id int64 | question string | solution list | final_answer list | context string | image_1 image | image_2 image | image_3 image | image_4 image | image_5 image | modality string | difficulty string | is_multiple_answer bool | unit string | answer_type string | error string | question_type string | subfield string | subject string | language string |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
2,926 | Let $T=5$. Compute the smallest positive integer $n$ such that there are at least $T$ positive integers in the domain of $f(x)=\sqrt{-x^{2}-2 x+n}$. | [
"Completing the square under the radical yields $\\sqrt{n+1-(x+1)^{2}}$. The larger zero of the radicand is $-1+\\sqrt{n+1}$, and the smaller zero is negative because $-1-\\sqrt{n+1}<0$, so the $T$ positive integers in the domain of $f$ must be $1,2,3, \\ldots, T$. Therefore $-1+\\sqrt{n+1} \\geq T$. Hence $\\sqrt{... | [
"35"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,927 | Let $T=35$. Compute the smallest positive real number $x$ such that $\frac{\lfloor x\rfloor}{x-\lfloor x\rfloor}=T$. | [
"If $\\frac{\\lfloor x\\rfloor}{x-\\lfloor x\\rfloor}=T$, the equation can be rewritten as follows:\n\n$$\n\\begin{aligned}\n\\frac{x-\\lfloor x\\rfloor}{\\lfloor x\\rfloor} & =\\frac{1}{T} \\\\\n\\frac{x}{\\lfloor x\\rfloor}-1 & =\\frac{1}{T} \\\\\n\\frac{x}{\\lfloor x\\rfloor} & =\\frac{T+1}{T} .\n\\end{aligned}\... | [
"$\\frac{36}{35}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,947 | Let set $S=\{1,2,3,4,5,6\}$, and let set $T$ be the set of all subsets of $S$ (including the empty set and $S$ itself). Let $t_{1}, t_{2}, t_{3}$ be elements of $T$, not necessarily distinct. The ordered triple $\left(t_{1}, t_{2}, t_{3}\right)$ is called satisfactory if either
(a) both $t_{1} \subseteq t_{3}$ and $t_{2} \subseteq t_{3}$, or
(b) $t_{3} \subseteq t_{1}$ and $t_{3} \subseteq t_{2}$.
Compute the number of satisfactory ordered triples $\left(t_{1}, t_{2}, t_{3}\right)$. | [
"Let $T_{1}=\\left\\{\\left(t_{1}, t_{2}, t_{3}\\right) \\mid t_{1} \\subseteq t_{3}\\right.$ and $\\left.t_{2} \\subseteq t_{3}\\right\\}$ and let $T_{2}=\\left\\{\\left(t_{1}, t_{2}, t_{3}\\right) \\mid t_{3} \\subseteq t_{1}\\right.$ and $\\left.t_{3} \\subseteq t_{2}\\right\\}$. Notice that if $\\left(t_{1}, t_... | [
"31186"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,948 | Let $A B C D$ be a parallelogram with $\angle A B C$ obtuse. Let $\overline{B E}$ be the altitude to side $\overline{A D}$ of $\triangle A B D$. Let $X$ be the point of intersection of $\overline{A C}$ and $\overline{B E}$, and let $F$ be the point of intersection of $\overline{A B}$ and $\overleftrightarrow{D X}$. If $B C=30, C D=13$, and $B E=12$, compute the ratio $\frac{A C}{A F}$. | [
"Extend $\\overline{A D}$ to a point $M$ such that $\\overline{C M} \\| \\overline{B E}$ as shown below.\n\n<img_3958>\n\nBecause $C D=A B=13$ and $B E=12=C M, A E=D M=5$. Then $A C=\\sqrt{35^{2}+12^{2}}=$ $\\sqrt{1369}=37$. Because $\\overline{E X} \\| \\overline{C M}, X E / C M=A E / A M=\\frac{1}{7}$. Thus $E X=... | [
"$\\frac{222}{13}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,949 | Compute the sum of all positive two-digit factors of $2^{32}-1$. | [
"Using the difference of squares, $2^{32}-1=\\left(2^{16}-1\\right)\\left(2^{16}+1\\right)$. The second factor, $2^{16}+1$, is the Fermat prime 65537 , so continue with the first factor:\n\n$$\n\\begin{aligned}\n2^{16}-1 & =\\left(2^{8}+1\\right)\\left(2^{8}-1\\right) \\\\\n2^{8}-1 & =\\left(2^{4}+1\\right)\\left(2... | [
"168"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,950 | Compute all ordered pairs of real numbers $(x, y)$ that satisfy both of the equations:
$$
x^{2}+y^{2}=6 y-4 x+12 \quad \text { and } \quad 4 y=x^{2}+4 x+12
$$ | [
"Rearrange the terms in the first equation to yield $x^{2}+4 x+12=6 y-y^{2}+24$, so that the two equations together yield $4 y=6 y-y^{2}+24$, or $y^{2}-2 y-24=0$, from which $y=6$ or $y=-4$. If $y=6$, then $x^{2}+4 x+12=24$, from which $x=-6$ or $x=2$. If $y=-4$, then $x^{2}+4 x+12=-16$, which has no real solutions... | [
"$(-6,6)$, $(2,6)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Algebra | Math | English |
2,951 | Define $\log ^{*}(n)$ to be the smallest number of times the log function must be iteratively applied to $n$ to get a result less than or equal to 1 . For example, $\log ^{*}(1000)=2$ since $\log 1000=3$ and $\log (\log 1000)=\log 3=0.477 \ldots \leq 1$. Let $a$ be the smallest integer such that $\log ^{*}(a)=3$. Compute the number of zeros in the base 10 representation of $a$. | [
"If $\\log ^{*}(a)=3$, then $\\log (\\log (\\log (a))) \\leq 1$ and $\\log (\\log (a))>1$. If $\\log (\\log (a))>1$, then $\\log (a)>10$ and $a>10^{10}$. Because the problem asks for the smallest such $a$ that is an integer, choose $a=10^{10}+1=10,000,000,001$, which has 9 zeros."
] | [
"9"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,952 | An integer $N$ is worth 1 point for each pair of digits it contains that forms a prime in its original order. For example, 6733 is worth 3 points (for 67,73 , and 73 again), and 20304 is worth 2 points (for 23 and 03). Compute the smallest positive integer that is worth exactly 11 points. [Note: Leading zeros are not allowed in the original integer.] | [
"If a number $N$ has $k$ base 10 digits, then its maximum point value is $(k-1)+(k-2)+\\cdots+1=$ $\\frac{1}{2}(k-1)(k)$. So if $k \\leq 5$, the number $N$ is worth at most 10 points. Therefore the desired number has at least six digits. If $100,000<N<101,000$, then $N$ is of the form $100 \\underline{A} \\underlin... | [
"100337"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,953 | The six sides of convex hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ are colored red. Each of the diagonals of the hexagon is colored either red or blue. Compute the number of colorings such that every triangle $A_{i} A_{j} A_{k}$ has at least one red side. | [
"Only two triangles have no sides that are sides of the original hexagon: $A_{1} A_{3} A_{5}$ and $A_{2} A_{4} A_{6}$. For each of these triangles, there are $2^{3}-1=7$ colorings in which at least one side is red, for a total of $7 \\cdot 7=49$ colorings of those six diagonals. The colorings of the three central d... | [
"392"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,954 | Compute the smallest positive integer $n$ such that $n^{n}$ has at least 1,000,000 positive divisors. | [
"Let $k$ denote the number of distinct prime divisors of $n$, so that $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdots p_{k}^{a_{k}}, a_{i}>0$. Then if $d(x)$ denotes the number of positive divisors of $x$,\n\n$$\nd\\left(n^{n}\\right)=\\left(a_{1} n+1\\right)\\left(a_{2} n+1\\right) \\cdots\\left(a_{k} n+1\\right) \\geq(n+1... | [
"84"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,955 | Given an arbitrary finite sequence of letters (represented as a word), a subsequence is a sequence of one or more letters that appear in the same order as in the original sequence. For example, $N, C T, O T T$, and CONTEST are subsequences of the word CONTEST, but NOT, ONSET, and TESS are not. Assuming the standard English alphabet $\{A, B, \ldots, Z\}$, compute the number of distinct four-letter "words" for which $E E$ is a subsequence. | [
"Divide into cases according to the number of $E$ 's in the word. If there are only two $E$ 's, then the word must have two non- $E$ letters, represented by ?'s. There are $\\left(\\begin{array}{l}4 \\\\ 2\\end{array}\\right)=6$ arrangements of two $E$ 's and two ?'s, and each of the ?'s can be any of 25 letters, s... | [
"3851"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,956 | Six solid regular tetrahedra are placed on a flat surface so that their bases form a regular hexagon $\mathcal{H}$ with side length 1 , and so that the vertices not lying in the plane of $\mathcal{H}$ (the "top" vertices) are themselves coplanar. A spherical ball of radius $r$ is placed so that its center is directly above the center of the hexagon. The sphere rests on the tetrahedra so that it is tangent to one edge from each tetrahedron. If the ball's center is coplanar with the top vertices of the tetrahedra, compute $r$. | [
"Let $O$ be the center of the sphere, $A$ be the top vertex of one tetrahedron, and $B$ be the center of the hexagon.\n\n<img_3299>\n\nThen $B O$ equals the height of the tetrahedron, which is $\\frac{\\sqrt{6}}{3}$. Because $A$ is directly above the centroid of the bottom face, $A O$ is two-thirds the length of th... | [
"$\\frac{\\sqrt{2}}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,957 | Derek starts at the point $(0,0)$, facing the point $(0,1)$, and he wants to get to the point $(1,1)$. He takes unit steps parallel to the coordinate axes. A move consists of either a step forward, or a $90^{\circ}$ right (clockwise) turn followed by a step forward, so that his path does not contain any left turns. His path is restricted to the square region defined by $0 \leq x \leq 17$ and $0 \leq y \leq 17$. Compute the number of ways he can get to $(1,1)$ without returning to any previously visited point. | [
"Divide into cases according to the number of right turns Derek makes.\n\n- There is one route involving only one turn: move first to $(0,1)$ and then to $(1,1)$.\n- If he makes two turns, he could move up to $(0, a)$ then to $(1, a)$ and then down to $(1,1)$. In order to do this, $a$ must satisfy $1<a \\leq 17$, l... | [
"529"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,958 | The equations $x^{3}+A x+10=0$ and $x^{3}+B x^{2}+50=0$ have two roots in common. Compute the product of these common roots. | [
"Let the roots of the first equation be $p, q, r$ and the roots of the second equation be $p, q, s$. Then $p q r=-10$ and $p q s=-50$, so $\\frac{s}{r}=5$. Also $p+q+r=0$ and $p+q+s=-B$, so $r-s=B$. Substituting yields $r-5 r=-4 r=B$, so $r=-\\frac{B}{4}$ and $s=-\\frac{5 B}{4}$. From the second given equation, $p ... | [
"$5 \\sqrt[3]{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,960 | Let $N$ be a perfect square between 100 and 400 , inclusive. What is the only digit that cannot appear in $N$ ? | [
"When the perfect squares between 100 and 400 inclusive are listed out, every digit except 7 is used. Note that the perfect squares 100, 256, 289, 324 use each of the other digits."
] | [
"7"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,961 | Let $T=7$. Let $A$ and $B$ be distinct digits in base $T$, and let $N$ be the largest number of the form $\underline{A} \underline{B} \underline{A}_{T}$. Compute the value of $N$ in base 10 . | [
"To maximize $\\underline{A} \\underline{B} \\underline{A}_{T}$ with $A \\neq B$, let $A=T-1$ and $B=T-2$. Then $\\underline{A} \\underline{B}^{A} \\underline{A}_{T}=$ $(T-1) \\cdot T^{2}+(T-2) \\cdot T^{1}+(T-1) \\cdot T^{0}=T^{3}-T-1$. With $T=7$, the answer is 335 ."
] | [
"335"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,962 | Let T be an integer. Given a nonzero integer $n$, let $f(n)$ denote the sum of all numbers of the form $i^{d}$, where $i=\sqrt{-1}$, and $d$ is a divisor (positive or negative) of $n$. Compute $f(2 T+1)$. | [
"Let $n=2^{m} r$, where $r$ is odd. If $m=0$, then $n$ is odd, and for each $d$ that divides $n$, $i^{d}+i^{-d}=i^{d}+\\frac{i^{d}}{\\left(i^{2}\\right)^{d}}=0$, hence $f(n)=0$ when $n$ is odd. If $m=1$, then for each $d$ that divides $n, i^{d}+i^{-d}$ equals 0 if $d$ is odd, and -2 if $d$ is even. Thus when $n$ is... | [
"0"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,963 | Let $T=0$. Compute the real value of $x$ for which there exists a solution to the system of equations
$$
\begin{aligned}
x+y & =0 \\
x^{3}-y^{3} & =54+T .
\end{aligned}
$$ | [
"$\\quad$ Plug $y=-x$ into the second equation to obtain $x=\\sqrt[3]{\\frac{54+T}{2}}$. With $T=0, x=\\sqrt[3]{27}=3$."
] | [
"3"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,964 | Let $T=3$. In $\triangle A B C, A C=T^{2}, \mathrm{~m} \angle A B C=45^{\circ}$, and $\sin \angle A C B=\frac{8}{9}$. Compute $A B$. | [
"From the Law of Sines, $\\frac{A B}{\\sin \\angle A C B}=\\frac{A C}{\\sin \\angle A B C}$. Thus $A B=\\frac{8}{9} \\cdot \\frac{T^{2}}{1 / \\sqrt{2}}=\\frac{8 \\sqrt{2}}{9} \\cdot T^{2}$. With $T=3, A B=\\mathbf{8} \\sqrt{\\mathbf{2}}$."
] | [
"$8 \\sqrt{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,966 | Let $T=9$. The sequence $a_{1}, a_{2}, a_{3}, \ldots$ is an arithmetic progression, $d$ is the common difference, $a_{T}=10$, and $a_{K}=2010$, where $K>T$. If $d$ is an integer, compute the value of $K$ such that $|K-d|$ is minimal. | [
"Note that $a_{T}=a_{1}+(T-1) d$ and $a_{K}=a_{1}+(K-1) d$, hence $a_{K}-a_{T}=(K-T) d=2010-10=$ 2000. Thus $K=\\frac{2000}{d}+T$, and to minimize $\\left|T+\\frac{2000}{d}-d\\right|$, choose a positive integer $d$ such that $\\frac{2000}{d}$ is also an integer and $\\frac{2000}{d}-d$ is as close as possible to $-T... | [
"49"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,967 | Let $A$ be the number you will receive from position 7 , and let $B$ be the number you will receive from position 9 . There are exactly two ordered pairs of real numbers $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ that satisfy both $|x+y|=6(\sqrt{A}-5)$ and $x^{2}+y^{2}=B^{2}$. Compute $\left|x_{1}\right|+\left|y_{1}\right|+\left|x_{2}\right|+\left|y_{2}\right|$. | [
"Note that the graph of $x^{2}+y^{2}=B^{2}$ is a circle of radius $|B|$ centered at $(0,0)$ (as long as $\\left.B^{2}>0\\right)$. Also note that the graph of $|x+y|=6(\\sqrt{A}-5)$ is either the line $y=-x$ if $A=25$, or the graph consists of two parallel lines with slope -1 if $A>25$. In the former case, the\n\n\n... | [
"24"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,968 | Let $T=23$. In triangle $A B C$, the altitude from $A$ to $\overline{B C}$ has length $\sqrt{T}, A B=A C$, and $B C=T-K$, where $K$ is the real root of the equation $x^{3}-8 x^{2}-8 x-9=0$. Compute the length $A B$. | [
"Rewrite the equation as $x^{3}-1=8\\left(x^{2}+x+1\\right)$, so that $(x-1)\\left(x^{2}+x+1\\right)=8\\left(x^{2}+x+1\\right)$. Because $x^{2}+x+1$ has no real zeros, it can be canceled from both sides of the equation to obtain $x-1=8$ or $x=9$. Hence $B C=T-9$, and $A B^{2}=(\\sqrt{T})^{2}+\\left(\\frac{T-9}{2}\\... | [
"$6 \\sqrt{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,969 | Let $T=8$. A cube has volume $T-2$. The cube's surface area equals one-eighth the surface area of a $2 \times 2 \times n$ rectangular prism. Compute $n$. | [
"The cube's side length is $\\sqrt[3]{T}$, so its surface area is $6 \\sqrt[3]{T^{2}}$. The rectangular prism has surface area $2(2 \\cdot 2+2 \\cdot n+2 \\cdot n)=8+8 n$, thus $6 \\sqrt[3]{T^{2}}=1+n$. With $T=8, n=6 \\sqrt[3]{64}-1=\\mathbf{2 3}$."
] | [
"23"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,970 | Let $T=98721$, and let $K$ be the sum of the digits of $T$. Let $A_{n}$ be the number of ways to tile a $1 \times n$ rectangle using $1 \times 3$ and $1 \times 1$ tiles that do not overlap. Tiles of both types need not be used; for example, $A_{3}=2$ because a $1 \times 3$ rectangle can be tiled with three $1 \times 1$ tiles or one $1 \times 3$ tile. Compute the smallest value of $n$ such that $A_{n} \geq K$. | [
"Consider the rightmost tile of the rectangle. If it's a $1 \\times 1$ tile, then there are $A_{n-1}$ ways to tile the remaining $1 \\times(n-1)$ rectangle, and if it's a $1 \\times 3$ tile, then there are $A_{n-3}$ ways to tile the remaining $1 \\times(n-3)$ rectangle. Hence $A_{n}=A_{n-1}+A_{n-3}$ for $n>3$, and ... | [
"10"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,971 | Let $T=3$, and let $K=T+2$. Compute the largest $K$-digit number which has distinct digits and is a multiple of 63. | [
"Let $N_{K}$ be the largest $K$-digit number which has distinct digits and is a multiple of 63 . It can readily be verified that $N_{1}=0, N_{2}=63$, and $N_{3}=945$. For $K>3$, compute $N_{K}$ using the following strategy: start with the number $M_{0}=\\underline{9} \\underline{8} \\underline{7} \\ldots(10-K)$; le... | [
"98721"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,972 | Let $T\neq 0$. Suppose that $a, b, c$, and $d$ are real numbers so that $\log _{a} c=\log _{b} d=T$. Compute
$$
\frac{\log _{\sqrt{a b}}(c d)^{3}}{\log _{a} c+\log _{b} d}
$$ | [
"Note that $a^{T}=c$ and $b^{T}=d$, thus $(a b)^{T}=c d$. Further note that $(a b)^{3 T}=(\\sqrt{a b})^{6 T}=(c d)^{3}$, thus $\\log _{\\sqrt{a b}}(c d)^{3}=6 T$. Thus the given expression simplifies to $\\frac{6 T}{2 T}=\\mathbf{3}$ (as long as $T \\neq 0$ )."
] | [
"3"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,973 | Let $T=2030$. Given that $\mathrm{A}, \mathrm{D}, \mathrm{E}, \mathrm{H}, \mathrm{S}$, and $\mathrm{W}$ are distinct digits, and that $\underline{\mathrm{W}} \underline{\mathrm{A}} \underline{\mathrm{D}} \underline{\mathrm{E}}+\underline{\mathrm{A}} \underline{\mathrm{S}} \underline{\mathrm{H}}=T$, what is the largest possible value of $\mathrm{D}+\mathrm{E}$ ? | [
"First note that if $T \\geq 10000$, then $\\mathrm{W}=9$ and $\\mathrm{A} \\geq 5$. If $T<10000$ and $x$ is the leading digit of $T$, then either $\\mathrm{W}=x$ and $\\mathrm{A} \\leq 4$ or $\\mathrm{W}=x-1$ and $\\mathrm{A} \\geq 5$. With $T=2030$, either $\\underline{\\mathrm{W}} \\underline{\\mathrm{A}}=20$\n\... | [
"9"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,974 | Let $f(x)=2^{x}+x^{2}$. Compute the smallest integer $n>10$ such that $f(n)$ and $f(10)$ have the same units digit. | [
"The units digit of $f(10)$ is the same as the units digit of $2^{10}$. Because the units digits of powers of 2 cycle in groups of four, the units digit of $2^{10}$ is 4 , so the units digit of $f(10)$ is 4 . Note that $n$ must be even, otherwise, the units digit of $f(n)$ is odd. If $n$ is a multiple of 4 , then $... | [
"30"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,975 | In rectangle $P A U L$, point $D$ is the midpoint of $\overline{U L}$ and points $E$ and $F$ lie on $\overline{P L}$ and $\overline{P A}$, respectively such that $\frac{P E}{E L}=\frac{3}{2}$ and $\frac{P F}{F A}=2$. Given that $P A=36$ and $P L=25$, compute the area of pentagon $A U D E F$. | [
"For convenience, let $P A=3 x$ and let $P L=5 y$. Then the given equations involving ratios of segment lengths imply that $P E=3 y, E L=2 y, P F=2 x$, and $F A=x$. Then $[P A U L]=(3 x)(5 y)=15 x y$ and\n\n$$\n\\begin{aligned}\n{[A U D E F] } & =[P A U L]-[P E F]-[E L D] \\\\\n& =15 x y-\\frac{1}{2}(3 y)(2 x)-\\fr... | [
"630"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,976 | Rectangle $A R M L$ has length 125 and width 8. The rectangle is divided into 1000 squares of area 1 by drawing in gridlines parallel to the sides of $A R M L$. Diagonal $\overline{A M}$ passes through the interior of exactly $n$ of the 1000 unit squares. Compute $n$. | [
"Notice that 125 and 8 are relatively prime. Examining rectangles of size $a \\times b$ where $a$ and $b$ are small and relatively prime suggests an answer of $a+b-1$. To see that this is the case, note that other than the endpoints, the diagonal does not pass through any vertex of any unit square. After the first ... | [
"132"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,977 | Compute the least integer $n>1$ such that the product of all positive divisors of $n$ equals $n^{4}$. | [
"Note that every factor pair $d$ and $\\frac{n}{d}$ have product $n$. For the product of all such divisor pairs to equal $n^{4}$, there must be exactly 4 divisor pairs, or 8 positive integer divisors. A number has 8 positive integer divisors if it is of the form $a^{3} b^{1}$ or $a^{7}$ where $a$ and $b$ are distin... | [
"24"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,978 | Each of the six faces of a cube is randomly colored red or blue with equal probability. Compute the probability that no three faces of the same color share a common vertex. | [
"There are $2^{6}=64$ colorings of the cube. Let $r$ be the number of faces that are colored red. Define a monochromatic vertex to be a vertex of the cube for which the three faces meeting there have the same color. It is clear that a coloring without a monochromatic vertex is only possible in the cases $2 \\leq r ... | [
"$\\frac{9}{32}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,979 | Scalene triangle $A B C$ has perimeter 2019 and integer side lengths. The angle bisector from $C$ meets $\overline{A B}$ at $D$ such that $A D=229$. Given that $A C$ and $A D$ are relatively prime, compute $B C$. | [
"Let $B C=a, A C=b, A B=c$. Also, let $A D=e$ and $B D=f$. Then $a+b+e+f=2019$, the values $a, b$, and $e+f$ are integers, and by the Angle Bisector Theorem, $\\frac{e}{f}=\\frac{b}{a}$. So $b=\\frac{a e}{f}=\\frac{229 a}{f}$. Because 229 is prime and $\\operatorname{gcd}(b, e)=1$, conclude that $f$ must be an inte... | [
"888"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,980 | Given that $a$ and $b$ are positive and
$$
\lfloor 20-a\rfloor=\lfloor 19-b\rfloor=\lfloor a b\rfloor,
$$
compute the least upper bound of the set of possible values of $a+b$. | [
"Let the common value of the three expressions in the given equation be $N$. Maximizing $a+b$ involves making at least one of $a$ and $b$ somewhat large, which makes the first two expressions for $N$ small. So, to maximize $a+b$, look for the least possible value of $N$. One can show that $N=14$ is not possible bec... | [
"$\\frac{41}{5}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,981 | Compute the number of five-digit integers $\underline{M} \underline{A} \underline{R} \underline{T} \underline{Y}$, with all digits distinct, such that $M>A>R$ and $R<T<Y$. | [
"There are $\\left(\\begin{array}{c}10 \\\\ 5\\end{array}\\right)=252$ ways to choose the values of the digits $M, A, R, T, Y$, without restrictions. Because $R$ is fixed as the least of the digits and because $T<Y$, it suffices to find the number of ways to choose $M$ and $A$. Once $M$ and $A$ are chosen, the othe... | [
"1512"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,982 | In parallelogram $A R M L$, points $P$ and $Q$ are the midpoints of sides $\overline{R M}$ and $\overline{A L}$, respectively. Point $X$ lies on segment $\overline{P Q}$, and $P X=3, R X=4$, and $P R=5$. Point $I$ lies on segment $\overline{R X}$ such that $I A=I L$. Compute the maximum possible value of $\frac{[P Q R]}{[L I P]}$. | [
"Because $A I=L I$ and $A Q=L Q$, line $I Q$ is the perpendicular bisector of $\\overline{A L}$. Because $A R M L$ is a parallelogram, $\\overline{Q I} \\perp \\overline{R P}$. Note also that $\\mathrm{m} \\angle R X P=90^{\\circ}$. Thus $I$ is the orthocenter of triangle $P Q R$, from\n\n\n\nwhich it follows that ... | [
"$\\frac{4}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,983 | Given that $a, b, c$, and $d$ are positive integers such that
$$
a ! \cdot b ! \cdot c !=d ! \quad \text { and } \quad a+b+c+d=37
$$
compute the product $a b c d$. | [
"Without loss of generality, assume $a \\leq b \\leq c<d$. Note that $d$ cannot be prime, as none of $a$ !, $b$ !, or $c$ ! would have it as a factor. If $d=p+1$ for some prime $p$, then $c=p$ and $a ! b !=p+1$. The least possible values of $a ! b$ ! are $1,2,4,6,24,36,48,120,144,240$, so the case where $d=p+1$ is ... | [
"2240"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,984 | Compute the value of
$$
\sin \left(6^{\circ}\right) \cdot \sin \left(12^{\circ}\right) \cdot \sin \left(24^{\circ}\right) \cdot \sin \left(42^{\circ}\right)+\sin \left(12^{\circ}\right) \cdot \sin \left(24^{\circ}\right) \cdot \sin \left(42^{\circ}\right) \text {. }
$$ | [
"Let $S=\\left(1+\\sin 6^{\\circ}\\right)\\left(\\sin 12^{\\circ} \\sin 24^{\\circ} \\sin 42^{\\circ}\\right)$. It follows from a sum-to-product identity that $1+\\sin 6^{\\circ}=$ $\\sin 90^{\\circ}+\\sin 6^{\\circ}=2 \\sin 48^{\\circ} \\cos 42^{\\circ}$. Because the sine of an angle is the cosine of its complemen... | [
"$\\frac{1}{16}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,985 | Let $a=19, b=20$, and $c=21$. Compute
$$
\frac{a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a}{a+b+c}
$$ | [
"Note that the numerator of the given expression factors as $(a+b+c)^{2}$, hence the expression to be computed equals $a+b+c=19+20+21=\\mathbf{6 0}$."
] | [
"60"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,986 | Let $T=60$ . Lydia is a professional swimmer and can swim one-fifth of a lap of a pool in an impressive 20.19 seconds, and she swims at a constant rate. Rounded to the nearest integer, compute the number of minutes required for Lydia to swim $T$ laps. | [
"Lydia swims a lap in $5 \\cdot 20.19=100.95$ seconds. The number of minutes required for Lydia to swim $T$ laps is therefore $100.95 \\cdot T / 60$. With $T=60$, the desired number of minutes, rounded to the nearest integer, is 101"
] | [
"101"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,987 | Let $T=101$. In $\triangle A B C, \mathrm{~m} \angle C=90^{\circ}$ and $A C=B C=\sqrt{T-3}$. Circles $O$ and $P$ each have radius $r$ and lie inside $\triangle A B C$. Circle $O$ is tangent to $\overline{A C}$ and $\overline{B C}$. Circle $P$ is externally tangent to circle $O$ and to $\overline{A B}$. Given that points $C, O$, and $P$ are collinear, compute $r$. | [
"Let $A^{\\prime}$ and $B^{\\prime}$ be the respective feet of the perpendiculars from $O$ to $\\overline{A C}$ and $\\overline{B C}$. Let $H$ be the foot of the altitude from $C$ to $\\overline{A B}$. Because $\\triangle A B C$ is isosceles, it follows that $A^{\\prime} O B^{\\prime} C$ is a square, $\\mathrm{m} \... | [
"$3-\\sqrt{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,988 | Given that $p=6.6 \times 10^{-27}$, then $\sqrt{p}=a \times 10^{b}$, where $1 \leq a<10$ and $b$ is an integer. Compute $10 a+b$ rounded to the nearest integer. | [
"Note that $p=6.6 \\times 10^{-27}=66 \\times 10^{-28}$, so $a=\\sqrt{66}$ and $b=-14$. Note that $\\sqrt{66}>\\sqrt{64}=8$. Because $8.1^{2}=65.61$ and $8.15^{2}=66.4225>66$, conclude that $81<10 \\sqrt{66}<81.5$, hence $10 a$ rounded to the nearest integer is 81 , and the answer is $81-14=\\mathbf{6 7}$."
] | [
"67"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,989 | Let $T=67$. A group of children and adults go to a rodeo. A child's admission ticket costs $\$ 5$, and an adult's admission ticket costs more than $\$ 5$. The total admission cost for the group is $\$ 10 \cdot T$. If the number of adults in the group were to increase by $20 \%$, then the total cost would increase by $10 \%$. Compute the number of children in the group. | [
"Suppose there are $x$ children and $y$ adults in the group and each adult's admission ticket costs $\\$ a$. The given information implies that $5 x+a y=10 T$ and $5 x+1.2 a y=11 T$. Subtracting the first equation from the second yields $0.2 a y=T \\rightarrow a y=5 T$, so from the first equation, $5 x=5 T \\righta... | [
"67"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,990 | Let $T=67$. Rectangles $F A K E$ and $F U N K$ lie in the same plane. Given that $E F=T$, $A F=\frac{4 T}{3}$, and $U F=\frac{12}{5}$, compute the area of the intersection of the two rectangles. | [
"Without loss of generality, let $A, U$, and $N$ lie on the same side of $\\overline{F K}$. Applying the Pythagorean Theorem to triangle $A F K$, conclude that $F K=\\frac{5 T}{3}$. Comparing the altitude to $\\overline{F K}$ in triangle $A F K$ to $\\overline{U F}$, note that the intersection of the two rectangles... | [
"262"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,991 | Elizabeth is in an "escape room" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to "flip" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room.
Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute).
For convenience, assume the $n$ light switches are numbered 1 through $n$.
Compute the $E(6,1)$ | [
"$E(6,1)=6$. Note that at least six minutes are required because exactly one switch is flipped each minute. By flipping all six switches (in any order) in the first six minutes, the door will open in six minutes."
] | [
"6"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,992 | Elizabeth is in an "escape room" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to "flip" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room.
Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute).
For convenience, assume the $n$ light switches are numbered 1 through $n$.
Compute the $E(6,2)$ | [
"$E(6,2)=3$. The sequence $\\{1,2\\},\\{3,4\\},\\{5,6\\}$ will allow Elizabeth to escape the room in three minutes. It is not possible to escape the room in fewer than three minutes because every switch must be flipped, and that requires at least $\\frac{6}{2}=3$ minutes."
] | [
"3"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,993 | Elizabeth is in an "escape room" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to "flip" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room.
Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute).
For convenience, assume the $n$ light switches are numbered 1 through $n$.
Compute the $E(7,3)$ | [
"$E(7,3)=3$. First, note that $E(7,3) \\geq 3$, because after only two minutes, it is impossible to flip each switch at least once. It is possible to escape in three minutes with the sequence $\\{1,2,3\\},\\{1,4,5\\}$, and $\\{1,6,7\\}$."
] | [
"3"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,994 | Elizabeth is in an "escape room" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to "flip" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room.
Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute).
For convenience, assume the $n$ light switches are numbered 1 through $n$.
Compute the $E(9,5)$ | [
"$E(9,5)=3$. Notice that $E(9,5) \\neq 1$ because each switch must be flipped at least once, and only five switches can be flipped in one minute. Notice also that $E(9,5) \\neq 2$ because after two minutes, there have been 10 flips, but in order to escape the room, each switch must be flipped at least once, and thi... | [
"3"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,995 | Elizabeth is in an "escape room" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to "flip" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room.
Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute).
For convenience, assume the $n$ light switches are numbered 1 through $n$.
Find the following in terms of $n$. $E(n, 2)$ for positive even integers $n$ | [
"If $n$ is even, then $E(n, 2)=\\frac{n}{2}$. This is the minimum number of minutes required to flip each switch at least once, and Elizabeth can clearly escape in $\\frac{n}{2}$ minutes by flipping each switch exactly once."
] | [
"$\\frac{n}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Combinatorics | Math | English |
2,996 | Elizabeth is in an "escape room" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to "flip" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room.
Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute).
For convenience, assume the $n$ light switches are numbered 1 through $n$.
Find the following in terms of $n$. $E(n, n-2)$ for $n \geq 5$ | [
"If $n \\geq 5$, then $E(n, n-2)=3$. Note that Elizabeth cannot flip every switch in one minute, and after two minutes, some switch (in fact, many switches) must be flipped exactly twice. However, Elizabeth can escape in three minutes using the sequence $\\{1,4,5, \\ldots, n\\},\\{2,4,5, \\ldots, n\\},\\{3,4,5, \\l... | [
"3"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,000 | Elizabeth is in an "escape room" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to "flip" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room.
Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute).
For convenience, assume the $n$ light switches are numbered 1 through $n$.
Find the $E(2020,1993)$ | [
"First, we prove that if $n$ is even and $k$ is odd, then $E(n, k)=E(n,n-k)$.\n\n\nBecause $n$ is even, and because each switch must be flipped an odd number of times in order to escape, the total number of flips is even. Because $k$ must be odd, $E(n, k)$ must be even. To show this, consider the case where $E(n, k... | [
"76"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,001 | Elizabeth is in an "escape room" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to "flip" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room.
Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute).
For convenience, assume the $n$ light switches are numbered 1 through $n$.
Find the $E(2001,501)$ | [
"$E(2001,501)=5$. First, note that three minutes is not enough time to flip each switch once. In four minutes, Elizabeth can flip each switch once, but has three flips left over. Because there are an odd number of leftover flips to distribute among the 2001 switches, some switch must get an odd number of leftover f... | [
"5"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,004 | Elizabeth is in an "escape room" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to "flip" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room.
Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute).
For convenience, assume the $n$ light switches are numbered 1 through $n$.
One might guess that in most cases, $E(n, k) \approx \frac{n}{k}$. In light of this guess, define the inefficiency of the ordered pair $(n, k)$, denoted $I(n, k)$, as
$$
I(n, k)=E(n, k)-\frac{n}{k}
$$
if $E(n, k) \neq \infty$. If $E(n, k)=\infty$, then by convention, $I(n, k)$ is undefined.
Compute $I(6,3)$. | [
"$I(6,3)=0$. By definition, $I(6,3)=E(6,3)-\\frac{6}{3}$. Because $3 \\mid 6, E(6,3)=\\frac{6}{3}=2$, and so $I(6,3)=2-2=0$."
] | [
"0"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,006 | Regular tetrahedra $J A N E, J O H N$, and $J O A N$ have non-overlapping interiors. Compute $\tan \angle H A E$. | [
"First note that $\\overline{J N}$ is a shared edge of all three pyramids, and that the viewpoint for the figure below is from along the line that is the extension of edge $\\overline{J N}$.\n\n<img_3460>\n\nLet $h$ denote the height of each pyramid. Let $X$ be the center of pyramid JOAN, and consider the plane pas... | [
"$\\frac{5 \\sqrt{2}}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,007 | Each positive integer less than or equal to 2019 is written on a blank sheet of paper, and each of the digits 0 and 5 is erased. Compute the remainder when the product of the remaining digits on the sheet of paper is divided by 1000 . | [
"Count the digits separately by position, noting that 1 is irrelevant to the product. There are a total of 20 instances of the digit 2 in the thousands place. The digit 0 only occurs in the hundreds place if the thousands digit is 2 , so look at the numbers 1 through 1999. Each non-zero digit contributes an equal n... | [
"976"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
3,008 | Compute the third least positive integer $n$ such that each of $n, n+1$, and $n+2$ is a product of exactly two (not necessarily distinct) primes. | [
"Define a positive integer $n$ to be a semiprime if it is a product of exactly two (not necessarily distinct) primes. Define a lucky trio to be a sequence of three consecutive integers, $n, n+1, n+2$, each of which is a semiprime. Note that a lucky trio must contain exactly one multiple of 3. Also note that the mid... | [
"93"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
3,009 | The points $(1,2,3)$ and $(3,3,2)$ are vertices of a cube. Compute the product of all possible distinct volumes of the cube. | [
"The distance between points $A(1,2,3)$ and $B(3,3,2)$ is $A B=\\sqrt{(3-1)^{2}+(3-2)^{2}+(2-3)^{2}}=\\sqrt{6}$. Denote by $s$ the side length of the cube. Consider three possibilities.\n\n- If $\\overline{A B}$ is an edge of the cube, then $A B=s$, so one possibility is $s_{1}=\\sqrt{6}$.\n- If $\\overline{A B}$ i... | [
"216"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,010 | Eight students attend a Harper Valley ARML practice. At the end of the practice, they decide to take selfies to celebrate the event. Each selfie will have either two or three students in the picture. Compute the minimum number of selfies so that each pair of the eight students appears in exactly one selfie. | [
"The answer is 12 . To give an example in which 12 selfies is possible, consider regular octagon $P_{1} P_{2} P_{3} P_{4} P_{5} P_{6} P_{7} P_{8}$. Each vertex of the octagon represents a student and each of the diagonals and sides of the octagon represents a pair of students. Construct eight triangles $P_{1} P_{2}... | [
"12"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,011 | $\quad$ Compute the least positive value of $t$ such that
$$
\operatorname{Arcsin}(\sin (t)), \operatorname{Arccos}(\cos (t)), \operatorname{Arctan}(\tan (t))
$$
form (in some order) a three-term arithmetic progression with a nonzero common difference. | [
"For $0 \\leq t<\\pi / 2$, all three values are $t$, so the desired $t$ does not lie in this interval.\n\nFor $\\pi / 2<t<\\pi$,\n\n$$\n\\begin{aligned}\n\\operatorname{Arcsin}(\\sin (t)) & =\\pi-t \\in(0, \\pi / 2) \\\\\n\\operatorname{Arccos}(\\cos (t)) & =t \\quad \\in(\\pi / 2, \\pi) \\\\\n\\operatorname{Arctan... | [
"$\\frac{3 \\pi}{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,012 | In non-right triangle $A B C$, distinct points $P, Q, R$, and $S$ lie on $\overline{B C}$ in that order such that $\angle B A P \cong \angle P A Q \cong \angle Q A R \cong \angle R A S \cong \angle S A C$. Given that the angles of $\triangle A B C$ are congruent to the angles of $\triangle A P Q$ in some order of correspondence, compute $\mathrm{m} \angle B$ in degrees. | [
"Let $\\theta=\\frac{1}{5} \\mathrm{~m} \\angle A$. Because $\\mathrm{m} \\angle P A Q=\\theta<5 \\theta=\\mathrm{m} \\angle A$, it follows that either $\\mathrm{m} \\angle B=\\theta$ or $\\mathrm{m} \\angle C=\\theta$. Thus there are two cases to consider.\n\nIf $\\mathrm{m} \\angle C=\\theta$, then it follows tha... | [
"$\\frac{45}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,013 | Consider the system of equations
$$
\begin{aligned}
& \log _{4} x+\log _{8}(y z)=2 \\
& \log _{4} y+\log _{8}(x z)=4 \\
& \log _{4} z+\log _{8}(x y)=5 .
\end{aligned}
$$
Given that $x y z$ can be expressed in the form $2^{k}$, compute $k$. | [
"Note that for $n>0, \\log _{4} n=\\log _{64} n^{3}$ and $\\log _{8} n=\\log _{64} n^{2}$. Adding together the three given equations and using both the preceding facts and properties of logarithms yields\n\n$$\n\\begin{aligned}\n& \\log _{4}(x y z)+\\log _{8}\\left(x^{2} y^{2} z^{2}\\right)=11 \\\\\n\\Longrightarro... | [
"$\\frac{66}{7}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,014 | A complex number $z$ is selected uniformly at random such that $|z|=1$. Compute the probability that $z$ and $z^{2019}$ both lie in Quadrant II in the complex plane. | [
"For convenience, let $\\alpha=\\pi / 4038$. Denote by\n\n$$\n0 \\leq \\theta<2 \\pi=8076 \\alpha\n$$\n\nthe complex argument of $z$, selected uniformly at random from the interval $[0,2 \\pi)$. Then $z$ itself lies in Quadrant II if and only if\n\n$$\n2019 \\alpha=\\frac{\\pi}{2}<\\theta<\\pi=4038 \\alpha\n$$\n\nO... | [
"$\\frac{505}{8076}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,015 | Compute the least positive integer $n$ such that the sum of the digits of $n$ is five times the sum of the digits of $(n+2019)$. | [
"Let $S(n)$ denote the sum of the digits of $n$, so that solving the problem is equivalent to solving $S(n)=5 S(n+2019)$. Using the fact that $S(n) \\equiv n(\\bmod 9)$ for all $n$, it follows that\n\n$$\n\\begin{aligned}\nn & \\equiv 5(n+2019) \\equiv 5(n+3)(\\bmod 9) \\\\\n4 n & \\equiv-15(\\bmod 9) \\\\\nn & \\e... | [
"7986"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
3,016 | $\quad$ Compute the greatest real number $K$ for which the graphs of
$$
(|x|-5)^{2}+(|y|-5)^{2}=K \quad \text { and } \quad(x-1)^{2}+(y+1)^{2}=37
$$
have exactly two intersection points. | [
"The graph of the second equation is simply the circle of radius $\\sqrt{37}$ centered at $(1,-1)$. The first graph is more interesting, and its behavior depends on $K$.\n\n- For small values of $K$, the first equation determines a set of four circles of radius $\\sqrt{K}$ with centers at $(5,5),(5,-5),(-5,5)$, and... | [
"29"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,017 | To morph a sequence means to replace two terms $a$ and $b$ with $a+1$ and $b-1$ if and only if $a+1<b-1$, and such an operation is referred to as a morph. Compute the least number of morphs needed to transform the sequence $1^{2}, 2^{2}, 3^{2}, \ldots, 10^{2}$ into an arithmetic progression. | [
"Call the original sequence of ten squares $T=\\left(1^{2}, 2^{2}, \\ldots, 10^{2}\\right)$. A morphed sequence is one that can be obtained by morphing $T$ a finite number of times.\n\nThis solution is divided into three steps. In the first step, a characterization of the possible final morphed sequences is given. ... | [
"56"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,018 | Triangle $A B C$ is inscribed in circle $\omega$. The tangents to $\omega$ at $B$ and $C$ meet at point $T$. The tangent to $\omega$ at $A$ intersects the perpendicular bisector of $\overline{A T}$ at point $P$. Given that $A B=14, A C=30$, and $B C=40$, compute $[P B C]$. | [
"To begin, denote by $R$ the radius of $\\omega$. The semiperimeter of triangle $A B C$ is 42 , and then applying Heron's formula yields\n\n$$\n[A B C]=\\frac{14 \\cdot 30 \\cdot 40}{4 R}=\\sqrt{42 \\cdot 28 \\cdot 12 \\cdot 2}=168\n$$\n\nfrom which it follows that $R=\\frac{14 \\cdot 30 \\cdot 40}{4 \\cdot 168}=25... | [
"$\\frac{800}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,019 | Given that $a, b, c$, and $d$ are integers such that $a+b c=20$ and $-a+c d=19$, compute the greatest possible value of $c$. | [
"Adding the two given equations yields $b c+c d=c(b+d)=39$. The greatest possible value of $c$ therefore occurs when $c=\\mathbf{3 9}$ and $b+d=1$."
] | [
"39"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,020 | Let $T$ = 39. Emile randomly chooses a set of $T$ cards from a standard deck of 52 cards. Given that Emile's set contains no clubs, compute the probability that his set contains three aces. | [
"Knowing that 13 of the cards are not in Emile's set, there are $\\left(\\begin{array}{c}39 \\\\ T\\end{array}\\right)$ ways for him to have chosen a set of $T$ cards. Given that Emile's set contains no clubs, the suits of the three aces are fixed (i.e., diamonds, hearts, and spades). The number of possible sets of... | [
"1"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,021 | Let $T=1$. In parallelogram $A B C D, \frac{A B}{B C}=T$. Given that $M$ is the midpoint of $\overline{A B}$ and $P$ and $Q$ are the trisection points of $\overline{C D}$, compute $\frac{[A B C D]}{[M P Q]}$. | [
"Let $C D=3 x$ and let $h$ be the length of the altitude between bases $\\overline{A B}$ and $\\overline{C D}$. Then $[A B C D]=3 x h$ and $[M P Q]=\\frac{1}{2} x h$. Hence $\\frac{[A B C D]}{[M P Q]}=\\mathbf{6}$. Both the position of $M$ and the ratio $\\frac{A B}{B C}=T$ are irrelevant."
] | [
"6"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,022 | Let $T=6$. Compute the value of $x$ such that $\log _{T} \sqrt{x-7}+\log _{T^{2}}(x-2)=1$. | [
"It can readily be shown that $\\log _{a} b=\\log _{a^{2}} b^{2}$. Thus it follows that $\\log _{T} \\sqrt{x-7}=\\log _{T^{2}}(x-7)$. Hence the left-hand side of the given equation is $\\log _{T^{2}}(x-7)(x-2)$ and the equation is equivalent to $(x-7)(x-2)=T^{2}$, which is equivalent to $x^{2}-9 x+14-T^{2}=0$. With... | [
"11"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,023 | Let $T=11$. Let $p$ be an odd prime and let $x, y$, and $z$ be positive integers less than $p$. When the trinomial $(p x+y+z)^{T-1}$ is expanded and simplified, there are $N$ terms, of which $M$ are always multiples of $p$. Compute $M$. | [
"A general term in the expansion of $(p x+y+z)^{T-1}$ has the form $K(p x)^{a} y^{b} z^{c}$, where $a, b$, and $c$ are nonnegative integers such that $a+b+c=T-1$. Using the \"stars and bars\" approach, the number of nonnegative integral solutions to $a+b+c=T-1$ is the number of arrangements of $T-1$ stars and 2 bar... | [
"55"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
3,024 | Let $T=55$. Compute the value of $K$ such that $20, T-5, K$ is an increasing geometric sequence and $19, K, 4 T+11$ is an increasing arithmetic sequence. | [
"The condition that $20, T-5, K$ is an increasing geometric sequence implies that $\\frac{T-5}{20}=\\frac{K}{T-5}$, hence $K=\\frac{(T-5)^{2}}{20}$. The condition that $19, K, 4 T+11$ is an increasing arithmetic sequence implies that $K-19=4 T+11-K$, hence $K=2 T+15$. With $T=55$, each of these equations implies th... | [
"125"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,025 | Let $T=125$. Cube $\mathcal{C}_{1}$ has volume $T$ and sphere $\mathcal{S}_{1}$ is circumscribed about $\mathcal{C}_{1}$. For $n \geq 1$, the sphere $\mathcal{S}_{n}$ is circumscribed about the cube $\mathcal{C}_{n}$ and is inscribed in the cube $\mathcal{C}_{n+1}$. Let $k$ be the least integer such that the volume of $\mathcal{C}_{k}$ is at least 2019. Compute the edge length of $\mathcal{C}_{k}$. | [
"In general, let cube $\\mathcal{C}_{n}$ have edge length $x$. Then the diameter of sphere $\\mathcal{S}_{n}$ is the space diagonal of $\\mathcal{C}_{n}$, which has length $x \\sqrt{3}$. This in turn is the edge length of cube $\\mathcal{C}_{n+1}$. Hence the edge lengths of $\\mathcal{C}_{1}, \\mathcal{C}_{2}, \\ld... | [
"15"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,026 | Square $K E N T$ has side length 20 . Point $M$ lies in the interior of $K E N T$ such that $\triangle M E N$ is equilateral. Given that $K M^{2}=a-b \sqrt{3}$, where $a$ and $b$ are integers, compute $b$. | [
"Let $s$ be the side length of square $K E N T$; then $M E=s$. Let $J$ be the foot of the altitude from $M$ to $\\overline{K E}$. Then $\\mathrm{m} \\angle J E M=30^{\\circ}$ and $\\mathrm{m} \\angle E M J=60^{\\circ}$. Hence $M J=\\frac{s}{2}, J E=\\frac{s \\sqrt{3}}{2}$, and $K J=K E-J E=s-\\frac{s \\sqrt{3}}{2}$... | [
"400"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,027 | Let $T$ be a rational number. Let $a, b$, and $c$ be the three solutions of the equation $x^{3}-20 x^{2}+19 x+T=0$. Compute $a^{2}+b^{2}+c^{2}$. | [
"According to Vieta's formulas, $a+b+c=-(-20)=20$ and $a b+b c+c a=19$. Noting that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)$, it follows that $a^{2}+b^{2}+c^{2}=20^{2}-2 \\cdot 19=\\mathbf{3 6 2}$. The value of $T$ is irrelevant."
] | [
"362"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,028 | Let $T=362$ and let $K=\sqrt{T-1}$. Compute $\left|(K-20)(K+1)+19 K-K^{2}\right|$. | [
"The expression inside the absolute value bars simplifies to $K^{2}-19 K-20+19 K-K^{2}=-20$. Hence the answer is $\\mathbf{2 0}$ and the value of $K(=\\sqrt{361}=19)$ is not needed."
] | [
"20"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,029 | Let $T=20$. In $\triangle L E O, \sin \angle L E O=\frac{1}{T}$. If $L E=\frac{1}{n}$ for some positive real number $n$, then $E O=$ $n^{3}-4 n^{2}+5 n$. As $n$ ranges over the positive reals, compute the least possible value of $[L E O]$. | [
"Note that $[L E O]=\\frac{1}{2}(\\sin \\angle L E O) \\cdot L E \\cdot E O=\\frac{1}{2} \\cdot \\frac{1}{T} \\cdot \\frac{1}{n} \\cdot\\left(n^{3}-4 n^{2}+5 n\\right)=\\frac{n^{2}-4 n+5}{2 T}$. Because $T$ is a constant, the least possible value of $[L E O]$ is achieved when the function $f(n)=n^{2}-4 n+5$ is mini... | [
"$\\frac{1}{40}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,030 | Let $T=\frac{1}{40}$. Given that $x, y$, and $z$ are real numbers such that $x+y=5, x^{2}-y^{2}=\frac{1}{T}$, and $x-z=-7$, compute $x+z$ | [
"Note that $x^{2}-y^{2}=(x+y)(x-y)=5(x-y)$, hence $x-y=\\frac{1}{5 T}$. Then $x+z=(x+y)+(x-y)+(z-x)=$ $5+\\frac{1}{5 T}+7=12+\\frac{1}{5 T}$. With $T=\\frac{1}{40}$, the answer is thus $12+8=\\mathbf{2 0}$."
] | [
"20"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,031 | Let $T=20$. The product of all positive divisors of $2^{T}$ can be written in the form $2^{K}$. Compute $K$. | [
"When $n$ is a nonnegative integer, the product of the positive divisors of $2^{n}$ is $2^{0} \\cdot 2^{1} \\cdot \\ldots \\cdot 2^{n-1} \\cdot 2^{n}=$ $2^{0+1+\\cdots+(n-1)+n}=2^{n(n+1) / 2}$. Because $T=20$ is an integer, it follows that $K=\\frac{T(T+1)}{2}=\\mathbf{2 1 0}$."
] | [
"210"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
3,032 | Let $T=210$. At the Westward House of Supper ("WHS"), a dinner special consists of an appetizer, an entrée, and dessert. There are 7 different appetizers and $K$ different entrées that a guest could order. There are 2 dessert choices, but ordering dessert is optional. Given that there are $T$ possible different orders that could be placed at the WHS, compute $K$. | [
"Because dessert is optional, there are effectively $2+1=3$ dessert choices. Hence, by the Multiplication Principle, it follows that $T=7 \\cdot K \\cdot 3$, thus $K=\\frac{T}{21}$. With $T=210$, the answer is 10 ."
] | [
"10"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,033 | Let $S=15$ and let $M=10$ . Sam and Marty each ride a bicycle at a constant speed. Sam's speed is $S \mathrm{~km} / \mathrm{hr}$ and Marty's speed is $M \mathrm{~km} / \mathrm{hr}$. Given that Sam and Marty are initially $100 \mathrm{~km}$ apart and they begin riding towards one another at the same time, along a straight path, compute the number of kilometers that Sam will have traveled when Sam and Marty meet. | [
"In km/hr, the combined speed of Sam and Marty is $S+M$. Thus one can determine the total time they traveled and use this to determine the number of kilometers that Sam traveled. However, this is not needed, and there is a simpler approach. Suppose that Marty traveled a distance of $d$. Then because Sam's speed is ... | [
"60"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,034 | Compute the $2011^{\text {th }}$ smallest positive integer $N$ that gains an extra digit when doubled. | [
"Let $S$ be the set of numbers that gain an extra digit when doubled. First notice that the numbers in $S$ are precisely those whose first digit is at least 5 . Thus there are five one-digit numbers in $S, 50$ two-digit numbers in $S$, and 500 three-digit numbers in $S$. Therefore 5000 is the $556^{\\text {th }}$ s... | [
"6455"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
3,035 | In triangle $A B C, C$ is a right angle and $M$ is on $\overline{A C}$. A circle with radius $r$ is centered at $M$, is tangent to $\overline{A B}$, and is tangent to $\overline{B C}$ at $C$. If $A C=5$ and $B C=12$, compute $r$. | [
"Let $N$ be the point of tangency of the circle with $\\overline{A B}$ and draw $\\overline{M B}$, as shown below.\n\n<img_3520>\n\nBecause $\\triangle B M C$ and $\\triangle B M N$ are right triangles sharing a hypotenuse, and $\\overline{M N}$ and $\\overline{M C}$ are radii, $\\triangle B M C \\cong \\triangle B... | [
"$\\frac{12}{5}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,036 | The product of the first five terms of a geometric progression is 32 . If the fourth term is 17 , compute the second term. | [
"Let $a$ be the third term of the geometric progression, and let $r$ be the common ratio. Then the product of the first five terms is\n\n$$\n\\left(a r^{-2}\\right)\\left(a r^{-1}\\right)(a)(a r)\\left(a r^{2}\\right)=a^{5}=32\n$$\n\nso $a=2$. Because the fourth term is $17, r=\\frac{17}{a}=\\frac{17}{2}$. The seco... | [
"$\\frac{4}{17}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
3,037 | Polygon $A_{1} A_{2} \ldots A_{n}$ is a regular $n$-gon. For some integer $k<n$, quadrilateral $A_{1} A_{2} A_{k} A_{k+1}$ is a rectangle of area 6 . If the area of $A_{1} A_{2} \ldots A_{n}$ is 60 , compute $n$. | [
"Because $A_{1} A_{2} A_{k} A_{k+1}$ is a rectangle, $n$ must be even, and moreover, $k=\\frac{n}{2}$. Also, the rectangle's diagonals meet at the center $O$ of the circumscribing circle. $O$ is also the center of the $n$-gon. The diagram below shows the case $n=16$.\n\n\n\n<img_3867>\n\nThen $\\left[A_{1} A_{2} O\... | [
"40"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,038 | A bag contains 20 lavender marbles, 12 emerald marbles, and some number of orange marbles. If the probability of drawing an orange marble in one try is $\frac{1}{y}$, compute the sum of all possible integer values of $y$. | [
"Let $x$ be the number of orange marbles. Then the probability of drawing an orange marble is $\\frac{x}{x+20+12}=\\frac{x}{x+32}$. If this probability equals $\\frac{1}{y}$, then $y=\\frac{x+32}{x}=1+\\frac{32}{x}$. This expression represents an integer only when $x$ is a factor of 32 , thus $x \\in\\{1,2,4,8,16,3... | [
"69"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,039 | Compute the number of ordered quadruples of integers $(a, b, c, d)$ satisfying the following system of equations:
$$
\left\{\begin{array}{l}
a b c=12,000 \\
b c d=24,000 \\
c d a=36,000
\end{array}\right.
$$ | [
"From the first two equations, conclude that $d=2 a$. From the last two, $3 b=2 a$. Thus all solutions to the system will be of the form $(3 K, 2 K, c, 6 K)$ for some integer $K$. Substituting these expressions into the system, each equation now becomes $c K^{2}=2000=2^{4} \\cdot 5^{3}$. So $K^{2}$ is of the form $... | [
"12"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,040 | Let $n$ be a positive integer such that $\frac{3+4+\cdots+3 n}{5+6+\cdots+5 n}=\frac{4}{11}$. Compute $\frac{2+3+\cdots+2 n}{4+5+\cdots+4 n}$. | [
"In simplifying the numerator and denominator of the left side of the equation, notice that\n\n$$\n\\begin{aligned}\nk+(k+1)+\\cdots+k n & =\\frac{1}{2}(k n(k n+1)-k(k-1)) \\\\\n& =\\frac{1}{2}(k(n+1)(k n-k+1))\n\\end{aligned}\n$$\n\nThis identity allows the given equation to be transformed:\n\n$$\n\\begin{aligned}... | [
"$\\frac{27}{106}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,041 | The quadratic polynomial $f(x)$ has a zero at $x=2$. The polynomial $f(f(x))$ has only one real zero, at $x=5$. Compute $f(0)$. | [
"Let $f(x)=a(x-b)^{2}+c$. The graph of $f$ is symmetric about $x=b$, so the graph of $y=f(f(x))$ is also symmetric about $x=b$. If $b \\neq 5$, then $2 b-5$, the reflection of 5 across $b$, must be a zero of $f(f(x))$. Because $f(f(x))$ has exactly one zero, $b=5$.\n\nBecause $f(2)=0$ and $f$ is symmetric about $x=... | [
"$-\\frac{32}{9}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,042 | The Local Area Inspirational Math Exam comprises 15 questions. All answers are integers ranging from 000 to 999, inclusive. If the 15 answers form an arithmetic progression with the largest possible difference, compute the largest possible sum of those 15 answers. | [
"Let $a$ represent the middle $\\left(8^{\\text {th }}\\right)$ term of the sequence, and let $d$ be the difference. Then the terms of the sequence are $a-7 d, a-6 d, \\ldots, a+6 d, a+7 d$, their sum is $15 a$, and the difference between the largest and the smallest terms is $14 d$. The largest $d$ such that $14 d... | [
"7530"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,043 | Circle $\omega_{1}$ has center $O$, which is on circle $\omega_{2}$. The circles intersect at points $A$ and $C$. Point $B$ lies on $\omega_{2}$ such that $B A=37, B O=17$, and $B C=7$. Compute the area of $\omega_{1}$. | [
"The points $O, A, B, C$ all lie on $\\omega_{2}$ in some order. There are two possible cases to consider: either $B$ is outside circle $\\omega_{1}$, or it is inside the circle, as shown below.\n\n\n\n<img_3962>\n\nThe following argument shows that the first case is impossible. By the Triangle Inequality on $\\tri... | [
"$548 \\pi$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,044 | Compute the number of integers $n$ for which $2^{4}<8^{n}<16^{32}$. | [
"$8^{n}=2^{3 n}$ and $16^{32}=2^{128}$. Therefore $4<3 n<128$, and $2 \\leq n \\leq 42$. Thus there are 41 such integers $n$."
] | [
"41"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,045 | Let $T=41$. Compute the number of positive integers $b$ such that the number $T$ has exactly two digits when written in base $b$. | [
"If $T$ has more than one digit when written in base $b$, then $b \\leq T$. If $T$ has fewer than three digits when written in base $b$, then $b^{2}>T$, or $b>\\sqrt{T}$. So the desired set of bases $b$ is $\\{b \\mid \\sqrt{T}<b \\leq T\\}$. When $T=41,\\lfloor\\sqrt{T}\\rfloor=6$ and so $6<b \\leq 41$. There are ... | [
"35"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
3,046 | Let $T=35$. Triangle $A B C$ has a right angle at $C$, and $A B=40$. If $A C-B C=T-1$, compute $[A B C]$, the area of $\triangle A B C$. | [
"Let $A C=b$ and $B C=a$. Then $a^{2}+b^{2}=1600$ and $|a-b|=T-1$. Squaring the second equation yields $a^{2}+b^{2}-2 a b=(T-1)^{2}$, so $1600-2 a b=(T-1)^{2}$. Hence the area of the triangle is $\\frac{1}{2} a b=\\frac{1600-(T-1)^{2}}{4}=400-\\frac{(T-1)^{2}}{4}$ or $400-\\left(\\frac{T-1}{2}\\right)^{2}$, which f... | [
"111"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,047 | Let $x$ be a positive real number such that $\log _{\sqrt{2}} x=20$. Compute $\log _{2} \sqrt{x}$. | [
"The identity $\\log _{b^{n}} x=\\frac{1}{n} \\log _{b} x$ yields $\\log _{2} x=10$. Then $\\log _{2} \\sqrt{x}=\\log _{2} x^{1 / 2}=\\frac{1}{2} \\log _{2} x=5$.",
"Use the definition of $\\log$ to obtain $x=(\\sqrt{2})^{20}=\\left(2^{1 / 2}\\right)^{20}=2^{10}$. Thus $\\log _{2} \\sqrt{x}=\\log _{2} 2^{5}=\\mat... | [
"5"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,048 | Let $T=5$. Hannah flips two fair coins, while Otto flips $T$ fair coins. Let $p$ be the probability that the number of heads showing on Hannah's coins is greater than the number of heads showing on Otto's coins. If $p=q / r$, where $q$ and $r$ are relatively prime positive integers, compute $q+r$. | [
"Because Hannah has only two coins, the only ways she can get more heads than Otto are if she gets 1 (and he gets 0 ), or she gets 2 (and he gets either 1 or 0 ).\n\nThe probability of Hannah getting exactly one head is $\\frac{1}{2}$. The probability of Otto getting no heads is $\\frac{1}{2^{T}}$. So the probabili... | [
"17"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,049 | Let $T=17$. In ARMLovia, the unit of currency is the edwah. Janet's wallet contains bills in denominations of 20 and 80 edwahs. If the bills are worth an average of $2 T$ edwahs each, compute the smallest possible value of the bills in Janet's wallet. | [
"Let $x$ be the number of twenty-edwah bills and $y$ be the number of eighty-edwah bills. Then\n\n$$\n\\begin{aligned}\n\\frac{20 x+80 y}{x+y} & =2 T \\\\\n20 x+80 y & =2 T x+2 T y \\\\\n(80-2 T) y & =(2 T-20) x\n\\end{aligned}\n$$\n\n\n\nIn the case where $T=17$ (and hence $2 T=34$ ), this equation reduces to $46 ... | [
"1020"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,075 | Spheres centered at points $P, Q, R$ are externally tangent to each other, and are tangent to plane $\mathcal{M}$ at points $P^{\prime}, Q^{\prime}, R^{\prime}$, respectively. All three spheres are on the same side of the plane. If $P^{\prime} Q^{\prime}=Q^{\prime} R^{\prime}=12$ and $P^{\prime} R^{\prime}=6$, compute the area of $\triangle P Q R$. | [
"Let the radii be $p, q, r$ respectively. Looking at a cross-section of the spheres through $\\overline{P Q}$ perpendicular to the plane, the points $P^{\\prime}, P, Q, Q^{\\prime}$ form a right trapezoid with $\\overline{P^{\\prime} P} \\perp \\overline{P^{\\prime} Q^{\\prime}}$ and $\\overline{Q^{\\prime} Q} \\pe... | [
"$18 \\sqrt{6}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,076 | Let $f(x)=x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\cdots$. Compute the coefficient of $x^{10}$ in $f(f(x))$. | [
"By the definition of $f$,\n\n$$\nf(f(x))=f(x)+(f(x))^{2}+(f(x))^{4}+(f(x))^{8}+\\cdots\n$$\n\nConsider this series term by term. The first term, $f(x)$, contains no $x^{10}$ terms, so its contribution is 0 . The second term, $(f(x))^{2}$, can produce terms of $x^{10}$ in two ways: as $x^{2} \\cdot x^{8}$ or as $x^... | [
"40"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,077 | Compute $\left\lfloor 100000(1.002)^{10}\right\rfloor$. | [
"Consider the expansion of $(1.002)^{10}$ as $(1+0.002)^{10}$. Using the Binomial Theorem yields the following:\n\n$$\n(1+0.002)^{10}=1+\\left(\\begin{array}{c}\n10 \\\\\n1\n\\end{array}\\right)(0.002)+\\left(\\begin{array}{c}\n10 \\\\\n2\n\\end{array}\\right)(0.002)^{2}+\\left(\\begin{array}{c}\n10 \\\\\n3\n\\end{... | [
"102018"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.