leq 2003$. Since $\phi(n)=n-1$ implies that $n$ is prime, we must only check for those $m$ relatively prime to 2001, except for 2002, which is relatively prime to $2002^{2}-1$. So $\varphi^{2001}\left(2002^{2}-1\right)=\varphi(2001)+1=\varphi(3 \cdot 23 \cdot 29)+1=$ $(3-1)(23-1)(29-1)+1=1233$. | 1233 | HMMT_2 |
[
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Let $\ell$ and $m$ be two non-coplanar lines in space, and let $P_{1}$ be a point on $\ell$. Let $P_{2}$ be the point on $m$ closest to $P_{1}, P_{3}$ be the point on $\ell$ closest to $P_{2}, P_{4}$ be the point on $m$ closest to $P_{3}$, and $P_{5}$ be the point on $\ell$ closest to $P_{4}$. Given that $P_{1} P_{2}=5, P_{2} P_{3}=3$, and $P_{3} P_{4}=2$, compute $P_{4} P_{5}$. | Let $a$ be the answer. By taking the $z$-axis to be the cross product of these two lines, we can let the lines be on the planes $z=0$ and $z=h$, respectively. Then, by projecting onto the $xy$-plane, we get the above diagram. The projected lengths of the first four segments are $\sqrt{25-h^{2}}, \sqrt{9-h^{2}}$, and $\sqrt{4-h^{2}}$, and $\sqrt{a^{2}-h^{2}}$. By similar triangles, these lengths must form a geometric progression. Therefore, $25-h^{2}$, $9-h^{2}, 4-h^{2}, a^{2}-h^{2}$ is a geometric progression. By taking consecutive differences, $16,5,4-a^{2}$ is a geometric progression. Hence, $4-a^{2}=\frac{25}{16} \Longrightarrow a=\frac{\sqrt{39}}{4}$. | \frac{\sqrt{39}}{4} | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Decimal Operations -> Other"
] | 4 | Compute the decimal expansion of \sqrt{\pi}$. Your score will be \min (23, k)$, where $k$ is the number of consecutive correct digits immediately following the decimal point in your answer. | For this problem, it is useful to know the following square root algorithm that allows for digit-by-digit extraction of \sqrt{x}$ and gives one decimal place of \sqrt{x}$ for each two decimal places of $x$. We will illustrate how to extract the second digit after the decimal point of \sqrt{\pi}$, knowing that \pi=3.1415 \cdots$ and \sqrt{\pi}=1.7 \cdots$. Let $d$ be the next decimal digit. Then $d$ should be the largest digit such that $(1.7+0.01 d)^{2}<\pi$, which in this case we will treat as $(1.7+0.01 d)^{2}<3.1415$. Expanding this, we get $2.89+0.034 d+0.0001 d^{2}<$ 3.1415, from which we get the value of $d$ to be approximately \left\lfloor\frac{3.1415-2.89}{0.034}\right\rfloor=\left\lfloor\frac{0.2515}{0.034}\right\rfloor=7$, since the $0.0001 d^{2}$ term is negligible. Indeed, 7 is the largest such digit, and so $d=7$ is the second digit of \sqrt{\pi}$. Because we are constantly subtracting the square of our extracted answer so far, we can record the difference in a manner similar to long division, which yields a quick method of extracting square roots by hand. | 1.77245385090551602729816 \ldots | HMMT_2 |
[
"Mathematics -> Algebra -> Prealgebra -> Absolute Values -> Other"
] | 5.25 | How many real solutions are there to the equation $|||| x|-2|-2|-2|=|||| x|-3|-3|-3|$? | 6. The graphs of the two sides of the equation can be graphed on the same plot to reveal six intersection points. | 6 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Bob Barker went back to school for a PhD in math, and decided to raise the intellectual level of The Price is Right by having contestants guess how many objects exist of a certain type, without going over. The number of points you will get is the percentage of the correct answer, divided by 10, with no points for going over (i.e. a maximum of 10 points). Let's see the first object for our contestants...a table of shape (5,4,3,2,1) is an arrangement of the integers 1 through 15 with five numbers in the top row, four in the next, three in the next, two in the next, and one in the last, such that each row and each column is increasing (from left to right, and top to bottom, respectively). For instance: \begin{tabular}{lcccc} 1 & 2 & 3 & 4 & 5 \\ 6 & 7 & 8 & 9 & \\ 10 & 11 & 12 & & \\ 13 & 14 & & & \\ 15 & & & & \end{tabular} is one table. How many tables are there? | $15!/\left(3^{4} \cdot 5^{3} \cdot 7^{2} \cdot 9\right)=292864$. These are Standard Young Tableaux. | 292864 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 3.5 | Find all numbers $n$ with the following property: there is exactly one set of 8 different positive integers whose sum is $n$. | The sum of 8 different positive integers is at least $1+2+3+\cdots+8=36$, so we must have $n \geq 36$. Now $n=36$ satisfies the desired property, since in this case we must have equality - the eight numbers must be $1, \ldots, 8$. And if $n=37$ the eight numbers must be $1,2, \ldots, 7,9$ : if the highest number is 8 then the sum is $36<n$, while if the highest number is more than 9 the sum is $>1+2+\cdots+7+9=37=n$. So the highest number must be 9 , and then the remaining numbers must be $1,2, \ldots, 7$. Thus $n=37$ also has the desired property. However, no other values of $n$ work: if $n>37$ then $\{1,2,3, \ldots, 7, n-28\}$ and $\{1,2, \ldots, 6,8, n-29\}$ are both sets of 8 distinct positive integers whose sum is $n$. So $n=36,37$ are the only solutions. | 36, 37 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 3.5 | Points $A, B, C$ in the plane satisfy $\overline{A B}=2002, \overline{A C}=9999$. The circles with diameters $A B$ and $A C$ intersect at $A$ and $D$. If $\overline{A D}=37$, what is the shortest distance from point $A$ to line $B C$? | $\angle A D B=\angle A D C=\pi / 2$ since $D$ lies on the circles with $A B$ and $A C$ as diameters, so $D$ is the foot of the perpendicular from $A$ to line $B C$, and the answer is the given 37. | 37 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 4.5 | A $5 \times 5$ square grid has the number -3 written in the upper-left square and the number 3 written in the lower-right square. In how many ways can the remaining squares be filled in with integers so that any two adjacent numbers differ by 1, where two squares are adjacent if they share a common edge (but not if they share only a corner)? | 250 If the square in row $i$, column $j$ contains the number $k$, let its 'index' be $i+j-k$. The constraint on adjacent squares now says that if a square has index $r$, the squares to its right and below it each have index $r$ or $r+2$. The upper-left square has index 5, and the lower-right square has index 7, so every square must have index 5 or 7. The boundary separating the two types of squares is a path consisting of upward and rightward steps; it can be extended along the grid's border so as to obtain a path between the lower-left and upper-right corners. Conversely, any such path uniquely determines each square's index and hence the entire array of numbers - except that the two paths lying entirely along the border of the grid fail to separate the upper-left from the lower-right square and thus do not create valid arrays (since these two squares should have different indices). Each path consists of 5 upward and 5 rightward steps, so there are $\binom{10}{5}=252$ paths, but two are impossible, so the answer is 250. | 250 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 5 | Cyclic quadrilateral $A B C D$ has side lengths $A B=1, B C=2, C D=3$ and $D A=4$. Points $P$ and $Q$ are the midpoints of $\overline{B C}$ and $\overline{D A}$. Compute $P Q^{2}$. | Construct $\overline{A C}, \overline{A Q}, \overline{B Q}, \overline{B D}$, and let $R$ denote the intersection of $\overline{A C}$ and $\overline{B D}$. Because $A B C D$ is cyclic, we have that $\triangle A B R \sim \triangle D C R$ and $\triangle A D R \sim \triangle B C R$. Thus, we may write $A R=4 x, B R=2 x, C R=6 x, D R=12 x$. Now, Ptolemy applied to $A B C D$ yields $140 x^{2}=1 \cdot 3+2 \cdot 4=11$. Now $\overline{B Q}$ is a median in triangle $A B D$. Hence, $B Q^{2}=\frac{2 B A^{2}+2 B D^{2}-A D^{2}}{4}$. Likewise, $C Q^{2}=\frac{2 C A^{2}+2 C D^{2}-D A^{2}}{4}$. But $P Q$ is a median in triangle $B Q C$, so $P Q^{2}=\frac{2 B Q^{2}+2 C Q^{2}-B C^{2}}{4}=\frac{A B^{2}+B D^{2}+C D^{2}+C A^{2}-B C^{2}-A D^{2}}{4}=$ $\frac{(196+100) x^{2}+1^{2}+3^{2}-2^{2}-4^{2}}{4}=\frac{148 x^{2}-5}{2}=\frac{148 \cdot \frac{11}{140}-5}{2}=\frac{116}{35}$. | \frac{116}{35} | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Cubic Functions -> Other"
] | 5 | The unknown real numbers $x, y, z$ satisfy the equations $$\frac{x+y}{1+z}=\frac{1-z+z^{2}}{x^{2}-x y+y^{2}} ; \quad \frac{x-y}{3-z}=\frac{9+3 z+z^{2}}{x^{2}+x y+y^{2}}$$ Find $x$. | $\sqrt[3]{14}$ Cross-multiplying in both equations, we get, respectively, $x^{3}+y^{3}=$ $1+z^{3}, x^{3}-y^{3}=27-z^{3}$. Now adding gives $2 x^{3}=28$, or $x=\sqrt[3]{14}$. | \sqrt[3]{14} | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 3.5 | Suppose $x$ and $y$ are real numbers such that $-1<x<y<1$. Let $G$ be the sum of the geometric series whose first term is $x$ and whose ratio is $y$, and let $G^{\prime}$ be the sum of the geometric series whose first term is $y$ and ratio is $x$. If $G=G^{\prime}$, find $x+y$. | We note that $G=x /(1-y)$ and $G^{\prime}=y /(1-x)$. Setting them equal gives $x /(1-y)=$ $y /(1-x) \Rightarrow x^{2}-x=y^{2}-x \Rightarrow(x+y-1)(x-y)=0$, so we get that $x+y-1=0 \Rightarrow x+y=1$. | 1 | HMMT_2 |
[
"Mathematics -> Number Theory -> Factorization",
"Mathematics -> Algebra -> Equations and Inequalities -> Other"
] | 4 | How many pairs of integers $(a, b)$, with $1 \leq a \leq b \leq 60$, have the property that $b$ is divisible by $a$ and $b+1$ is divisible by $a+1$? | The divisibility condition is equivalent to $b-a$ being divisible by both $a$ and $a+1$, or, equivalently (since these are relatively prime), by $a(a+1)$. Any $b$ satisfying the condition is automatically $\geq a$, so it suffices to count the number of values $b-a \in$ $\{1-a, 2-a, \ldots, 60-a\}$ that are divisible by $a(a+1)$ and sum over all $a$. The number of such values will be precisely $60 /[a(a+1)]$ whenever this quantity is an integer, which fortunately happens for every $a \leq 5$; we count: $a=1$ gives 30 values of $b ;$ $a=2$ gives 10 values of $b ;$ $a=3$ gives 5 values of $b$; $a=4$ gives 3 values of $b$; $a=5$ gives 2 values of $b$; $a=6$ gives 2 values ($b=6$ or 48); any $a \geq 7$ gives only one value, namely $b=a$, since $b>a$ implies $b \geq a+a(a+1)>60$. Adding these up, we get a total of 106 pairs. | 106 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | FemtoPravis is walking on an $8 \times 8$ chessboard that wraps around at its edges (so squares on the left edge of the chessboard are adjacent to squares on the right edge, and similarly for the top and bottom edges). Each femtosecond, FemtoPravis moves in one of the four diagonal directions uniformly at random. After 2012 femtoseconds, what is the probability that FemtoPravis is at his original location? | We note the probability that he ends up in the same row is equal to the probability that he ends up in the same column by symmetry. Clearly these are independent, so we calculate the probability that he ends up in the same row. Now we number the rows $0-7$ where 0 and 7 are adjacent. Suppose he starts at row 0 . After two more turns, the probability he is in row 2 (or row 6 ) is \frac{1}{4}$, and the probability he is in row 0 again is \frac{1}{2}$. Let $a_{n}, b_{n}, c_{n}$ and $d_{n}$ denote the probability he is in row $0,2,4,6$ respectively after $2 n$ moves. We have $a_{0}=1$, and for $n \geq 0$ we have the following equations: $$ \begin{aligned} & a_{n+1}=\frac{1}{2} a_{n}+\frac{1}{4} b_{n}+\frac{1}{4} d_{n} \\ & b_{n+1}=\frac{1}{2} b_{n}+\frac{1}{4} a_{n}+\frac{1}{4} c_{n} \\ & c_{n+1}=\frac{1}{2} c_{n}+\frac{1}{4} b_{n}+\frac{1}{4} d_{n} \\ & d_{n+1}=\frac{1}{2} d_{n}+\frac{1}{4} a_{n}+\frac{1}{4} c_{n} \end{aligned} $$ From which we get the following equations: $$ \begin{gathered} a_{n}+c_{n}=\frac{1}{2} \\ x_{n}=a_{n}-c_{n}=\frac{1}{2}\left(a_{n-1}-c_{n-1}\right)=\frac{x_{n-1}}{2} \end{gathered} $$ So $$ \begin{gathered} a_{1006}+c_{1006}=\frac{1}{2} \\ x_{0}=1, x_{1006}=\frac{1}{2^{1006}} \\ a_{1006}=\frac{1+2^{1005}}{2^{1007}} \end{gathered} $$ And thus the answer is \left(\frac{1+2^{1005}}{2^{1007}}\right)^{2}$. | \left(\frac{1+2^{1005}}{2^{1007}}\right)^{2} | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Geometry -> Plane Geometry -> Circles"
] | 5 | Tessa has a figure created by adding a semicircle of radius 1 on each side of an equilateral triangle with side length 2, with semicircles oriented outwards. She then marks two points on the boundary of the figure. What is the greatest possible distance between the two points? | Note that both points must be in different semicircles to reach the maximum distance. Let these points be $M$ and $N$, and $O_{1}$ and $O_{2}$ be the centers of the two semicircles where they lie respectively. Then $$M N \leq M O_{1}+O_{1} O_{2}+O_{2} N$$ Note that the the right side will always be equal to 3 ($M O_{1}=O_{2} N=1$ from the radius condition, and $O_{1} O_{2}=1$ from being a midline of the equilateral triangle), hence $M N$ can be at most 3. Finally, if the four points are collinear (when $M$ and $N$ are defined as the intersection of line $O_{1} O_{2}$ with the two semicircles), then equality will hold. Therefore, the greatest possible distance between $M$ and $N$ is 3. | 3 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 4.5 | $A B C$ is a triangle with points $E, F$ on sides $A C, A B$, respectively. Suppose that $B E, C F$ intersect at $X$. It is given that $A F / F B=(A E / E C)^{2}$ and that $X$ is the midpoint of $B E$. Find the ratio $C X / X F$. | Let $x=A E / E C$. By Menelaus's theorem applied to triangle $A B E$ and line $C X F$, $$1=\frac{A F}{F B} \cdot \frac{B X}{X E} \cdot \frac{E C}{C A}=\frac{x^{2}}{x+1}$$ Thus, $x^{2}=x+1$, and $x$ must be positive, so $x=(1+\sqrt{5}) / 2$. Now apply Menelaus to triangle $A C F$ and line $B X E$, obtaining $$1=\frac{A E}{E C} \cdot \frac{C X}{X F} \cdot \frac{F B}{B A}=\frac{C X}{X F} \cdot \frac{x}{x^{2}+1}$$ so $C X / X F=\left(x^{2}+1\right) / x=\left(2 x^{2}-x\right) / x=2 x-1=\sqrt{5}$. | \sqrt{5} | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5.5 | Let $A B C$ be an equilateral triangle with side length 1. Points $D, E, F$ lie inside triangle $A B C$ such that $A, E, F$ are collinear, $B, F, D$ are collinear, $C, D, E$ are collinear, and triangle $D E F$ is equilateral. Suppose that there exists a unique equilateral triangle $X Y Z$ with $X$ on side $\overline{B C}, Y$ on side $\overline{A B}$, and $Z$ on side $\overline{A C}$ such that $D$ lies on side $\overline{X Z}, E$ lies on side $\overline{Y Z}$, and $F$ lies on side $\overline{X Y}$. Compute $A Z$. | First, note that point $X$ can be constructed from intersection of $\odot(D O F)$ and side $\overline{B C}$. Thus, if there is a unique equilateral triangle, then we must have that $\odot(D O F)$ is tangent to $\overline{B C}$. Furthermore, $\odot(D O F)$ is tangent to $D E$, so by equal tangents, we have $C D=C X$. We now compute the answer. Let $x=A Z=C X=C D=B F$. Then, by power of point, $$B F \cdot B D=B X^{2} \Longrightarrow B D=\frac{(1-x)^{2}}{x}$$ Thus, by law of cosine on $\triangle B D C$, we have that $$\begin{aligned} x^{2}+\left(\frac{(1-x)^{2}}{x}\right)^{2}+x \cdot \frac{(1-x)^{2}}{x} & =1 \\ x^{2}+\frac{(1-x)^{4}}{x^{2}}+(1-x)^{2} & =1 \\ \frac{(1-x)^{4}}{x^{2}} & =2x(1-x) \\ \frac{1-x}{x} & =\sqrt[3]{2} \\ x & =\frac{1}{1+\sqrt[3]{2}} \end{aligned}$$ | \frac{1}{1+\sqrt[3]{2}} | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 4 | The real function $f$ has the property that, whenever $a, b, n$ are positive integers such that $a+b=2^{n}$, the equation $f(a)+f(b)=n^{2}$ holds. What is $f(2002)$? | We know $f(a)=n^{2}-f\left(2^{n}-a\right)$ for any $a$, $n$ with $2^{n}>a$; repeated application gives $$f(2002)=11^{2}-f(46)=11^{2}-\left(6^{2}-f(18)\right)=11^{2}-\left(6^{2}-\left(5^{2}-f(14)\right)\right) =11^{2}-\left(6^{2}-\left(5^{2}-\left(4^{2}-f(2)\right)\right)\right)$$ But $f(2)=2^{2}-f(2)$, giving $f(2)=2$, so the above simplifies to $11^{2}-\left(6^{2}-\left(5^{2}-\left(4^{2}-\right.\right.\right.$ 2)) $=96$. | 96 | HMMT_2 |
[
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes"
] | 4.75 | Jarris is a weighted tetrahedral die with faces $F_{1}, F_{2}, F_{3}, F_{4}$. He tosses himself onto a table, so that the probability he lands on a given face is proportional to the area of that face. Let $k$ be the maximum distance any part of Jarris is from the table after he rolls himself. Given that Jarris has an inscribed sphere of radius 3 and circumscribed sphere of radius 10, find the minimum possible value of the expected value of $k$. | Since the maximum distance to the table is just the height, the expected value is equal to $\frac{\sum_{i=1}^{4} h_{i}\left[F_{i}\right]}{\sum_{i=1}^{4}\left[F_{i}\right]}$. Let $V$ be the volume of Jarris. Recall that $V=\frac{1}{3} h_{i}\left[F_{i}\right]$ for any $i$, but also $V=\frac{r}{3}\left(\sum_{i=1}^{4}\left[F_{i}\right]\right)$ where $r$ is the inradius (by decomposing into four tetrahedra with a vertex at the incenter). Therefore $\frac{\sum_{i=1}^{4} h_{i}\left[F_{i}\right]}{\sum_{i=1}^{4}\left[F_{i}\right]}=\frac{12 V}{3 V / r}=4 r=12$. | 12 | HMMT_2 |
[
"Mathematics -> Geometry -> Solid Geometry -> Volume"
] | 4.5 | Find the volume of the three-dimensional solid given by the inequality $\sqrt{x^{2}+y^{2}}+$ $|z| \leq 1$. | $2 \pi / 3$. The solid consists of two cones, one whose base is the circle $x^{2}+y^{2}=1$ in the $x y$-plane and whose vertex is $(0,0,1)$, and the other with the same base but vertex $(0,0,-1)$. Each cone has a base area of $\pi$ and a height of 1, for a volume of $\pi / 3$, so the answer is $2 \pi / 3$. | 2 \pi / 3 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 4 | Count how many 8-digit numbers there are that contain exactly four nines as digits. | There are $\binom{8}{4} \cdot 9^{4}$ sequences of 8 numbers with exactly four nines. A sequence of digits of length 8 is not an 8-digit number, however, if and only if the first digit is zero. There are $\binom{7}{4} 9^{3}$ 8-digit sequences that are not 8-digit numbers. The answer is thus $\binom{8}{4} \cdot 9^{4}-\binom{7}{4} 9^{3}=433755$. | 433755 | HMMT_2 |
[
"Mathematics -> Algebra -> Abstract Algebra -> Field Theory"
] | 4.5 | Find the smallest positive integer $k$ such that $z^{10}+z^{9}+z^{6}+z^{5}+z^{4}+z+1$ divides $z^{k}-1$. | Let $Q(z)$ denote the polynomial divisor. We need that the roots of $Q$ are $k$-th roots of unity. With this in mind, we might observe that solutions to $z^{7}=1$ and $z \neq 1$ are roots of $Q$, which leads to its factorization. Alternatively, we note that $$(z-1) Q(z)=z^{11}-z^{9}+z^{7}-z^{4}+z^{2}-1=\left(z^{4}-z^{2}+1\right)\left(z^{7}-1\right)$$ Solving for the roots of the first factor, $z^{2}=\frac{1+i \sqrt{3}}{2}= \pm \operatorname{cis} \pi / 3$ (we use the notation $\operatorname{cis}(x)=\cos (x)+i \sin (x))$ so that $z= \pm \operatorname{cis}( \pm \pi / 6)$. These are primitive 12 -th roots of unity. The other roots of $Q(z)$ are the primitive 7 -th roots of unity (we introduced $z=1$ by multiplication.) It follows that the answer is $\operatorname{lcm}[12,7]=84$. | 84 | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers"
] | 5 | Let $$\begin{aligned} & A=(1+2 \sqrt{2}+3 \sqrt{3}+6 \sqrt{6})(2+6 \sqrt{2}+\sqrt{3}+3 \sqrt{6})(3+\sqrt{2}+6 \sqrt{3}+2 \sqrt{6})(6+3 \sqrt{2}+2 \sqrt{3}+\sqrt{6}) \\ & B=(1+3 \sqrt{2}+2 \sqrt{3}+6 \sqrt{6})(2+\sqrt{2}+6 \sqrt{3}+3 \sqrt{6})(3+6 \sqrt{2}+\sqrt{3}+2 \sqrt{6})(6+2 \sqrt{2}+3 \sqrt{3}+\sqrt{6}) \end{aligned}$$ Compute the value of $A / B$. | Note that $$\begin{aligned} & A=((1+2 \sqrt{2})(1+3 \sqrt{3}))((2+\sqrt{3})(1+3 \sqrt{2}))((3+\sqrt{2})(1+2 \sqrt{3}))((3+\sqrt{3})(2+\sqrt{2})) \\ & B=((1+3 \sqrt{2})(1+2 \sqrt{3}))((2+\sqrt{2})(1+3 \sqrt{3}))((3+\sqrt{3})(1+2 \sqrt{2}))((2+\sqrt{3})(3+\sqrt{2})) \end{aligned}$$ It is not difficult to check that they have the exact same set of factors, so $A=B$ and thus the ratio is 1. | 1 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Circles"
] | 5.25 | Two circles have radii 13 and 30, and their centers are 41 units apart. The line through the centers of the two circles intersects the smaller circle at two points; let $A$ be the one outside the larger circle. Suppose $B$ is a point on the smaller circle and $C$ a point on the larger circle such that $B$ is the midpoint of $A C$. Compute the distance $A C$. | $12 \sqrt{13}$ Call the large circle's center $O_{1}$. Scale the small circle by a factor of 2 about $A$; we obtain a new circle whose center $O_{2}$ is at a distance of $41-13=28$ from $O_{1}$, and whose radius is 26. Also, the dilation sends $B$ to $C$, which thus lies on circles $O_{1}$ and $O_{2}$. So points $O_{1}, O_{2}, C$ form a 26-28-30 triangle. Let $H$ be the foot of the altitude from $C$ to $O_{1} O_{2}$; we have $C H=24$ and $H O_{2}=10$. Thus, $H A=36$, and $A C=\sqrt{24^{2}+36^{2}}=12 \sqrt{13}$. | 12 \sqrt{13} | HMMT_2 |
[
"Mathematics -> Algebra -> Prealgebra -> Other"
] | 4 | The expression $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. Find the value of $$\left\lfloor\frac{2002!}{2001!+2000!+1999!+\cdots+1!}\right\rfloor.$$ | 2000 We break up 2002! = 2002(2001)! as $$2000(2001!)+2 \cdot 2001(2000!)=2000(2001!)+2000(2000!)+2002 \cdot 2000(1999!) >2000(2001!+2000!+1999!+\cdots+1!)$$ On the other hand, $$2001(2001!+2000!+\cdots+1!)>2001(2001!+2000!)=2001(2001!)+2001!=2002!$$ Thus we have $2000<2002!/(2001!+\cdots+1!)<2001$, so the answer is 2000. | 2000 | HMMT_2 |
[
"Mathematics -> Number Theory -> Prime Numbers"
] | 5 | Call a positive integer $n$ weird if $n$ does not divide $(n-2)$!. Determine the number of weird numbers between 2 and 100 inclusive. | We claim that all the weird numbers are all the prime numbers and 4. Since no numbers between 1 and $p-2$ divide prime $p,(p-2)$! will not be divisible by $p$. We also have $2!=2$ not being a multiple of 4. Now we show that all other numbers are not weird. If $n=p q$ where $p \neq q$ and $p, q \geq 2$, then since $p$ and $q$ both appear in $1,2, \ldots, n-2$ and are distinct, we have $p q \mid(n-2)$!. This leaves the only case of $n=p^{2}$ for prime $p \geq 3$. In this case, we can note that $p$ and $2 p$ are both less than $p^{2}-2$, so $2 p^{2} \mid(n-2)$! and we are similarly done. Since there are 25 prime numbers not exceeding 100, there are $25+1=26$ weird numbers. | 26 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Probability -> Other",
"Mathematics -> Geometry -> Plane Geometry -> Other"
] | 5 | In the $x-y$ plane, draw a circle of radius 2 centered at $(0,0)$. Color the circle red above the line $y=1$, color the circle blue below the line $y=-1$, and color the rest of the circle white. Now consider an arbitrary straight line at distance 1 from the circle. We color each point $P$ of the line with the color of the closest point to $P$ on the circle. If we pick such an arbitrary line, randomly oriented, what is the probability that it contains red, white, and blue points? | Let $O=(0,0), P=(1,0)$, and $H$ the foot of the perpendicular from $O$ to the line. If $\angle P O H$ (as measured counterclockwise) lies between $\pi / 3$ and $2 \pi / 3$, the line will fail to contain blue points; if it lies between $4 \pi / 3$ and $5 \pi / 3$, the line will fail to contain red points. Otherwise, it has points of every color. Thus, the answer is $1-\frac{2 \pi}{3} / 2 \pi=\frac{2}{3}$. | \frac{2}{3} | HMMT_2 |
[
"Mathematics -> Number Theory -> Other"
] | 4 | Call a positive integer 'mild' if its base-3 representation never contains the digit 2. How many values of $n(1 \leq n \leq 1000)$ have the property that $n$ and $n^{2}$ are both mild? | 7 Such a number, which must consist entirely of 0's and 1's in base 3, can never have more than one 1. Indeed, if $n=3^{a}+3^{b}+$ higher powers where $b>a$, then $n^{2}=3^{2 a}+2 \cdot 3^{a+b}+$ higher powers which will not be mild. On the other hand, if $n$ does just have one 1 in base 3, then clearly $n$ and $n^{2}$ are mild. So the values of $n \leq 1000$ that work are $3^{0}, 3^{1}, \ldots, 3^{6}$; there are 7 of them. | 7 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 3.5 | A regular decagon $A_{0} A_{1} A_{2} \cdots A_{9}$ is given in the plane. Compute $\angle A_{0} A_{3} A_{7}$ in degrees. | Put the decagon in a circle. Each side subtends an arc of $360^{\circ} / 10=36^{\circ}$. The inscribed angle $\angle A_{0} A_{3} A_{7}$ contains 3 segments, namely $A_{7} A_{8}, A_{8} A_{9}, A_{9} A_{0}$, so the angle is $108^{\circ} / 2=54^{\circ}$. | 54^{\circ} | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers",
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5 | The sequence $\left(z_{n}\right)$ of complex numbers satisfies the following properties: $z_{1}$ and $z_{2}$ are not real. $z_{n+2}=z_{n+1}^{2} z_{n}$ for all integers $n \geq 1$. $\frac{z_{n+3}}{z_{n}^{2}}$ is real for all integers $n \geq 1$. $\left|\frac{z_{3}}{z_{4}}\right|=\left|\frac{z_{4}}{z_{5}}\right|=2$ Find the product of all possible values of $z_{1}$. | All complex numbers can be expressed as $r(\cos \theta+i \sin \theta)=r e^{i \theta}$. Let $z_{n}$ be $r_{n} e^{i \theta_{n}}$. $\frac{z_{n+3}}{z_{n}^{2}}=\frac{z_{n+2}^{2} z_{n+1}}{z_{n}^{2}}=\frac{z_{n+1}^{5} z_{n}^{2}}{z_{n}^{2}}=z_{n+1}^{5}$ is real for all $n \geq 1$, so $\theta_{n}=\frac{\pi k_{n}}{5}$ for all $n \geq 2$, where $k_{n}$ is an integer. $\theta_{1}+2 \theta_{2}=\theta_{3}$, so we may write $\theta_{1}=\frac{\pi k_{1}}{5}$ with $k_{1}$ an integer. $\frac{r_{3}}{r_{4}}=\frac{r_{4}}{r_{5}} \Rightarrow r_{5}=\frac{r_{4}^{2}}{r_{3}}=r_{4}^{2} r_{3}$, so $r_{3}=1 . \frac{r_{3}}{r_{4}}=2 \Rightarrow r_{4}=\frac{1}{2}, r_{4}=r_{3}^{2} r_{2} \Rightarrow r_{2}=\frac{1}{2}$, and $r_{3}=r_{2}^{2} r_{1} \Rightarrow r_{1}=4$. Therefore, the possible values of $z_{1}$ are the nonreal roots of the equation $x^{10}-4^{10}=0$, and the product of the eight possible values is $\frac{4^{10}}{4^{2}}=4^{8}=65536$. For these values of $z_{1}$, it is not difficult to construct a sequence which works, by choosing $z_{2}$ nonreal so that $\left|z_{2}\right|=\frac{1}{2}$. | 65536 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 5 | Knot is on an epic quest to save the land of Hyruler from the evil Gammadorf. To do this, he must collect the two pieces of the Lineforce, then go to the Temple of Lime. As shown on the figure, Knot starts on point $K$, and must travel to point $T$, where $O K=2$ and $O T=4$. However, he must first reach both solid lines in the figure below to collect the pieces of the Lineforce. What is the minimal distance Knot must travel to do so? | Let $l_{1}$ and $l_{2}$ be the lines as labeled in the above diagram. First, suppose Knot visits $l_{1}$ first, at point $P_{1}$, then $l_{2}$, at point $P_{2}$. Let $K^{\prime}$ be the reflection of $K$ over $l_{1}$, and let $T^{\prime}$ be the reflection of $T$ over $l_{2}$. The length of Knot's path is at least $$ K P_{1}+P_{1} P_{2}+P_{2} T=K^{\prime} P_{1}+P_{1} P_{2}+P_{2} T^{\prime} \geq K^{\prime} T^{\prime} $$ by the Triangle Inequality (This bound can be achieved by taking $P_{1}, P_{2}$ to be the intersections of $K^{\prime} T^{\prime}$ with $l_{1}, l_{2}$, respectively.) Also, note that \measuredangle K^{\prime} O T^{\prime}=90^{\circ}$, so that $K^{\prime} T^{\prime}=2 \sqrt{5}$. Now, suppose Knot instead visits $l_{2}$ first, at point $Q_{2}$, then $l_{1}$, at point $Q_{1}$. Letting $K^{\prime \prime}$ be the reflection of $K$ over $l_{2}$ and $T^{\prime \prime}$ be the reflection of $T$ over $l_{1}$, by similar logic to before the length of his path is at least the length of $K^{\prime \prime} T^{\prime \prime}$. However, by inspection $K^{\prime \prime} T^{\prime \prime}>K^{\prime} T^{\prime}$, so our answer is $2 \sqrt{5}$. | 2 \sqrt{5} | HMMT_2 |
[
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes"
] | 5 | A conical flask contains some water. When the flask is oriented so that its base is horizontal and lies at the bottom (so that the vertex is at the top), the water is 1 inch deep. When the flask is turned upside-down, so that the vertex is at the bottom, the water is 2 inches deep. What is the height of the cone? | $\frac{1}{2}+\frac{\sqrt{93}}{6}$. Let $h$ be the height, and let $V$ be such that $V h^{3}$ equals the volume of the flask. When the base is at the bottom, the portion of the flask not occupied by water forms a cone similar to the entire flask, with a height of $h-1$; thus its volume is $V(h-1)^{3}$. When the base is at the top, the water occupies a cone with a height of 2, so its volume is $V \cdot 2^{3}$. Since the water's volume does not change, $$V h^{3}-V(h-1)^{3}=8 V \Rightarrow 3 h^{2}-3 h+1=h^{3}-(h-1)^{3}=8 \Rightarrow 3 h^{2}-3 h-7=0$$ Solving via the quadratic formula and taking the positive root gives $h=\frac{1}{2}+\frac{\sqrt{93}}{6}$. | \frac{1}{2}+\frac{\sqrt{93}}{6} | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5.25 | For a string of $P$ 's and $Q$ 's, the value is defined to be the product of the positions of the $P$ 's. For example, the string $P P Q P Q Q$ has value $1 \cdot 2 \cdot 4=8$. Also, a string is called antipalindromic if writing it backwards, then turning all the $P$ 's into $Q$ 's and vice versa, produces the original string. For example, $P P Q P Q Q$ is antipalindromic. There are $2^{1002}$ antipalindromic strings of length 2004. Find the sum of the reciprocals of their values. | Consider the product $$ \left(\frac{1}{1}+\frac{1}{2004}\right)\left(\frac{1}{2}+\frac{1}{2003}\right)\left(\frac{1}{3}+\frac{1}{2002}\right) \cdots\left(\frac{1}{1002}+\frac{1}{1003}\right) $$ This product expands to $2^{1002}$ terms, and each term gives the reciprocal of the value of a corresponding antipalindromic string of $P$ 's and $Q$ 's as follows: if we choose the term $1 / n$ for the $n$th factor, then our string has a $P$ in position $n$ and $Q$ in position $2005-n$; if we choose the term $1 /(2005-n)$, then we get a $Q$ in position $n$ and $P$ in position $2005-n$. Conversely, each antipalindromic string has its value represented by exactly one of our $2^{1002}$ terms. So the value of the product is the number we are looking for. But when we simplify this product, the $n$th factor becomes $1 / n+1 /(2005-n)=2005 / n(2005-n)$. Multiplying these together, we get 1002 factors of 2005 in the numerator and each integer from 1 to 2004 exactly once in the denominator, for a total of $2005^{1002} / 2004$ !. | 2005^{1002} / 2004! | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 4 | Reimu has 2019 coins $C_{0}, C_{1}, \ldots, C_{2018}$, one of which is fake, though they look identical to each other (so each of them is equally likely to be fake). She has a machine that takes any two coins and picks one that is not fake. If both coins are not fake, the machine picks one uniformly at random. For each $i=1,2, \ldots, 1009$, she puts $C_{0}$ and $C_{i}$ into the machine once, and machine picks $C_{i}$. What is the probability that $C_{0}$ is fake? | Let $E$ denote the event that $C_{0}$ is fake, and let $F$ denote the event that the machine picks $C_{i}$ over $C_{0}$ for all $i=1,2, \ldots 1009$. By the definition of conditional probability, $P(E \mid F)=\frac{P(E \cap F)}{P(F)}$. Since $E$ implies $F$, $P(E \cap F)=P(E)=\frac{1}{2019}$. Now we want to compute $P(F)$. If $C_{0}$ is fake, $F$ is guaranteed to happen. If $C_{i}$ is fake for some $1 \leq i \leq 1009$, then $F$ is impossible. Finally, if $C_{i}$ is fake for some $1010 \leq i \leq 2018$, then $F$ occurs with probability $2^{-1009}$, since there is a $\frac{1}{2}$ probability for each machine decision. Therefore, $P(F)=\frac{1}{2019} \cdot 1+\frac{1009}{2019} \cdot 0+\frac{1009}{2019} \cdot 2^{-1009}=\frac{2^{1009}+1009}{2019 \cdot 2^{1009}}$. Therefore, $P(E \mid F)=\frac{2^{1009}}{2^{1009}+1009} \cdot$ | \frac{2^{1009}}{2^{1009}+1009} | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5.25 | Let $A B C$ be a triangle with $A B=6, A C=7, B C=8$. Let $I$ be the incenter of $A B C$. Points $Z$ and $Y$ lie on the interior of segments $A B$ and $A C$ respectively such that $Y Z$ is tangent to the incircle. Given point $P$ such that $$\angle Z P C=\angle Y P B=90^{\circ}$$ find the length of $I P$. | Solution 1. Let $P U, P V$ tangent from $P$ to the incircle. We will invoke the dual of the Desargues Involution Theorem, which states the following: Given a point $P$ in the plane and four lines $\ell_{1}, \ell_{2}, \ell_{3}, \ell_{4}$, consider the set of conics tangent to all four lines. Then we define a function on the pencil of lines through $P$ by mapping one tangent from $P$ to each conic to the other. This map is well defined and is a projective involution, and in particular maps $P A \rightarrow P D, P B \rightarrow P E, P C \rightarrow P F$, where $A B C D E F$ is the complete quadrilateral given by the pairwise intersections of $\ell_{1}, \ell_{2}, \ell_{3}, \ell_{4}$. An overview of the projective background behind the (Dual) Desargues Involution Theorem can be found here: https://www.scribd.com/document/384321704/Desargues-Involution-Theorem, and a proof can be found at https://www2.washjeff.edu/users/mwoltermann/Dorrie/63.pdf. Now, we apply this to the point $P$ and the lines $A B, A C, B C, Y Z$, to get that the pairs $$(P U, P V),(P Y, P B),(P Z, P C)$$ are swapped by some involution. But we know that the involution on lines through $P$ which rotates by $90^{\circ}$ swaps the latter two pairs, thus it must also swap the first one and $\angle U P V=90$. It follows by equal tangents that $I U P V$ is a square, thus $I P=r \sqrt{2}$ where $r$ is the inradius of $A B C$. Since $r=\frac{2 K}{a+b+c}=\frac{21 \sqrt{15} / 2}{21}=\frac{\sqrt{15}}{2}$, we have $I P=\frac{\sqrt{30}}{2}$. Solution 2. Let $H$ be the orthocenter of $A B C$. Lemma. $H I^{2}=2 r^{2}-4 R^{2} \cos (A) \cos (B) \cos (C)$, where $r$ is the inradius and $R$ is the circumradius. Proof. This follows from barycentric coordinates or the general result that for a point $X$ in the plane, $$a X A^{2}+b X B^{2}+c X C^{2}=(a+b+c) X I^{2}+a A I^{2}+b B I^{2}+c C I^{2}$$ which itself is a fact about vectors that follows from barycentric coordinates. This can also be computed directly using trigonometry. Let $E=B H \cap A C, F=C H \cap A B$, then note that $B, P, E, Y$ are concyclic on the circle of diameter $B Y$, and $C, P, F, Z$ are concyclic on the circle of diameter $C Z$. Let $Q$ be the second intersection of these circles. Since $B C Y Z$ is a tangential quadrilateral, the midpoints of $B Y$ and $C Z$ are collinear with $I$ (this is known as Newton's theorem), which implies that $I P=I Q$ by symmetry. Note that as $B H \cdot H E=C H \cdot H F, H$ lies on the radical axis of the two circles, which is $P Q$. Thus, if $I P=I Q=x$, $B H \cdot H E$ is the power of $H$ with respect to the circle centered at $I$ with radius $x$, which implies $B H \cdot H E=x^{2}-H I^{2}$. As with the first solution, we claim that $x=r \sqrt{2}$, which by the lemma is equivalent to $B H \cdot H E=$ $4 R^{2} \cos (A) \cos (B) \cos (C)$. Then note that $$B H \cdot H E=B H \cdot C H \cos (A)=(2 R \cos (B))(2 R \cos (C)) \cos (A)$$ so our claim holds and we finish as with the first solution. Note. Under the assumption that the problem is well-posed (the answer does not depend on the choice of $Y, Z$, or $P$ ), then here is an alternative method to obtain $I P=r \sqrt{2}$ by making convenient choices. Let $U$ be the point where $Y Z$ is tangent to the incircle, and choose $U$ so that $I U \| B C$ (and therefore $Y Z \perp B C)$. Note that $Y Z \cap B C$ is a valid choice for $P$, so assume that $P$ is the foot from $U$ to $B C$. If $D$ is the point where $B C$ is tangent to the incircle, then $I U P D$ is a square so $I P=r \sqrt{2}$. (This disregards the condition that $Y$ and $Z$ are in the interior of segments $A C$ and $A B$, but there is no reason to expect that this condition is important.) | \frac{\sqrt{30}}{2} | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics",
"Mathematics -> Number Theory -> Congruences"
] | 5 | Determine the number of subsets $S$ of $\{1,2, \ldots, 1000\}$ that satisfy the following conditions: - $S$ has 19 elements, and - the sum of the elements in any non-empty subset of $S$ is not divisible by 20 . | First we prove that each subset must consist of elements that have the same residue mod 20. Let a subset consist of elements $a_{1}, \ldots, a_{19}$, and consider two lists of partial sums $$\begin{aligned} & a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \ldots, a_{1}+a_{2}+\cdots+a_{19} \\ & a_{2}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \ldots, a_{1}+a_{2}+\cdots+a_{19} \end{aligned}$$ The residues mod 20 of the partial sums in each list must be pairwise distinct, otherwise subtracting the sum with less terms from the sum with more terms yields a subset whose sum of elements is 0 $(\bmod 20)$. Since the residues must also be nonzero, each list forms a complete nonzero residue class $\bmod 20$. Since the latter 18 sums in the two lists are identical, $a_{1} \equiv a_{2}(\bmod 20)$. By symmetric arguments, $a_{i} \equiv a_{j}(\bmod 20)$ for any $i, j$. Furthermore this residue $1 \leq r \leq 20$ must be relatively prime to 20, because if $d=\operatorname{gcd}(r, 20)>1$ then any $20 / d$ elements of the subset will sum to a multiple of 20. Hence there are $\varphi(20)=8$ possible residues. Since there are 50 elements in each residue class, the answer is $\binom{50}{19}$. We can see that any such subset whose elements are a relatively prime residue $r(\bmod 20)$ works because the sum of any $1 \leq k \leq 19$ elements will be $k r \neq 0(\bmod 20)$ | 8 \cdot\binom{50}{19} | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Math Word Problems",
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 3.5 | The mathematician John is having trouble remembering his girlfriend Alicia's 7-digit phone number. He remembers that the first four digits consist of one 1, one 2, and two 3s. He also remembers that the fifth digit is either a 4 or 5. While he has no memory of the sixth digit, he remembers that the seventh digit is 9 minus the sixth digit. If this is all the information he has, how many phone numbers does he have to try if he is to make sure he dials the correct number? | There are $\frac{4!}{2!}=12$ possibilities for the first four digits. There are two possibilities for the fifth digit. There are 10 possibilities for the sixth digit, and this uniquely determines the seventh digit. So he has to dial $12 \cdot 2 \cdot 10=240$ numbers. | 240 | HMMT_2 |
[
"Mathematics -> Algebra -> Prealgebra -> Integers",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Find all the integers $n>1$ with the following property: the numbers $1,2, \ldots, n$ can be arranged in a line so that, of any two adjacent numbers, one is divisible by the other. | $2,3,4,6$ The values $n=2,3,4,6$ work, as shown by respective examples 1,$2 ; 2,1,3 ; 2,4,1,3 ; 3,6,2,4,1,5$. We shall show that there are no other possibilities. If $n=2 k+1$ is odd, then none of the numbers $k+1, k+2, \ldots, 2 k+1$ can divide any other, so no two of these numbers are adjacent. This is only possible if they occupy the 1st, 3rd, $\ldots,(2 k+1)$th positions in the line, which means every number $\leq k$ is adjacent to two of these and hence divides two of them. But $k$ only divides one of these numbers when $k \geq 2$. Thus no odd $n \geq 5$ works. If $n=2 k$ is even, the numbers $k+1, k+2, \ldots, 2 k$ again must be mutually nonadjacent, but now this means we can have up to two numbers $\leq k$ each of which is adjacent to only one number $>k$, and if there are two such numbers, they must be adjacent. If $k \geq 4$, then each of $k-1, k$ divides only one of the numbers $k+1, \ldots, 2 k$, so $k-1, k$ must be adjacent, but this is impossible. Thus no even $k \geq 8$ works, and we are done. | 2, 3, 4, 6 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Let $S$ be a set of size 3. How many collections $T$ of subsets of $S$ have the property that for any two subsets $U \in T$ and $V \in T$, both $U \cap V$ and $U \cup V$ are in $T$ ? | Let us consider the collections $T$ grouped based on the size of the set $X=\bigcup_{U \in T} U$, which we can see also must be in $T$ as long as $T$ contains at least one set. This leads us to count the number of collections on a set of size at most 3 satisfying the desired property with the additional property that the entire set must be in the collection. Let $C_{n}$ denote that number of such collections on a set of size $n$. Our answer will then be $1+\binom{3}{0} C_{0}+\binom{3}{1} C_{1}+\binom{3}{2} C_{2}+\binom{3}{3} C_{3}$, with the additional 1 coming from the empty collection. Now for such a collection $T$ on a set of $n$ elements, consider the set $I=\bigcap_{U \in T} U$. Suppose this set has size $k$. Then removing all these elements from consideration gives us another such collection on a set of size $n-k$, but now containing the empty set. We can see that for each particular choice of $I$, this gives a bijection to the collections on the set $S$ to the collections on the set $S-I$. This leads us to consider the further restricted collections that must contain both the entire set and the empty set. It turns out that such restricted collections are a well-studied class of objects called topological spaces. Let $T_{n}$ be the number of topological spaces on $n$ elements. Our argument before shows that $C_{n}=$ $\sum_{k=0}^{n}\binom{n}{k} T_{k}$. It is relatively straightforward to see that $T_{0}=1, T_{1}=1$, and $T_{2}=4$. For a set of size 3 , there are the following spaces. The number of symmetric versions is shown in parentheses. - $\emptyset,\{a, b, c\}(1)$ - $\emptyset,\{a, b\},\{a, b, c\}(3)$ - $\emptyset,\{a\},\{a, b, c\}(3)$ - $\emptyset,\{a\},\{a, b\},\{a, b, c\}$ (6) - $\emptyset,\{a\},\{b, c\},\{a, b, c\}$ - $\emptyset,\{a\},\{a, b\},\{a, c\},\{a, b, c\}(3)$ - $\emptyset,\{a\},\{b\},\{a, b\} .\{a, b, c\}(3)$ - $\emptyset,\{a\},\{b\},\{a, b\},\{a, c\},\{a, b, c\}(6)$ - $\emptyset,\{a\},\{b\},\{c\},\{a, b\},\{a, c\},\{b, c\},\{a, b, c\}$ which gives $T_{3}=29$. Tracing back our reductions, we have that $C_{0}=\binom{0}{0} T_{0}=1, C_{1}=\binom{1}{0} T_{0}+\binom{1}{1} T_{1}=$ 2, $C_{2}=\binom{2}{0} T_{0}+\binom{2}{1} T_{1}+\binom{2}{2} T_{2}=7, C_{3}=\binom{3}{0} T_{0}+\binom{3}{1} T_{1}+\binom{3}{2} T_{2}+\binom{3}{3} T_{3}=45$, and then our answer is $1+\binom{3}{0} C_{0}+\binom{3}{1} C_{1}+\binom{3}{2} C_{2}+\binom{3}{3} C_{3}=1+1+6+21+45=74$. | 74 | HMMT_2 |
[
"Mathematics -> Number Theory -> Factorization",
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5.25 | Let $S$ be the set of all positive factors of 6000. What is the probability of a random quadruple $(a, b, c, d) \in S^{4}$ satisfies $$\operatorname{lcm}(\operatorname{gcd}(a, b), \operatorname{gcd}(c, d))=\operatorname{gcd}(\operatorname{lcm}(a, b), \operatorname{lcm}(c, d)) ?$$ | For each prime factor, let the greatest power that divides $a, b, c, d$ be $p, q, r, s$. WLOG assume that $p \leq q$ and $r \leq s$, and further WLOG assume that $p \leq r$. Then we need $r=\min (q, s)$. If $q=r$ then we have $p \leq q=r \leq s$, and if $r=s$ then we have $p \leq r=s \leq q$, and in either case the condition reduces to the two 'medians' among $p, q, r, s$ are equal. (It is not difficult to see that this condition is also sufficient.) Now we compute the number of quadruples $(p, q, r, s)$ of integers between 0 and $n$ inclusive that satisfy the above condition. If there are three distinct numbers then there are $\binom{n+1}{3}$ ways to choose the three numbers and $4!/ 2=12$ ways to assign them (it must be a $1-2-1$ split). If there are two distinct numbers then there are $\binom{n+1}{2}$ ways to choose the numbers and $4+4=8$ ways to assign them (it must be a $3-1$ or a 1-3 split). If there is one distinct number then there are $n+1$ ways to assign. Together we have $12\binom{n+1}{3}+8\binom{n+1}{2}+(n+1)=2(n+1) n(n-1)+4(n+1) n+(n+1)=(n+1)(2 n(n+1)+1)$ possible quadruples. So if we choose a random quadruple then the probability that it satisfies the condition is $\frac{(n+1)(2 n(n+1)+1)}{(n+1)^{4}}=\frac{2 n(n+1)+1}{(n+1)^{3}}$. Since $6000=2^{4} \cdot 5^{3} \cdot 3^{1}$ and the power of different primes are independent, we plug in $n=4,3,1$ to get the overall probability to be $$\frac{41}{125} \cdot \frac{25}{64} \cdot \frac{5}{8}=\frac{41}{512}$$ | \frac{41}{512} | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Exponential Functions"
] | 4 | In the Year 0 of Cambridge there is one squirrel and one rabbit. Both animals multiply in numbers quickly. In particular, if there are $m$ squirrels and $n$ rabbits in Year $k$, then there will be $2 m+2019$ squirrels and $4 n-2$ rabbits in Year $k+1$. What is the first year in which there will be strictly more rabbits than squirrels? | In year $k$, the number of squirrels is $$2(2(\cdots(2 \cdot 1+2019)+2019)+\cdots)+2019=2^{k}+2019 \cdot\left(2^{k-1}+2^{k-2}+\cdots+1\right)=2020 \cdot 2^{k}-2019$$ and the number of rabbits is $$4(4(\cdots(4 \cdot 1-2)-2)-\cdots)-2=4^{k}-2 \cdot\left(4^{k-1}+4^{k-2}+\cdots+1\right)=\frac{4^{k}+2}{3}$$ For the number of rabbits to exceed that of squirrels, we need $$4^{k}+2>6060 \cdot 2^{k}-6057 \Leftrightarrow 2^{k}>6059$$ Since $2^{13}>6059>2^{12}, k=13$ is the first year for which there are more rabbits than squirrels. | 13 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other"
] | 5 | Milan has a bag of 2020 red balls and 2021 green balls. He repeatedly draws 2 balls out of the bag uniformly at random. If they are the same color, he changes them both to the opposite color and returns them to the bag. If they are different colors, he discards them. Eventually the bag has 1 ball left. Let $p$ be the probability that it is green. Compute $\lfloor 2021 p \rfloor$. | The difference between the number of green balls and red balls in the bag is always 1 modulo 4. Thus the last ball must be green and $p=1$. | 2021 | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers"
] | 4.5 | Complex numbers $a, b, c$ form an equilateral triangle with side length 18 in the complex plane. If $|a+b+c|=36$, find $|b c+c a+a b|$. | Using basic properties of vectors, we see that the complex number $d=\frac{a+b+c}{3}$ is the center of the triangle. From the given, $|a+b+c|=36 \Longrightarrow|d|=12$. Then, let $a^{\prime}=a-d, b^{\prime}=b-d$, and $c^{\prime}=c-d$. Due to symmetry, $\left|a^{\prime}+b^{\prime}+c^{\prime}\right|=0$ and $\left|b^{\prime} c^{\prime}+c^{\prime} a^{\prime}+a^{\prime} b^{\prime}\right|=0$. Finally, we compute $$\begin{aligned} |b c+c a+a b| & =\left|\left(b^{\prime}+d\right)\left(c^{\prime}+d\right)+\left(c^{\prime}+d\right)\left(a^{\prime}+d\right)+\left(a^{\prime}+d\right)\left(b^{\prime}+d\right)\right| \\ & =\left|b^{\prime} c^{\prime}+c^{\prime} a^{\prime}+a^{\prime} b^{\prime}+2 d\left(a^{\prime}+b^{\prime}+c^{\prime}\right)+3 d^{2}\right| \\ & =\left|3 d^{2}\right|=3 \cdot 12^{2}=432 . \end{aligned}$$ | 432 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Geometry -> Plane Geometry -> Triangulations",
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 5.25 | A regular hexagon $A B C D E F$ has side length 1 and center $O$. Parabolas $P_{1}, P_{2}, \ldots, P_{6}$ are constructed with common focus $O$ and directrices $A B, B C, C D, D E, E F, F A$ respectively. Let $\chi$ be the set of all distinct points on the plane that lie on at least two of the six parabolas. Compute $$\sum_{X \in \chi}|O X|$$ (Recall that the focus is the point and the directrix is the line such that the parabola is the locus of points that are equidistant from the focus and the directrix.) | Recall the focus and the directrix are such that the parabola is the locus of points equidistant from the focus and the directrix. We will consider pairs of parabolas and find their points of intersections (we label counterclockwise): (1): $P_{1} \cap P_{2}$, two parabolas with directrices adjacent edges on the hexagon (sharing vertex $A$ ). The intersection inside the hexagon can be found by using similar triangles: by symmetry this $X$ must lie on $O A$ and must have that its distance from $A B$ and $F A$ are equal to $|O X|=x$, which is to say $$\sin 60^{\circ}=\frac{\sqrt{3}}{2}=\frac{x}{|O A|-x}=\frac{x}{1-x} \Longrightarrow x=2 \sqrt{3}-3$$ By symmetry also, the second intersection point, outside the hexagon, must lie on $O D$. Furthermore, $X$ must have that its distance $A B$ and $F A$ are equal to $|O X|$. Then again by similar triangles $$\sin 60^{\circ}=\frac{\sqrt{3}}{2}=\frac{x}{|O A|+x}=\frac{x}{1+x} \Longrightarrow x=2 \sqrt{3}+3$$ (2): $P_{1} \cap P_{3}$, two parabolas with directrices edges one apart on the hexagon, say $A B$ and $C D$. The intersection inside the hexagon is clearly immediately the circumcenter of triangle $B O C$ (equidistance condition), which gives $$x=\frac{\sqrt{3}}{3}$$ Again by symmetry the $X$ outside the hexagon must lie on the lie through $O$ and the midpoint of $E F$; then one can either observe immediately that $x=\sqrt{3}$ or set up $$\sin 30^{\circ}=\frac{1}{2}=\frac{x}{x+\sqrt{3}} \Longrightarrow x=\sqrt{3}$$ where we notice $\sqrt{3}$ is the distance from $O$ to the intersection of $A B$ with the line through $O$ and the midpoint of $B C$. (3): $P_{1} \cap P_{4}$, two parabolas with directrices edges opposite on the hexagon, say $A B$ and $D E$. Clearly the two intersection points are both inside the hexagon and must lie on $C F$, which gives $$x=\frac{\sqrt{3}}{2}$$ These together give that the sum desired is $$6(2 \sqrt{3}-3)+6(2 \sqrt{3}+3)+6\left(\frac{\sqrt{3}}{3}\right)+6(\sqrt{3})+6\left(\frac{\sqrt{3}}{2}\right)=35 \sqrt{3}$$ | 35 \sqrt{3} | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 4.5 | Let $x<y$ be positive real numbers such that $\sqrt{x}+\sqrt{y}=4$ and $\sqrt{x+2}+\sqrt{y+2}=5$. Compute $x$. | Adding and subtracting both equations gives $$\begin{aligned} & \sqrt{x+2}+\sqrt{x}+\sqrt{y+2}+\sqrt{y}=9 \\ & \sqrt{x+2}-\sqrt{x}+\sqrt{y+2}-\sqrt{y}=1 \end{aligned}$$ Substitute $a=\sqrt{x}+\sqrt{x+2}$ and $b=\sqrt{y}+\sqrt{y+2}$. Then since $(\sqrt{x+2}+\sqrt{x})(\sqrt{x+2}-\sqrt{x})=2$, we have $$\begin{gathered} a+b=9 \\ \frac{2}{a}+\frac{2}{b}=1 \end{gathered}$$ Dividing the first equation by the second one gives $$ab=18, a=3, b=6$$ Lastly, $\sqrt{x}=\frac{\sqrt{x+2}+\sqrt{x}-(\sqrt{x+2}-\sqrt{x})}{2}=\frac{3-\frac{2}{3}}{2}=\frac{7}{6}$, so $x=\frac{49}{36}$. | \frac{49}{36} | HMMT_2 |
[
"Mathematics -> Precalculus -> Functions"
] | 4 | Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be a function such that for any integers $x, y$, we have $f\left(x^{2}-3 y^{2}\right)+f\left(x^{2}+y^{2}\right)=2(x+y) f(x-y)$. Suppose that $f(n)>0$ for all $n>0$ and that $f(2015) \cdot f(2016)$ is a perfect square. Find the minimum possible value of $f(1)+f(2)$. | Plugging in $-y$ in place of $y$ in the equation and comparing the result with the original equation gives $(x-y) f(x+y)=(x+y) f(x-y)$. This shows that whenever $a, b \in \mathbb{Z}-\{0\}$ with $a \equiv b(\bmod 2)$, we have $\frac{f(a)}{a}=\frac{f(b)}{b}$ which implies that there are constants $\alpha=f(1) \in \mathbb{Z}_{>0}, \beta=f(2) \in \mathbb{Z}_{>0}$ for which $f$ satisfies the equation $(*)$: $f(n)= \begin{cases}n \cdot \alpha & \text { when } 2 \nmid n \\ \frac{n}{2} \cdot \beta & \text { when } 2 \mid n\end{cases}$. Therefore, $f(2015) f(2016)=2015 \alpha \cdot 1008 \beta=2^{4} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 13 \cdot 31 \alpha \beta$, so $\alpha \beta=5 \cdot 7 \cdot 13 \cdot 31 \cdot t^{2}$ for some $t \in \mathbb{Z}_{>0}$. We claim that $(\alpha, \beta, t)=(5 \cdot 31,7 \cdot 13,1)$ is a triple which gives the minimum $\alpha+\beta$. In particular, we claim $\alpha+\beta \geq 246$. Consider the case $t \geq 2$ first. We have, by AM-GM, $\alpha+\beta \geq 2 \cdot \sqrt{\alpha \beta} \geq 4 \cdot \sqrt{14105}>246$. Suppose $t=1$. We have $\alpha \cdot \beta=5 \cdot 7 \cdot 13 \cdot 31$. Because $(\alpha+\beta)^{2}-(\alpha-\beta)^{2}=4 \alpha \beta$ is fixed, we want to have $\alpha$ as close as $\beta$ as possible. This happens when one of $\alpha, \beta$ is $5 \cdot 31$ and the other is $7 \cdot 13$. In this case, $\alpha+\beta=91+155=246$. Finally, we note that the equality $f(1)+f(2)=246$ can be attained. Consider $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(n)=91 n$ for every odd $n \in \mathbb{Z}$ and $f(n)=\frac{155}{2} n$ for every even $n \in \mathbb{Z}$. It can be verified that $f$ satisfies the condition in the problem and $f(1)+f(2)=246$ as claimed. | 246 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes"
] | 5 | Determine the number of unordered triples of distinct points in the $4 \times 4 \times 4$ lattice grid $\{0,1,2,3\}^{3}$ that are collinear in $\mathbb{R}^{3}$ (i.e. there exists a line passing through the three points). | Define a main plane to be one of the $x y, y z, z x$ planes. Define a space diagonal to be a set of collinear points not parallel to a main plane. We classify the lines as follows: (a) Lines parallel to two axes (i.e. orthogonal to a main plane). Notice that given a plane of the form $v=k$, where $v \in\{x, y, z\}, k \in\{0,1,2,3\}$, there are 8 such lines, four in one direction and four in a perpendicular direction. There are $4 \times 3=12$ such planes. However, each line lies in two of these $(v, k)$ planes, so there are $\frac{8 \times 4 \times 3}{2}=48$ such lines. Each of these lines has 4 points, so there are 4 possible ways to choose 3 collinear points, giving $4 \times 48=192$ triplets. (b) Diagonal lines containing four points parallel to some main plane. Consider a plane of the form $(v, k)$, as defined above. These each have 2 diagonals that contain 4 collinear points. Each of these diagonals uniquely determines $v, k$ so these diagonals are each counted once. There are 12 possible $(v, k)$ pairs, yielding $12 \times 2 \times 4=96$ triplets. (c) Diagonal lines containing three points parallel to some main plane. Again, consider a plane $(v, k)$. By inspection, there are four such lines and one way to choose the triplet of points for each of these lines. This yields $4 \times 12=48$ triplets. (d) Main diagonals. There are four main diagonals, each with 4 collinear points, yielding $4 \times 4=16$ triplets. (e) Space diagonals containing three points. Choose one of the points in the set $\{1,2\}^{3}$ to be the midpoint of the line. Since these 8 possibilities are symmetric, say we take the point $(1,1,1)$. There are four space diagonals passing through this point, but one is a main diagonal. So each of the 8 points has 3 such diagonals with 3 points each, yielding $8 \times 3=24$ ways. Adding all these yields $192+96+48+16+24=376$. | 376 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 3.5 | Find the number of pentominoes (5-square polyominoes) that span a 3-by-3 rectangle, where polyominoes that are flips or rotations of each other are considered the same polyomino. | By enumeration, the answer is 6. | 6 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5 | Given two distinct points $A, B$ and line $\ell$ that is not perpendicular to $A B$, what is the maximum possible number of points $P$ on $\ell$ such that $A B P$ is an isosceles triangle? | In an isosceles triangle, one vertex lies on the perpendicular bisector of the opposite side. Thus, either $P$ is the intersection of $A B$ and $\ell$, or $P$ lies on the circle centered at $A$ with radius $A B$, or $P$ lies on the circle centered at $B$ with radius $A B$. Each circle-line intersection has at most two solutions, and the line-line intersection has at most one, giving 5. This can be easily constructed by taking any $\overline{A B}$, and taking $\ell$ that isn't a diameter but intersects both relevant circles twice. | 5 | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Other",
"Mathematics -> Number Theory -> Congruences"
] | 5 | Let $f: \mathbb{Z}^{2} \rightarrow \mathbb{Z}$ be a function such that, for all positive integers $a$ and $b$, $$f(a, b)= \begin{cases}b & \text { if } a>b \\ f(2 a, b) & \text { if } a \leq b \text { and } f(2 a, b)<a \\ f(2 a, b)-a & \text { otherwise }\end{cases}$$ Compute $f\left(1000,3^{2021}\right)$. | Note that $f(a, b)$ is the remainder of $b$ when divided by $a$. If $a>b$ then $f(a, b)$ is exactly $b$ $\bmod a$. If instead $a \leq b$, our "algorithm" doubles our $a$ by $n$ times until we have $a \times 2^{n}>b$. At this point, we subtract $a^{\overline{2 n-1}}$ from $f\left(a \cdot 2^{n}, b\right)$ and iterate back down until we get $a>b-a \cdot k>0$ and $f(a, b)=b-a \cdot k$ for some positive integer $k$. This expression is equivalent to $b-a \cdot k \bmod a$, or $b \bmod a$. Thus, we want to compute $3^{2021} \bmod 1000$. This is equal to $3 \bmod 8$ and $78 \bmod 125$. By CRT, this implies that the answer is 203. | 203 | HMMT_2 |
[
"Mathematics -> Algebra -> Number Theory -> Other"
] | 3.5 | I have written a strictly increasing sequence of six positive integers, such that each number (besides the first) is a multiple of the one before it, and the sum of all six numbers is 79 . What is the largest number in my sequence? | If the fourth number is \geq 12, then the last three numbers must sum to at least $12+$ $2 \cdot 12+2^{2} \cdot 12=84>79$. This is impossible, so the fourth number must be less than 12. Then the only way we can have the required divisibilities among the first four numbers is if they are $1,2,4,8$. So the last two numbers now sum to $79-15=64$. If we call these numbers $8 a, 8 a b(a, b>1)$ then we get $a(1+b)=a+a b=8$, which forces $a=2, b=3$. So the last two numbers are 16,48. | 48 | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Other"
] | 5 | Compute the number of real solutions $(x, y, z, w)$ to the system of equations: $$\begin{array}{rlrl} x & =z+w+z w x & z & =x+y+x y z \\ y & =w+x+w x y & w & =y+z+y z w \end{array}$$ | The first equation rewrites as $x=\frac{w+z}{1-w z}$, which is a fairly strong reason to consider trigonometric substitution. Let $x=\tan (a), y=\tan (b), z=\tan (c)$, and $w=\tan (d)$, where $-90^{\circ}<a, b, c, d<90^{\circ}$. Under modulo $180^{\circ}$, we find $a \equiv c+d ; b \equiv$ $d+a ; c \equiv a+b ; d \equiv b+c$. Adding all of these together yields $a+b+c+d \equiv 0$. Then $a \equiv c+d \equiv-a-b$ so $b \equiv-2 a$. Similarly, $c \equiv-2 b ; d \equiv-2 c ; d \equiv-2 a$. Hence, $c \equiv-2 b \equiv 4 a, d \equiv-2 c \equiv-8 a$, and $a \equiv-2 d \equiv 16 a$, so the only possible solutions are $(a, b, c, d) \equiv(t,-2 t, 4 t,-8 t)$ where $15 t \equiv 0$. Checking, these, we see that actually $5 t \equiv 0$, which yields 5 solutions. Our division by $1-y z$ is valid since $1-y z=0$ iff $y z=1$, but $x=y+z+x y z$ so $y=-z$, which implies that $y z \leq 0<1$, which is impossible. (The solutions we have computed are in fact $(0,0,0,0)$ and the cyclic permutations of $\left.\left(\tan \left(36^{\circ}\right), \tan \left(-72^{\circ}\right), \tan \left(-36^{\circ}\right), \tan \left(72^{\circ}\right)\right).\right)$ | 5 | HMMT_2 |
[
"Mathematics -> Algebra -> Prealgebra -> Arithmetic Sequences -> Other"
] | 5 | Determine the value of $$1 \cdot 2-2 \cdot 3+3 \cdot 4-4 \cdot 5+\cdots+2001 \cdot 2002$$ | 2004002. Rewrite the expression as $$2+3 \cdot(4-2)+5 \cdot(6-4)+\cdots+2001 \cdot(2002-2000)$$ $$=2+6+10+\cdots+4002$$ This is an arithmetic progression with $(4002-2) / 4+1=1001$ terms and average 2002, so its sum is $1001 \cdot 2002=2004002$. | 2004002 | HMMT_2 |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 5 | A convex quadrilateral is drawn in the coordinate plane such that each of its vertices $(x, y)$ satisfies the equations $x^{2}+y^{2}=73$ and $x y=24$. What is the area of this quadrilateral? | The vertices all satisfy $(x+y)^{2}=x^{2}+y^{2}+2 x y=73+2 \cdot 24=121$, so $x+y= \pm 11$. Similarly, $(x-y)^{2}=x^{2}+y^{2}-2 x y=73-2 \cdot 24=25$, so $x-y= \pm 5$. Thus, there are four solutions: $(x, y)=(8,3),(3,8),(-3,-8),(-8,-3)$. All four of these solutions satisfy the original equations. The quadrilateral is therefore a rectangle with side lengths of $5 \sqrt{2}$ and $11 \sqrt{2}$, so its area is 110. | 110 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other",
"Mathematics -> Algebra -> Intermediate Algebra -> Logarithmic Functions"
] | 5.25 | Yannick picks a number $N$ randomly from the set of positive integers such that the probability that $n$ is selected is $2^{-n}$ for each positive integer $n$. He then puts $N$ identical slips of paper numbered 1 through $N$ into a hat and gives the hat to Annie. Annie does not know the value of $N$, but she draws one of the slips uniformly at random and discovers that it is the number 2. What is the expected value of $N$ given Annie's information? | Let $S$ denote the value drawn from the hat. The probability that 2 is picked is $\frac{1}{n}$ if $n \geq 2$ and 0 if $n=1$. Thus, the total probability $X$ that 2 is picked is $$P(S=2)=\sum_{k=2}^{\infty} \frac{2^{-k}}{k}$$ By the definition of conditional probability, $P(N=n \mid S=2)=\frac{P(N=n, S=2)}{P(S=2)}=\frac{2^{-n} / n}{X}$ if $n \geq 2$ and 0 if $n=1$. Thus the conditional expectation of $N$ is $$\mathbb{E}[N \mid S=2]=\sum_{n=1}^{\infty} n \cdot P(N=n \mid S=2)=\sum_{n=2}^{\infty} n \cdot \frac{2^{-n} / n}{X}=\frac{1}{X} \sum_{n=2}^{\infty} 2^{-n}=\frac{1}{2 X}$$ It remains to compute $X$. Note that $\sum_{k=0}^{\infty} x^{k}=\frac{1}{1-x}$ for $|x|<1$. Integrating both sides with respect to $x$ yields $$\sum_{k=1}^{\infty} \frac{x^{k}}{k}=-\ln (1-x)+C$$ for some constant $C$, and plugging in $x=0$ shows that $C=0$. Plugging in $x=\frac{1}{2}$ shows that $\sum_{k=1}^{\infty} \frac{2^{-k}}{k}=\ln 2$. Note that $X$ is exactly this summation but without the first term. Thus, $X=\ln 2-\frac{1}{2}$, so $\frac{1}{2 X}=\frac{1}{2 \ln 2-1}$. | \frac{1}{2 \ln 2-1} | HMMT_2 |
[
"Mathematics -> Algebra -> Abstract Algebra -> Field Theory",
"Mathematics -> Number Theory -> Congruences"
] | 4 | For integers $a, b, c, d$, let $f(a, b, c, d)$ denote the number of ordered pairs of integers $(x, y) \in \{1,2,3,4,5\}^{2}$ such that $a x+b y$ and $c x+d y$ are both divisible by 5. Find the sum of all possible values of $f(a, b, c, d)$. | Standard linear algebra over the field $\mathbb{F}_{5}$ (the integers modulo 5). The dimension of the solution set is at least 0 and at most 2, and any intermediate value can also be attained. So the answer is $1+5+5^{2}=31$. This also can be easily reformulated in more concrete equation/congruence-solving terms, especially since there are few variables/equations. | 31 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 5.5 | Let $AD, BE$, and $CF$ be segments sharing a common midpoint, with $AB < AE$ and $BC < BF$. Suppose that each pair of segments forms a $60^{\circ}$ angle, and that $AD=7, BE=10$, and $CF=18$. Let $K$ denote the sum of the areas of the six triangles $\triangle ABC, \triangle BCD, \triangle CDE, \triangle DEF, \triangle EFA$, and $\triangle FAB$. Compute $K \sqrt{3}$. | Let $M$ be the common midpoint, and let $x=7, y=10, z=18$. One can verify that hexagon $ABCDEF$ is convex. We have $[ABC]=[ABM]+[BCM]-[ACM]=\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{x}{2} \cdot \frac{y}{2}+\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{y}{2} \cdot \frac{z}{2}-\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{x}{2} \cdot \frac{z}{2}=\frac{\sqrt{3}(xy+yz-zx)}{16}$. Summing similar expressions for all 6 triangles, we have $$K=\frac{\sqrt{3}(2xy+2yz+2zx)}{16}$$ Substituting $x, y, z$ gives $K=47 \sqrt{3}$, for an answer of 141. | 141 | HMMT_2 |
[
"Mathematics -> Number Theory -> Greatest Common Divisors (GCD)"
] | 5.25 | Two integers are relatively prime if they don't share any common factors, i.e. if their greatest common divisor is 1. Define $\varphi(n)$ as the number of positive integers that are less than $n$ and relatively prime to $n$. Define $\varphi_{d}(n)$ as the number of positive integers that are less than $d n$ and relatively prime to $n$. What is the least $n$ such that $\varphi_{x}(n)=64000$, where $x=\varphi_{y}(n)$, where $y=\varphi(n)$? | For fixed $n$, the pattern of integers relatively prime to $n$ repeats every $n$ integers, so $\varphi_{d}(n)=d \varphi(n)$. Therefore the expression in the problem equals $\varphi(n)^{3}$. The cube root of 64000 is $40 . \varphi(p)=p-1$ for any prime $p$. Since 40 is one less than a prime, the least $n$ such that $\varphi(n)=40$ is 41. | 41 | HMMT_2 |
Subsets and Splits