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1. Let $x_{i} \in\{0,1\}(i=1,2, \cdots, n)$. If the function $f=f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ takes values only 0 or 1, then $f$ is called an $n$-ary Boolean function, and we denote
$$
D_{n}(f)=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=0\right\} \text {.... | 1. (1) The total number of all possible values of $x_{1}, x_{2}, \cdots, x_{n}$ is $2^{n}$, and each corresponding function value can be either 0 or 1. Therefore, the number of all different $n$-ary Boolean functions is $2^{2^{n}}$.
(2) Let $\mid D_{10}(g)$ | denote the number of elements in the set $D_{10}(g)$. Below,... | 1817 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. For a finite set
$$
A=\left\{a_{i} \mid 1 \leqslant i \leqslant n, i \in \mathbf{Z}_{+}\right\}\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
let $S=\sum_{i=1}^{n} a_{i}$, then $S$ is called the "sum" of set $A$, denoted as
$|A|$. Given the set $P=\{2 n-1 \mid n=1,2, \cdots, 10\}$,
all the subsets of $P$ containin... | 8. 3600.
Since $1+3+\cdots+19=100$, and each element in $1,3, \cdots, 19$ appears in the three-element subsets of set $P$ a number of times equal to $\mathrm{C}_{9}^{2}=36$, therefore, $\sum_{i=1}^{k}\left|P_{i}\right|=3600$. | 3600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. ${ }^{[1]}$ (2016, National High School Mathematic... | Let the graph that satisfies the conditions be $G(V, E)$.
First, we prove a lemma.
Lemma In any $n(n \leqslant 5)$-order subgraph $G^{\prime}$ of graph $G(V, E)$, there can be at most five edges.
Proof It suffices to prove the case when $n=5$.
If there exists a vertex $A$ in $G^{\prime}$ with a degree of 4, then no edg... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given 2015 circles of radius 1 in the plane. Prove: Among these 2015 circles, there exists a subset $S$ of 27 circles such that any two circles in $S$ either both have a common point or both do not have a common point. ${ }^{[1]}$
(The 28th Korean Mathematical Olympiad) | Proof that there do not exist 27 circles such that any two circles have a common point.
Select a line $l$ such that $l$ is neither parallel to the line connecting the centers of any two of the 2015 circles nor perpendicular to these lines.
Let this line be the $x$-axis, and the 2015 circles be denoted as $C_{1}, C_{2... | 27 | Combinatorics | proof | Yes | Yes | cn_contest | false |
The third question: In a $33 \times 33$ grid, each cell is colored with one of three colors, such that the number of cells of each color is equal. If two adjacent cells have different colors, their common edge is called a "separating edge." Find the minimum number of separating edges.
Translate the above text into Eng... | Assume the number of separating edges is no more than 55, and denote the three colors as $A$, $B$, and $C$.
If a separating edge corresponds to two cells of colors $A$ and $B$ or $A$ and $C$, then it is called an $A$-colored separating edge. Similarly, define $B$-colored and $C$-colored separating edges.
Since $55 < ... | 56 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $n$ is a positive integer, such that there exist positive integers $x_{1}$, $x_{2}, \cdots, x_{n}$ satisfying
$$
x_{1} x_{2} \cdots x_{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right)=100 n .
$$
Find the maximum possible value of $n$.
(Lin Jin, problem contributor) | 2. The maximum possible value of $n$ is 9702.
Obviously, from the given equation, we have $\sum_{i=1}^{n} x_{i} \geqslant n$. Therefore, $\prod_{i=1}^{n} x_{i} \leqslant 100$.
Since equality cannot hold, then $\prod_{i=1}^{n} x_{i} \leqslant 99$.
$$
\begin{array}{l}
\text { and } \prod_{i=1}^{n} x_{i}=\prod_{i=1}^{n}\... | 9702 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. A magician and his assistant have a deck of cards, all of which have the same back, and the front is one of 2017 colors (each color has 1000000 cards). The magic trick is: the magician first leaves the room, the audience arranges $n$ face-up cards in a row on the table, the magician's assistant then flips $n-1$ of t... | 4. When $n=2018$, the magician and the assistant can agree that for $i=1,2, \cdots, 2017$, if the assistant retains the $i$-th card face up, the magician will guess that the 2018-th card is of the $i$-th color. This way, the magic trick can be completed.
Assume for some positive integer $n \leqslant 2017$, the magic t... | 2018 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
6. On a $200 \times 200$ chessboard, some cells contain a red or blue piece, while others are empty. If two pieces are in the same row or column, we say one piece can "see" the other. Assume each piece can see exactly five pieces of the opposite color (it may also see some pieces of the same color). Find the maximum nu... | 6. First, give an example with 3800 pieces.
The intersections of rows 1 to 5 and columns 11 to 200, as well as the intersections of columns 1 to 5 and rows 11 to 200, are all placed with red pieces; the intersections of rows 6 to 10 and columns 11 to 200, as well as the intersections of columns 6 to 10 and rows 11 to ... | 3800 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. If the digits $a_{i}(i=1,2, \cdots, 9)$ satisfy
$$
a_{9}a_{4}>\cdots>a_{1} \text {, }
$$
then the nine-digit positive integer $\overline{a_{9} a_{8} \cdots a_{1}}$ is called a "nine-digit peak number", for example 134698752. Then, the number of all nine-digit peak numbers is . $\qquad$ | 4. 11875 .
From the conditions, we know that the middle number of the nine-digit mountain number can only be 9, 8, 7, 6, 5.
When the middle number is 9, there are $\mathrm{C}_{8}^{4} \mathrm{C}_{9}^{4}$ nine-digit mountain numbers; when the middle number is 8, there are $\mathrm{C}_{7}^{4} \mathrm{C}_{8}^{4}$ nine-dig... | 11875 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that the 2017 roots of the equation $x^{2017}=1$ are 1, $x_{1}, x_{2}, \cdots, x_{2016}$. Then $\sum_{k=1}^{2016} \frac{1}{1+x_{k}}=$ $\qquad$ . | 5.1008 .
Given $x_{k}=\mathrm{e}^{\frac{2 \pi m}{2017} \mathrm{i}}(k=1,2, \cdots, 2016)$, we know
$$
\begin{array}{l}
\overline{x_{k}}=\mathrm{e}^{\frac{-2 k \pi}{2017} \mathrm{i}}=\mathrm{e}^{\frac{2(2017-k) \pi \mathrm{i}}{2017} \mathrm{i}}=x_{2017-k} . \\
\text { Then } \frac{1}{1+x_{k}}+\frac{1}{1+x_{2017-k}} \\
=... | 1008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In a country, some cities have direct two-way flights between them. It is known that one can fly from any city to any other city with no more than 100 flights, and also one can fly from any city to any other city with an even number of flights. Find the smallest positive integer $d$ such that it is guaranteed that f... | Prompt Answer: $d=200$.
First, construct an example, then prove that $d=200$ is feasible. For any two cities $A$ and $B$, consider the shortest path connecting them that has an even length, let its length be $2k$. Use proof by contradiction combined with the extremal principle to show that $k \leqslant 100$. | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Figure 1 is a road map of a city, where $A$, $B, \cdots, I$ represent nine bus stops. A bus departs from station $A$, travels along the roads, reaches each bus stop exactly once, and finally returns to station $A$. The number of different sequences of stops it can pass through is
| 8. 32.
The bus route can be considered as a circle $\Gamma$ (direction is not considered for now), and each station on $\Gamma$ has exactly two adjacent stations.
Let the two stations adjacent to $I$ on $\Gamma$ be $\alpha$ and $\beta$. There are two cases:
(1) $\{\alpha, \beta\} \cap \{E, F, G, H\} \neq \varnothing$... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given real numbers $x, y$ satisfy $x+y=3, \frac{1}{x+y^{2}}+\frac{1}{x^{2}+y}=\frac{1}{2}$.
Find the value of $x^{5}+y^{5}$. [3]
(2017, National Junior High School Mathematics League) | From the given equation, we have
$$
\begin{array}{l}
2 x^{2}+2 y+2 x+2 y^{2}=x^{3}+x y+x^{2} y^{2}+y^{3} \\
\Rightarrow 2(x+y)^{2}-4 x y+2(x+y) \\
\quad=(x+y)\left((x+y)^{2}-3 x y\right)+x y+x^{2} y^{2} .
\end{array}
$$
Substituting $x+y=3$ into the above equation, we get
$$
\begin{array}{l}
(x y)^{2}-4 x y+3=0 \\
\Ri... | 123 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given $a+b+c=5$,
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}=15, a^{3}+b^{3}+c^{3}=47 . \\
\text { Find }\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)
\end{array}
$$
the value. ${ }^{[4]}$
$(2016$, National Junior High School Mathematics League (B Volume)) | From formula (1), we know
$$
\begin{array}{l}
2(a b+b c+c a) \\
=(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=10 \\
\Rightarrow a b+b c+c a=5 .
\end{array}
$$
From formula (2), we know
$$
47-3 a b c=5(15-5) \Rightarrow a b c=-1 \text {. }
$$
And $a^{2}+a b+b^{2}$
$$
\begin{array}{l}
=(a+b)(a+b+c)-(a b+b c+c a) \\
=5(5-... | 625 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given $a+b+c=1$,
$$
\frac{1}{a+1}+\frac{1}{b+3}+\frac{1}{c+5}=0 \text {. }
$$
Find the value of $(a+1)^{2}+(b+3)^{2}+(c+5)^{2}$. (2017, National Junior High School Mathematics League (Grade 8)) | Let $x=a+1, y=b+3, z=c+5$.
Then the given equations can be transformed into
$$
\begin{array}{l}
x+y+z=10, \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \Rightarrow x y+y z+z x=0 .
\end{array}
$$
From formula (1), we get
$$
\begin{array}{l}
(a+1)^{2}+(b+3)^{2}+(c+5)^{2} \\
=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x y+y z+z x) \\
=1... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let planar vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ satisfy $|\boldsymbol{\alpha}+2 \boldsymbol{\beta}|=3,|2 \boldsymbol{\alpha}+3 \boldsymbol{\beta}|=4$. Then the minimum value of $\boldsymbol{\alpha} \cdot \boldsymbol{\beta}$ is $\qquad$ . | 5. -170 .
Let $\alpha+2 \beta=u, 2 \alpha+3 \beta=v,|u|=3,|v|=4$.
From $\alpha=2 v-3 u, \beta=2 u-v$, we get
$$
\boldsymbol{\alpha} \cdot \boldsymbol{\beta}=-6|u|^{2}-2|\boldsymbol{v}|^{2}+7 u \cdot v \text {. }
$$
When $\boldsymbol{u} \cdot \boldsymbol{v}=-12$,
$\boldsymbol{\alpha} \cdot \boldsymbol{\beta}=-54-32-84... | -170 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $P_{1}, P_{2}, \cdots, P_{100}$ as 100 points on a plane, satisfying that no three points are collinear. For any three of these points, if their indices are in increasing order and they form a clockwise orientation, then the triangle with these three points as vertices is called "clockwise". Question: Is it po... | 4. Suppose $P_{1}, P_{2}, \cdots, P_{100}$ are arranged counterclockwise on a circle. At this point, the number of clockwise triangles is 0.
Now, move these points (not necessarily along the circumference). When a point $P_{i}$ crosses the line $P_{j} P_{k}$ during the movement, the orientation (clockwise or countercl... | 2017 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $AB$ is a line segment of length 8, and point $P$ is at a distance of 3 from the line containing $AB$. Then the minimum value of $AP \cdot PB$ is $\qquad$. | $-, 1.24$
Draw a perpendicular from point $P$ to the line $AB$, with the foot of the perpendicular being $H$. Then
$$
\begin{array}{l}
\frac{1}{2} AB \cdot PH = S_{\triangle PAB} = \frac{1}{2} AP \cdot PB \sin \angle APB \\
\Rightarrow AP \cdot PB = \frac{24}{\sin \angle APB} \geqslant 24 .
\end{array}
$$
When $\angle... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the set
$$
A=\{n|n \in \mathbf{N}, 11| S(n), 11 \mid S(n+1)\},
$$
where $S(m)$ denotes the sum of the digits of the natural number $m$. Then the smallest number in set $A$ is $\qquad$ . | 5.2899999.
Let the smallest number in $A$ be $n=\overline{a_{1} a_{2} \cdots a_{t}}$,
$$
S(n)=a_{1}+a_{2}+\cdots+a_{t} \text {. }
$$
If the unit digit of $n$, $a_{t} \neq 9$, then
$$
\begin{array}{l}
n+1=\overline{a_{1} a_{2} \cdots a_{t-1} a_{t}^{\prime}}\left(a_{t}^{\prime}=a_{t}+1\right) . \\
\text { Hence } S(n+1... | 2899999 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange all positive odd numbers in ascending order, take the first number as $a_{1}$, take the sum of the next two numbers as $a_{2}$, then take the sum of the next three numbers as $a_{3}$, and so on, to get the sequence $\left\{a_{n}\right\}$, that is,
$$
a_{1}=1, a_{2}=3+5, a_{3}=7+9+11, \cdots \cdots
$$
Then $... | 4. 44100 .
Notice that, $\sum_{i=1}^{20} a_{i}$ is the sum of the first $\sum_{i=1}^{20} i=210$ odd numbers. Therefore, $210^{2}=44100$. | 44100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $M=\{1,2, \cdots, 2017\}$ be the set of the first 2017 positive integers. If one element is removed from the set $M$, and the sum of the remaining elements is exactly a perfect square, then the removed element is $\qquad$ . | 8. 1677 .
Notice,
$$
\begin{array}{l}
1+2+\cdots+2017=\frac{2017 \times 2018}{2} \\
>[\sqrt{2017 \times 1009}]^{2}=1426^{2} \\
=2033476 . \\
\text { Then } \frac{2017 \times 2018}{2}-2033476=1677 .
\end{array}
$$ | 1677 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. The sum $\sum_{i=1}^{k} a_{m+i}$ is called the sum of $k$ consecutive terms of the sequence $a_{1}, a_{2}, \cdots, a_{n}$, where $m, k \in \mathbf{N}, k \geqslant 1, m+k \leqslant n$. The number of groups of consecutive terms in the sequence $1,2, \cdots, 100$ whose sum is a multiple of 11 is $\qquad$. | 7.801.
Let $S_{k}=\sum_{i=1}^{k} i=\frac{k(k+1)}{2}$.
Notice,
$$
\begin{array}{l}
S_{k+11}=\frac{(k+11)(k+12)}{2} \\
\equiv \frac{k(k+1)}{2}=S_{k}(\bmod 11),
\end{array}
$$
and the remainders of $S_{1}, S_{2}, \cdots, S_{11}$ modulo 11 are $1,3,6$, $10,4,10,6,3,1,0,0$.
Since $100=9 \times 11+1$, thus, among the rema... | 801 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $x_{k} 、 y_{k} \geqslant 0(k=1,2,3)$. Calculate:
$$
\begin{array}{l}
\sqrt{\left(2018-y_{1}-y_{2}-y_{3}\right)^{2}+x_{3}^{2}}+\sqrt{y_{3}^{2}+x_{2}^{2}}+ \\
\sqrt{y_{2}^{2}+x_{1}^{2}}+\sqrt{y_{1}^{2}+\left(x_{1}+x_{2}+x_{3}\right)^{2}}
\end{array}
$$
the minimum value is | 6. 2018.
Let $O(0,0), A(0,2018)$,
$$
\begin{array}{l}
P_{1}\left(x_{1}+x_{2}+x_{3}, y_{1}\right), P_{2}\left(x_{2}+x_{3}, y_{1}+y_{2}\right), \\
P_{3}\left(x_{3}, y_{1}+y_{2}+y_{3}\right) .
\end{array}
$$
The required is
$$
\begin{array}{l}
\left|\overrightarrow{A P_{3}}\right|+\left|\overrightarrow{P_{3} P_{2}}\righ... | 2018 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. (15 points) In the sequence $\left\{a_{n}\right\}$,
$$
a_{n}=2^{n} a+b n-80\left(a 、 b \in \mathbf{Z}_{+}\right) \text {. }
$$
It is known that the minimum value of the sum of the first $n$ terms $S_{n}$ is obtained only when $n=6$, and $7 \mid a_{36}$. Find the value of $\sum_{i=1}^{12}\left|a_{i}\right|$. | Three, 13. Notice that, $\left\{a_{n}\right\}$ is an increasing sequence.
From the given, $a_{6}0$, that is,
$$
64 a+6 b-800 \text {. }
$$
Combining $a, b \in \mathbf{Z}_{+}$, we get
$$
a=1, b=1 \text { or } 2 \text {. }
$$
Also, $a_{36}=2^{36}+36 b-80$
$$
\equiv 1+b-3 \equiv 0(\bmod 7) .
$$
Thus, $b=2$.
Therefore, ... | 8010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Given non-zero complex numbers $x, y$ satisfy $y^{2}\left(x^{2}-x y+y^{2}\right)+x^{3}(x-y)=0$.
Find the value of $\sum_{m=0}^{29} \sum_{n=0}^{29} x^{18 m n} y^{-18 m n}$. | 11. Divide both sides of the known equation by $y^{4}$,
$$
\left(\frac{x}{y}\right)^{4}-\left(\frac{x}{y}\right)^{3}+\left(\frac{x}{y}\right)^{2}-\left(\frac{x}{y}\right)+1=0 \text {. }
$$
Let $\frac{x}{y}=\omega$, then
$$
\begin{array}{l}
\omega^{4}-\omega^{3}+\omega^{2}-\omega+1=0 \\
\Rightarrow \omega^{5}=-1 \Right... | 180 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
25. Among the integers between 100 and 999, there are ( ) numbers that have the property: the digits of the number can be rearranged to form a number that is a multiple of 11 and is between 100 and 999 (for example, 121 and 211 both have this property).
(A) 226
(B) 243
( C) 270
(D) 469
(E) 486 | 25. A.
Let a three-digit number be $\overline{A C B}$. Then
11. $\overline{A C B} \Leftrightarrow 11 \mathrm{I}(A+B-C)$
$\Leftrightarrow A+B=C$ or $A+B=C+11$.
We will discuss the following scenarios.
Note that, $A$ and $B$ are of equal status, so we can assume $A \geqslant B$ (the case for $A < B$ is similar).
(1) $A+... | 226 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
In a box, there are 10 red cards and 10 blue cards, each set of cards containing one card labeled with each of the numbers $1, 3, 3^{2}, \cdots, 3^{9}$. The total sum of the numbers on the cards of both colors is denoted as $S$. For a given positive integer $n$, if it is possible to select several cards from the box su... | Let the maximum sum of the labels of two-color cards marked as $1,3,3^{2}, \cdots, 3^{k}$ be denoted as $S_{k}$. Then,
$$
S_{k}=2 \sum_{n=0}^{k} 3^{n}=3^{k+1}-1<3^{k+1} \text {. }
$$
In the sequence $1,3,3^{2}, \cdots, 3^{k}$, the sum of any subset of these numbers is not equal to $3^{m}$. Therefore, the number of way... | 6423 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the sum of the digits of a positive integer $m$ be denoted as $S(m)$, for example, $S(2017)=2+0+1+7=10$. Now, from the 2017 positive integers $1,2, \cdots$, 2017, any $n$ different numbers are taken. It is always possible to find eight different numbers $a_{1}, a_{2}, \cdots, a_{8}$ among these $n$ numbers such ... | 6. A.
Notice that, among $1,2, \cdots, 2017$, the minimum sum of digits is 1, and the maximum sum is 28.
It is easy to see that the numbers with a digit sum of 1 are $1, 10, 100, 1000$; the numbers with a digit sum of $2,3, \cdots, 26$ are no less than eight; the numbers with a digit sum of 27 are only 999, $1899, 19... | 185 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
2. Given $x_{1}=1, x_{2}=2, x_{3}=3$ are zeros of the function
$$
f(x)=x^{4}+a x^{3}+b x^{2}+c x+d
$$
then $f(0)+f(4)=$ $\qquad$ | 2. 24 .
Let $f(x)=(x-1)(x-2)(x-3)(x-k)$. Then $f(0)+f(4)=6k+6(4-k)=24$. | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the sequence $\left\{a_{n}\right\}$ with the first term being 2, and satisfying
$$
6 S_{n}=3 a_{n+1}+4^{n}-1 \text {. }
$$
Then the maximum value of $S_{n}$ is $\qquad$. | 8. 35 .
According to the problem, we have
$$
\left\{\begin{array}{l}
6 S_{n}=3 a_{n+1}+4^{n}-1 \\
6 S_{n-1}=3 a_{n}+4^{n-1}-1
\end{array}\right.
$$
Subtracting the two equations and simplifying, we get
$$
\begin{array}{l}
a_{n+1}=3 a_{n}-4^{n-1} \\
\Rightarrow a_{n+1}+4^{n}=3 a_{n}-4^{n-1}+4^{n} \\
\quad=3\left(a_{n}... | 35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Given that the angle between vector $\boldsymbol{a}$ and $\boldsymbol{b}$ is $120^{\circ}$, and $|a|=2,|b|=5$. Then $(2 a-b) \cdot a=$ $\qquad$ | \begin{array}{l}\text { II.13. 13. } \\ (2 a-b) \cdot a=2|a|^{2}-a \cdot b=13 \text {. }\end{array} | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Given the parabola $y^{2}=a x(a>0)$ and the line $x=1$ enclose a closed figure with an area of $\frac{4}{3}$. Then, the coefficient of the $x^{-18}$ term in the expansion of $\left(x+\frac{a}{x}\right)^{20}$ is | 14. 20 .
According to the problem, we know $2 \int_{0}^{1} \sqrt{a x} \mathrm{~d} x=\frac{4}{3} \Rightarrow a=1$. Therefore, the term containing $x^{-18}$ is $\mathrm{C}_{20}^{19} x\left(\frac{1}{x}\right)^{19}$. | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If the expansion of $(a+2 b)^{n}$ has three consecutive terms whose binomial coefficients form an arithmetic sequence, then the largest three-digit positive integer $n$ is $\qquad$ | 3. 959 .
Let the binomial coefficients $\mathrm{C}_{n}^{k-1}, \mathrm{C}_{n}^{k}, \mathrm{C}_{n}^{k+1}(1 \leqslant k \leqslant n-1)$ of three consecutive terms in the expansion of $(a+2 b)^{n}$ satisfy
$$
\begin{array}{l}
2 \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k+1} . \\
\text { Then } n^{2}-(4 k+1) ... | 959 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. In the sequence $\left\{a_{n}\right\}$, for $1 \leqslant n \leqslant 5$, we have $a_{n}=n^{2}$, and for all positive integers $n$, we have
$$
a_{n+5}+a_{n+1}=a_{n+4}+a_{n} \text {. }
$$
Then $a_{2023}=$ . $\qquad$ | 3. 17 .
For all positive integers $n$, we have
$$
a_{n+5}+a_{n+1}=a_{n+4}+a_{n}=\cdots=a_{5}+a_{1}=26 \text {. }
$$
Then $a_{n}=26-a_{n+4}=26-\left(26-a_{n+8}\right)=a_{n+8}$, which means $\left\{a_{n}\right\}$ is a sequence with a period of 8.
Therefore, $a_{2023}=a_{7}=26-a_{3}=26-9=17$. | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In a certain social event, it was originally planned that each pair of people would shake hands exactly once, but four people each shook hands twice and then left. As a result, there were a total of 60 handshakes during the entire event. Then the number of people who initially participated in the event is $\qquad$ | 5. 15 .
Let the number of people participating in the activity be $n+4$, among which, the number of handshakes among the four people who quit is $x\left(0 \leqslant x \leqslant \mathrm{C}_{4}^{2}=6\right)$.
From the problem, we have $\mathrm{C}_{n}^{2}+4 \times 2=60+x$, which simplifies to $n(n-1)=104+2 x$.
Given $0 \... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. If the function
$$
f(x)=\left(x^{2}-1\right)\left(x^{2}+a x+b\right)
$$
satisfies $f(x)=f(4-x)$ for any $x \in \mathbf{R}$, then the minimum value of $f(x)$ is $\qquad$ . | 2. -16 .
Notice that, $f(1)=f(-1)=0$.
Also, $f(x)=f(4-x)$, so $f(3)=f(5)=0$.
Therefore, $f(x)=(x^{2}-1)(x-3)(x-5)$ $=(x^{2}-4x+3)(x^{2}-4x-5)$.
Let $t=x^{2}-4x+4 \geqslant 0$, then $f(x)=(t-1)(t-9)=(t-5)^{2}-16$.
Thus, the minimum value of $f(x)$ is -16. | -16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { 6. Let } a_{n}=1+2+\cdots+n\left(n \in \mathbf{Z}_{+}\right) , \\
S_{m}=a_{1}+a_{2}+\cdots+a_{m}(m=1,2, \cdots) \text {. }
\end{array}
$$
Then among $S_{1}, S_{2}, \cdots, S_{2017}$, the numbers that are divisible by 2 but not by 4 are $\qquad$ in number.
$$ | 6. 252 .
Notice that, $S_{m}=\frac{m(m+1)(m+2)}{6}$.
Thus $S_{m} \equiv 2(\bmod 4)$
$$
\begin{array}{l}
\Leftrightarrow m(m+1)(m+2) \equiv 4(\bmod 8) \\
\Leftrightarrow m \equiv 3(\bmod 8) .
\end{array}
$$
Therefore, among $S_{1}, S_{2}, \cdots, S_{2017}$, the numbers that are divisible by 2 but not by 4 are $\left[\... | 252 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a positive geometric sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{6}+a_{5}+a_{4}-a_{3}-a_{2}-a_{1}=49 \text {. }
$$
Then the minimum value of $a_{9}+a_{8}+a_{7}$ is $\qquad$ | 2. 196.
Let the common ratio be $q$. From the condition, we have
$$
\left(q^{3}-1\right)\left(a_{3}+a_{2}+a_{1}\right)=49.
$$
Clearly, $q^{3}-1>0$.
Then $a_{3}+a_{2}+a_{1}=\frac{49}{q^{3}-1}$.
Thus, $a_{9}+a_{8}+a_{7}=q^{6}\left(a_{3}+a_{2}+a_{1}\right)$
$$
\begin{array}{l}
=\frac{49 q^{6}}{q^{3}-1}=49\left(\sqrt{q^{... | 196 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let the function be
$$
f(x)=x^{3}+a x^{2}+b x+c \quad (x \in \mathbf{R}),
$$
where $a, b, c$ are distinct non-zero integers, and
$$
f(a)=a^{3}, f(b)=b^{3} \text {. }
$$
Then $a+b+c=$ $\qquad$ | 3. 18 .
Let $g(x)=f(x)-x^{3}=a x^{2}+b x+c$.
From the problem, we have $g(a)=g(b)=0$.
Thus, $g(x)=a(x-a)(x-b)$
$$
\begin{array}{l}
\Rightarrow b=-a(a+b), c=a^{2} b \\
\Rightarrow b=-\frac{a^{2}}{a+1}=1-a-\frac{1}{a+1} .
\end{array}
$$
Since $b$ is an integer, we have $a+1= \pm 1$.
Also, $a \neq 0$, so $a=-2, b=4, c=1... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Arrange all positive integers that are coprime with 70 in ascending order. The 2017th term of this sequence is $\qquad$ . | 10. 5881.
It is easy to know that the number of positive integers not exceeding 70 and coprime with 70 is
$$
35-(7+5)+1=24 \text{.}
$$
Let the sequence of all positive integers coprime with 70, arranged in ascending order, be $\left\{a_{n}\right\}$. Then
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, a_{3}=9, \cdots, a_{24}=6... | 5881 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. In two regular tetrahedrons $A-OBC$ and $D-OBC$ with their bases coinciding, $M$ and $N$ are the centroids of $\triangle ADC$ and $\triangle BDC$ respectively. Let $\overrightarrow{OA}=\boldsymbol{a}, \overrightarrow{OB}=\boldsymbol{b}, \overrightarrow{OC}=\boldsymbol{c}$. If point $P$ satisfies $\overrightarrow{OP}... | Take $O$ as the origin and the line $O B$ as the $x$-axis, establishing a spatial rectangular coordinate system as shown in Figure 13.
Let $B(1,0,0)$. Then
$$
\begin{array}{l}
C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right), \\
A\left(\frac{1}{2}, \frac{\sqrt{3}}{6}, \frac{\sqrt{6}}{3}\right), \\
D\left(\frac{1}{2}, \... | 439 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. For the positive integer $n$, define $a_{n}$ as the unit digit of $n^{(n+1)^{n-1}}$. Then $\sum_{n=1}^{2018} a_{n}=$ $\qquad$ . | 5. 5857 .
When $n \equiv 0,1,5,6(\bmod 10)$, $a_{n} \equiv n^{(n+1)^{n+2}} \equiv n(\bmod 10)$; when $n \equiv 2,4,8(\bmod 10)$, $(n+1)^{n+2} \equiv 1(\bmod 4)$, then $a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k+1} \equiv n(\bmod 10)$; when $n \equiv 3,7,9(\bmod 10)$, $(n+1)^{n+2} \equiv 0(\bmod 4)$, then $a_{n} \equiv n^{(n+... | 5857 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A five-digit number $\overline{a b c d e}$ satisfies:
$$
ac>d, dd, b>e \text {. }
$$
For example, 34 201, 49 412. If the digits of the number change in a pattern similar to the monotonicity of a sine function over one period, then the five-digit number is said to follow the "sine rule." The number of five-digit num... | 6. 2892.
From the problem, we know that $b$ and $d$ are the maximum and minimum numbers, respectively, and $2 \leqslant b-d \leqslant 9$.
Let $b-d=k$, at this point, there are $10-k$ ways to choose $(b, d)$, and $a$, $c$, and $e$ each have $k-1$ ways to be chosen, i.e., $(a, c, e)$ has $(k-1)^{3}$ combinations. There... | 2892 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
One, (40 points) Let $S$ be a set of positive integers with the property: for any $x \in S$, the arithmetic mean of the remaining numbers in $S$ after removing $x$ is a positive integer, and it satisfies $1 \in S$, 2016 is the largest element in $S$. Find the maximum value of $|S|$. | Let the elements of set $S$ be
$$
1=x_{1}<x_{2}<\cdots<x_{n}=2016 \text {. }
$$
Then for $1 \leqslant j \leqslant n$, we have
$$
y_{j}=\frac{\sum_{i=1}^{n} x_{i}-x_{j}}{n-1} \in \mathbf{Z}_{+} \text {. }
$$
Thus, for $1 \leqslant i<j \leqslant n$, we have
$$
\begin{array}{l}
y_{i}-y_{j}=\frac{x_{j}-x_{i}}{n-1} \in \m... | 32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. In 1993, American mathematician F. Smarandache proposed many number theory problems, attracting the attention of scholars both at home and abroad. One of these is the famous Smarandache function. The Smarandache function of a positive integer \( n \) is defined as
\[
S(n)=\min \left\{m \left| m \in \mathbf{Z}_{+}, ... | 14. (1) It is easy to know that $16=2^{4}$. Then $S(16)=6$.
From $2016=2^{5} \times 3^{2} \times 7$, we know
$S(2016)=\max \left\{S\left(2^{5}\right), S\left(3^{2}\right), S(7)\right\}$.
Also, $S(7)=7, S\left(3^{2}\right)=6, S\left(2^{5}\right)=8$, so $S(2016)=8$.
(2) From $S(n)=7$, we know $n \mid 7!$.
Thus, $n \leq... | 5040 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. In a chess tournament, $n$ players participate in a round-robin competition. After players A and B each played two games, they withdrew from the competition due to certain reasons. It is known that a total of 81 games were ultimately played. Then $n=$ $\qquad$ | 4. 15 .
If there is no match between A and B, then
$$
\mathrm{C}_{n-2}^{2}+2 \times 2=81 \text{, }
$$
this equation has no positive integer solution.
If A and B have a match, then
$$
\mathrm{C}_{n-2}^{2}+3=81 \Rightarrow n=15 .
$$ | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $G$ be a simple graph of order 100. It is known that for any vertex $u$, there exists another vertex $v$ such that $u$ and $v$ are adjacent, and there is no vertex adjacent to both $u$ and $v$. Find the maximum possible number of edges in graph $G$. (Provided by Yunhao Fu) | 2. Let $G=(V, E)$.
For $u v \in E$, if there is no other vertex adjacent to both $u$ and $v$, then $u v$ is called a "good edge".
Let $E_{0}$ be the set of all good edges, and $G_{0}=\left(V, E_{0}\right)$.
From the problem statement, we know that each vertex in graph $G$ has at least one good edge, i.e., $G_{0}$ has ... | 3822 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $A_{1}, A_{2}, \cdots, A_{n}$ be binary subsets of the set $\{1,2, \cdots, 2018\}$, such that the sets $A_{i}+A_{j}(1 \leqslant i \leqslant j \leqslant n)$ are all distinct, where,
$$
A+B=\{a+b \mid a \in A, b \in B\} \text {. }
$$
Find the maximum possible value of $n$.
(Qiu Zhenhua) | 4. The maximum value of $n$ is 4033.
If there exists $A_{i}=A_{j}(i \neq j)$, then $A_{i}+A_{i}=A_{j}+A_{j}$, which contradicts the condition.
Thus, $A_{1}, A_{2}, \cdots, A_{n}$ are all distinct.
For a binary set $A=\{a, b\}(a<b)$, let $S_{i}=\{x-y \mid x, y \in M_{i}, x>y\}$.
If there exists $i \neq j$ such that $S_... | 4033 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. If the cube of a three-digit positive integer is an eight-digit number of the form $\overline{A B C D C D A B}$, then such a three-digit number is | 5. 303 | 303 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Arrange all positive integers whose sum of digits is 10 in ascending order to form the sequence $\left\{a_{n}\right\}$. If $a_{n}=2017$, then $n=$ . $\qquad$ | 8. 120.
It is easy to know that there are 9 two-digit numbers $\overline{a b}$ whose digits sum to 10.
For three-digit numbers $\overline{a b c}$ whose digits sum to 10, the first digit $a$ can take any value in $\{1,2, \cdots, 9\}$. Once $a$ is determined, $b$ can take any value in $\{0,1, \cdots, 10-a\}$, which give... | 120 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Let the three-digit number $n=\overline{a b c}$, where the lengths $a, b, c$ can form an isosceles (including equilateral) triangle. Then the number of such three-digit numbers $n$ is. $\qquad$ | 4. 165.
When $a=b=c$, there are 9 cases in total.
When $a=b \neq c, a=c \neq b, b=c \neq a$, the number of cases for each situation is the same. Taking $a=b \neq c$ as an example:
(1) When $a=b \geqslant 5$, $a$ and $b$ have 5 cases, and $c$ has 8 cases, making a total of 40 cases;
(2) When $a=b=4$, $c$ can be any one... | 165 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $p(x)=a x^{3}+b x^{2}+c x+d$ is a cubic polynomial, satisfying
$$
p\left(\frac{1}{2}\right)+p\left(-\frac{1}{2}\right)=1000 p(0) \text {. }
$$
Let $x_{1} 、 x_{2} 、 x_{3}$ be the three roots of $p(x)=0$. Then the value of $\frac{1}{x_{1} x_{2}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{1} x_{3}}$ is $\qquad$ . | 6. 1996.
From equation (1) we get
$$
\frac{1}{2} b+2 d=1000 d \Rightarrow \frac{b}{d}=1996 \text {. }
$$
By Vieta's formulas we get
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=-\frac{b}{a}, x_{1} x_{2} x_{3}=-\frac{d}{a} . \\
\text { Therefore } \frac{1}{x_{1} x_{2}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{1} x_{3}}=\frac{x_{1}+... | 1996 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Let the set $M=\{1,2, \cdots, 10\}$,
$$
\begin{aligned}
A= & \{(x, y, z) \mid x, y, z \in M, \text { and } \\
& \left.9 \mid\left(x^{3}+y^{3}+z^{3}\right)\right\} .
\end{aligned}
$$
Then the number of elements in set $A$ is $\qquad$ . | 12.243.
Notice that,
when $x \equiv 1(\bmod 3)$, $x^{3} \equiv 1(\bmod 9)$;
when $x \equiv 2(\bmod 3)$, $x^{3} \equiv-1(\bmod 9)$;
when $x \equiv 0(\bmod 3)$, $x^{3} \equiv 0(\bmod 9)$.
Thus, for $x \in \mathbf{Z}$, we have
$$
x^{3} \equiv 0,1,-1(\bmod 9) \text {. }
$$
To make $9 \mid\left(x^{3}+y^{3}+z^{3}\right)$, ... | 243 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Set $a_{n}=\sum_{k=1}^{n}\left[\frac{n}{k}\right]$. Then the number of even numbers in $a_{1}, a_{2}, \cdots, a_{2018}$ is . $\qquad$ | 8. 1028.
It is easy to see that if and only if $k \mid n$,
$$
\left[\frac{n}{k}\right]-\left[\frac{n-1}{k}\right]=1 \text{. }
$$
Otherwise, $\left[\frac{n}{k}\right]-\left[\frac{n-1}{k}\right]=0$.
Thus, $a_{n}-a_{n-1}$
$$
\begin{array}{l}
=\sum_{k=1}^{n}\left[\frac{n}{k}\right]-\sum_{k=1}^{n-1}\left[\frac{n-1}{k}\rig... | 1028 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given an increasing sequence composed of powers of 3 or the sum of several different powers of 3: $1,3,4,9,10,12,13, \cdots$. Then the 100th term of this sequence is $\qquad$ | 6. 981.
The terms of the sequence are given by $\sum_{i=0}^{n} 3^{i} a_{i}$, where,
$$
a_{i} \in\{0,1\}(i=1,2, \cdots, n) \text {. }
$$
When $n=5$, there are $2^{6}-1=63$ numbers that can be formed, and the 64th term is $3^{6}=729$.
Starting from the 65th term, there are $2^{5}-1=31$ terms that do not contain $3^{5}... | 981 | Number Theory | 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 |
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 |
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 |
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 |
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.
$$
\begin{array}{l}
\text { Therefore, }(\sqrt{3}+i)^{10}=\left((-2 i)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)\right)^{10} \\
=(-2 i)^{10}\left(-\frac{1}{2}+\frac{\sqrt{3}}... | 512 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. As shown in Figure 2, two equal circles with a radius of 5 are externally tangent to each other, and both are internally tangent to a larger circle with a radius of 13, with the points of tangency being $A$ and $B$. Let $AB = \frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)\right. = 1)$. Then the value of $m+n$ is (... | 15. D.
As shown in Figure 5, let the centers of the three circles be $X$, $Y$, and $Z$.
Then points $A$ and $B$ are on the extensions of $XY$ and $XZ$, respectively, satisfying
$$
\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 |
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 |
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 |
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}
$$
When $a=... | 343 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given four positive integers
$a, b, c, d$ satisfy:
$$
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 |
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
$$
a_{k}=\left\{\begin{array}{ll}
k, & S_{k-1}<k ; \\
-k, & S_{k-1} \geqslant k
\end{array}\left(1 \leqslant k \leqslant n, k 、 n \in \mathbf{Z}_... | 10. Let the indices $n$ that satisfy $S_{n}=0$ be arranged in ascending order, denoted as the sequence $\left\{b_{n}\right\}$, then $b_{1}=0$.
To find the recurrence relation that $\left\{b_{n}\right\}$ should satisfy.
In fact, without loss of generality, assume $S_{b_{k}}=0$.
Thus, by Table 1, it is easy to prove by m... | 1092 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $x, y, z \in \mathbf{R}_{+}$, satisfying $x+y+z=x y z$. Then the function
$$
\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 |
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 |
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 |
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 |
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