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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
AMC | 0.205451 | 0.023824 | 0.187673 | AMC12 | 12A | 2,012 | N/A | 3 | A box $2$ centimeters high, $3$ centimeters wide, and $5$ centimeters long can hold $40$ grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold $n$ grams of clay. What is $n$? | 240 | The first box has volume $2\times3\times5=30\text{ cm}^3$, and the second has volume $(2\times2)\times(3\times3)\times(5)=180\text{ cm}^3$. The second has a volume that is $6$ times greater, so it holds $6\times40=\fbox{240}$ grams. | AMC12 First Half | AMC12 A | 83.82 | 2.015947 | 0.148423 | 1.5 | 2 | false |
HMMT | 0.612865 | 0.055438 | 0.726541 | HMMT-Nov | team | 2,017 | Nov | 6 | Consider five-dimensional Cartesian space \[ \mathbb{R}^{5}=\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right) \mid x_{i} \in \mathbb{R}\right\} \] and consider the hyperplanes with the following equations: \begin{itemize} $x_{i}=x_{j}$ for every $1 \leq i<j \leq 5$; $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=-1$; $x_{1}+x_{2}+... | 480 | (Joint with Junyao Peng) Note that given a set of plane equations $P_{i}\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=0$, for $i=1,2, \ldots, n$, each region that the planes separate the space into correspond to a $n$-tuple of -1 and 1 , representing the sign of $P_{1}, P_{2}, \ldots P_{n}$ for all points in that regi... | HMMT Nov Team | HMMT-Nov Team | 4.225352 | 4.554144 | 0.345382 | 4 | 5.5 | false |
AMC | 0.194194 | 0.041502 | 0.16805 | AMC10 | 10A | 2,016 | N/A | 2 | For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$? | 3 | We can rewrite $10^{x}\cdot 100^{2x}=1000^{5}$ as $10^{5x}=10^{15}$: \[\begin{split} 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ 10^x\cdot10^{4x} & =(10^3)^5 \\ 10^{5x} & =10^{15} \end{split}\] Since the bases are equal, we can set the exponents equal, giving us $5x=15$. Solving the equation gives us $x = \fbox{3}.$ | AMC10 First Half | AMC10 A | 62.87 | 1.945815 | 0.258557 | 1 | 2 | false |
HMMT | 0.582458 | 0.082624 | 0.698491 | HMMT-Nov | guts | 2,017 | Nov | 35 | Rebecca has twenty-four resistors, each with resistance $1 \mathrm{ohm}$. Every minute, she chooses any two resistors with resistance of $a$ and $b$ ohms respectively, and combine them into one by one of the following methods: \begin{itemize} Connect them in series, which produces a resistor with resistance of $a+b$ oh... | 1015080877 | This is the same problem as in OEIS A153588. It is helpful to see (or guess) that neither the numerator or the denominator of the final resistance exceed the $(n+1)$-th Fibonacci number, which in this case is $F_{25}=75025$, using concepts on the line of continued fractions. So $75025^{2} \approx 5.6 \times 10^{9}$ is ... | HMMT Nov Guts | HMMT-Nov Guts | 1.408451 | 4.364709 | 0.514749 | 3.5 | 6 | false |
HMMT | 0.945273 | 0.106344 | 0.998616 | HMMT-Feb | alg | 2,010 | Feb | 10 | Let $p(x)$ and $q(x)$ be two cubic polynomials such that $p(0)=-24, q(0)=30$, and \[ p(q(x))=q(p(x)) \] for all real numbers $x$. Find the ordered pair $(p(3), q(6))$. | (3,-24) | Note that the polynomials $f(x)=a x^{3}$ and $g(x)=-a x^{3}$ commute under composition. Let $h(x)=x+b$ be a linear polynomial, and note that its inverse $h^{-1}(x)=x-b$ is also a linear polynomial. The composite polynomials $h^{-1} f h$ and $h^{-1} g h$ commute, since function composition is associative, and these poly... | HMMT Feb Hard | HMMT-Feb Algebra | 0 | 6.625052 | 0.662526 | 5.5 | 6.5 | false |
AIME | 0.497252 | 0.049183 | 0.59673 | AIME | I | 2,016 | N/A | 5 | Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she comp... | 53 | Let $d$ be the number of days Anh reads for. Because the difference between the number of pages and minutes Anh reads for each day stays constant and is an integer, $d$ must be a factor of the total difference, which is $374-319=55$. Also note that the number of pages Anh reads is $dn+\frac{d(d-1)}{2}$. Similarly, t... | Easy AIME Problems | AIME | 84.21 | 3.833869 | 0.306409 | 3 | 3.5 | false |
AMC | 0.288558 | 0.09768 | 0.378868 | AMC10 | 10B | 2,018 | N/A | 24 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$? | \frac {15}{32}\sqrt{3} | [asy] /* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y);... | AMC10 Final Problems | AMC10 B | 5.08 | 2.533704 | 0.608548 | 3.5 | 4.5 | false |
HMMT | 0.721608 | 0.040236 | 0.820881 | HMMT-Feb | comb | 2,018 | Feb | 3 | A $4 \times 4$ window is made out of 16 square windowpanes. How many ways are there to stain each of the windowpanes, red, pink, or magenta, such that each windowpane is the same color as exactly two of its neighbors? Two different windowpanes are neighbors if they share a side. | 24 | For the purpose of explaining this solution, let's label the squares as \[ \begin{array}{llll} 11 & 12 & 13 & 14 \\ 21 & 22 & 23 & 24 \\ 31 & 32 & 33 & 34 \\ 41 & 42 & 43 & 44 \end{array} \] Note that since the corner squares $11,14,41,44$ each only have two neighbors, each corner square is the same color as both of it... | HMMT Feb Easy | HMMT-Feb Combinatorics | 43.026706 | 5.231614 | 0.250674 | 4.5 | 5.5 | false |
HMMT | 0.521526 | 0.073852 | 0.630692 | HMMT-Nov | team | 2,018 | Nov | 3 | For how many positive integers $n \leq 100$ is it true that $10 n$ has exactly three times as many positive divisors as $n$ has? | 28 | Let $n=2^{a} 5^{b} c$, where $2,5 \nmid c$. Then, the ratio of the number of divisors of $10 n$ to the number of divisors of $n$ is $\frac{a+2}{a+1} \frac{b+2}{b+1}=3$. Solving for $b$, we find that $b=\frac{1-a}{2 a+1}$. This forces $(a, b)=(0,1),(1,0)$. Therefore, the answers are of the form $2 k$ and $5 k$ whenever ... | HMMT Nov Team | HMMT-Nov Team | 47.019868 | 3.9851 | 0.460099 | 4 | 5.5 | false |
AMC | 0.224672 | 0.0828 | 0.220881 | AMC10 | 10B | 2,007 | N/A | 3 | A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip? | 24 | The trip was $240$ miles long and took $\dfrac{120}{30}+\dfrac{120}{20}=4+6=10$ gallons. Therefore, the average mileage was $\dfrac{240}{10}= \fbox{24}$ | AMC10 First Half | AMC10 B | 30.4 | 2.135692 | 0.515846 | 1 | 2 | false |
HMMT | 0.507951 | 0.041508 | 0.610314 | HMMT-Nov | gen | 2,009 | Nov | 6 | Find the maximum value of $x+y$, given that $x^{2}+y^{2}-3 y-1=0$. | \frac{\sqrt{26}+3}{2} | We can rewrite $x^{2}+y^{2}-3 y-1=0$ as $x^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{13}{4}$. We then see that the set of solutions to $x^{2}-y^{2}-3 y-1=0$ is the circle of radius $\frac{\sqrt{13}}{2}$ and center $\left(0, \frac{3}{2}\right)$. This can be written as $x=\frac{\sqrt{13}}{2} \cos (\theta)$ and $y=\frac{\s... | HMMT Nov Hard | HMMT-Nov General | 8.333333 | 3.900525 | 0.258596 | 3.5 | 4.5 | false |
HMMT | 0.539687 | 0.05302 | 0.650566 | HMMT-Nov | guts | 2,021 | Nov | 28 | Find the smallest positive integer $n$ such that the divisors of $n$ can be partitioned into three sets with equal sums. | 120 | Solution: I claim the answer is 120 . First, note that $120=2^{3} \cdot 3 \cdot 5$, so the sum of divisors is $(1+2+4+8)(1+3)(1+5)=15 \cdot 4 \cdot 6=360$. Thus, we need to split the divisors into groups summing to 120 . But then we can just take $\{120\},\{20,40,60\},\{1,2,3,4,5,6,8,10,12,15,24,30\}$. Thus, 120 works.... | HMMT Nov Guts | HMMT-Nov Guts | 5.504587 | 4.098244 | 0.330315 | 3.5 | 6 | false |
AIME | 0.599589 | 0.064728 | 0.712704 | AIME | I | 2,012 | N/A | 10 | Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$. Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$, where $x$ is in $\mathcal{S}$. In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number ... | 170 | It is apparent that for a perfect square $s^2$ to satisfy the constraints, we must have $s^2 - 256 = 1000n$ or $(s+16)(s-16) = 1000n.$ Now in order for $(s+16)(s-16)$ to be a multiple of $1000,$ at least one of $s+16$ and $s-16$ must be a multiple of $5,$ and since $s+16 \not\equiv s-16 \pmod{5},$ one term must have al... | Hard AIME Problems | AIME | 15.59 | 4.471436 | 0.403257 | 5 | 5.5 | false |
AMC | 0.346481 | 0.034124 | 0.469434 | AMC12 | 12A | 2,015 | N/A | 22 | For each positive integer $n$, let $S(n)$ be the number of sequences of length $n$ consisting solely of the letters $A$ and $B$, with no more than three $A$s in a row and no more than three $B$s in a row. What is the remainder when $S(2015)$ is divided by $12$? | 8 | We can start off by finding patterns in $S(n)$. When we calculate a few values we realize either from performing the calculation or because the calculation was performed in the exact same way that $S(n) = 2^n - 2((n_4)- (n_5) \dots (n_n))$. Rearranging the expression we realize that the terms aside from $2^{2015}$ are ... | AMC12 Final Problems | AMC12 A | 4.79 | 2.894561 | 0.212594 | 3 | 5.5 | true |
HMMT | 0.74697 | 0.07556 | 0.850314 | HMMT-Feb | guts | 2,020 | Feb | 8 | Tessa picks three real numbers $x, y, z$ and computes the values of the eight expressions of the form $\pm x \pm y \pm z$. She notices that the eight values are all distinct, so she writes the expressions down in increasing order. For example, if $x=2, y=3, z=4$, then the order she writes them down is \[ -x-y-z,+x-y-z,... | 96 | Solution: There are $2^{3}=8$ ways to choose the sign for each of $x, y$, and $z$. Furthermore, we can order $|x|,|y|$, and $|z|$ in $3 !=6$ different ways. Now assume without loss of generality that $0<x<y<z$. Then there are only two possible orders depending on the sign of $x+y-z$ : \[ \begin{aligned} & -x-y-z,+x-y-z... | HMMT Feb Guts | HMMT-Feb Guts | 25.925926 | 5.389624 | 0.470738 | 4 | 6.5 | false |
AMC | 0.287319 | 0.09846 | 0.376352 | AMC10 | 10B | 2,010 | N/A | 23 | The entries in a $3 \times 3$ array include all the digits from $1$ through $9$, arranged so that the entries in every row and column are in increasing order. How many such arrays are there? | 42 | Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right. Case 1: Center 4 \[\begin{tabular}{|c|c|... | AMC10 Final Problems | AMC10 B | 5.28 | 2.525983 | 0.613408 | 3.5 | 4.5 | false |
HMMT | 0.605583 | 0.09568 | 0.719245 | HMMT-Nov | guts | 2,023 | Nov | 36 | Isabella writes the expression $\sqrt{d}$ for each positive integer $d$ not exceeding 8 ! on the board. Seeing that these expressions might not be worth points on HMMT, Vidur simplifies each expression to the form $a \sqrt{b}$, where $a$ and $b$ are integers such that $b$ is not divisible by the square of a prime numbe... | 534810086 | Solution: Let $\sqrt{n}$ simplifies to $a_{n} \sqrt{b_{n}}$, and replace 8 ! by $x$. First, notice that $\sum_{n \leq x} a_{n}$ is small $\left(O\left(x^{3 / 2}\right)\right.$ in particular) because each term cannot exceed $\sqrt{x}$. On the other hand, $\sum_{n \leq x} b_{n}$ will be large; we have $b_{n}=n$ when $n$ ... | HMMT Nov Guts | HMMT-Nov Guts | 0.663717 | 4.508776 | 0.596091 | 3.5 | 6 | false |
AMC | 0.279419 | 0.046827 | 0.348931 | AMC10 | 10A | 2,003 | N/A | 17 | The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? | \frac{3\sqrt{3}}{\pi} | Let $s$ be the length of a side of the equilateral triangle and let $r$ be the radius of the circle. In a circle with a radius $r$, the side of an inscribed equilateral triangle is $r\sqrt{3}$. So $s=r\sqrt{3}$. The perimeter of the triangle is $3s=3r\sqrt{3}$ The area of the circle is $\pi r^{2}$ So: $\pi r^{2} = 3r\s... | AMC10 Second Half | AMC10 A | 9.33 | 2.476769 | 0.291734 | 2 | 3 | false |
AMC | 0.344222 | 0.035955 | 0.46717 | AMC12 | 12A | 2,013 | N/A | 23 | $ABCD$ is a square of side length $\sqrt{3} + 1$. Point $P$ is on $\overline{AC}$ such that $AP = \sqrt{2}$. The square region bounded by $ABCD$ is rotated $90^{\circ}$ counterclockwise with center $P$, sweeping out a region whose area is $\frac{1}{c} (a \pi + b)$, where $a$, $b$, and $c$ are positive integers and $\te... | 19 | We first note that diagonal $\overline{AC}$ is of length $\sqrt{6} + \sqrt{2}$. It must be that $\overline{AP}$ divides the diagonal into two segments in the ratio $\sqrt{3}$ to $1$. It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimens... | AMC12 Final Problems | AMC12 A | 5.14 | 2.880493 | 0.224003 | 3 | 5.5 | false |
HMMT | 0.813262 | 0.035253 | 0.921006 | HMMT-Feb | geo | 2,013 | Feb | 7 | Let $A B C$ be an obtuse triangle with circumcenter $O$ such that $\angle A B C=15^{\circ}$ and $\angle B A C>90^{\circ}$. Suppose that $A O$ meets $B C$ at $D$, and that $O D^{2}+O C \cdot D C=O C^{2}$. Find $\angle C$. | 35 | Let the radius of the circumcircle of $\triangle A B C$ be $r$. \[ \begin{gathered} O D^{2}+O C \cdot C D=O C^{2} \\ O C \cdot C D=O C^{2}-O D^{2} \\ O C \cdot C D=(O C+O D)(O C-O D) \\ O C \cdot C D=(r+O D)(r-O D) \end{gathered} \] By the power of the point at $\mathrm{D}$, \[ \begin{gathered} O C \cdot C D=B D \cdot ... | HMMT Feb Hard | HMMT-Feb Geometry | 6.336088 | 5.802621 | 0.21963 | 5.5 | 6.5 | true |
AMC | 0.080635 | 0.023923 | 0.024654 | AMC8 | 8 | 2,005 | N/A | 4 | A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.1 cm, 8.2 cm and 9.7 cm. What is the area of the square in square centimeters? | 36 | The perimeter of the triangle is $6.1+8.2+9.7=24$ cm. A square's perimeter is four times its sidelength, since all its sidelengths are equal. If the square's perimeter is $24$, the sidelength is $24/4=6$, and the area is $6^2=\fbox{36}$. | AMC8 First Half | AMC8 | 56.1 | 1.238338 | 0.149044 | 1 | 1.25 | false |
AMC | 0.259101 | 0.118243 | 0.300377 | AMC10 | 10B | 2,012 | N/A | 23 | A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface fa... | \frac{2\sqrt{3}}{3} | This tetrahedron has the 4 vertices in these positions: on a corner (lets call this $A$) of the cube, and the other three corners ($B$, $C$, and $D$) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron ... | AMC10 Final Problems | AMC10 B | 12.35 | 2.350182 | 0.736659 | 3.5 | 4.5 | false |
AMC | 0.089607 | 0.052132 | 0.033082 | AMC8 | 8 | 2,011 | N/A | 21 | Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$, and $49$. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb? | 37 | If at least half the guesses are too low, then Norb's age must be greater than $36.$ If two of the guesses are off by one, then his age is in between two guesses whose difference is $2.$ It could be $31,37,$ or $48,$ but because his age is greater than $36$ it can only be $37$ or $48.$ Lastly, Norb's age is a prime num... | AMC8 Second Half | AMC8 | 48.76 | 1.294233 | 0.324784 | 1.5 | 2 | false |
HMMT | 0.663351 | 0.125896 | 0.760629 | HMMT-Nov | guts | 2,018 | Nov | 32 | Over all real numbers $x$ and $y$, find the minimum possible value of \[ (x y)^{2}+(x+7)^{2}+(2 y+7)^{2} \] | 45 | Solution 1: Rewrite the given expression as $\left(x^{2}+4\right)\left(1+y^{2}\right)+14(x+2 y)+94$. By Cauchy-Schwartz, this is at least $(x+2 y)^{2}+14(x+2 y)+94=(x+2 y+7)^{2}+45$. The minimum is 45 , attained when $x y=2, x+2 y=-7$. Solution 2: Let $z=2 y, s=x+z, p=x z$. We seek to minimize \[ \begin{aligned} \left(... | HMMT Nov Guts | HMMT-Nov Guts | 0 | 4.86867 | 0.784335 | 3.5 | 6 | false |
HMMT | 0.437895 | 0.143924 | 0.532327 | HMMT-Nov | guts | 2,009 | Nov | 3 | Consider a square, inside which is inscribed a circle, inside which is inscribed a square, inside which is inscribed a circle, and so on, with the outermost square having side length 1 . Find the difference between the sum of the areas of the squares and the sum of the areas of the circles. | 2-\frac{\pi}{2} | The ratio of the area of each square and the circle immediately inside it is $\frac{4}{\pi}$. The total sum of the areas of the squares is $1+\frac{1}{2}+\frac{1}{4}+\ldots=2$. Difference in area is then $2-2 \cdot \frac{4}{\pi}$. $\fbox{2-\frac{\pi}{2}}$. | HMMT Nov Guts | HMMT-Nov Guts | 62.295082 | 3.464076 | 0.89665 | 3.5 | 6 | false |
HMMT | 0.622138 | 0.105192 | 0.731572 | HMMT-Nov | thm | 2,017 | Nov | 7 | On a blackboard a stranger writes the values of $s_{7}(n)^{2}$ for $n=0,1, \ldots, 7^{20}-1$, where $s_{7}(n)$ denotes the sum of digits of $n$ in base 7. Compute the average value of all the numbers on the board. | 3680 | Solution 1: We solve for 0 to $b^{n}-1$ and $s_{b}(n)^{2}$ (i.e. base $b$ ). Let $n=d_{1} \ldots d_{n}$ in base $b$, where there may be leading zeros. Then $s_{b}(n)=d_{1}+\cdots+d_{n}$, regardless of the leading zeros. \[ \mathbb{E}\left[s_{d}(n)^{2}\right]=\mathbb{E}\left[\left(d_{1}+\cdots+d_{n}\right)^{2}\right]=\s... | HMMT Nov Hard | HMMT-Nov Theme | 0.121212 | 4.611914 | 0.655351 | 3.5 | 4.5 | false |
AMC | 0.185405 | 0.038327 | 0.154969 | AMC10 | 10B | 2,021 | N/A | 1 | How many integer values of $x$ satisfy $|x|<3\pi$? | 19 | Since $3\pi\approx9.42$, we multiply $9$ by $2$ for the integers from $1$ to $9$ and the integers from $-1$ to $-9$ and add $1$ to account for $0$ to get $\fbox{19}$. | AMC10 First Half | AMC10 B | 61.35 | 1.89106 | 0.23878 | 1 | 2 | false |
HMMT | 0.511951 | 0.044271 | 0.615849 | HMMT-Nov | gen | 2,012 | Nov | 8 | Let $n$ be the 200th smallest positive real solution to the equation $x-\frac{\pi}{2}=\tan x$. Find the greatest integer that does not exceed $\frac{n}{2}$. | 314 | Drawing the graphs of the functions $y=x-\frac{\pi}{2}$ and $y=\tan x$, we may observe that the graphs intersect exactly once in each of the intervals $\left(\frac{(2 k-1) \pi}{2}, \frac{(2 k+1) \pi}{2}\right)$ for each $k=1,2, \cdots$. Hence, the 200th intersection has $x$ in the range $\left(\frac{399 \pi}{2}, \frac{... | HMMT Nov Hard | HMMT-Nov General | 7.38255 | 3.925448 | 0.275811 | 3.5 | 4.5 | false |
HMMT | 0.72197 | 0.026157 | 0.821635 | HMMT-Feb | guts | 2,014 | Feb | 16 | Suppose that $x$ and $y$ are positive real numbers such that $x^{2}-x y+2 y^{2}=8$. Find the maximum possible value of $x^{2}+x y+2 y^{2}$. | \frac{72+32 \sqrt{2}}{7} | Let $u=x^{2}+2 y^{2}$. By AM-GM, $u \geq \sqrt{8} x y$, so $x y \leq \frac{u}{\sqrt{8}}$. If we let $x y=k u$ where $k \leq \frac{1}{\sqrt{8}}$, then we have \[ \begin{gathered} u(1-k)=8 \\ u(1+k)=x^{2}+x y+2 y^{2} \end{gathered} \] that is, $u(1+k)=8 \cdot \frac{1+k}{1-k}$. It is not hard to see that the maximum value... | HMMT Feb Guts | HMMT-Feb Guts | 44.318182 | 5.233869 | 0.162957 | 4 | 6.5 | false |
HMMT | 0.9252 | 0.216693 | 0.983899 | HMMT-Feb | guts | 2,016 | Feb | 27 | Find the smallest possible area of an ellipse passing through $(2,0),(0,3),(0,7)$, and $(6,0)$. | \frac{56 \pi \sqrt{3}}{9} | Let $\Gamma$ be an ellipse passing through $A=(2,0), B=(0,3), C=(0,7), D=(6,0)$, and let $P=(0,0)$ be the intersection of $A D$ and $B C$. Area of $\Gamma$ As is inchanged under an affine transformation, so we just have to minimize this quantity over situations where $\Gamma$ is a circle and $\frac{P A}{P D}=\frac{1}{3... | HMMT Feb Guts | HMMT-Feb Guts | 0 | 6.5 | 1.35 | 4 | 6.5 | false |
AMC | 0.118674 | 0.028786 | 0.076226 | AMC8 | 8 | 2,017 | N/A | 16 | In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD$? [asy]draw((0,0)--(4,0)--(0,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,0), ESE); label("$C$", (0, 3), N); label("$3$", (0, 1.5), W); label("$4$", (2, ... | \frac{12}{5} | Because $\overline{BD} + \overline{CD} = 5,$ we can see that when we draw a line from point $B$ to imaginary point $D$ that line applies to both triangles. Let us say that $x$ is that line. Perimeter of $\triangle{ABD}$ would be $\overline{AD} + 4 + x$, while the perimeter of $\triangle{ACD}$ would be $\overline{AD} + ... | AMC8 Second Half | AMC8 | 26.8 | 1.47532 | 0.17934 | 1.5 | 2 | false |
HMMT | 0.768364 | 0.079825 | 0.874717 | HMMT-Feb | geo | 2,023 | Feb | 5 | Let $A B C$ be a triangle with $A B=13, B C=14$, and $C A=15$. Suppose $P Q R S$ is a square such that $P$ and $R$ lie on line $B C, Q$ lies on line $C A$, and $S$ lies on line $A B$. Compute the side length of this square. | 42 \sqrt{2} | \section*{Solution:} Let $A^{\prime}$ be the reflection of $A$ across $B C$. Since $Q$ and $S$ are symmetric across $B C$, we get that $Q \in B A^{\prime}$, $S \in C A^{\prime}$. Now, let $X$ and $M$ be the midpoints of $A A^{\prime}$ and $P R$. Standard altitude computation gives $B X=5, C X=9, A X=12$. Moreover, from... | HMMT Feb Easy | HMMT-Feb Geometry | 22.828283 | 5.522906 | 0.497312 | 4.5 | 5.5 | false |
HMMT | 0.806625 | 0.030041 | 0.915723 | HMMT-Feb | calc | 2,010 | Feb | 8 | Let $f(n)=\sum_{k=2}^{\infty} \frac{1}{k^{n} \cdot k !}$. Calculate $\sum_{n=2}^{\infty} f(n)$. | 3-e | \[ \begin{aligned} \sum_{n=2}^{\infty} f(n) & =\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{k^{n} \cdot k !} \\ & =\sum_{k=2}^{\infty} \frac{1}{k !} \sum_{n=2}^{\infty} \frac{1}{k^{n}} \\ & =\sum_{k=2}^{\infty} \frac{1}{k !} \cdot \frac{1}{k(k-1)} \\ & =\sum_{k=2}^{\infty} \frac{1}{(k-1) !} \cdot \frac{1}{k^{2}(k-1... | HMMT Feb Hard | HMMT-Feb Calculus | 8.695652 | 5.761275 | 0.187159 | 5.5 | 6.5 | false |
AMC | 0.267826 | 0.118269 | 0.318742 | AMC10 | 10A | 2,019 | N/A | 9 | What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers? | 996 | The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$, and we want this not to be a divisor of $n!$ (the product of the first $n$ positive integers). Notice that if and only if $n+1$ were composite, all of its factors would be less than or equal to $n$, which means they would be able to cancel with the fac... | AMC10 First Half | AMC10 A | 13.09 | 2.404544 | 0.736822 | 1 | 2 | false |
HMMT | 0.789875 | 0.022691 | 0.897358 | HMMT-Feb | comb | 2,018 | Feb | 7 | A tourist is learning an incorrect way to sort a permutation $\left(p_{1}, \ldots, p_{n}\right)$ of the integers $(1, \ldots, n)$. We define a fix on two adjacent elements $p_{i}$ and $p_{i+1}$, to be an operation which swaps the two elements if $p_{i}>p_{i+1}$, and does nothing otherwise. The tourist performs $n-1$ ro... | 1009 ! \cdot 1010 ! | Note that the given algorithm is very similar to the well-known Bubble Sort algorithm for sorting an array. The exception is that in the $i$-th round through the array, the first $i-1$ pairs are not checked. We claim a necessary and sufficient condition for the array to be sorted after the tourist's process is: for all... | HMMT Feb Hard | HMMT-Feb Combinatorics | 7.418398 | 5.656919 | 0.141368 | 5.5 | 6.5 | false |
AIME | 0.566375 | 0.023038 | 0.680755 | AIME | II | 2,014 | N/A | 6 | Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes... | 167 | The probability that he rolls a six twice when using the fair die is $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$. The probability that he rolls a six twice using the biased die is $\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}=\frac{16}{36}$. Given that Charles rolled two sixes, we can see that it is $16$ times more lik... | Intermediate AIME Problems | AIME | 35.49 | 4.264509 | 0.143529 | 4 | 4.5 | true |
AMC | 0.155282 | 0.028524 | 0.128805 | AMC10 | 10A | 2,011 | N/A | 2 | A small bottle of shampoo can hold $35$ milliliters of shampoo, whereas a large bottle can hold $500$ milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy? | 15 | To find how many small bottles we need, we can simply divide $500$ by $35$. This simplifies to $\frac{100}{7}=14 \frac{2}{7}.$ Since the answer must be an integer greater than $14$, we have to round up to $15$ bottles, or $\fbox{15}$ | AMC10 First Half | AMC10 A | 85.88 | 1.703392 | 0.177704 | 1 | 2 | false |
AMC | 0.235603 | 0.018452 | 0.241258 | AMC10 | 10B | 2,008 | N/A | 11 | Suppose that $(u_n)$ is a sequence of real numbers satifying $u_{n+2}=2u_{n+1}+u_n$, and that $u_3=9$ and $u_6=128$. What is $u_5$? | 53 | If we plug in $n=4$, we get $128=2u_5+u_4.$ By plugging in $n=3$, we get $u_5=2u_4+9.$ This is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives $128=5u_4+18 \Longrightarrow u_4=22$, therefore $u_5=\frac{128-22}{2}=53 \longrightarrow \textbf... | AMC10 Second Half | AMC10 B | 23.37 | 2.203793 | 0.114953 | 2 | 3 | true |
AMC | 0.22857 | 0.086868 | 0.227547 | AMC10 | 10B | 2,003 | N/A | 9 | Find the value of $x$ that satisfies the equation $25^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot 25^{17/x}}.$ | 3 | Manipulate the powers of $5$ in order to get a clean expression. \[\frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 25^{\frac{17}{x}}} = \frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 5^{\frac{34}{x}}} = 5^{\frac{48}{x}-(\frac{26}{x}+\frac{34}{x})} = 5^{-\frac{12}{x}}\] \[25^{-2} = (5^2)^{-2} = 5^{-4}\] \[5^{-4} = 5^{-\f... | AMC10 First Half | AMC10 B | 27.76 | 2.159973 | 0.541187 | 1 | 2 | true |
HMMT | 0.516768 | 0.030516 | 0.62327 | HMMT-Nov | guts | 2,021 | Nov | 25 | Let $x, y, z$ be real numbers satisfying \[ \begin{aligned} 2 x+y+4 x y+6 x z & =-6 \\ y+2 z+2 x y+6 y z & =4 \\ x-z+2 x z-4 y z & =-3 \end{aligned} \] Find $x^{2}+y^{2}+z^{2}$. | 29 | Solution: We multiply the first, second, and third equations by $\frac{1}{2},-\frac{1}{2}$, and -1 , respectively, then add the three resulting equations. This gives $x y+x z+y z=-2$. Doing the same with the coefficients $-1,2$, and 3 gives $x+y+z=5$, from which $(x+y+z)^{2}=25$. So $x^{2}+y^{2}+z^{2}=25-2 \cdot-2=29$.... | HMMT Nov Guts | HMMT-Nov Guts | 11.009174 | 3.955457 | 0.190114 | 3.5 | 6 | false |
AMC | 0.095931 | 0.034079 | 0.040252 | AMC8 | 8 | 2,005 | N/A | 3 | What is the minimum number of small squares that must be colored black so that a line of symmetry lies on the diagonal $\overline{BD}$ of square $ABCD$? [asy] defaultpen(linewidth(1)); for ( int x = 0; x < 5; ++x ) { draw((0,x)--(4,x)); draw((x,0)--(x,4)); } fill((1,0)--(2,0)--(2,1)--(1,1)--cycle); fill((0,... | 4 | Rotating square $ABCD$ counterclockwise $45^\circ$ so that the line of symmetry $BD$ is a vertical line makes it easier to see that $\fbox{4}$ squares need to be colored to match its corresponding square. | AMC8 First Half | AMC8 | 43.6 | 1.333629 | 0.212314 | 1 | 1.25 | true |
HMMT | 0.437586 | 0.171576 | 0.532075 | HMMT-Nov | team | 2,014 | Nov | 4 | How many ways are there to color the vertices of a triangle red, green, blue, or yellow such that no two vertices have the same color? Rotations and reflections are considered distinct. | 24 | There are 4 ways to color the first vertex, then 3 ways to color the second vertex to be distinct from the first, and finally 2 ways to color the third vertex to be distinct from the earlier two vertices. Multiplying gives 24 ways. $\fbox{24}$. | HMMT Nov Team | HMMT-Nov Team | 93.333333 | 3.462153 | 1.068924 | 4 | 5.5 | false |
HMMT | 0.71574 | 0.019281 | 0.813082 | HMMT-Feb | guts | 2,024 | Feb | 11 | Let $A B C D$ be a rectangle such that $A B=20$ and $A D=24$. Point $P$ lies inside $A B C D$ such that triangles $P A C$ and $P B D$ have areas 20 and 24, respectively. Compute all possible areas of triangle $P A B$. | 98,118,122,142 | Solution: There are four possible locations of $P$ as shown in the diagram. Let $O$ be the center. Then, $[P A O]=10$ and $[P B O]=12$. Thus, $[P A B]=[A O B] \pm[P A O] \pm[P B O]=120 \pm 10 \pm 12$, giving the four values $98,118,122$, and 142 . $\fbox{98,118,122,142}$. | HMMT Feb Guts | HMMT-Feb Guts | 49.411765 | 5.195056 | 0.120119 | 4 | 6.5 | false |
AIME | 0.486331 | 0.042033 | 0.582138 | AIME | I | 2,020 | N/A | 2 | There is a unique positive real number $x$ such that the three numbers $\log_8{2x}$, $\log_4{x}$, and $\log_2{x}$, in that order, form a geometric progression with positive common ratio. The number $x$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 17 | Since these form a geometric series, $\frac{\log_2{x}}{\log_4{x}}$ is the common ratio. Rewriting this, we get $\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2$ by base change formula. Therefore, the common ratio is 2. Now $\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\l... | Easy AIME Problems | AIME | 88.42 | 3.765831 | 0.261865 | 3 | 3.5 | false |
AMC | 0.245382 | 0.072294 | 0.264654 | AMC12 | 12B | 2,006 | N/A | 11 | Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee? | \frac 76 | Joe has 2 ounces of cream, as stated in the problem. JoAnn had 14 ounces of liquid, and drank $\frac{1}{7}$ of it. Therefore, she drank $\frac{1}{7}$ of her cream, leaving her $2*\frac{6}{7}$. $\frac{2}{2*\frac{6}{7}}=\frac{7}{6} \Rightarrow \fbox{\frac 76}$ | AMC12 Second Half | AMC12 B | 46.55 | 2.264714 | 0.450393 | 2.5 | 3.5 | false |
AMC | 0.313597 | 0.020914 | 0.422893 | AMC12 | 12A | 2,015 | N/A | 18 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$? | 16 | The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers. The quadratic formula gives the roots of the quadratic equation: $x=\frac{a\pm\sqrt{a^2-8a}}{2}$ As long as the numerator is an even integer, the roots are both integers. But first of all, the... | AMC12 Second Half | AMC12 A | 12.91 | 2.689695 | 0.130297 | 2.5 | 3.5 | false |
AMC | 0.209678 | 0.029962 | 0.195975 | AMC12 | 12B | 2,016 | N/A | 3 | Let $x=-2016$. What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$? | 4032 | By: dragonfly First of all, lets plug in all of the $x$'s into the equation. $\bigg|$ $||-2016|-(-2016)|-|-2016|$ $\bigg|$ $-(-2016)$ Then we simplify to get $\bigg|$ $|2016+2016|-2016$ $\bigg|$ $+2016$ which simplifies into $\bigg|$ $2016$ $\bigg|$ $+2016$ and finally we get $\fbox{4032}$ | AMC12 First Half | AMC12 B | 73.79 | 2.042276 | 0.186663 | 1.5 | 2 | false |
HMMT | 0.9252 | 0.216693 | 0.983899 | HMMT-Feb | guts | 2,018 | Feb | 27 | There are 2018 frogs in a pool and there is 1 frog on the shore. In each time-step thereafter, one random frog moves position. If it was in the pool, it jumps to the shore, and vice versa. Find the expected number of time-steps before all frogs are in the pool for the first time. | 2^{2018}-1 | Consider the general case of $n$ frogs. Let $E_{i}$ be the expected time for all frogs to enter the pool when $i$ frogs are on the shore and $n-i$ frogs are in the pool. We have $E_{0}=0, E_{n}=1+E_{n-1}$, and \[ E_{i}=\frac{i}{n} E_{i-1}+\frac{n-i}{n} E_{i+1}+1 \] for $0<i<n$. Define $f_{i}$ so that \[ E_{i}=\frac{f_{... | HMMT Feb Guts | HMMT-Feb Guts | 0 | 6.5 | 1.35 | 4 | 6.5 | false |
HMMT | 0.847503 | 0.055511 | 0.949434 | HMMT-Feb | alg | 2,020 | Feb | 7 | Find the sum of all positive integers $n$ for which \[ \frac{15 \cdot n !^{2}+1}{2 n-3} \] is an integer. | 90 | Solution: It is clear that $n=1$ and $n=2$ work so assume that $n>2$. If $2 n-3$ is composite then its smallest prime factor is at most $\frac{2 n-3}{2}<n$ so will be coprime to $15 \cdot n !^{2}+1$. Therefore assume that $2 n-3=p$ is prime. We can rewrite the numerator as \[ (-1)^{n} \cdot 15 \cdot\left(1 \cdot 2 \cdo... | HMMT Feb Hard | HMMT-Feb Algebra | 2.427184 | 6.015942 | 0.345835 | 5.5 | 6.5 | false |
AMC | 0.327716 | 0.044931 | 0.443019 | AMC12 | 12B | 2,014 | N/A | 23 | The number $2017$ is prime. Let $S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}$. What is the remainder when $S$ is divided by $2017?$ | 1024 | Note that $2014\equiv -3 \mod2017$. We have for $k\ge1$ \[\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017\] \[\equiv (-1)^k\dbinom{k+2}{k} \mod 2017\] Therefore \[\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017\] This is simply an alternating seri... | AMC12 Final Problems | AMC12 B | 5.5 | 2.777657 | 0.279922 | 3 | 5.5 | true |
AMC | 0.327857 | 0.031592 | 0.44327 | AMC12 | 12A | 2,011 | N/A | 14 | Suppose $a$ and $b$ are single-digit positive integers chosen independently and at random. What is the probability that the point $(a,b)$ lies above the parabola $y=ax^2-bx$? | \frac{19}{81} | If $(a,b)$ lies above the parabola, then $b$ must be greater than $y(a)$. We thus get the inequality $b>a^3-ba$. Solving this for $b$ gives us $b>\frac{a^3}{a+1}$. Now note that $\frac{a^3}{a+1}$ constantly increases when $a$ is positive. Then since this expression is greater than $9$ when $a=4$, we can deduce that $a$... | AMC12 Second Half | AMC12 A | 8.49 | 2.778537 | 0.196819 | 2.5 | 3.5 | false |
AMC | 0.283332 | 0.084919 | 0.364528 | AMC12 | 12A | 2,021 | Nov | 8 | Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,32,33,34,35,36,37,38,39,$ and $40.$ What is the value of $\frac{N}{M}?$ | 74 | By the definition of least common mutiple, we take the greatest powers of the prime numbers of the prime factorization of all the numbers, that we are taking the $\text{lcm}$ of. In this case, \[M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.\] Now, using the same logic, we fi... | AMC12 First Half | AMC12 A | 28.61 | 2.501147 | 0.529048 | 1.5 | 2 | false |
HMMT | 0.67169 | 0.103876 | 0.770566 | HMMT-Feb | guts | 2,023 | Feb | 6 | Let $A, E, H, L, T$, and $V$ be chosen independently and at random from the set $\left\{0, \frac{1}{2}, 1\right\}$. Compute the probability that $\lfloor T \cdot H \cdot E\rfloor=L \cdot A \cdot V \cdot A$. | \frac{55}{81} | Solution: There are $3^{3}-2^{3}=19$ ways to choose $L, A$, and $V$ such that $L \cdot A \cdot V \cdot A=0$, since at least one of $\{L, A, V\}$ must be 0 , and $3^{3}-1=26$ ways to choose $T, H$, and $E$ such that $\lfloor T \cdot H \cdot E\rfloor=0$, since at least one of $\{T, H, E\}$ must not be 1 , for a total of ... | HMMT Feb Guts | HMMT-Feb Guts | 80.597015 | 4.920622 | 0.647152 | 4 | 6.5 | false |
AMC | 0.189465 | 0.036288 | 0.16 | AMC10 | 10A | 2,006 | N/A | 4 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display? | 23 | From the greedy algorithm, we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\fbox{23}$ | AMC10 First Half | AMC10 A | 66.42 | 1.916349 | 0.226075 | 1 | 2 | false |
HMMT | 0.492247 | 0.035271 | 0.588679 | HMMT-Nov | guts | 2,022 | Nov | 21 | Let $P(x)$ be a quadratic polynomial with real coefficients. Suppose that $P(1)=20, P(-1)=22$, and $P(P(0))=400$. Compute the largest possible value of $P(10)$. | 2486 | Solution: Let $P(x)=a x^{2}+b x+c$. The given equations give us: \[ \begin{aligned} & a+b+c=20 \\ & a-b+c=22 \end{aligned} \] Hence $b=-1, a+c=21$, and so the final equation gives us $a c^{2}=400$. Substituting $a=21-c$ and solving the cubic in $c$, we get $c=-4,5,20$. Of these, the smallest value $c=-4$ (and hence $\l... | HMMT Nov Guts | HMMT-Nov Guts | 21.686747 | 3.802693 | 0.219738 | 3.5 | 6 | false |
AMC | 0.331522 | 0.034138 | 0.449057 | AMC12 | 12A | 2,019 | N/A | 14 | For a certain complex number $c$, the polynomial \[P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)\]has exactly 4 distinct roots. What is $|c|$? | \sqrt{10} | The polynomial can be factored further broken down into $P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)$ by using the quadratic formula on each of the quadratic factors. Since the first four roots are all distinct, the term $(x^2 - cx + 4)$ must be a product of any combination of two (not n... | AMC12 Second Half | AMC12 A | 7.6 | 2.801367 | 0.212683 | 2.5 | 3.5 | false |
HMMT | 0.430864 | 0.157983 | 0.525031 | HMMT-Nov | guts | 2,018 | Nov | 7 | At Easter-Egg Academy, each student has two eyes, each of which can be eggshell, cream, or cornsilk. It is known that $30 \%$ of the students have at least one eggshell eye, $40 \%$ of the students have at least one cream eye, and $50 \%$ of the students have at least one cornsilk eye. What percentage of the students a... | 80 \% | For the purposes of this solution, we abbreviate "eggshell" by "egg", and "cornsilk" by "corn". We know that there are only six combinations of eye color possible: egg-cream, egg-corn, egg-egg, creamcorn, cream-cream, corn-corn. If we let the proportions for each of these be represented by $a, b, c, d$, $e$, and $f$ re... | HMMT Nov Guts | HMMT-Nov Guts | 67.549669 | 3.420271 | 0.984235 | 3.5 | 6 | false |
AIME | 0.57138 | 0.027014 | 0.686541 | AIME | II | 2,018 | N/A | 9 | Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$, and partition the octagon... | 184 | We represent octagon $ABCDEFGH$ in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that $A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\frac{23}{2})$. Recall that the centroid is $\frac{1}{3}$ way up each median in t... | Intermediate AIME Problems | AIME | 31.82 | 4.295693 | 0.168296 | 4 | 4.5 | false |
AMC | 0.206879 | 0.049323 | 0.19044 | AMC10 | 10B | 2,011 | N/A | 14 | A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot? | 62 | Let the sides of the rectangular parking lot be $a$ and $b$. Then $a^2 + b^2 = 625$ and $ab = 168$. Add the two equations together, then factor. \begin{align} a^2 + 2ab + b^2 &= 625 + 168 \times 2\\ (a + b)^2 &= 961\\ a + b &= 31 \end{align} The perimeter of a rectangle is $2 (a + b) = 2 (31) = \fbox{62}$ | AMC10 Second Half | AMC10 B | 43.94 | 2.024838 | 0.307281 | 2 | 3 | false |
AMC | 0.28146 | 0.050255 | 0.359497 | AMC10 | 10B | 2,020 | N/A | 20 | Let $B$ be a right rectangular prism (box) with edges lengths $1,$ $3,$ and $4$, together with its interior. For real $r\geq0$, let $S(r)$ be the set of points in $3$-dimensional space that lie within a distance $r$ of some point in $B$. The volume of $S(r)$ can be expressed as $ar^{3} + br^{2} + cr +d$, where $a,$ $b,... | 19 | Split $S(r)$ into 4 regions: 1. The rectangular prism itself 2. The extensions of the faces of $B$ 3. The quarter cylinders at each edge of $B$ 4. The one-eighth spheres at each corner of $B$ Region 1: The volume of $B$ is $1 \cdot 3 \cdot 4 = 12$, so $d=12$. Region 2: This volume is equal to the surface area of $B$ ti... | AMC10 Second Half | AMC10 B | 6.33 | 2.489484 | 0.313092 | 2 | 3 | true |
AMC | 0.21516 | 0.03416 | 0.20327 | AMC12 | 12B | 2,008 | N/A | 5 | A class collects $50$ dollars to buy flowers for a classmate who is in the hospital. Roses cost $3$ dollars each, and carnations cost $2$ dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly $50$ dollars? | 9 | The class could send $25$ carnations and no roses, $22$ carnations and $2$ roses, $19$ carnations and $4$ roses, and so on, down to $1$ carnation and $16$ roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), $\Rightarrow \fbox{9}$ | AMC12 First Half | AMC12 B | 70.16 | 2.07643 | 0.212815 | 1.5 | 2 | false |
AMC | 0.289691 | 0.053961 | 0.380629 | AMC10 | 10A | 2,006 | N/A | 20 | Six distinct positive integers are randomly chosen between $1$ and $2006$, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of $5$? | 1 | For two numbers to have a difference that is a multiple of $5$, the numbers must be congruent $\bmod{5}$ (their remainders after division by $5$ must be the same). $0, 1, 2, 3, 4$ are the possible values of numbers in $\bmod{5}$. Since there are only $5$ possible values in $\bmod{5}$ and we are picking $6$ numbers, by ... | AMC10 Second Half | AMC10 A | 6.84 | 2.54076 | 0.33618 | 2 | 3 | false |
HMMT | 0.502753 | 0.019398 | 0.60327 | HMMT-Nov | guts | 2,009 | Nov | 24 | Penta chooses 5 of the vertices of a unit cube. What is the maximum possible volume of the figure whose vertices are the 5 chosen points? | \frac{1}{2} | Label the vertices of the cube $A, B, C, D, E, F, G, H$, such that $A B C D$ is the top face of the cube, $E$ is directly below $A, F$ is directly below $B, G$ is directly below $C$, and $H$ is directly below $D$. We can obtain a volume of $\frac{1}{2}$ by taking the vertices $A, B, C, F$, and $H$. To compute the volum... | HMMT Nov Guts | HMMT-Nov Guts | 16.393443 | 3.868141 | 0.120847 | 3.5 | 6 | false |
AMC | 0.310088 | 0.021969 | 0.418616 | AMC12 | 12B | 2,021 | N/A | 18 | Let $z$ be a complex number satisfying $12|z|^2=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\frac 6z?$ | -2 | Using the fact $z\bar{z}=|z|^2$, the equation rewrites itself as \begin{align} 12z\bar{z}&=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31 \\ -12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32&=0 \\ \left((z^2+2z\bar{z}+\bar{z}^2)+4(z+\bar{z})+4\right)+\left(z^2\bar{z}^2-12z\bar{z}+36\right)&=0 \\ (z+\bar{z... | AMC12 Second Half | AMC12 B | 9.41 | 2.667835 | 0.136867 | 2.5 | 3.5 | true |
AMC | 0.368395 | 0.018352 | 0.48805 | AMC12 | 12A | 2,013 | N/A | 24 | Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area? | \frac{223}{286} | Suppose $p$ is the answer. We calculate $1-p$. Assume that the circumradius of the 12-gon is $1$, and the 6 different lengths are $a_1$, $a_2$, $\cdots$, $a_6$, in increasing order. Then $a_k = 2\sin ( \frac{k\pi}{12} )$. So $a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5$, $a_2=1$, $a_3=\sqrt{2}\approx 1.4$, $a_4=\sqrt{3}\appr... | AMC12 Final Problems | AMC12 A | 2.39 | 3.031086 | 0.114332 | 3 | 5.5 | false |
AMC | 0.252227 | 0.028672 | 0.28327 | AMC10 | 10B | 2,015 | N/A | 19 | In $\triangle{ABC}$, $\angle{C} = 90^{\circ}$ and $AB = 12$. Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X, Y, Z$, and $W$ lie on a circle. What is the perimeter of the triangle? | 12+12\sqrt{2} | The center of the circle lies on the intersection between the perpendicular bisectors of chords $ZW$ and $YX$. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be $O$. Draw perpendiculars to $ZW$ and $YX$ from $O$, and connect $OZ$ and $OY$. $OY^2=6^2+12^2=180$. Let... | AMC10 Second Half | AMC10 B | 15.01 | 2.307359 | 0.17863 | 2 | 3 | false |
HMMT | 0.790781 | 0.103563 | 0.898113 | HMMT-Feb | comb | 2,017 | Feb | 5 | Kelvin the Frog likes numbers whose digits strictly decrease, but numbers that violate this condition in at most one place are good enough. In other words, if $d_{i}$ denotes the $i$ th digit, then $d_{i} \leq d_{i+1}$ for at most one value of $i$. For example, Kelvin likes the numbers 43210, 132, and 3, but not the nu... | 14034 | Suppose first that no digit violates the constraint; i.e. the digits are in strictly decreasing order. There are $\left(\begin{array}{c}10 \\ 5\end{array}\right)$ ways to choose the digits of the number, and each set of digits can be arranged in exactly one way, so there are $\left(\begin{array}{c}10 \\ 5\end{array}\ri... | HMMT Feb Easy | HMMT-Feb Combinatorics | 7.216495 | 5.662563 | 0.645199 | 4.5 | 5.5 | false |
AMC | 0.160165 | 0.020111 | 0.132075 | AMC10 | 10B | 2,003 | N/A | 2 | Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs $$$1$ more than a pink pill, and Al's pills cost a total of $\textdollar 546$ for the two weeks. How much does one green pill cost? | 20 | Because there are $14$ days in two weeks, Al spends $546/14 = 39$ dollars per day for the cost of a green pill and a pink pill. If the green pill costs $x$ dollars and the pink pill $x-1$ dollars, the sum of the two costs $2x-1$ should equal $39$ dollars. Then the cost of the green pill $x$ is $\fbox{20}$. | AMC10 First Half | AMC10 B | 78.44 | 1.733811 | 0.125295 | 1 | 2 | false |
AIME | 0.64546 | 0.118947 | 0.744654 | AIME | II | 2,014 | N/A | 13 | Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer $k<5$, no collection of $k$ pairs made by the child contains the shoes from exactly... | 28 | Label the left shoes be $L_1,\dots, L_{10}$ and the right shoes $R_1,\dots, R_{10}$. Notice that there are $10!$ possible pairings. Let a pairing be "bad" if it violates the stated condition. We would like a better condition to determine if a given pairing is bad. Note that, in order to have a bad pairing, there must e... | Very Hard AIME Problems | AIME | 3.93 | 4.757211 | 0.741045 | 6 | 7 | true |
HMMT | 0.512243 | 0.087655 | 0.616604 | HMMT-Nov | team | 2,021 | Nov | 2 | Joey wrote a system of equations on a blackboard, where each of the equations was of the form $a+b=c$ or $a \cdot b=c$ for some variables or integers $a, b, c$. Then Sean came to the board and erased all of the plus signs and multiplication signs, so that the board reads: \[ \begin{array}{ll} x & z=15 \\ x & y=12 \\ x ... | 2037 | Solution: The bottom line gives $x=-6, x=6$ or $x=18$. If $x=-6, y$ can be -2 or 18 and $z$ must be 21 , so the possible values for $100 x+10 y+z$ are -599 and -399 . If $x=6, y$ can be 2 or 6 and $z$ must be 9 , so the possible values are 629 and 669 . If $x=18, y$ must be -6 and $z$ must be -3 , so the only possible ... | HMMT Nov Team | HMMT-Nov Team | 54.62963 | 3.927265 | 0.54609 | 4 | 5.5 | false |
AMC | 0.103538 | 0.039091 | 0.050314 | AMC8 | 8 | 2,019 | N/A | 7 | Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ , $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests? | 48 | We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. \[\frac{76+94+87+x+y}{5} = 81,\] \[\frac{257+x+y}{5} = 81.\] We can now cross multiply to get rid of the denominator. \[257... | AMC8 First Half | AMC8 | 37.58 | 1.381025 | 0.243537 | 1 | 1.25 | false |
AMC | 0.143795 | 0.021085 | 0.119497 | AMC8 | 8 | 2,018 | N/A | 25 | How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive? | 58 | We compute $2^8+1=257$. We're all familiar with what $6^3$ is, namely $216$, which is too small. The smallest cube greater than it is $7^3=343$. $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^6)^3=64^3$, which therefore will clearly be the largest cube less than $2^{18}+1$. So, the required number ... | AMC8 Second Half | AMC8 | 13.82 | 1.631826 | 0.131363 | 1.5 | 2 | false |
AMC | 0.232238 | 0.047863 | 0.23522 | AMC12 | 12B | 2,009 | N/A | 9 | Triangle $ABC$ has vertices $A = (3,0)$, $B = (0,3)$, and $C$, where $C$ is on the line $x + y = 7$. What is the area of $\triangle ABC$? | 6 | Because the line $x + y = 7$ is parallel to $\overline {AB}$, the area of $\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$. In that case the triangle has base $AC = 4$ and altitude $3$, so its area is $\frac 12 \cdot 4 \cdot 3 = \fbox{6}$. | AMC12 First Half | AMC12 B | 57.29 | 2.182831 | 0.29819 | 1.5 | 2 | true |
AMC | 0.246392 | 0.019061 | 0.268428 | AMC10 | 10A | 2,005 | N/A | 11 | A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n$? | 4 | Since there are $n^2$ little faces on each face of the big wooden cube, there are $6n^2$ little faces painted red. Since each unit cube has $6$ faces, there are $6n^3$ little faces total. Since one-fourth of the little faces are painted red, $\frac{6n^2}{6n^3}=\frac{1}{4}$ $\frac{1}{n}=\frac{1}{4}$ $n=\fbox{4}$ | AMC10 Second Half | AMC10 A | 23.35 | 2.271006 | 0.118753 | 2 | 3 | false |
AMC | 0.228116 | 0.025959 | 0.226667 | AMC10 | 10B | 2,006 | N/A | 13 | Joe and JoAnn each bought $12$ ounces of coffee in a $16$ ounce cup. Joe drank $2$ ounces of his coffee and then added $2$ ounces of cream. JoAnn added $2$ ounces of cream, stirred the coffee well, and then drank $2$ ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee? | \frac{7}{6} | After drinking and adding cream, Joe's cup has $2$ ounces of cream. After adding cream to her cup, JoAnn's cup had $14$ ounces of liquid. By stirring and then drinking $2$ ounces out of the $14$ ounces of liquid, she drank $\frac{2}{14}=\frac{1}{7}$th of the cream. So there are $2\cdot\frac{6}{7}=\frac{12}{7}$ ounces o... | AMC10 Second Half | AMC10 B | 28.06 | 2.157147 | 0.161727 | 2 | 3 | false |
AMC | 0.358388 | 0.022733 | 0.48 | AMC12 | 12B | 2,004 | N/A | 23 | The polynomial $x^3 - 2004 x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $n$ are possible? | 250,500 | Let the roots be $r,s,r + s$, and let $t = rs$. Then $(x - r)(x - s)(x - (r + s))$ $= x^3 - (r + s + r + s) x^2 + (rs + r(r + s) + s(r + s))x - rs(r + s) = 0$ and by matching coefficients, $2(r + s) = 2004 \Longrightarrow r + s = 1002$. Then our polynomial looks like \[x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0\] and we... | AMC12 Final Problems | AMC12 B | 2.08 | 2.968744 | 0.141628 | 3 | 5.5 | false |
AMC | 0.190725 | 0.037664 | 0.16327 | AMC10 | 10A | 2,013 | N/A | 10 | A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations? | 70 | Let the total amount of flowers be $x$. Thus, the number of pink flowers is $0.6x$, and the number of red flowers is $0.4x$. The number of pink carnations is $\frac{2}{3}(0.6x) = 0.4x$ and the number of red carnations is $\frac{3}{4}(0.4x) = 0.3x$. Summing these, the total number of carnations is $0.4x+0.3x=0.7x$. ... | AMC10 First Half | AMC10 A | 65.49 | 1.924201 | 0.234651 | 1 | 2 | false |
AMC | 0.117262 | 0.047931 | 0.070692 | AMC8 | 8 | 2,022 | N/A | 3 | When three positive integers $a$, $b$, and $c$ are multiplied together, their product is $100$. Suppose $a < b < c$. In how many ways can the numbers be chosen? | 4 | The positive divisors of $100$ are \[1,2,4,5,10,20,25,50,100.\] It is clear that $10\leq c\leq50,$ so we apply casework to $c:$ If $c=10,$ then $(a,b,c)=(2,5,10).$ If $c=20,$ then $(a,b,c)=(1,5,20).$ If $c=25,$ then $(a,b,c)=(1,4,25).$ If $c=50,$ then $(a,b,c)=(1,2,50).$ Together, the numbers $a,b,$ and $c$ can be chos... | AMC8 First Half | AMC8 | 27.72 | 1.466523 | 0.298612 | 1 | 1.25 | false |
HMMT | 0.743148 | 0.048585 | 0.846792 | HMMT-Feb | comb | 2,024 | Feb | 6 | In each cell of a $4 \times 4$ grid, one of the two diagonals is drawn uniformly at random. Compute the probability that the resulting 32 triangular regions can be colored red and blue so that any two regions sharing an edge have different colors. | \frac{1}{512} | Solution: Give each cell coordinates from $(1,1)$ to $(4,4)$. Claim. The grid has a desired coloring if and only if every vertex not on the boundary meets an even number of edges and diagonals. Proof. If this were not the case, the odd number of regions around the vertex would have to alternate between the two colors, ... | HMMT Feb Hard | HMMT-Feb Combinatorics | 27.118644 | 5.365811 | 0.302688 | 5.5 | 6.5 | false |
AMC | 0.089132 | 0.029566 | 0.032453 | AMC8 | 8 | 2,018 | N/A | 3 | Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle? | \text{Dan} | The five numbers which cause people to leave the circle are $7, 14, 17, 21,$ and $27.$ The most straightforward way to do this would be to draw out the circle with the people, and cross off people as you count. Assuming the six people start with $1$, Arn counts $7$ so he leaves first. Then, Cyd counts $14$ as there are... | AMC8 First Half | AMC8 | 49.15 | 1.291274 | 0.1842 | 1 | 1.25 | false |
AMC | 0.340783 | 0.041939 | 0.46239 | AMC12 | 12B | 2,004 | N/A | 17 | For some real numbers $a$ and $b$, the equation \[8x^3 + 4ax^2 + 2bx + a = 0\] has three distinct positive roots. If the sum of the base-$2$ logarithms of the roots is $5$, what is the value of $a$? | -256 | Let the three roots be $x_1,x_2,x_3$. \[\log_2 x_1 + \log_2 x_2 + \log_2 x_3 = \log_2 x_1x_2x_3= 5 \Longrightarrow x_1x_2x_3 = 32\] By Vieta’s formulas, \[8(x-x_1)(x-x_2)(x-x_3) = 8x^3 + 4ax^2 + 2bx + a\] gives us that $a = -8x_1x_2x_3 = -256 $. $\fbox{-256}$. | AMC12 Second Half | AMC12 B | 3.65 | 2.859065 | 0.261283 | 2.5 | 3.5 | true |
HMMT | 0.515346 | 0.028869 | 0.620377 | HMMT-Nov | guts | 2,009 | Nov | 30 | Regular hexagon $A B C D E F$ has side length 2. A laser beam is fired inside the hexagon from point $A$ and hits $\overline{B C}$ at point $G$. The laser then reflects off $\overline{B C}$ and hits the midpoint of $\overline{D E}$. Find $B G$. | \frac{2}{5} | Look at the diagram below, in which points $J, K, M, T$ and $X$ have been defined. $M$ is the midpoint of $\overline{D E}, B C J K$ is a rhombus with $J$ lying on the extension of $\overline{C D}, T$ is the intersection of lines $\overline{C D}$ and $\overline{G M}$ when extended, and $X$ is on $\overline{J T}$ such th... | HMMT Nov Guts | HMMT-Nov Guts | 11.47541 | 3.946597 | 0.179856 | 3.5 | 6 | false |
AMC | 0.374883 | 0.017898 | 0.490818 | AMC12 | 12A | 2,005 | N/A | 25 | Let $S$ be the set of all points with coordinates $(x,y,z)$, where $x$, $y$, and $z$ are each chosen from the set $\{0,1,2\}$. How many equilateral triangles all have their vertices in $S$? | 80 | For this solution, we will just find as many solutions as possible, until it becomes intuitive that there are no more size of triangles left. First, try to make three of its vertices form an equilateral triangle. This we find is possible by taking any vertex, and connecting the three adjacent vertices into a triangle. ... | AMC12 Final Problems | AMC12 A | 1.94 | 3.071507 | 0.111503 | 3 | 5.5 | true |
AMC | 0.122808 | 0.025684 | 0.085535 | AMC8 | 8 | 2,020 | N/A | 19 | A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$ | 4 | A number is divisible by $15$ precisely if it is divisible by $3$ and $5$. The latter means the last digit must be either $5$ or $0$, and the former means the sum of the digits must be divisible by $3$. If the last digit is $0$, the first digit would be $0$ (because the digits alternate), which is not possible. Hence t... | AMC8 Second Half | AMC8 | 24.22 | 1.501077 | 0.160011 | 1.5 | 2 | false |
HMMT | 0.665499 | 0.112961 | 0.767296 | HMMT-Feb | guts | 2,023 | Feb | 1 | Suppose $a$ and $b$ are positive integers such that $a^{b}=2^{2023}$. Compute the smallest possible value of $b^{a}$. | 1 | Solution: By taking $a=2^{2023}$ and $b=1$, we get $b^{a}=1$, which is clearly the minimum. $\fbox{1}$. | HMMT Feb Guts | HMMT-Feb Guts | 83.58209 | 4.882056 | 0.703748 | 4 | 6.5 | true |
HMMT | 0.590809 | 0.08747 | 0.704654 | HMMT-Nov | guts | 2,014 | Nov | 35 | Ten points are equally spaced on a circle. A graph is a set of segments (possibly empty) drawn between pairs of points, so that every two points are joined by either zero or one segments. Two graphs are considered the same if we can obtain one from the other by rearranging the points. Let $N$ denote the number of graph... | 11716571 | The question asks for the number of isomorphism classes of connected graphs on 10 vertices. This is enumerated in \href{http://oeis.org/A001349}{http://oeis.org/A001349} the answer is 11716571. In fact, of the $2^{45} \approx 3.51 \cdot 10^{13} \approx 3 \cdot 10^{13}$ graphs on 10 labelled vertices, virtually all (abo... | HMMT Nov Guts | HMMT-Nov Guts | 1.074074 | 4.416733 | 0.544941 | 3.5 | 6 | false |
AMC | 0.103163 | 0.040984 | 0.049811 | AMC8 | 8 | 2,000 | N/A | 15 | Triangles $ABC$, $ADE$, and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$, respectively. If $AB = 4$, what is the perimeter of figure $ABCDEFG$? [asy] pair A,B,C,D,EE,F,G; A = (4,0); B = (0,0); C = (2,2*sqrt(3)); D = (3,sqrt(3)); EE = (5,sqrt(3)); F = (5.5,sqrt(3)/2)... | 15 | The large triangle $ABC$ has sides of length $4$. The medium triangle has sides of length $2$. The small triangle has sides of length $1$. There are $3$ segment sizes, and all segments depicted are one of these lengths. Starting at $A$ and going clockwise, the perimeter is: $AB + BC + CD + DE + EF + FG + GA$ $4 + 4 ... | AMC8 Second Half | AMC8 | 37.87 | 1.378685 | 0.255333 | 1.5 | 2 | false |
AMC | 0.088061 | 0.028854 | 0.031447 | AMC8 | 8 | 2,012 | N/A | 11 | The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$? | 11 | We can eliminate answer choices ${\textbf{(A)}\ 5}$ and ${\textbf{(C)}\ 7}$, because of the above statement. Now we need to test the remaining answer choices. Case 1: $x = 6$ Mode: $6$ Median: $6$ Mean: $\frac{37}{7}$ Since the mean does not equal the median or mode, ${\textbf{(B)}\ 6}$ can also be eliminated. Case 2: ... | AMC8 First Half | AMC8 | 50.03 | 1.2846 | 0.179763 | 1 | 1.25 | false |
AMC | 0.31095 | 0.148331 | 0.420126 | AMC10 | 10B | 2,004 | N/A | 4 | A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$? | 12 | The product of all six numbers is $6!=720$. The products of numbers that can be visible are $720/1$, $720/2$, ..., $720/6$. The answer to this problem is their greatest common divisor -- which is $720/L$, where $L$ is the least common multiple of $\{1,2,3,4,5,6\}$. Clearly $L=60$ and the answer is $720/60=\fbox{12}$. | AMC10 First Half | AMC10 B | 2.5 | 2.673207 | 0.924103 | 1 | 2 | false |
HMMT | 0.850924 | 0.177274 | 0.952956 | HMMT-Feb | guts | 2,014 | Feb | 29 | Natalie has a copy of the unit interval $[0,1]$ that is colored white. She also has a black marker, and she colors the interval in the following manner: at each step, she selects a value $x \in[0,1]$ uniformly at random, and (a) If $x \leq \frac{1}{2}$ she colors the interval $\left[x, x+\frac{1}{2}\right]$ with her ma... | 5 | The first choice always wipes out half the interval. So we calculate the expected value of the amount of time needed to wipe out the other half. Solution 1 (non-calculus): We assume the interval has $2 n$ points and we start with the last $n$ colored black. We let $f(k)$ be the expected value of the number of turns we ... | HMMT Feb Guts | HMMT-Feb Guts | 1.136364 | 6.037257 | 1.104424 | 4 | 6.5 | false |
AIME | 0.570321 | 0.106436 | 0.685535 | AIME | II | 2,019 | N/A | 4 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 187 | Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes). Case 1 (easy): Four 5's are rolled. This has probability $\frac{1}{6^4}$ of occurring. Case 2: Two 5's are rolled. Cas... | Easy AIME Problems | AIME | 32.58 | 4.289093 | 0.663096 | 3 | 3.5 | true |
HMMT | 0.724877 | 0.032311 | 0.823899 | HMMT-Feb | guts | 2,020 | Feb | 11 | Find the number of ordered pairs of positive integers $(x, y)$ with $x, y \leq 2020$ such that $3 x^{2}+10 x y+$ $3 y^{2}$ is the power of some prime. | 29 | Solution: We can factor as $(3 x+y)(x+3 y)$. If $x \geq y$, we need $\frac{3 x+y}{x+3 y} \in\{1,2\}$ to be an integer. So we get the case where $x=y$, in which we need both to be a power of 2 , or the case $x=5 y$, in which case we need $y$ to be a power of 2 . This gives us $11+9+9=29$ solutions, where we account for ... | HMMT Feb Guts | HMMT-Feb Guts | 41.975309 | 5.251979 | 0.201298 | 4 | 6.5 | false |
HMMT | 0.644955 | 0.137934 | 0.744151 | HMMT-Feb | guts | 2,011 | Feb | 1 | Let $A B C$ be a triangle with area 1. Let points $D$ and $E$ lie on $A B$ and $A C$, respectively, such that $D E$ is parallel to $B C$ and $D E / B C=1 / 3$. If $F$ is the reflection of $A$ across $D E$, find the area of triangle $F B C$. | \frac{1}{3} | Let $A F$ intersect $B C$ at $H$. Since $D E / B C=1 / 3$ and $F$ and $A$ are equidistant from $D E$, we have $A F=\frac{2}{3} A H$ and $F H=A H-A F=\frac{1}{3} A H$. Furthermore, since $A F$ is perpendicular to $D E$, we have $A H$ and $F H$ are the altitudes of triangles $A B C$ and $F B C$ respectively. Therefore th... | HMMT Feb Guts | HMMT-Feb Guts | 90.909091 | 4.754063 | 0.859332 | 4 | 6.5 | true |
HMMT | 0.854817 | 0.059754 | 0.959497 | HMMT-Feb | geo | 2,015 | Feb | 9 | Let $A B C D$ be a regular tetrahedron with side length 1 . Let $X$ be the point in triangle $B C D$ such that $[X B C]=2[X B D]=4[X C D]$, where $[\varpi]$ denotes the area of figure $\varpi$. Let $Y$ lie on segment $A X$ such that $2 A Y=Y X$. Let $M$ be the midpoint of $B D$. Let $Z$ be a point on segment $A M$ such... | \frac{4}{7} | We apply three-dimensional barycentric coordinates with reference tetrahedron $A B C D$. The given conditions imply that \[ \begin{aligned} X & =(0: 1: 2: 4) \\ Y & =(14: 1: 2: 4) \\ M & =(0: 1: 0: 1) \\ Z & =(t: 1: 0: 1) \end{aligned} \] for some real number $t$. Normalizing, we obtain $Y=\left(\frac{14}{21}, \frac{1}... | HMMT Feb Hard | HMMT-Feb Geometry | 1.697313 | 6.061507 | 0.372267 | 5.5 | 6.5 | false |
AMC | 0.297259 | 0.019967 | 0.396352 | AMC12 | 12B | 2,021 | Nov | 17 | A bug starts at a vertex of a grid made of equilateral triangles of side length $1$. At each step the bug moves in one of the $6$ possible directions along the grid lines randomly and independently with equal probability. What is the probability that after $5$ moves the bug never will have been more than $1$ unit away ... | \frac{13}{108} | Let $S(n)$ be the number of paths of $n$ moves such that the bug never will have been more than $1$ unit away from the starting position. Clearly, by symmetry, there are two possible states here, the bug being on the center and the bug being on one of the vertices of the unit hexagon around the center. Let $C(n)$ be th... | AMC12 Second Half | AMC12 B | 13.67 | 2.587907 | 0.124397 | 2.5 | 3.5 | false |
AMC | 0.295159 | 0.057546 | 0.392201 | AMC10 | 10A | 2,018 | N/A | 14 | What is the greatest integer less than or equal to \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?\] | 80 | We write \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot\frac{3^{100}}{3^{96}}+\frac{2^{96}}{3^{96}+2^{96}}\cdot\frac{2^{100}}{2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot 81+\frac{2^{96}}{3^{96}+2^{96}}\cdot 16.\] Hence we see that our number is a weighted average of 81 and 16, extremely heavily... | AMC10 Second Half | AMC10 A | 5.78 | 2.574828 | 0.358514 | 2 | 3 | true |
AMC | 0.209281 | 0.026453 | 0.194969 | AMC12 | 12A | 2,012 | N/A | 5 | A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of $280$ pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad? | 64 | So let the number of blueberries be $b,$ the number of raspberries be $r,$ the number of grapes be $g,$ and finally the number of cherries be $c.$ Observe that since there are $280$ pieces of fruit, \[b+r+g+c=280.\] Since there are twice as many raspberries as blueberries, \[2b=r.\] The fact that there are three times ... | AMC12 First Half | AMC12 A | 82.04 | 2.039803 | 0.164802 | 1.5 | 2 | false |
HMMT | 0.50655 | 0.017893 | 0.608805 | HMMT-Nov | guts | 2,009 | Nov | 14 | Let $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$ be a polynomial whose roots are all negative integers. If $a+b+c+d=2009$, find $d$. | 528 | Call the roots $-x_{1},-x_{2},-x_{3}$, and $-x_{4}$. Then $f(x)$ must factor as $\left(x+x_{1}\right)\left(x+x_{2}\right)(x+$ $\left.x_{3}\right)\left(x+x_{4}\right)$. If we evaluate $f$ at 1 , we get $\left(1+x_{1}\right)\left(1+x_{2}\right)\left(1+x_{3}\right)\left(1+x_{4}\right)=a+b+c+d+1=$ $2009+1=2010$. $2010=2 \c... | HMMT Nov Guts | HMMT-Nov Guts | 14.754098 | 3.891797 | 0.111472 | 3.5 | 6 | true |
AMC | 0.066423 | 0.01749 | 0.014591 | AMC8 | 8 | 2,008 | N/A | 4 | In the figure, the outer equilateral triangle has area $16$, the inner equilateral triangle has area $1$, and the three trapezoids are congruent. What is the area of one of the trapezoids? [asy] size((70)); draw((0,0)--(7.5,13)--(15,0)--(0,0)); draw((1.88,3.25)--(9.45,3.25)); draw((11.2,0)--(7.5,6.5)); draw((9.4,9.7)--... | 5 | The area outside the small triangle but inside the large triangle is $16-1=15$. This is equally distributed between the three trapezoids. Each trapezoid has an area of $15/3 = \fbox{5}$. | AMC8 First Half | AMC8 | 67.09 | 1.149795 | 0.108965 | 1 | 1.25 | false |
AMC | 0.143821 | 0.021102 | 0.119748 | AMC8 | 8 | 2,003 | N/A | 20 | What is the measure of the acute angle formed by the hands of the clock at 4:20 PM? | 10 | Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the hour. The central angle formed between $4$ and $5$ is $30$ degrees (since it is 1/12 of a full circle, 360). By $4:20$, the hour hand would have moved $\frac{1}{3}$ way from 4 to 5 since $\frac{20}{60}$ is reducible to $\frac{1}{3}$... | AMC8 Second Half | AMC8 | 13.81 | 1.631985 | 0.131468 | 1.5 | 2 | false |
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