problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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9. The function
$$
f(x)=\sqrt{2 x-7}+\sqrt{12-x}+\sqrt{44-x}
$$
has a maximum value of $\qquad$ | 9.11.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(\sqrt{2 x-7}+\sqrt{12-x}+\sqrt{44-x})^{2} \\
\leqslant(3+2+6)\left(\frac{2 x-7}{3}+\frac{12-x}{2}+\frac{44-x}{6}\right) \\
=11^{2},
\end{array}
$$
The equality holds if and only if $\frac{9}{2 x-7}=\frac{4}{12-x}=\frac{36}{44-x}$, which is when $x=8$... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) Given the function
$$
f(x)=2 \cos x(\cos x+\sqrt{3} \sin x)-1(x \in \mathbf{R}) \text {. }
$$
(1) Find the intervals where the function $f(x)$ is monotonically increasing;
(2) Let points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots$, $P_{n}\left(x_{n}, y_{n}\right), \cdots$ a... | (1) From the problem, we have
$$
f(x)=\cos 2 x+\sqrt{3} \sin 2 x=2 \sin \left(2 x+\frac{\pi}{6}\right) \text {. }
$$
Therefore, the monotonic increasing interval of $f(x)$ is
$$
\left[-\frac{\pi}{3}+k \pi, \frac{\pi}{6}+k \pi\right](k \in \mathbf{Z}) \text {. }
$$
(2) Let $t_{n}=2 x_{n}+\frac{\pi}{6}, t_{1}=2 x_{1}+\f... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Choose any two numbers from $2, 4, 6, 7, 8, 11, 12, 13$ to form a fraction. Then, there are $\qquad$ irreducible fractions among these fractions. | 4. 36 .
Among 7, 11, 13, choose one number and among $2, 4, 6, 8, 12$, choose one number to form a reduced fraction, there are $2 \mathrm{C}_{3}^{1} \mathrm{C}_{5}^{1}$ $=30$ kinds; among $7, 11, 13$, choose two numbers to form a reduced fraction, there are $\mathrm{A}_{3}^{2}=6$ kinds.
There are a total of 36 differe... | 36 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $f(x)=\lg (x+1)-\frac{1}{2} \log _{3} x$. Then the set
$$
M=\left\{n \mid f\left(n^{2}-8 n-2018\right) \geqslant 0, n \in \mathbf{Z}\right\}
$$
the number of subsets of $M$ is $\qquad$. | 2. 1.
For any $0< x_2 < x_1$, we have
$$
\begin{array}{l}
f\left(x_{1}\right)-f\left(x_{2}\right)=\lg \frac{x_{1}}{x_{2}}-\frac{\lg \frac{x_{1}}{x_{2}}}{\lg 9}>0 .
\end{array}
$$
Thus, $f(x)$ is a decreasing function on the interval $(0,+\infty)$. Note that, $f(9)=0$.
Therefore, when $x>9$, $f(x)f(9)=0$.
Hence, $f(x)... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 1, let $P\left(x_{p}, y_{p}\right)$ be a point on the graph of the inverse proportion function $y=\frac{2}{x}$ in the first quadrant of the Cartesian coordinate system $x O y$. Draw lines parallel to the $x$-axis and $y$-axis through point $P$, intersecting the graph of $y=\frac{10}{x}$ in the fir... | 3. B.
Connect $O P$.
Then $S_{\triangle A O B}=S_{\triangle A O P}+S_{\triangle P O B}+S_{\triangle A P B}$
$$
\begin{aligned}
= & \frac{1}{2}\left(\frac{10}{y_{p}}-x_{p}\right) y_{p}+\frac{1}{2}\left(\frac{10}{x_{p}}-y_{p}\right) x_{p}+ \\
& \frac{1}{2}\left(\frac{10}{y_{p}}-x_{p}\right)\left(\frac{10}{x_{p}}-y_{p}\r... | 24 | Algebra | MCQ | Yes | Yes | cn_contest | false |
5. Arrange natural numbers whose digits sum to 11 in ascending order to form a sequence. The $m$-th number is 2018. Then $m$ is ( ).
(A) 134
(B) 143
(C) 341
(D) 413 | 5. A.
Among single-digit numbers, there are no numbers whose digit sum is 11.
Among two-digit numbers, there are 8 numbers: $29, 38, 47, 56, 65, 74, 83, 92$.
For three-digit numbers $\overline{x y z}$, when $x=1$, $y$ can take 9 numbers: $1, 2, \cdots, 9$, and the corresponding $z$ takes $9, 8, \cdots, 1$, a total of... | 134 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
4. Given that $x_{1}, x_{2}, \cdots, x_{n}$ where $x_{i}(i=1,2, \cdots, n)$ can only take one of the values $-2, 0, 1$, and satisfy
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=-17, \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=37 .
\end{array}
$$
Then $\left(x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}\right)^{2}$ is $\qquad$ | 4. 5041.
Let $x_{1}, x_{2}, \cdots, x_{n}$ have $p$ values of $x_{i}$ equal to 1, $q$ values of $x_{i}$ equal to -2, and the rest of the $x_{i}$ equal to 0. We can obtain
$$
\left\{\begin{array} { l }
{ p - 2 q = - 1 7 } \\
{ p + 4 q = 3 7 }
\end{array} \Rightarrow \left\{\begin{array}{l}
p=1, \\
q=9 .
\end{array}\ri... | 5041 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Among the $n$ positive integers from 1 to $n$, those with the most positive divisors are called the "prosperous numbers" among these $n$ positive integers. For example, among the positive integers from 1 to 20, the numbers with the most positive divisors are $12, 18, 20$, so $12, 18, 20$ are all prosperous numbers a... | 5. 10080.
First, in the prime factorization of the first 100 positive integers, the maximum number of different prime factors is three. This is because the product of the smallest four primes is $2 \times 3 \times 5 \times 7=210$, which exceeds 100.
Second, to maximize the number of divisors, the prime factors should... | 10080 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Divide the set of positive even numbers $\{2,4, \cdots\}$ into groups in ascending order, with the $n$-th group containing $3 n-2$ numbers:
$$
\{2\},\{4,6,8,10\},\{12,14, \cdots, 24\}, \cdots \text {. }
$$
Then 2018 is in the group. | - 1. 27 .
Let 2018 be in the $n$-th group. Since 2018 is the 1009th positive even number and according to the problem, we have
$$
\begin{array}{l}
\sum_{i=1}^{n-1}(3 i-2)<1009 \leqslant \sum_{i=1}^{n}(3 i-2) \\
\Rightarrow \frac{3(n-1)^{2}-(n-1)}{2}<1009 \leqslant \frac{3 n^{2}-n}{2} \\
\Rightarrow n=27 .
\end{array}
... | 27 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms $S_{n}$ satisfies $2 S_{n}-n a_{n}=n\left(n \in \mathbf{Z}_{+}\right)$, and $a_{2}=3$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{1}{a_{n} \sqrt{a_{n+1}}+a_{n+1} \sqrt{a_{n}}}$, and $T_{n}$ ... | (1) From $2 S_{n}-n a_{n}=n$, we get
$$
2 S_{n+1}-(n+1) a_{n+1}=n+1 \text {. }
$$
Subtracting the above two equations yields
$$
2 a_{n+1}-(n+1) a_{n+1}+n a_{n}=1 \text {. }
$$
Thus, $n a_{n}-(n-1) a_{n+1}=1$,
$$
(n+1) a_{n+1}-n a_{n+2}=1 \text {. }
$$
Subtracting (2) from (1) and rearranging gives
$$
a_{n}+a_{n+2}=2 ... | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Let $M$ be a set composed of a finite number of positive integers
$$
\begin{array}{l}
\text { such that, } M=\bigcup_{i=1}^{20} A_{i}=\bigcup_{i=1}^{20} B_{i}, \\
A_{i} \neq \varnothing, B_{i} \neq \varnothing(i=1,2, \cdots, 20),
\end{array}
$$
and satisfies:
(1) For any $1 \leqslant i<j \leqslant 20$,
$$
A_{i} \... | 15. Let $\min _{1 \leqslant i \leqslant 20}\left\{\left|A_{i}\right|,\left|B_{i}\right|\right\}=t$.
Assume $\left|A_{1}\right|=t$,
$$
\begin{array}{l}
A_{1} \cap B_{i} \neq \varnothing(i=1,2, \cdots, k) ; \\
A_{1} \cap B_{j}=\varnothing(j=k+1, k+2, \cdots, 20) .
\end{array}
$$
Let $a_{i} \in A_{1} \cap B_{i}(i=1,2, \... | 180 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. In $\triangle A B C$, $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ respectively, satisfying $a^{2}+b^{2}=4-\cos ^{2} C, a b=2$. Then $S_{\triangle A B C}=$ $\qquad$ | $-1.1$
From the problem, we have $(a-b)^{2}+\cos ^{2} C=0$.
Solving, we get $a=b=\sqrt{2}, \cos C=0$.
Therefore, $S_{\triangle A B C}=1$. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $x, y>0$. If
$$
f(x, y)=\left(x^{2}+y^{2}+2\right)\left(\frac{1}{x+y}+\frac{1}{x y+1}\right) \text {, }
$$
then the minimum value of $f(x, y)$ is | 3.4.
By completing the square, we get
$$
x^{2}+y^{2}+2 \geqslant(x+y)+(x y+1) \text {. }
$$
Then $f(x, y)$
$$
\begin{array}{l}
\geqslant((x+y)+(x y+1))\left(\frac{1}{x+y}+\frac{1}{x y+1}\right) \\
\geqslant(1+1)^{2}=4 .
\end{array}
$$
Equality holds if and only if $x=y=1$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 In an $8 \times 8$ chessboard, how many ways are there to select 56 squares such that: all the black squares are selected, and each row and each column has exactly seven squares selected? ? ${ }^{[1]}$
(2014, Irish Mathematical Olympiad) | The problem is equivalent to selecting eight white squares on the chessboard, with exactly one square selected from each row and each column.
The white squares on the chessboard are formed by the intersections of rows $1, 3, 5, 7$ and columns $1, 3, 5, 7$, resulting in a $4 \times 4$ submatrix, as well as the intersec... | 576 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For the four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b 、 c$ 、 $d \leqslant 9)$ : if $a>b, bd$, then $\overline{a b c d}$ is called a $P$ class number; if $ac, c<d$, then $\overline{a b c d}$ is called a $Q$ class number. Let $N(P)$ and $N(Q)$ represent the number of $P$ class numbers and ... | Let the set of all numbers of type $P$ and type $Q$ be denoted as $A$ and $B$, respectively. Further, let the set of all numbers of type $P$ that end in zero be denoted as $A_{0}$, and the set of all numbers of type $P$ that do not end in zero be denoted as $A_{1}$.
For any four-digit number $\overline{a b c d} \in A_... | 285 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $f(x)=\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{2013!}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$). For an integer $n$, if the equation $f(x)=n$ has a real solution, then $n$ is called a "good number". Find the number of good numbers... | First, point out two obvious conclusions:
(1) If $m \in \mathbf{Z}_{+}, x \in \mathbf{R}$, then $\left[\frac{x}{m}\right]=\left[\frac{[x]}{m}\right]$;
(2) For any integer $l$ and positive even number $m$, we have
$$
\left[\frac{2 l+1}{m}\right]=\left[\frac{2 l}{m}\right] \text {. }
$$
In (1), let $m=k!(k=1,2, \cdots, ... | 587 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Equation
$$
x^{2}-31 x+220=2^{x}\left(31-2 x-2^{x}\right)
$$
The sum of the squares of all real roots is $\qquad$ . | 2. 25 .
Let $y=x+2^{x}$. Then the original equation is equivalent to
$$
\begin{array}{l}
y^{2}-31 y+220=0 \\
\Rightarrow y_{1}=11, y_{2}=20 \\
\Rightarrow x_{1}+2^{x_{1}}=11 \text { and } x_{2}+2^{x_{2}}=20 .
\end{array}
$$
Since $f(x)=x+2^{x}$ is a monotonically increasing function, each equation has at most one rea... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In the plane, there are 200 points, no three of which are collinear, and each point is labeled with one of the numbers $1, 2, 3$. All pairs of points labeled with different numbers are connected by line segments, and each line segment is labeled with a number 1, 2, or 3, which is different from the numbers at its en... | 5.199.
Let the points labeled with $1, 2, 3$ be $a, b, c$ respectively.
Thus, $a+b+c=200$, and the number of line segments labeled with $1, 2, 3$ are $bc, ca, ab$ respectively.
Then $n=a+bc=b+ca=c+ab$.
Therefore, $(a+bc)-(b+ca)=(a-b)(1-c)=0$.
Similarly, $(b-c)(1-a)=(c-a)(1-b)=0$.
If at least two of $a, b, c$ are not 1... | 199 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The sequence $\left\{a_{n}\right\}$ has nine terms, $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in\left\{2,1,-\frac{1}{2}\right\}$. The number of such sequences is $\qquad$
(2013, National High School Mathematics League Competition) | Let $b_{i}=\frac{a_{i+1}}{a_{i}}(1 \leqslant i \leqslant 8)$, mapping each sequence $\left\{a_{n}\right\}$ that meets the conditions to a unique eight-term sequence $\left\{b_{n}\right\}$, where $\prod_{i=1}^{8} b_{i}=\frac{a_{9}}{a_{1}}=1$, and $b_{i} \in\left\{2,1,-\frac{1}{2}\right\}(1 \leqslant i \leqslant 8)$.
Fro... | 491 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $n$ be a three-digit positive integer without the digit 0. If the digits of $n$ in the units, tens, and hundreds places are permuted arbitrarily, the resulting three-digit number is never a multiple of 4. Find the number of such $n$.
(54th Ukrainian Mathematical Olympiad) | Hint: Classify by the number of even digits (i.e., $2,4,6,8$) appearing in the three-digit code of $n$. It is known from the discussion that the number of $n$ satisfying the condition is $125+150+0+8=283$.
| 283 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $x \in\left(0, \frac{\pi}{2}\right)$. Then the minimum value of the function $y=\frac{1}{\sin ^{2} x}+\frac{12 \sqrt{3}}{\cos x}$ is $\qquad$ . | 7. 28 .
Notice that,
$$
\begin{array}{l}
y=16\left(\frac{1}{16 \sin ^{2} x}+\sin ^{2} x\right)+16\left(\frac{3 \sqrt{3}}{4 \cos x}+\cos ^{2} x\right)-16 \\
\geqslant 16 \times \frac{1}{2}+16\left(\frac{3 \sqrt{3}}{8 \cos x}+\frac{3 \sqrt{3}}{8 \cos x}+\cos ^{2} x\right)-16 \\
\geqslant 16 \times \frac{1}{2}+16 \times ... | 28 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Find the smallest positive integer $k$ such that for any $k$-element subset $A$ of the set $S=\{1,2, \cdots, 2012\}$, there exist three distinct elements $a, b, c$ in $S$ such that $a+b, b+c, c+a$ are all in the subset $A$.
(2012, China Mathematical Olympiad) | 【Analysis】The condition of the problem is equivalent to being able to find three numbers $x, y, z$ in the subset $A$ such that
$$
\frac{-x+y+z}{2}, \frac{x-y+z}{2}, \frac{x+y-z}{2}
$$
are three distinct positive integers. For this, $x, y, z$ must be distinct, satisfy the triangle inequality, and their sum must be even... | 1008 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
A=\{1,2, \cdots, 99\}, \\
B=\{2 x \mid x \in A\}, \\
C=\{x \mid 2 x \in A\} .
\end{array}
$$
Then the number of elements in $B \cap C$ is $\qquad$ | 1. 24 .
From the conditions, we have
$$
\begin{array}{l}
B \cap C \\
=\{2,4, \cdots, 198\} \cap\left\{\frac{1}{2}, 1, \frac{3}{2}, 2, \cdots, \frac{99}{2}\right\} \\
=\{2,4, \cdots, 48\} .
\end{array}
$$
Therefore, the number of elements in $B \cap C$ is 24 . | 24 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Let the integer sequence $a_{1}, a_{2}, \cdots, a_{10}$ satisfy:
$$
a_{10}=3 a_{1}, a_{2}+a_{8}=2 a_{5} \text {, }
$$
and $a_{i+1} \in\left\{1+a_{i}, 2+a_{i}\right\}(i=1,2, \cdots, 9)$. Then the number of such sequences is $\qquad$ | 8. 80 .
$$
\begin{array}{l}
\text { Let } b_{i}=a_{i+1}-a_{i} \in\{1,2\}(i=1,2, \cdots, 9) \text {. } \\
\text { Then } 2 a_{1}=a_{10}-a_{1}=b_{1}+b_{2}+\cdots+b_{9}, \\
b_{2}+b_{3}+b_{4}=a_{5}-a_{2}=a_{8}-a_{5} \\
=b_{5}+b_{6}+b_{7} .
\end{array}
$$
Let $t$ represent the number of terms with value 2 among $b_{2}, b_{... | 80 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let non-zero real numbers $a, b$ satisfy $a^{2}+b^{2}=25$. If the function $y=\frac{a x+b}{x^{2}+1}$ has a maximum value $y_{1}$ and a minimum value $y_{2}$, then $y_{1}-y_{2}=$ $\qquad$. | 3.5.
From
$$
\begin{aligned}
y & =\frac{a x+b}{x^{2}+1} \Rightarrow y x^{2}-a x+y-b=0 \\
& \Rightarrow \Delta=a^{2}-4 y(y-b) \geqslant 0 .
\end{aligned}
$$
Thus, $y_{2}$ and $y_{1}$ are the two roots of $a^{2}-4 y(y-b)=0$, at this point,
$$
\Delta_{1}=16 b^{2}+16 a^{2}=400>0 \text {. }
$$
Therefore, $y_{2}$ and $y_{... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the side length of rhombus $A_{1} A_{2} A_{3} A_{4}$ be $1, \angle A_{1} A_{2} A_{3}=$ $\frac{\pi}{6}, P$ be a point in the plane of rhombus $A_{1} A_{2} A_{3} A_{4}$. Then the minimum value of $\sum_{1 \leqslant i<j \leqslant 4} \overrightarrow{P A_{i}} \cdot \overrightarrow{P A_{j}}$ is $\qquad$ | 6. -1 .
Let the center of the rhombus be $O$. Then
$$
\begin{aligned}
& \sum_{1 \leqslant i<j \leqslant 4} \overrightarrow{P A_{i}} \cdot \overrightarrow{P A_{j}} \\
= & \mathrm{C}_{4}^{2}|\overrightarrow{P Q}|^{2}+\overrightarrow{P O} \cdot 3 \sum_{1 \leqslant i \leqslant 4} \overrightarrow{O A_{i}}+\sum_{1 \leqslant... | -1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $P(x)=x^{5}-x^{2}+1$ have five roots $r_{1}$, $r_{2}, \cdots, r_{5}$, and $Q(x)=x^{2}+1$. Then
$$
\begin{array}{l}
Q\left(r_{1}\right) Q\left(r_{2}\right) Q\left(r_{3}\right) Q\left(r_{4}\right) Q\left(r_{5}\right) \\
=
\end{array}
$$ | 7.5.
Given that $P(x)=\prod_{j=1}^{5}\left(x-r_{j}\right)$.
$$
\begin{array}{l}
\text { Then } \prod_{j=1}^{5} Q\left(r_{j}\right)=\left(\prod_{j=1}^{5}\left(r_{j}+\mathrm{i}\right)\right)\left(\prod_{j=1}^{5}\left(r_{j}-\mathrm{i}\right)\right) \\
=P(\mathrm{i}) P(-\mathrm{i}) \\
=\left(\mathrm{i}^{5}-\mathrm{i}^{2}+... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let set $S \subset\{1,2, \cdots, 200\}, S$ such that the difference between any two elements is not 4, 5, or 9. Find the maximum value of $|S|$.
untranslated portion:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note to the translator and should not be included in the translated text. Here is ... | Hint Consider the maximum number of numbers from $S$ that can be contained in any sequence of 13 consecutive numbers. Answer: 64. | 64 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a large regular tetrahedron with an edge length of 6, a smaller regular tetrahedron is placed inside it. If the smaller tetrahedron can rotate freely within the larger one, the maximum edge length of the smaller tetrahedron is . $\qquad$ | 2. 2 .
Given that a smaller regular tetrahedron can rotate freely inside a larger regular tetrahedron, the maximum edge length of the smaller tetrahedron occurs when it is inscribed in the insphere of the larger tetrahedron.
Let the circumradius of the larger tetrahedron be $R$, and the circumradius of the smaller te... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. In the expansion of $(\sqrt{3}+i)^{10}$, the sum of all odd terms is $\qquad$ . | 3. 512 .
It is known that in the expansion of $(\sqrt{3}+\mathrm{i})^{10}$, the sum of all odd terms is the real part of the complex number.
$$
\begin{array}{l}
\text { Therefore, }(\sqrt{3}+i)^{10}=\left((-2 i)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)\right)^{10} \\
=(-2 i)^{10}\left(-\frac{1}{2}+\frac{\sqrt{3}}... | 512 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If $(2 \dot{x}+4)^{2 n}=\sum_{i=0}^{2 n} a_{i} x^{i}\left(n \in \mathbf{Z}_{+}\right)$, then the remainder of $\sum_{i=1}^{n} a_{2 i}$ when divided by 3 is $\qquad$ | 6.1.
Let $x=0$, we get $a_{0}=4^{2 n}$.
By setting $x=1$ and $x=-1$ respectively, and adding the two resulting equations, we get
$$
\begin{array}{l}
\sum_{i=0}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right) . \\
\text { Therefore, } \sum_{i=1}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right)-4^{2 n} \\
\equiv(... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Divide a circle into a group of $n$ equal parts and color each point either red or blue. Starting from any point, record the colors of $k(k \leqslant n)$ consecutive points in a counterclockwise direction, which is called a “$k$-order color sequence” of the circle. Two $k$-order color sequences are considered differ... | 8. 8 .
In a 3rd-order color sequence, since each point has two color choices, there are $2 \times 2 \times 2=8$ kinds of 3rd-order color sequences.
Given that $n$ points can form $n$ 3rd-order color sequences, we know $n \leqslant 8$.
Thus, $n=8$ can be achieved.
For example, determining the colors of eight points in ... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find the smallest positive integer $n$ such that the polynomial $(x+1)^{n}-1$ modulo 3 is divisible by $x^{2}+1$.
$(2015$, Harvard-MIT Mathematics Tournament) | 【Analysis】More explicitly, the polynomial $(x+1)^{n}-1$ needs to satisfy the equivalent condition:
There exist integer-coefficient polynomials $P$ and $Q$ such that
$$
(x+1)^{n}-1=\left(x^{2}+1\right) P(x)+3 Q(x) \text {. }
$$
Assume without loss of generality that
$$
R(x)=(x+1)^{n}-1-P(x)\left(x^{2}+1\right) \text {.... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. As shown in Figure 2, two equal circles with a radius of 5 are externally tangent to each other, and both are internally tangent to a larger circle with a radius of 13, with the points of tangency being $A$ and $B$. Let $AB = \frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)\right. = 1)$. Then the value of $m+n$ is (... | 15. D.
As shown in Figure 5, let the centers of the three circles be $X$, $Y$, and $Z$.
Then points $A$ and $B$ are on the extensions of $XY$ and $XZ$, respectively, satisfying
$$
\begin{array}{l}
XY = XZ \\
= 13 - 5 = 8, \\
YZ = 5 + 5 = 10, \\
XA = XB = 13.
\end{array}
$$
Since $YZ \parallel AB \Rightarrow \triangl... | 69 | Geometry | MCQ | Yes | Yes | cn_contest | false |
7. Let $x, y, z$ be complex numbers, and
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=x y+y z+z x, \\
|x+y+z|=21,|x-y|=2 \sqrt{3},|x|=3 \sqrt{3} .
\end{array}
$$
Then $|y|^{2}+|z|^{2}=$ . $\qquad$ | 7. 132.
It is easy to see that the figures corresponding to $x$, $y$, and $z$ on the complex plane form an equilateral triangle, and note that
\[
\begin{array}{l}
|x-y|^{2}+|y-z|^{2}+|z-x|^{2}+|x+y+z|^{2} \\
=3\left(|x|^{2}+|y|^{2}+|z|^{2}\right),
\end{array}
\]
Combining the conditions and $|x-y|=|y-z|=|z-x|$, we ca... | 132 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
In $1,2, \cdots, 100$ these 100 positive integers, remove 50 so that in the remaining positive integers, any two different $a, b$ have $a \nmid b$. Find the maximum possible value of the sum of all removed positive integers. | Let the remaining numbers be $a_{i}=2^{3} t_{i}$, where $i \in \{1,2, \cdots, 50\}$, $s_{i}$ is a natural number, and $t_{i}$ is an odd number.
Since for any $1 \leqslant i \neq j \leqslant 50$, we have $a_{i} \nmid a_{j}$, it follows that $t_{i} \neq t_{j}$, meaning $t_{1}, t_{2}, \cdots, t_{50}$ are 50 distinct odd ... | 2165 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Color each cell of a $5 \times 5$ grid with one of five colors, such that the number of cells of each color is the same. If two adjacent cells have different colors, their common edge is called a "separating edge." Find the minimum number of separating edges.
Color each cell of a $5 \times 5$ grid wi... | As shown in Figure 4, a $5 \times 5$ grid is divided into five parts, each colored with one of 5 colors, and at this point, there are 16 dividing edges.
Below is the proof that the number of dividing edges is at least 16.
First, let the 5 colors be denoted as $1, 2, \cdots, 5$, and let the number of rows and columns oc... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Let positive real numbers $x, y$ satisfy
$$
x^{2}+y^{2}+\frac{1}{x}+\frac{1}{y}=\frac{27}{4} \text {. }
$$
Then the minimum value of $P=\frac{15}{x}-\frac{3}{4 y}$ is | 7.6.
By the AM-GM inequality for three terms, we have
$$
\begin{array}{l}
x^{2}+\frac{1}{x}=\left(x^{2}+\frac{8}{x}+\frac{8}{x}\right)-\frac{15}{x} \geqslant 12-\frac{15}{x} \\
y^{2}+\frac{1}{y}=\left(y^{2}+\frac{1}{8 y}+\frac{1}{8 y}\right)+\frac{3}{4 y} \geqslant \frac{3}{4}+\frac{3}{4 y} .
\end{array}
$$
Adding th... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let the general term formula of the sequence $\left\{a_{n}\right\}$ be $a_{n}=n^{3}-n$ $\left(n \in \mathbf{Z}_{+}\right)$, and the terms in this sequence whose unit digit is 0, arranged in ascending order, form the sequence $\left\{b_{n}\right\}$. Then the remainder when $b_{2} 018$ is divided by 7 is $\qquad$ . | 8. 4 .
Since $a_{n}=n^{3}-n=n(n-1)(n+1)$, therefore, $a_{n}$ has a units digit of 0 if and only if the units digit of $n$ is $1, 4, 5, 6, 9, 0$. Hence, in any consecutive 10 terms of the sequence $\left\{a_{n}\right\}$, there are 6 terms whose units digit is 0.
Since $2018=336 \times 6+2,336 \times 10=3360$, the rema... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. If the function
$$
f(x)=x^{2}-2 a x-2 a|x-a|+1
$$
has exactly three zeros, then the value of the real number $a$ is $\qquad$. | 2. 1.
Let $t=|x-a|(t \geqslant 0)$.
Then the original problem is equivalent to the equation $t^{2}-2 a t+1-a^{2}=0$ having two roots $t_{1}=0, t_{2}>0$.
Upon verification, $a=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms as $S_{n}$, and
$$
a_{1}=3, S_{n}=2 a_{n}+\frac{3}{2}\left((-1)^{n}-1\right) \text {. }
$$
If $\left\{a_{n}\right\}$ contains three terms $a_{1} 、 a_{p} 、 a_{q}(p 、 q \in$ $\left.\mathbf{Z}_{+}, 1<p<q\right)$ that form an arithmetic seque... | 4. 1 .
Given $S_{n}=2 a_{n}+\frac{3}{2}\left((-1)^{n}-1\right)$
$$
\Rightarrow S_{n-1}=2 a_{n-1}+\frac{3}{2}\left((-1)^{n-1}-1\right)(n \geqslant 2) \text {. }
$$
Subtracting the two equations yields $a_{n}=2 a_{n-1}-3(-1)^{n}(n \geqslant 2)$.
Let $b_{n}=\frac{a_{n}}{(-1)^{n}}$. Then
$$
\begin{array}{l}
b_{n}=-2 b_{n... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. In the Cartesian coordinate system, color the set of points
$$
\left\{(m, n) \mid m, n \in \mathbf{Z}_{+}, 1 \leqslant m, n \leqslant 6\right\}
$$
red or blue. Then the number of different coloring schemes where each unit square has exactly two red vertices is $\qquad$ kinds. | 8. 126 .
Dye the first row (points with a y-coordinate of 6), there are $2^{6}$ ways to do this, which can be divided into two cases.
(1) No two same-colored points are adjacent (i.e., red and blue alternate), there are 2 ways, and the second row can only be dyed in 2 ways, each row has only 2 ways, totaling $2^{6}$ w... | 126 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. A meeting was attended by 24 representatives, and between any two representatives, they either shook hands once or did not shake hands at all. After the meeting, it was found that there were a total of 216 handshakes, and for any two representatives $P$ and $Q$ who shook hands, among the remaining 22 representatives... | 7. Let the 24 representatives be $v_{1}, v_{2}, \cdots, v_{24}$, and for $i=1,2, \cdots, 24$, let $d_{i}$ denote the number of people who have shaken hands with $v_{i}$.
Define the set $E=\left\{\left\{v_{i}, v_{j}\right\} \mid v_{i}\right.$ has shaken hands with $v_{j} \}$.
For any $e=\left\{v_{i}, v_{j}\right\} \in E... | 864 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let integers $x, y$ satisfy $x^{2}+y^{2}4$. Then the maximum value of $x^{2}-2 x y-3 y$ is
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. 3 .
Using a TI calculator, we can obtain the region satisfying
$$
\left\{\begin{array}{l}
x^{2}+y^{2}4
\end{array},\right.
$$
and since we need to find the maximum value of $x^{2}-2 x y-3 y$, the integer point is in the third quadrant.
Substituting $(-2,-3)$ and $(-3,-2)$ into $x^{2}-2 x y-3 y$ and comparing the ... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
11. Arrange 10 flowers in a row using red, yellow, and blue flowers (assuming there are plenty of each color), and yellow flowers cannot be adjacent. How many different arrangements are there (the 10 flowers can be of one color or two colors)? | 11. Let $x_{n}$ be the number of different arrangements of $n$ flowers that meet the requirements. Then
$$
x_{1}=3, x_{2}=3^{2}-1=8 \text {. }
$$
When $n \geqslant 3$, let the arrangement of $n$ flowers be $a_{1}, a_{2}, \cdots, a_{n}$. If $a_{1}$ is a red or blue flower, then $a_{2}, a_{3}, \cdots, a_{n}$ is an arran... | 24960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $x, y$ satisfy $x^{2}+y^{2}=20$. Then the maximum value of $x y+8 x+y$ is $\qquad$ . | 4. 42 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(x y+8 x+y)^{2} \\
\leqslant\left(x^{2}+8^{2}+y^{2}\right)\left(y^{2}+x^{2}+1^{2}\right) \\
=84 \times 21=42^{2} .
\end{array}
$$
Therefore, the maximum value sought is 42. | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the inequality $\left|a x^{2}+b x+a\right| \leqslant x$ holds for $x \in$ $[1,2]$. Then the maximum value of $3 a+b$ is $\qquad$ | 6. 3 .
From the problem, we know that $\left|a\left(x+\frac{1}{x}\right)+b\right| \leqslant 1$.
Given $x \in[1,2]$, we have $t=x+\frac{1}{x} \in\left[2, \frac{5}{2}\right]$.
Thus, $|2 a+b| \leqslant 1$, and $\left|\frac{5}{2} a+b\right| \leqslant 1$.
Therefore, $3 a+b=2\left(\frac{5}{2} a+b\right)-(2 a+b) \leqslant 3$... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
7. In the sequence
$$
\left[\frac{1^{2}}{2019}\right],\left[\frac{2^{2}}{2019}\right], \cdots,\left[\frac{2019^{2}}{2019}\right]
$$
there are $\qquad$ distinct integers ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 7.1515.
Let the $k$-th term of the known sequence be $\left[\frac{k^{2}}{2019}\right]$. Then, when $(k+1)^{2}-k^{2} \leqslant 2019$, i.e., $k \leqslant 1009$,
$$
\begin{array}{l}
\frac{(k+1)^{2}}{2019}=\frac{k^{2}}{2019}+\frac{2 k+1}{2019} \leqslant \frac{k^{2}}{2019}+1 \\
\Rightarrow\left[\frac{(k+1)^{2}}{2019}\right... | 1515 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. On each face of a cube, randomly fill in one of the numbers 1, 2, $\cdots$, 6 (the numbers on different faces are distinct). Then number the eight vertices such that the number assigned to each vertex is the product of the numbers on the three adjacent faces. The maximum value of the sum of the numbers assigned to t... | 8. 343 .
Let the numbers on the six faces be $a, b, c, d, e, f$, and $(a, b), (c, d), (e, f)$ be the numbers on the opposite faces. Thus, the sum of the numbers at the eight vertices is
$$
\begin{array}{l}
(a+b)(c+d)(e+f) \\
\leqslant\left(\frac{(a+b)+(c+d)+(e+f)}{3}\right)^{3} \\
=7^{3}=343 .
\end{array}
$$
When $a=... | 343 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) A planet has 1000 cities $c_{1}$, $c_{2}, \cdots, c_{1000}$, and three airlines $X$, $Y$, and $Z$ provide flights between these cities. For any $1 \leqslant i<j \leqslant 1000$, exactly one airline operates a one-way flight from city $c_{i}$ to city $c_{j}$. Find the largest positive integer $n$, such... | Four, the maximum value of $n$ is 9.
For each city $c_{i}(i=1,2, \cdots, 1000)$, define a triplet of non-negative integers $\left(x_{i}, y_{i}, z_{i}\right)$ according to the following rules:
If there are no flights from company $X$ arriving at city $c_{i}$, set $x_{i}=0$; otherwise, there exists a largest positive in... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given four positive integers
$a, b, c, d$ satisfy:
$$
a^{2}=c(d+20), b^{2}=c(d-18) \text {. }
$$
Then the value of $d$ is $\qquad$ | 4. 180 .
Let $(a, b)=t, a=t a_{1}, b=t b_{1}$.
Then $\frac{d+20}{d-18}=\frac{c(d+20)}{c(d-18)}=\frac{a^{2}}{b^{2}}=\frac{a_{1}^{2}}{b_{1}^{2}}$ (simplest fraction).
Let $d+20=k a_{1}^{2}, d-18=k b_{1}^{2}$.
Eliminating $d$ yields
$$
\begin{array}{l}
k\left(a_{1}+b_{1}\right)\left(a_{1}-b_{1}\right)=2 \times 19 \\
\Rig... | 180 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a, b, c, d$ are positive integers, and $\log _{a} b=\frac{3}{2}, \log _{c} d=\frac{5}{4}, a-c=9$.
Then $a+b+c+d=$ $\qquad$ | 3. 198 .
Given $a=x^{2}, b=x^{3}, c=y^{4}, d=y^{5}$.
$$
\begin{array}{l}
\text { Given } a-c=x^{2}-y^{4}=9 \\
\Rightarrow\left(x+y^{2}\right)\left(x-y^{2}\right)=9 \\
\Rightarrow x+y^{2}=9, x-y^{2}=1 \\
\Rightarrow x=5, y^{2}=4 \\
\Rightarrow a=25, b=125, c=16, d=32 \\
\Rightarrow a+b+c+d=198 .
\end{array}
$$ | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, $\angle A B C=120^{\circ}$, the angle bisector of $\angle A B C$ intersects $A C$ at point $D$, and $B D=1$. Then the minimum value of $4 a+c$ is $\qquad$ | 4.9.
From the problem, we know that $S_{\triangle A B C}=S_{\triangle A B D}+S_{\triangle B C D}$.
By the angle bisector property and the formula for the area of a triangle, we get
$$
\begin{array}{l}
\frac{1}{2} a c \sin 120^{\circ} \\
=\frac{1}{2} a \times 1 \times \sin 60^{\circ}+\frac{1}{2} c \times 1 \times \sin ... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) In the sequence $\left\{a_{n}\right\}$, let $S_{n}=\sum_{i=1}^{n} a_{i}$ $\left(n \in \mathbf{Z}_{+}\right)$, with the convention: $S_{0}=0$. It is known that
$$
a_{k}=\left\{\begin{array}{ll}
k, & S_{k-1}<k ; \\
-k, & S_{k-1} \geqslant k
\end{array}\left(1 \leqslant k \leqslant n, k 、 n \in \mathbf{Z}_... | 10. Let the indices $n$ that satisfy $S_{n}=0$ be arranged in ascending order, denoted as the sequence $\left\{b_{n}\right\}$, then $b_{1}=0$.
To find the recurrence relation that $\left\{b_{n}\right\}$ should satisfy.
In fact, without loss of generality, assume $S_{b_{k}}=0$.
Thus, by Table 1, it is easy to prove by m... | 1092 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (40 points) Find the smallest integer $c$, such that there exists a sequence of positive integers $\left\{a_{n}\right\}(n \geqslant 1)$ satisfying:
$$
a_{1}+a_{2}+\cdots+a_{n+1}<c a_{n}
$$
for all $n \geqslant 1$. | Given the problem, we have
$$
c>\frac{a_{1}+a_{2}+\cdots+a_{n+1}}{a_{n}}.
$$
For any \( n \geqslant 1 \), we have
$$
\begin{array}{l}
nc > \frac{a_{1}+a_{2}}{a_{1}} + \frac{a_{1}+a_{2}+a_{3}}{a_{2}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n+1}}{a_{n}} \\
= n + \frac{a_{2}}{a_{1}} + \left(\frac{a_{1}}{a_{2}} + \frac{a_{... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. If three angles $\alpha, \beta, \gamma$ form an arithmetic sequence with a common difference of $\frac{\pi}{3}$, then $\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \gamma \cdot \tan \alpha$ $\qquad$ | 4. -3 .
According to the problem, $\alpha=\beta-\frac{\pi}{3}, \gamma=\beta+\frac{\pi}{3}$. Therefore, $\tan \alpha=\frac{\tan \beta-\sqrt{3}}{1+\sqrt{3} \tan \beta}, \tan \gamma=\frac{\tan \beta+\sqrt{3}}{1-\sqrt{3} \tan \beta}$.
Then, $\tan \alpha \cdot \tan \beta=\frac{\tan ^{2} \beta-\sqrt{3} \tan \beta}{1+\sqrt{3... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $x, y, z \in \mathbf{R}_{+}$, satisfying $x+y+z=x y z$. Then the function
$$
\begin{array}{l}
f(x, y, z) \\
=x^{2}(y z-1)+y^{2}(z x-1)+z^{2}(x y-1)
\end{array}
$$
has the minimum value of $\qquad$ | 5. 18 .
According to the conditions, we have
$$
y+z=x(y z-1) \Rightarrow y z-1=\frac{y+z}{x} \text{. }
$$
Similarly, $z x-1=\frac{z+x}{y}, x y-1=\frac{x+y}{z}$.
From $x y z=x+y+z \geqslant 3 \sqrt[3]{x y z} \Rightarrow x y z \geqslant 3 \sqrt{3}$, thus
$$
\begin{array}{l}
f(x, y, z)=2(x y+y z+z x) \\
\geqslant 2 \tim... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. The sequence of positive integers $\left\{a_{n}\right\}: a_{n}=3 n+2$ and $\left\{b_{n}\right\}$ $b_{n}=5 n+3(n \in \mathbf{N})$ have a common number of terms in $M=\{1,2, \cdots, 2018\}$ which is $\qquad$ | 6. 135.
It is known that 2018 is the largest common term of the two sequences within $M$. Excluding this common term, subtract 2018 from the remaining terms of $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively, to get
$$
\left\{\overline{a_{n}}\right\}=\{3,6,9, \cdots, 2016\},
$$
which are all multiples... | 135 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. For a positive integer $n$, let the sum of its digits be denoted as $s(n)$, and the product of its digits as $p(n)$. If $s(n) +$ $p(n) = n$ holds, then $n$ is called a "coincidence number". Therefore, the sum of all coincidence numbers is | 8.531.
Let \( n = \overline{a_{1} a_{2} \cdots a_{k}} \left(a_{1} \neq 0\right) \).
From \( n - s(n) = p(n) \), we get
\[
\begin{array}{l}
a_{1}\left(10^{k-1}-1\right) + a_{2}\left(10^{k-2}-1\right) + \cdots + a_{k-1}(10-1) \\
= a_{1} a_{2} \cdots a_{k},
\end{array}
\]
which simplifies to \( a_{1}\left(10^{k-1}-1-a_{... | 531 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the function $f(x)=x+\frac{2}{x}$ on the interval $[1,4]$, the maximum value is $M$, and the minimum value is $m$. Then the value of $M-m$ is $\qquad$ | ニ.7.4.
Since $f(x)$ is monotonically decreasing on the interval $[1,3]$ and monotonically increasing on the interval $[3,4]$, the minimum value of $f(x)$ is $f(3)=6$.
Also, $f(1)=10, f(4)=\frac{25}{4}$, so the maximum value of $f(x)$ is $f(1)=10$.
Therefore, $M-m=10-6=4$. | 4 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
11. Let the line $y=k x+b$ intersect the curve $y=x^{3}-x$ at three distinct points $A, B, C$, and $|A B|=|B C|=2$. Then the value of $k$ is | 11. 1.
Given that the curve is symmetric about the point $(0,0)$, and
$$
|A B|=|B C|=2 \text {, }
$$
we know that the line $y=k x+b$ must pass through the origin.
Thus, $b=0$.
Let $A(x, y)$. Then
$$
\begin{array}{l}
y=k x, y=x^{3}-x, \sqrt{x^{2}+y^{2}}=2 \\
\Rightarrow x=\sqrt{k+1}, y=k \sqrt{k+1} . \\
\text { Substi... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $f(x)=\frac{10}{x+1}-\frac{\sqrt{x}}{3}$. Then the number of elements in the set $M=\left\{n \in \mathbf{Z} \mid f\left(n^{2}-1\right) \geqslant 0\right\}$ is $\qquad$. | ,- 1.6 .
From the problem, we know that $f(x)$ is monotonically decreasing on the interval $[0,+\infty)$, and $f(9)=0$.
Then $f\left(n^{2}-1\right) \geqslant f(9) \Rightarrow 1 \leqslant n^{2} \leqslant 10$.
Thus, the number of elements in set $M$ is 6. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $a_{k}$ as the number of integer terms in $\log _{2} k, \log _{3} k, \cdots, \log _{2018} k$. Then $\sum_{k=1}^{2018} a_{k}=$ $\qquad$ | 8.4102 .
Let $b_{m}$ be the number of integer terms in $\log _{m} 1, \log _{m} 2, \cdots, \log _{m} 2018$.
Then, $\sum_{k=1}^{2018} a_{k}=\sum_{m=2}^{2018} b_{m}$.
Notice that, $\log _{m} t$ is an integer if and only if $t$ is a power of $m$.
\[
\begin{array}{l}
\text { Then } b_{2}=11, b_{3}=7, b_{4}=6, b_{5}=b_{6}=5... | 4102 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Let
$$
\begin{array}{l}
P(z)=z^{4}-(6 \mathrm{i}+6) z^{3}+24 \mathrm{iz}^{2}- \\
(18 \mathrm{i}-18) z-13 .
\end{array}
$$
Find the area of the convex quadrilateral formed by the four points in the complex plane corresponding to the four roots of $P(z)=0$. | II. 9. Notice that, $P(1)=0$.
Then $P(z)=(z-1)\left(z^{3}-(6 \mathrm{i}+6) z^{2}+24 \mathrm{iz}^{2}+\right.$ $(18 \mathrm{i}-5) z+13)$.
Let $Q(z)=z^{3}-(6 \mathrm{i}+5) z^{2}+(18 \mathrm{i}-5) z+13$.
Then $Q(\mathrm{i})=0$.
Hence $Q(z)=(z-\mathrm{i})\left(z^{2}-(5 \mathrm{i}+5) z+13 \mathrm{i}\right)$.
Using the quadr... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that the three sides of a triangle are consecutive natural numbers. If the largest angle is twice the smallest angle, then the perimeter of the triangle is $\qquad$ . | -,1.15.
Assuming the three sides of a triangle are $n-1$, $n$, and $n+1$, with the largest angle being $2 \theta$ and the smallest angle being $\theta$.
Then, by the Law of Sines, we have
$$
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \cos \theta=\frac{n+1}{2(n-1)} \text {. }
$$
By the Law of Cosines... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. The integer sequence $\left\{a_{i, j}\right\}(i, j \in \mathbf{N})$, where,
$$
\begin{array}{l}
a_{1, n}=n^{n}\left(n \in \mathbf{Z}_{+}\right), \\
a_{i, j}=a_{i-1, j}+a_{i-1, j+1}(i, j \geqslant 1) .
\end{array}
$$
Then the unit digit of the value taken by $a_{128,1}$ is | 8. 4 .
By the recursive relation, we have
$$
\begin{array}{l}
a_{1,1}=1, a_{2, n}=n^{n}+(n+1)^{n+1}, \\
a_{3, n}=n^{n}+2(n+1)^{n+1}+(n+2)^{n+2} .
\end{array}
$$
Accordingly, by induction, we get
$$
\begin{array}{l}
a_{n, m}=\sum_{k=0}^{m-1} \mathrm{C}_{m-1}^{k}(n+k)^{n+k} \\
=\sum_{k \geqslant 0} \mathrm{C}_{m-1}^{k}... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. There are five cities in a line connected by semi-circular roads, as shown in Figure 1. Each segment of the journey is from one city to another along a semi-circle. If the journey can be repeated, the total number of possible ways to start from city 5 and return to city 5 after four segments is $\qquad$.
保留源文本的换行和格... | ,- 1.80 .
After four segments, there are five possible ways to start from city 5 and return to city 5:
$$
\begin{array}{l}
5 \rightarrow 1 \rightarrow 5 \rightarrow 1 \rightarrow 5, \\
5 \rightarrow 1 \rightarrow 5 \rightarrow 2 \rightarrow 5, \\
5 \rightarrow 2 \rightarrow 5 \rightarrow 2 \rightarrow 5, \\
5 \rightarr... | 80 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let positive integers $m, n$ satisfy
$$
m(n-m)=-11 n+8 \text {. }
$$
Then the sum of all possible values of $m-n$ is $\qquad$ | 2. 18 .
From the problem, we have
$$
n=\frac{m^{2}+8}{m+11}=m-11+\frac{129}{m+11} \in \mathbf{Z}_{+} \text {. }
$$
Then $(m+11) \mid 129$
$$
\Rightarrow m+11=1,3,43,129 \text {. }
$$
Also, $m \in \mathbf{Z}_{+}$, checking we find that when $m=32,118$, the corresponding $n$ is a positive integer.
$$
\text { Hence }(m... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Person A tosses a fair coin twice, and Person B tosses the same coin three times. If the probability that they end up with the same number of heads is written as a simplified fraction, the sum of the numerator and the denominator is $\qquad$ . (Romania) | 3.21.
Let the outcomes of a coin landing heads up and tails up be denoted as $\mathrm{H}$ and $\mathrm{T}$, respectively.
Jia has four equally probable outcomes: HH, HT, TH, TT;
Yi has eight equally probable outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
The outcomes that match $\mathrm{HH}$ are $\mathrm{HHT}$, $... | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. The product $1!\cdot 2!\cdot 3!\cdot \cdots \cdot 99!\cdot 100!$ ends with $\qquad$ consecutive 0s.
(Hong Kong, China, provided) | 6. 1124.
Since in the product $1!\cdot 2!\cdot 3!\cdots \cdots 99!\cdot 100!$, there are a large number of factor 2s, the number of consecutive 0s at the end is determined by the number of factor 5s. The number of factor 5s in each factorial $x$! is shown in Table 1.
Thus, the total number of factor 5s is
$$
\begin{ar... | 1124 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? ${ }^{[3]}$
(2012, Girls' Mathematical Olympiad) | Notice that, $2012=4 \times 503$, where $p=503$ is a prime number, and
$$
\mathrm{C}_{2012}^{k}=\mathrm{C}_{4 p}^{k}=\frac{(4 p)!}{k!\cdot(4 p-k)!}=\frac{4 p}{k} \mathrm{C}_{4 p-1}^{k-1} .
$$
If $p \nmid k$, then $p \mid \mathrm{C}_{2012}^{k}$;
If $p \mid k$, then $k \in\{0, p, 2 p, 3 p, 4 p\}$.
Notice that, $\mathrm{... | 1498 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are real coefficients. Assume that
$$
P(1)=7, P(2)=52, P(3)=97 \text {, }
$$
then $\frac{P(9)+P(-5)}{4}=$ $\qquad$ . (Vietnam) | 7. 1202.
Notice that, $52-7=97-52=45$,
$$
\begin{array}{l}
7=45 \times 1-38,52=45 \times 2-38, \\
97=45 \times 3-38 .
\end{array}
$$
Let $Q(x)=P(x)-45 x+38$. Then $Q(x)$ is a fourth-degree polynomial with a leading coefficient of 1, and
$$
Q(1)=Q(2)=Q(3)=0 \text {. }
$$
Thus, for some $r$,
$$
\begin{array}{l}
Q(x)=(... | 1202 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given that $a$ and $b$ are real numbers, satisfying:
$$
\sqrt[3]{a}-\sqrt[3]{b}=12, \quad a b=\left(\frac{a+b+8}{6}\right)^{3} \text {. }
$$
Then $a-b=$ $\qquad$ (Proposed by Thailand) | 11. 468 .
Let $x=\sqrt[3]{a}, y=\sqrt[3]{b}$. Then $x-y=12, 6xy=x^{3}+y^{3}+8$.
Thus, $x^{3}+y^{3}+2^{3}-3 \times 2xy=0$
$$
\Rightarrow(x+y+2)\left(x^{2}+y^{2}+2^{2}-xy-2x-2y\right)=0 \text {. }
$$
Therefore, $x+y+2=0$, or
$$
x^{2}-xy+y^{2}-2x-2y+4=0 \text {. }
$$
Equation (1) can be rewritten as
$$
\frac{1}{2}(x-y)... | 468 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the equation $\left|x^{2}-2 a x+b\right|=8$ has exactly three real roots, and they are the side lengths of a right triangle. Find the value of $a+b$.
(Bulgaria) | II. 1. Note that, for the equation
$$
x^{2}-2 a x+b-8=0
$$
the discriminant $\Delta_{1}=4\left(a^{2}-b+8\right)$; for the equation
$$
x^{2}-2 a x+b+8=0
$$
the discriminant $\Delta_{2}=4\left(a^{2}-b-8\right)$.
According to the problem, the original equation has exactly three real roots, so one of the discriminants is... | 264 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Color the six vertices of a regular pentagonal pyramid using at most six different colors, such that the two endpoints of any edge are colored differently. If one coloring method can be obtained from another by rotation, they are considered the same method. How many different coloring methods are there?
(Romania) | 2. Since the problem does not specify the number of colors to choose from, we will calculate the coloring methods based on the actual number of colors used.
If fewer than four colors are used, the bottom face can have at most two colors. Since 5 is an odd number, the colors of the bottom face vertices cannot be altern... | 288 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. For what positive integer $k$ does $\frac{20^{k}+18^{k}}{k!}$ achieve its maximum value?
(Hong Kong, China, Contributed) | 5.19.
Let $A_{k}=\frac{20^{k}+18^{k}}{k!}$. Then
$A_{k+1}=\frac{20^{k+1}+18^{k+1}}{(k+1)!}$,
$\frac{A_{k+1}}{A_{k}}=\frac{20^{k+1}+18^{k+1}}{(k+1)\left(20^{k}+18^{k}\right)}$.
Notice that,
$$
\begin{array}{l}
\frac{A_{k+1}}{A_{k}}>1 \Leftrightarrow \frac{20^{k+1}+18^{k+1}}{(k+1)\left(20^{k}+18^{k}\right)}>1 \\
\Leftri... | 19 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In the square in Figure 1, 10000 points are randomly thrown. Then the estimated number of points that fall into the shaded area (the equation of curve $C$ is $x^{2}-y=0$) is ( ).
(A) 5000
(B) 6667
(C) 7500
(D) 7854 | 5. B.
According to the problem, the area of the blank region is
$$
\int_{0}^{1} x^{2} \mathrm{~d} x=\left.\frac{1}{3} x^{3}\right|_{0} ^{1}=\frac{1}{3} \text {. }
$$
Therefore, the area of the shaded part is $\frac{2}{3}$, and the ratio of the area of the shaded part to the area of the blank part is $2: 1$.
Thus, the... | 6667 | Geometry | MCQ | Yes | Yes | cn_contest | false |
16. Nine consecutive positive integers are arranged in an increasing sequence $a_{1}, a_{2}, \cdots, a_{9}$. If $a_{1}+a_{3}+a_{5}+a_{7}+a_{9}$ is a perfect square, and $a_{2}+a_{4}+a_{6}+a_{8}$ is a perfect cube, then the minimum value of $a_{1}+a_{2}+\cdots+a_{9}$ is $\qquad$. | 16. 18000 .
Let these nine numbers be $a-4, a-3, \cdots, a+4$. Then $5 a=m^{2}, 4 a=n^{3}, S=9 a$. Thus, $a=\frac{m^{2}}{5}=\frac{n^{3}}{4} \Rightarrow 4 m^{2}=5 n^{3}$.
Clearly, $m$ and $n$ are both multiples of 10.
Let $m=10 m_{1}, n=10 n_{1}$, then $4 m_{1}^{2}=50 n_{1}^{3}$.
Take $m_{1}=10, n_{1}=2$. At this point... | 18000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Laura has 2010 lamps and 2010 switches in front of her, with different switches controlling different lamps. She wants to find the correspondence between the switches and the lamps. For this, Charlie operates the switches. Each time Charlie presses some switches, and the number of lamps that light up is the same as ... | (1) Charlie selects a pair of switches $(A, B)$, each operation involves pressing them simultaneously or leaving them untouched, while other switches can be chosen arbitrarily, resulting in $2^{2009}$ different operations, but Laura cannot determine which lights are controlled by switches $A$ and $B$.
If Charlie perfo... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) Mr. Wanda often forgets the numbers he should remember, such as his friends' phone numbers, the password of the safe, etc. For this reason, the manufacturer specially designed a password lock for his office safe, with keys labeled $0 \sim 9$. It is known that the password of the safe is a three-digit ... | Four, 50 times.
First, introduce several notations. Let
$E=\{1,3,5,7,9\}, F=\{0,2,4,6,8\}$
$\Phi=E \cup F=\{0,1, \cdots, 9\}$,
$\Omega=\{x y z \mid x, y, z \in \Phi\}$ (the set of all three-digit codes),
$\Omega_{11}=\{x a b \in \Omega \mid x \in \Phi, a, b \in E\}$,
$\Omega_{12}=\{a y b \in \Omega \mid y \in \Phi, a, ... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a regular 2019-gon, then, the maximum number of diagonals such that any two of them are either perpendicular or do not intersect except at endpoints. | 7.2016 .
On one hand, no two diagonals of the regular 2019-gon $\Gamma$ are perpendicular.
Assume there exist two diagonals of the regular 2019-gon $\Gamma$ that are perpendicular, say $AB \perp CD$, and $A, B, C, D$ are four vertices of $\Gamma$. Let $E$ be a point on the perpendicular bisector of $AB$, and let $E'$... | 2016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ with the general terms $a_{n}=2^{n}, b_{n}=5 n-2$. Then the sum of all elements in the set
$$
\left\{a_{1}, a_{2}, \cdots, a_{2019}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{2019}\right\}
$$
is $\qquad$ | 8.2184 .
Let the elements of $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ be arranged in ascending order to form the sequence $\left\{c_{n}\right\}$.
Then the problem is reduced to finding the number of solutions to the equation
$$
2^{n}=5 m-2\left(n, m \in \mathbf{Z}_{+}... | 2184 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Periodical cicadas are insects with very long larval periods and brief adult lives. For each species of periodical cicada with larval period of 17 years, there is a similar species with a larval period of 13 years. If both the 17 -year and 13 -year species emerged in a particular location in 1900, when will they nex... | 5. 2121 | 2121 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
14. Find the least common multiple of each of the following pairs of integers
a) 8,12
d) 111,303
b) 14,15
e) 256,5040
c) 28,35
f) 343,999 . | 14. a) 24 b) 210 c) 140 d) 11211 e) 80640 f) 342657 | 140 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
10. How many ways can change be made for one dollar using
a) dimes and quarters
b) nickels. dimes, and quarters
c) pennies, nickels, dimes, and quarters? | 10. a) 3
b) 29
c) 242 | 242 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
1. Find the values of the following sums
a) $\sum_{j=1}^{10} 2$
c) $\sum_{j=1}^{10} j^{2}$
b) $\sum_{j=1}^{10} j$
d) $\sum_{j=1}^{10} 2^{j}$. | 1. a) 20 b) 55 c) 385 d) 2046 | 2046 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
6. Find the smallest positive integer $n$ with $\tau(n)$ equal to
a) 1
d) 6
b) 2
e) 14
c) 3
f) 100 . | 6. a) 1
b) 2
c) 4
d) 12
e) 192
f) 45360 | 4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. Find the values of the following products
a) $\prod_{j=1}^{5} 2$
c) $\prod_{j=1}^{5} j^{2}$
b) $\prod_{j=1}^{5} j$
d) $\prod_{j=1}^{5} 2^{j}$. | 2. a) 32 b) 120 c) 14400 d) 32768 | 14400 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
3. What is the ciphertext that is produced when the RSA cipher with key $(e, n)=(3,2669)$ is used to encipher the message BEST WISHES? | 3. 12151224147100230116 | 12151224147100230116 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
3. Find $n$ ! for $n$ equal to each of the first ten positive integers. | 3. $1,2,6,24,120,720,5040,40320,362880,3628800$ | 3628800 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
12. Let $n=2^{t} \cdot p_{1}^{t} p_{2}^{t} \cdots p_{m}^{t}$ be the prime-power factorization of $n$. Let $a$ be an integer relatively prime to $n$. Let $r_{1}, r_{2}, \ldots, r_{m}$ be primitive roots of $p_{1}^{t_{1}}, p_{2}^{t_{2}}, \ldots, p_{m}^{t}$, respectively, and let $\gamma_{1}=\operatorname{ind}_{r_{1}} a\l... | 12. b) $(0,0,1,1),(0,0,1,4)$
d) $x \equiv 17(\bmod 60)$ | 17 | Number Theory | proof | Yes | Yes | number_theory | false |
1. Find $\lambda(n)$, the minimal universal exponent of $n$, for the following values of $n$
a) 100
e) $2^{4} \cdot 3^{3} \cdot 5^{2} \cdot 7$
b) 144
f) $2^{5} \cdot 3^{2} \cdot 5^{2} \cdot 7^{3} \cdot 11^{2} \cdot 13 \cdot 17 \cdot 19$
c) 222
g) 10 !
d) 884
h) 20 !. | a) 20
b) 12
c) 36
d) 48
e) 180
f) 388080
g) 8640
h) 125411328000 | 48 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1. Find the sequence of two-digit pseudo-random numbers generated using the middle-square method, taking 69 as the seed. | 1. $69,76,77,92,46,11,12,14,19,36,29,84,5,25,62,84,5,25,62, \ldots$ | 62 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. Find the first ten terms of the sequence of pseudo-random numbers generated by the linear congruential method with $x_{0}=6$ and $x_{n+1} \equiv 5 x_{n}+2(\bmod 19)$. What is the period length of this generator? | 2. $6.13,10,14,15,1,7,18,16,6,13, \ldots$ period length is 9 | 9 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the number of prime numbers not exceeding 100. | We take $N=100$, the prime numbers not exceeding $\sqrt{100}=10$ are $2,3,5,7$, so by formula (20) we get
$$\begin{aligned}
\pi(100)= & 4-1+100-\left\{\left[\frac{100}{2}\right]+\left[\frac{100}{3}\right]+\left[\frac{100}{5}\right]+\left[\frac{100}{7}\right]\right\} \\
& +\left\{\left[\frac{100}{2 \cdot 3}\right]+\left... | 25 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 (i) Find the number of integers between 1 and 500 that are not divisible by any of 5, 6, 8;
(ii) Find the number of integers between 1 and 500 that are divisible by 5, or 6, or 8. | Consider the sequence $A$ composed of integers from 1 to 500. Property $P_{1}$: divisible by 5, property $P_{2}$: divisible by 6, property $P_{3}$: divisible by 8. Thus, (i) is to find the number of $B^{(0)}$, and (ii) is to find the number of $A^{(0)}-B^{(0)}$.
The number of $A(1)$ is: 100; the number of $A(2)$ is: 83... | 299 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Let sequence $A$ be $1,3,2,5,4,3,5,7,6,11$. Property $P_{1}$ is that the element is divisible by 3, property $P_{2}$ is that the element leaves a remainder of 1 when divided by 3, and property $P_{3}$ is that the element is divisible by 2. Try to verify whether Theorem 7 holds. | $A(1)$ is $3,3,6; A(2)$ is $1,4,7; A(3)$ is $2,4,6. A(1,2)$ has no elements; $A(1,3)$ is $6; A(2,3)$ is $4. A(1,2,3)$ has no elements.$
$$B^{(0)}=B(0) \text { is } 5,5,11; B(1) \text { is } 3,3; B(2) \text { is } 1,7; B(3) \text { is } 2 \text {. }$$
$B(1,2)$ has no elements; $B(1,3)$ is $6; B(2,3)$ is $4. B(1,2,3)$ ha... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 Let sequence $A$ be $1,3,2,5,4,3,5,7,6,11$. Property $P_{1}$ is that the element is divisible by 3, Property $P_{2}$ is that the element leaves a remainder of 1 when divided by 3, Property $P_{3}$ is that the element is divisible by 2, Property $P_{4}$ is that the element is divisible by 3. Try to verify whet... | Solve $A(1)$ is $3,3,6 ; A(2)$ is $1,4,7 ; A(3)$ is $2,4,6 ; A(4)$ is $3,3,6. A(1,2)$ has no elements; $A(1,3)$ is $6 ; A(1,4)$ is $3,3,6 ; A(2,3)$ is $4 ; A(2,4)$ has no elements; $A(3,4)$ is $6. A(1,2,3)$ has no elements; $A(1,2,4)$ has no elements; $A(1,3,4)$ is $6 ; A(2,3,4)$ has no elements; $A(1,2,3,4)$ has no el... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
4. Let $n$ be any given positive integer. Find the value of $\mu(n) \mu(n+1) \mu(n+2) \mu(n+3)$. | 4. 0 , because $n, n+1, n+2, n+3$ must include one number that is divisible by 4. | 0 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5. Find the smallest positive integer $n$ that satisfies $\tau(n)=6$.
| 5. $n=p_{1}^{a_{1}} \cdots p_{s}^{a_{s}}, p_{1}<\cdots<p_{3} . \tau(n)=\left(\alpha_{1}+1\right) \cdots\left(\alpha_{s}+1\right)=6$. It must be that $\alpha_{1}=1$, $\alpha_{2}=2$; or $\alpha_{1}=2, \alpha_{2}=1$. Therefore, the smallest $n=2^{2} \cdot 3^{1}=12$. | 12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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