id
int64 | subfield
string | context
null | question
string | solution
list | final_answer
list | is_multiple_answer
bool | unit
string | answer_type
string | error
string | answer
string |
|---|---|---|---|---|---|---|---|---|---|---|
2,764
|
Number Theory
| null |
Compute the least integer value of the function
$$
f(x)=\frac{x^{4}-6 x^{3}+2 x^{2}-6 x+2}{x^{2}+1}
$$
whose domain is the set of all real numbers.
|
[
"$\\quad$ Use polynomial long division to rewrite $f(x)$ as\n\n$$\nf(x)=x^{2}-6 x+1+\\frac{1}{x^{2}+1}\n$$\n\nThe quadratic function $x^{2}-6 x+1=(x-3)^{2}-8$ has a minimum of -8 , achieved at $x=3$. The \"remainder term\" $\\frac{1}{x^{2}+1}$ is always positive. Thus $f(x)>-8$ for all $x$, so any integer value of $f(x)$ must be at least -7 .\n\nWhen $x=3$, the remainder term is less than 1 , so $f(3)$ is less than -7 . But $f(4)=-\\frac{34}{5}>-7$, so there must be some value of $x$ between 3 and 4 for which $f(x)=-7$, so the least integer value of $f(x)$ is $\\mathbf{- 7}$. The reader may note that $f(x)=-7$ when $x \\approx 2.097$ and $x \\approx 3.970$."
] |
[
"-7"
] | false
| null |
Numerical
| null |
\boxed{-7}
|
2,727
|
Geometry
| null |
Let $T=T N Y W R$. The parabola $y=x^{2}+T x$ is tangent to the parabola $y=-(x-2 T)^{2}+b$. Compute $b$.
|
[
"In this case, the two parabolas are tangent exactly when the system of equations has a unique solution. (Query: Is this the case for every pair of equations representing parabolas?) So set the right sides equal to each other: $x^{2}+T x=-(x-2 T)^{2}+b$. Then $x^{2}+T x=$ $-x^{2}+4 T x-4 T^{2}+b$, or equivalently, $2 x^{2}-3 T x+4 T^{2}-b=0$. The equation has a double root when the discriminant is 0 , so set $(-3 T)^{2}-4\\left(4 T^{2}-b\\right)(2)=0$ and solve: $9 T^{2}-32 T^{2}+8 b=0$ implies $-23 T^{2}+8 b=0$, or $b=23 T^{2} / 8$. Using $T=8$ yields $b=\\mathbf{1 8 4}$."
] |
[
"184"
] | false
| null |
Numerical
| null |
\boxed{184}
|
2,344
|
Algebra
| null |
The geometric sequence with $n$ terms $t_{1}, t_{2}, \ldots, t_{n-1}, t_{n}$ has $t_{1} t_{n}=3$. Also, the product of all $n$ terms equals 59049 (that is, $t_{1} t_{2} \cdots t_{n-1} t_{n}=59049$ ). Determine the value of $n$.
(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.)
|
[
"Suppose that the first term in the geometric sequence is $t_{1}=a$ and the common ratio in the sequence is $r$.\n\nThen the sequence, which has $n$ terms, is $a, a r, a r^{2}, a r^{3}, \\ldots, a r^{n-1}$.\n\nIn general, the $k$ th term is $t_{k}=a r^{k-1}$; in particular, the $n$th term is $t_{n}=a r^{n-1}$.\n\nSince $t_{1} t_{n}=3$, then $a \\cdot a r^{n-1}=3$ or $a^{2} r^{n-1}=3$.\n\nSince $t_{1} t_{2} \\cdots t_{n-1} t_{n}=59049$, then\n\n$$\n\\begin{aligned}\n(a)(a r) \\cdots\\left(a r^{n-2}\\right)\\left(a r^{n-1}\\right) & =59049 \\\\\na^{n} r r^{2} \\cdots r^{n-2} r^{n-1} & =59049 \\\\\na^{n} r^{1+2+\\cdots+(n-2)+(n-1)} & =59049 \\\\\na^{n} r^{\\frac{1}{2}(n-1)(n)} & =59049\n\\end{aligned}\n$$\n\n$$\na^{n} r r^{2} \\cdots r^{n-2} r^{n-1}=59049 \\quad \\text { (since there are } n \\text { factors of } a \\text { on the left side) }\n$$\n\nsince $1+2+\\cdots+(n-2)+(n-1)=\\frac{1}{2}(n-1)(n)$.\n\nSince $a^{2} r^{n-1}=3$, then $\\left(a^{2} r^{n-1}\\right)^{n}=3^{n}$ or $a^{2 n} r^{(n-1)(n)}=3^{n}$.\n\nSince $a^{n} r^{\\frac{1}{2}(n-1)(n)}=59049$, then $\\left(a^{n} r^{\\frac{1}{2}(n-1)(n)}\\right)^{2}=59049^{2}$ or $a^{2 n} r^{(n-1)(n)}=59049^{2}$.\n\nSince the left sides of these equations are the same, then $3^{n}=59049^{2}$.\n\nNow\n\n$$\n59049=3(19683)=3^{2}(6561)=3^{3}(2187)=3^{4}(729)=3^{5}(243)=3^{6}(81)=3^{6} 3^{4}=3^{10}\n$$\n\nSince $59049=3^{10}$, then $59049^{2}=3^{20}$ and so $3^{n}=3^{20}$, which gives $n=20$."
] |
[
"20"
] | false
| null |
Numerical
| null |
\boxed{20}
|
2,582
|
Combinatorics
| null |
Eight people, including triplets Barry, Carrie and Mary, are going for a trip in four canoes. Each canoe seats two people. The eight people are to be randomly assigned to the four canoes in pairs. What is the probability that no two of Barry, Carrie and Mary will be in the same canoe?
|
[
"Among a group of $n$ people, there are $\\frac{n(n-1)}{2}$ ways of choosing a pair of these people:\n\nThere are $n$ people that can be chosen first.\n\nFor each of these $n$ people, there are $n-1$ people that can be chosen second.\n\nThis gives $n(n-1)$ orderings of two people.\n\nEach pair is counted twice (given two people A and B, we have counted both the\n\npair $\\mathrm{AB}$ and the pair $\\mathrm{BA})$, so the total number of pairs is $\\frac{n(n-1)}{2}$.\n\nWe label the four canoes W, X, Y, and Z.\n\nFirst, we determine the total number of ways to put the 8 people in the 4 canoes.\n\nWe choose 2 people to put in W. There are $\\frac{8 \\cdot 7}{2}$ pairs. This leaves 6 people for the remaining 3 canoes.\n\nNext, we choose 2 people to put in X. There are $\\frac{6 \\cdot 5}{2}$ pairs. This leaves 4 people for the remaining 2 canoes.\n\nNext, we choose 2 people to put in Y. There are $\\frac{4 \\cdot 3}{2}$ pairs. This leaves 2 people for the remaining canoe.\n\nThere is now 1 way to put the remaining people in $\\mathrm{Z}$.\n\nTherefore, there are\n\n$$\n\\frac{8 \\cdot 7}{2} \\cdot \\frac{6 \\cdot 5}{2} \\cdot \\frac{4 \\cdot 3}{2}=\\frac{8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3}{2^{3}}=7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3\n$$\n\nways to put the 8 people in the 4 canoes.\n\nNow, we determine the number of ways in which no two of Barry, Carrie and Mary will be in the same canoe.\n\nThere are 4 possible canoes in which Barry can go.\n\nThere are then 3 possible canoes in which Carrie can go, because she cannot go in the same canoe as Barry.\n\nThere are then 2 possible canoes in which Mary can go, because she cannot go in the same canoe as Barry or Carrie.\n\nThis leaves 5 people left to put in the canoes.\n\nThere are 5 choices of the person that can go with Barry, and then 4 choices of the person that can go with Carrie, and then 3 choices of the person that can go with Mary.\n\nThe remaining 2 people are put in the remaining empty canoe.\n\nThis means that there are $4 \\cdot 3 \\cdot 2 \\cdot 5 \\cdot 4 \\cdot 3$ ways in which the 8 people can be put in 4 canoes so that no two of Barry, Carrie and Mary are in the same canoe.\n\nTherefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is $\\frac{4 \\cdot 3 \\cdot 2 \\cdot 5 \\cdot 4 \\cdot 3}{7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3}=\\frac{4 \\cdot 3 \\cdot 2}{7 \\cdot 6}=\\frac{24}{42}=\\frac{4}{7}$.",
"Let $p$ be the probability that two of Barry, Carrie and Mary are in the same canoe.\n\nThe answer to the original problem will be $1-p$.\n\nLet $q$ be the probability that Barry and Carrie are in the same canoe.\n\nBy symmetry, the probability that Barry and Mary are in the same canoe also equals $q$ as does the probability that Carrie and Mary are in the same canoe.\n\nThis means that $p=3 q$.\n\nSo we calculate $q$.\n\nTo do this, we put Barry in a canoe. Since there are 7 possible people who can go in the canoe with him, then the probability that Carrie is in the canoe with him equals $\\frac{1}{7}$. The other 6 people can be put in the canoes in any way.\n\nThis means that the probability that Barry and Carrie are in the same canoe is $q=\\frac{1}{7}$.\n\nTherefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is $1-3 \\cdot \\frac{1}{7}$ or $\\frac{4}{7}$."
] |
[
"$\\frac{4}{7}$"
] | false
| null |
Numerical
| null |
\boxed{\frac{4}{7}}
|
2,831
|
Geometry
| null |
Let $T=8$. Let $A=(1,5)$ and $B=(T-1,17)$. Compute the value of $x$ such that $(x, 3)$ lies on the perpendicular bisector of $\overline{A B}$.
|
[
"The midpoint of $\\overline{A B}$ is $\\left(\\frac{T}{2}, 11\\right)$, and the slope of $\\overleftrightarrow{A B}$ is $\\frac{12}{T-2}$. Thus the perpendicular bisector of $\\overline{A B}$ has slope $\\frac{2-T}{12}$ and passes through the point $\\left(\\frac{T}{2}, 11\\right)$. Thus the equation of the perpendicular bisector of $\\overline{A B}$ is $y=\\left(\\frac{2-T}{12}\\right) x+\\left(11-\\frac{2 T-T^{2}}{24}\\right)$. Plugging $y=3$ into this equation and solving for $x$ yields $x=\\frac{96}{T-2}+\\frac{T}{2}$. With $T=8$, it follows that $x=\\frac{96}{6}+\\frac{8}{2}=16+4=\\mathbf{2 0}$."
] |
[
"20"
] | false
| null |
Numerical
| null |
\boxed{20}
|
2,212
|
Algebra
| null |
Given a positive integer $n \geq 2$, determine the largest positive integer $N$ for which there exist $N+1$ real numbers $a_{0}, a_{1}, \ldots, a_{N}$ such that
(1) $a_{0}+a_{1}=-\frac{1}{n}$, and
(2) $\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ for $1 \leq k \leq N-1$.
|
[
"$\\left(a_{k}+a_{k-1}\\right)\\left(a_{k}+a_{k+1}\\right)=a_{k-1}-a_{k+1}$ is equivalent to $\\left(a_{k}+a_{k-1}+1\\right)\\left(a_{k}+a_{k+1}-1\\right)=-1$. Let $b_{k}=a_{k}+a_{k+1}$. Thus we need $b_{0}, b_{1}, \\ldots$ the following way: $b_{0}=-\\frac{1}{n}$ and $\\left(b_{k-1}+1\\right)\\left(b_{k}-1\\right)=-1$. There is a proper sequence $b_{0}, b_{1}, \\ldots, b_{N-1}$ if and only if there is proper sequence $a_{0}, a_{1}, \\ldots, a_{N}$, because from a a proper $\\left(a_{k}\\right)$ sequence we can get a proper $\\left(b_{k}\\right)$ sequence with $b_{k}=a_{k}+a_{k+1}$ for $k=0,1, \\ldots, N-1$ and from a proper $\\left(b_{k}\\right)$ sequence we can get a proper $\\left(a_{k}\\right)$ sequence by arbitrarily setting $a_{0}$ and then inductively defining $a_{k}=b_{k-1}-a_{k-1}$ for $k=1,2, \\ldots, N$.\n\nWe prove by induction that $b_{k}=-\\frac{1}{n-k}$ for $k<n$. This is true for $k=0$, as $b_{0}=-\\frac{1}{n}$ and\n\n$$\nb_{k}=1-\\frac{1}{b_{k-1}+1}=1-\\frac{1}{1-\\frac{1}{n-k+1}}=-\\frac{1}{n-k}\n$$\n\nfor $k<n$. Thus there is a proper sequence $b_{0}, b_{1}, \\ldots, b_{n-1}$, but it can't be continued, because $b_{n-1}+1=$ 0 so there is no $b_{n}$ for which $\\left(b_{n-1}+1\\right)\\left(b_{n}-1\\right)=-1$.\n\nTherefore the longest proper sequence $\\left(b_{k}\\right)$ is $n$-long, so the longest proper sequence $\\left(a_{k}\\right)$ is $n+1$ long, so $N=n$.",
"The required maximum is $N=n$.\n\nTo rule out the case $N \\geq n+1$, it is clearly sufficient to rule out the case $N=n+1$.\n\nAssume for contradiction that $a_{0}, a_{1}, \\ldots, a_{n+1}$ are real numbers satisfying both conditions in the statement. It is sufficient to show that $a_{k}+a_{k+1}=0$ for some $k \\leq n$, because then $a_{k-1}-a_{k+1}=0$ so $a_{k+1}=a_{k-1}$, therefore $a_{k-1}+a_{k}=0$ and so on, by backwards recursion we get that $a_{j}+a_{j+1}=0$ for all $0 \\leq j \\leq k$, but this is a contradiction with $a_{0}+a_{1}=-\\frac{1}{n}$.\n\nTo prove that $a_{k}+a_{k+1}=0$ for some $k \\leq n$, assume that $a_{k}+a_{k+1} \\neq 0$ for all $k \\leq n$, to rewrite the second condition in the statement in the form\n\n$$\n\\frac{1}{a_{k}+a_{k+1}}-\\frac{1}{a_{k-1}+a_{k}}=1, \\quad k=1, \\ldots, n\n$$\n\nand sum both sides over the full range from $k=1$ to $n$. This gives\n\n$$\n\\frac{1}{a_{n}+a_{n+1}}-\\frac{1}{a_{0}+a_{1}}=n\n$$\n\nAs $a_{0}+a_{1}=-\\frac{1}{n}$, this means that $\\frac{1}{a_{n}+a_{n+1}}=0$, which is a contradiction. Consequently, $N \\leq n$.\n\nTo provide $n+1$ real numbers satisfying both conditions in the statement, fix $a_{0}$ and go through the telescoping procedure above to obtain\n\n$$\na_{k}=(-1)^{k} a_{0}+\\sum_{j=1}^{k} \\frac{(-1)^{k-j+1}}{n-j+1}, \\quad k=1, \\ldots, n\n$$\n\nThis concludes the proof."
] |
[
"$N=n$"
] | false
| null |
Expression
| null |
\boxed{N=n}
|
3,013
|
Algebra
| null |
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$$\n\\begin{aligned}\n& \\log _{4}(x y z)+\\log _{8}\\left(x^{2} y^{2} z^{2}\\right)=11 \\\\\n\\Longrightarrow & \\log _{64}(x y z)^{3}+\\log _{64}(x y z)^{4}=11 \\\\\n\\Longrightarrow & \\log _{64}(x y z)^{7}=11 \\\\\n\\Longrightarrow & 7 \\log _{64}(x y z)=11 .\n\\end{aligned}\n$$\n\nThe last equation is equivalent to $x y z=64^{11 / 7}=2^{66 / 7}$, hence the desired value of $k$ is $\\frac{\\mathbf{6 6}}{\\mathbf{7}}$."
] |
[
"$\\frac{66}{7}$"
] | false
| null |
Numerical
| null |
\boxed{\frac{66}{7}}
|
2,778
|
Number Theory
| null |
Let $T=23$ . Compute the units digit of $T^{2023}+T^{20}-T^{23}$.
|
[
"Assuming that $T$ is a positive integer, because units digits of powers of $T$ cycle in groups of at most 4, the numbers $T^{2023}$ and $T^{23}$ have the same units digit, hence the number $T^{2023}-T^{23}$ has a units digit of 0 , and the answer is thus the units digit of $T^{20}$. With $T=23$, the units digit of $23^{20}$ is the same as the units digit of $3^{20}$, which is the same as the units digit of $3^{4}=81$, so the answer is $\\mathbf{1}$."
] |
[
"1"
] | false
| null |
Numerical
| null |
\boxed{1}
|
2,772
|
Number Theory
| null |
Compute the least integer greater than 2023 , the sum of whose digits is 17 .
|
[
"A candidate for desired number is $\\underline{2} \\underline{0} \\underline{X} \\underline{Y}$, where $X$ and $Y$ are digits and $X+Y=15$. To minimize this number, take $Y=9$. Then $X=6$, and the desired number is 2069 ."
] |
[
"2069"
] | false
| null |
Numerical
| null |
\boxed{2069}
|
2,238
|
Algebra
| null |
Determine all integer values of $x$ such that $\left(x^{2}-3\right)\left(x^{2}+5\right)<0$.
|
[
"Since $x^{2} \\geq 0$ for all $x, x^{2}+5>0$. Since $\\left(x^{2}-3\\right)\\left(x^{2}+5\\right)<0, x^{2}-3<0$, so $x^{2}<3$ or $-\\sqrt{3}<x<\\sqrt{3}$. Thus $x=-1,0,1$."
] |
[
"-1,0,1"
] | true
| null |
Numerical
| null |
\boxed{-1,0,1}
|
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