id string | answer string | question string | solution string |
|---|---|---|---|
1 | 103 | Compute the sum of the positive divisors (including $1$) of $9!$ that have units digit $1$. | The prime factorization of $9!$ is $2^7 \cdot 3^4 \cdot 5 \cdot 7$. Every divisor of $9!$ has prime factorization $2^a \cdot 3^b \cdot 5^c \cdot 7^d$, where $0 \le a \le 7$, $0 \le b \le 4$, $0 \le c \le 1$, and $0 \le d \le 1$. If the divisor has units digit 1, it cannot be divisible by 2 or 5, so $a=c=0$. Now take ca... |
2 | 3375 | Mark writes the expression $\sqrt{\underline{a b c d}}$ on the board, where $\underline{a b c d}$ is a four-digit number and $a \neq 0$. Derek, a toddler, decides to move the $a$, changing Mark's expression to $a \sqrt{\underline{b c d}}$. Surprisingly, these two expressions are equal. Compute the only possible four-di... | Let $x = \underline{bcd}$. Then, we rewrite the given condition $\sqrt{abcd} = a\sqrt{bcd}$ as $$ 1000a + x = a^2x, $$ which simplifies as $$ (a^2 - 1)x = 1000a. $$ In particular, $a^2 - 1$ divides $1000a$. Since $\gcd(a^2-1, a) = 1$, it follows that $a^2 - 1 \mid 1000$. The only $a \in \{1, 2, \dots, 9\}$ that satisfi... |
3 | \frac{1}{576} | Given that $x, y$, and $z$ are positive real numbers such that
$$
x^{\log _{2}(y z)}=2^{8} \cdot 3^{4}, \quad y^{\log _{2}(z x)}=2^{9} \cdot 3^{6}, \quad \text { and } \quad z^{\log _{2}(x y)}=2^{5} \cdot 3^{10}
$$
compute the smallest possible value of $x y z$. | Let $k = \log_2 3$ for brevity. Taking the base-2 log of each equation gives $(\log_2 x)(\log_2 y + \log_2 z) = 8 + 4k, \; (\log_2 y)(\log_2 z + \log_2 x) = 9 + 6k, \; (\log_2 z)(\log_2 x + \log_2 y) = 5 + 10k.$ Adding the first two equations and subtracting the third yields $2 \log_2 x \log_2 y = 12$, so $\log_2 x \lo... |
4 | -984 | Let $\lfloor z\rfloor$ denote the greatest integer less than or equal to $z$. Compute
$$
\sum_{j=-1000}^{1000}\left\lfloor\frac{2025}{j+0.5}\right\rfloor
$$ | The key idea is to pair up the terms $\lfloor \frac{2025}{-x} \rfloor$ and $\lfloor \frac{2025}{x} \rfloor$. There are 1000 such pairs and one lone term, $\lfloor \frac{2025}{1000.5} \rfloor = 2$. Thus, $$ \sum_{j=-1000}^{1000} \left\lfloor \frac{2025}{j+0.5} \right\rfloor = 2 + \sum_{x \in \{0.5, 1.5, \dots, 999.5\}} ... |
5 | 890 | Let $\mathcal{S}$ be the set of all nonconstant monic polynomials $P$ with integer coefficients satisfying $P(\sqrt{3}+\sqrt{2})=$ $P(\sqrt{3}-\sqrt{2})$. If $Q$ is an element of $\mathcal{S}$ with minimal degree, compute the only possible value of $Q(10)-Q(0)$. | First, note that the polynomial $x^4 - 10x^2 + 1$ has both $\sqrt{3}+\sqrt{2}$ and $\sqrt{3}-\sqrt{2}$ as roots. It suffices to check whether a polynomial of degree at most 3 belongs in $S$. Suppose $f(x) = ax^3 + bx^2 + cx + d \in S$. We compute $(\sqrt{3}+\sqrt{2})^3 - (\sqrt{3}-\sqrt{2})^3 = 22\sqrt{2}; \; (\sqrt{3}... |
6 | \frac{1311}{2017} | Let $r$ be the remainder when $2017^{2025!}-1$ is divided by 2025!. Compute $\frac{r}{2025!}$. (Note that $2017$ is prime.) | Solution: Let $N = 2017^{2025!}$. Let $p$ be a prime dividing $2025!$ other than $2017$. Let $p^k$ be the largest power of $p$ dividing $2025!$. Clearly, $\phi(p^k) = (p-1)p^{k-1}$ divides $2025!$ and $\gcd(2017, p^k) = 1$, so by Euler's Totient Theorem, $$ N \equiv 1 \pmod{p^k}. $$ Repeating for all such primes $p$, w... |
7 | \frac{9 \sqrt{23}}{23} | There exists a unique triple $(a, b, c)$ of positive real numbers that satisfies the equations
$$
2\left(a^{2}+1\right)=3\left(b^{2}+1\right)=4\left(c^{2}+1\right) \quad \text { and } \quad a b+b c+c a=1
$$
Compute $a+b+c$. | Solution 1: The crux of this problem is to apply the trigonometric substitutions $a = \cot \alpha$, $b = \cot \beta$, and $c = \cot \gamma$, with $0 < \alpha, \beta, \gamma < \pi/2$. Then, the given equations translate to $$ \frac{2}{\sin^2 \alpha} = \frac{3}{\sin^2 \beta} = \frac{4}{\sin^2 \gamma} \quad \text{and} \qu... |
8 | 1-\frac{2}{\pi} | Define $\operatorname{sgn}(x)$ to be $1$ when $x$ is positive, $-1$ when $x$ is negative, and $0$ when $x$ is $0$. Compute
$$
\sum_{n=1}^{\infty} \frac{\operatorname{sgn}\left(\sin \left(2^{n}\right)\right)}{2^{n}}
$$
(The arguments to sin are in radians.) | Note that each of following is equivalent to the next. $sgn(\sin(2^n)) = +1$. $0 < 2^n \pmod{2\pi} < \pi$. $0 < 2^n/\pi \pmod 2 < 1$. $The $n$th digit after the decimal point in the binary representation of $\frac{1}{\pi}$ is 0. Similarly, $sgn(\sin(2^n)) = -1$ if and only if the $n$-th digit after the decimal point in... |
9 | 1037 | Let $f$ be the unique polynomial of degree at most $2026$ such that for all $n \in\{1,2,3, \ldots, 2027\}$,
$$
f(n)= \begin{cases}1 & \text { if } n \text { is a perfect square } \\ 0 & \text { otherwise }\end{cases}
$$
Suppose that $\frac{a}{b}$ is the coefficient of $x^{2025}$ in $f$, where $a$ and $b$ are integers ... | Let $p=2027$. We work in $\mathbb{F}_p$ for the entire solution. Recall the well-known fact that $$ \sum_{x \in \mathbb{F}_p} x^k = \begin{cases} -1 & \text{if } k>0 \text{ and } p-1 \mid k, \\ 0 & \text{otherwise,} \end{cases} $$ assuming $0^0=1$. In particular, for any polynomial $g(x) = b_0 + b_1x + \dots + b_nx^n$,... |
10 | \frac{-1+\sqrt{17}}{2}, \frac{-1-\sqrt{17}}{2} | Let $a, b$, and $c$ be pairwise distinct complex numbers such that
$$
a^{2}=b+6, \quad b^{2}=c+6, \quad \text { and } \quad c^{2}=a+6
$$
Compute the two possible values of $a+b+c$. In your answer, list the two values in a comma-separated list of two valid \LaTeX expressions. | Notice that any of $a$, $b$, or $c$ being 3 or $-2$ implies $a=b=c$, which is invalid. Thus, $(a^2-9)(b^2-9)(c^2-9) = (b-3)(c-3)(a-3) \implies (a+3)(b+3)(c+3)=1; \; (a^2-4)(b^2-4)(c^2-4) = (b+2)(c+2)(a+2) \implies (a-2)(b-2)(c-2)=1.$ Therefore, 2 and $-3$ are roots of the polynomial $(x-a)(x-b)(x-c)+1$, and so there ex... |
11 | 56 | Compute the number of ways to arrange the numbers $1,2,3,4,5,6$, and $7$ around a circle such that the product of every pair of adjacent numbers on the circle is at most 20. (Rotations and reflections count as different arrangements.) | Fix the position of the number $7$. Note that the only numbers that can be next to $7$ are $1$ and $2$, so they must occupy the two slots adjacent to $7$. Now the number $6$ can only be adjacent to $1, 2,$ and $3$. As $6$ can no longer be adjacent to both $1$ and $2$, we conclude that $6$ must be adjacent to $3$ as wel... |
12 | 29 | Kevin the frog in on the bottom-left lily pad of a $3 \times 3$ grid of lily pads, and his home is at the topright lily pad. He can only jump between two lily pads which are horizontally or vertically adjacent. Compute the number of ways to remove $4$ of the lily pads so that the bottom-left and top-right lily pads bot... | We instead count the arrangements for which Kelvin can get home. Note that at minimum, Kelvin must use $5$ lily pads to get home, leaving $4$ lily pads that are not on the path. This means that if we were to remove $4$ lily pads and Kelvin can still get home, the non-removed lily pads form a shortest path from the bott... |
13 | 105 | Ben has $16$ balls labeled $1,2,3, \ldots, 16$, as well as $4$ indistinguishable boxes. Two balls are \emph{neighbors} if their labels differ by $1$. Compute the number of ways for him to put $4$ balls in each box such that each ball is in the same box as at least one of its neighbors. (The order in which the balls are... | Each box must either contain a single group of four consecutive balls (e.g., $5, 6, 7, 8$) or two groups of two consecutive balls (e.g., $5, 6, 9, 10$). Since all groups have even lengths, this means that $1$ and $2$ are in the same group, $3$ and $4$ are in the same group, and so on. We can think of each of these $8$ ... |
14 | 2304 | Sophie is at $(0,0)$ on a coordinate grid and would like to get to $(3,3)$. If Sophie is at $(x, y)$, in a single step she can move to one of $(x+1, y),(x, y+1),(x-1, y+1)$, or $(x+1, y-1)$. She cannot revisit any points along her path, and neither her $x$-coordinate nor her $y$-coordinate can ever be less than $0$ or ... | Let a lateral move refer to one which is either up or right. Then the lateral moves are the only ones which increase Sophie's sum of coordinates by $1$, while all other moves (diagonal) do not change the sum, so Kelvin must make $6$ of them, one to increase this sum from $i$ to $i+1$ for each $i \in [0, 5]$. We claim t... |
15 | 200 | In an $11 \times 11$ grid of cells, each pair of edge-adjacent cells is connected by a door. Karthik wants to walk a path in this grid. He can start in any cell, but he must end in the same cell he started in, and he cannot go through any door more than once (not even in opposite directions). Compute the maximum number... | This is simply asking for the longest circuit in the adjacency graph of this grid. Note that this grid has $4 \cdot 9 = 36$ cells of odd degree, 9 along each side. If we color the cells with checkerboard colors so that the corners are black, then $20$ of these $36$ odd-degree cells are white. An Eulerian circuit uses a... |
16 | 6300 | Compute the number of ways to pick two rectangles in a $5 \times 5$ grid of squares such that the edges of the rectangles lie on the lines of the grid and the rectangles do not overlap at their interiors, edges, or vertices. The order in which the rectangles are chosen does not matter. | A rectangle can be specified by two intervals, one specifying its horizontal extent (x-coordinates of left and right sides) and one specifying its vertical extent (y-coordinates of bottom and top sides). For the rectangles to not overlap, we need either the horizontal intervals or the vertical intervals to be disjoint ... |
17 | 2^{25} \cdot 26! | Compute the number of ways to arrange $3$ copies of each of the $26$ lowercase letters of the English alphabet such that for any two distinct letters $x_{1}$ and $x_{2}$, the number of $x_{2}$ 's between the first and second occurrences of $x_{1}$ equals the number of $x_{2}$ 's between the second and third occurrences... | First, we prove such a string can be divided into blocks where each block consists of the same substring written three times. We prove the following lemma.
Lemma1: For any letter $x_1$, the strings between the first and second occurrences of $x_1$ and between the second and third occurrences of $x_1$ are the same.
... |
18 | \frac{2025}{101} | Albert writes $2025$ numbers $a_{1}, \ldots, a_{2025}$ in a circle on a blackboard. Initially, each of the numbers is uniformly and independently sampled at random from the interval $[0,1]$. Then, each second, he \emph{simultaneously} replaces $a_{i}$ with $\max \left(a_{i-1}, a_{i}, a_{i+1}\right)$ for all $i=1,2, \ld... | Solution: We can assume that the initial numbers are all distinct, since this occurs with probability 1. For clarity, we denote the value of $a_i$ after $t$ seconds as $a_{i,t}$. The index $i$ is taken mod 2025.
In general, after $k < 1012$ seconds, we claim the expected number of distinct values remaining is $\frac{2... |
19 | \frac{4}{9} | Two points are selected independently and uniformly at random inside a regular hexagon. Compute the probability that a line passing through both of the points intersects a pair of opposite edges of the hexagon. | First, we compute the probability that the line through two random points in a triangle $ABC$ passes through segments $\overline{AB}$ and $\overline{AC}$. We can take an affine transform of the two random points and the triangle such that $ABC$ becomes equilateral. Since the distribution of the two points is still unif... |
20 | \frac{448}{3} | The circumference of a circle is divided into $45$ arcs, each of length $1$. Initially, there are $15$ snakes, each of length $1$, occupying every third arc. Every second, each snake independently moves either one arc left or one arc right, each with probability $\frac{1}{2}$. If two snakes ever touch, they merge to fo... | Solution: We solve the problem generally for $n$ snakes and $3n$ arcs. Without loss of generality, fix the two snakes $A$ and $B$ that will eventually form the ends of the last snake. Note that $A$ and $B$ must be consecutive in the initial configuration; assume $B$ lies immediately clockwise from $A$.
Let $d$ be the ... |
21 | 26 | Equilateral triangles $\triangle A B C$ and $\triangle D E F$ are drawn such that points $B, E, F$, and $C$ lie on a line in this order, and point $D$ lies inside triangle $\triangle A B C$. If $B E=14, E F=15$, and $F C=16$, compute $A D$. | Extend $DE$ to meet $AC$ at $X$. Observe that $ABEX$ and $DFCX$ are isosceles trapezoids (both with base angles of $60^{\circ}$), so we have
* $AX = BE = 14$,
* $DX = FC = 16$, and
* $\angle AXD = 120^{\circ}$.
By Law of Cosines on $\triangle ADX$, the answer is
$$ AD = \sqrt{AX^2 + DX^2 - 2\cos(120^{\circ}... |
22 | 63 | In a two-dimensional cave with a parallel floor and ceiling, two stalactites of lengths $16$ and $36$ hang perpendicularly from the ceiling, while two stalagmites of heights $25$ and $49$ grow perpendicularly from the ground. If the tips of these four structures form the vertices of a square in some order, compute the ... | Note that the difference in heights between the two stalactites does not equal the difference in heights between the stalagmites. This tells us that the two stalactites form a pair of opposite vertices of the square, and likewise for the stalagmites. As the midpoint of the pairs of structures must then coincide, we kno... |
23 | 8\sqrt{10} | Point $P$ lies inside square $A B C D$ such that the areas of $\triangle P A B, \triangle P B C, \triangle P C D$, and $\triangle P D A$ are 1, $2,3$, and $4$, in some order. Compute $P A \cdot P B \cdot P C \cdot P D$. | Let $h_1, h_2, h_3,$ and $h_4$ be the lengths of the altitudes from $P$ to sides $AB, BC, CD,$ and $DA$, respectively. Then, the problem statement implies that $\{h_1, h_2, h_3, h_4\} = \{x, 2x, 3x, 4x\}$ for some $x$. Furthermore, the area of the square is $1+2+3+4 = 10$, so we have
$$ h_1+h_3 = \sqrt{10} = h_2+h_4. ... |
24 | 20 | A semicircle is inscribed in another semicircle if the smaller semicircle's diameter is a chord of the larger semicircle, and the smaller semicircle's arc is tangent to the diameter of the larger semicircle.
Semicircle $S_{1}$ is inscribed in a semicircle $S_{2}$, which is inscribed in another semicircle $S_{3}$. The r... | Let $P, Q,$ and $R$ be the midpoints of the diameters (i.e., the center of the circular arcs) of $S_3, S_2,$ and $S_1$, respectively. Observe that if one fixes $S_3$, the location of $S_2$ is uniquely determined by the angle between the diameters of $S_2$ and $S_3$. The same holds for $S_2$ and $S_1$. Thus, the figures... |
25 | \sqrt{23}-2 \sqrt{3} | Let $\triangle A B C$ be an equilateral triangle with side length $6$. Let $P$ be a point inside triangle $\triangle A B C$ such that $\angle B P C=120^{\circ}$. The circle with diameter $\overline{A P}$ meets the circumcircle of $\triangle A B C$ again at $X \neq A$. Given that $A X=5$, compute $X P$. | Let $A'$ be the antipode of $A$. As $\angle AXA' = 90^{\circ}$, we have $X, P,$ and $A'$ are collinear. As $\angle BPC = 120^{\circ}$ and $\angle BA'C = 180^{\circ} - \angle BAC = 120^{\circ}$, it follows that $P$ lies on the circle with center $A'$ passing through $B$ and $C$, so $A'P = \frac{BC}{\sqrt{3}} = 2\sqrt{3}... |
26 | 9\sqrt{15} | Trapezoid $A B C D$, with $A B \| C D$, has side lengths $A B=11, B C=8, C D=19$, and $D A=4$. Compute the area of the convex quadrilateral whose vertices are the circumcenters of $\triangle A B C, \triangle B C D$, $\triangle C D A$, and $\triangle D A B$. | Let $O_A, O_B, O_C,$ and $O_D$ be the circumcenters of $\triangle BCD, \triangle CDA, \triangle DAB,$ and $\triangle ABC$, respectively. Note that $O_B O_C$ is the perpendicular bisector of $AD$. Similarly, $O_B O_D \perp AC$ and $O_C O_D \perp AB$. As $AB \parallel CD$, we have $O_C O_D \perp CD$. Then, $\triangle O_B... |
27 | \frac{7}{18} | Point $P$ is inside triangle $\triangle A B C$ such that $\angle A B P=\angle A C P$. Given that $A B=6, A C=8, B C=7$, and $\frac{B P}{P C}=\frac{1}{2}$, compute $\frac{[B P C]}{[A B C]}$.
(Here, $[X Y Z]$ denotes the area of $\triangle X Y Z$ ). | Let the internal and external bisectors of $\angle BAC$ meet $BC$ at $D$ and $E$. Similarly, let the internal and external bisectors of $\angle BPC$ meet $BC$ at $D'$ and $E'$. The angle condition implies that $AD \parallel PD'$ and $AE \parallel PE'$. Thus, triangles $ADE$ and $PD'E'$ are homothetic. Hence, the reques... |
28 | \sqrt{6} | Let $A B C D$ be an isosceles trapezoid such that $C D>A B=4$. Let $E$ be a point on line $C D$ such that $D E=2$ and $D$ lies between $E$ and $C$. Let $M$ be the midpoint of $\overline{A E}$. Given that points $A, B, C, D$, and $M$ lie on a circle with radius $5$, compute $M D$. | Let $D'$ be the reflection of $D$ across $M$. Then, $ADED'$ is a parallelogram. Hence, $D'A = 2$, so $D'B = 6$. Thus, if $D'M = MD = x$, then Power of a Point at $D'$ gives $x \cdot (2x) = 2 \cdot 6$, so $x = \boxed{\sqrt{6}}$. *Remark*. The radius of the circle is unnecessary. |
29 | 14+4\sqrt{37} | Let $A B C D$ be a rectangle with $B C=24$. Point $X$ lies inside the rectangle such that $\angle A X B=90^{\circ}$. Given that triangles $\triangle A X D$ and $\triangle B X C$ are both acute and have circumradii $13$ and $15$, respectively, compute $A B$. | Solution 1: Let $M$ be the midpoint of $AB$. Let $O_1$ and $O_2$ be the circumcenters of $\triangle AXD$ and $\triangle BXC$, respectively. Since $O_1M$ is the perpendicular bisector of $AX$ and $O_2M$ is the perpendicular bisector of $BX$, we get that $\angle O_1MO_2 = 90^\circ$. Let $P_1$ and $P_2$ be the projections... |
30 | \sqrt{\frac{95}{24}} | A plane $\mathcal{P}$ intersects a rectangular prism at a hexagon which has side lengths $45,66,63,55,54$, and 77, in that order. Compute the distance from the center of the rectangular prism to $\mathcal{P}$. | Translate $\mathcal{P}$ so that it contains the center. The intersection of the translated plane with the rectangular prism is a centrally symmetric hexagon. Let its side lengths be $a$, $b$, $c$, $a$, $b$, and $c$, in that order. Then, for some $t_a$, $t_b$, and $t_c$, the side lengths of the hexagon before the transl... |
YAML Metadata Warning:The task_categories "text2text-generation" is not in the official list: text-classification, token-classification, table-question-answering, question-answering, zero-shot-classification, translation, summarization, feature-extraction, text-generation, fill-mask, sentence-similarity, text-to-speech, text-to-audio, automatic-speech-recognition, audio-to-audio, audio-classification, audio-text-to-text, voice-activity-detection, depth-estimation, image-classification, object-detection, image-segmentation, text-to-image, image-to-text, image-to-image, image-to-video, unconditional-image-generation, video-classification, reinforcement-learning, robotics, tabular-classification, tabular-regression, tabular-to-text, table-to-text, multiple-choice, text-ranking, text-retrieval, time-series-forecasting, text-to-video, image-text-to-text, image-text-to-image, image-text-to-video, visual-question-answering, document-question-answering, zero-shot-image-classification, graph-ml, mask-generation, zero-shot-object-detection, text-to-3d, image-to-3d, image-feature-extraction, video-text-to-text, keypoint-detection, visual-document-retrieval, any-to-any, video-to-video, other
Dataset Summary
This dataset comprises the questions, answers, and solutions from HMMT February 2025, all of which were extracted by OCR, converted to LaTeX, and manually verified by FlagEval Team.
Data Fields
Below one can find the description of each field in the dataset.
id(str): Index of the problem in the competitionproblem(str): Full problem statementanswer(str): Ground-truth answer to the questionsolution(str): Ground-truth solution to the question
Licensing Information
This dataset is licensed under the Attribution-NonCommercial-ShareAlike 4.0 International (CC BY-NC-SA 4.0).
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