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1. Among the natural numbers from $1 \sim 10000$, the integers that are neither perfect squares nor perfect cubes are $\qquad$ in number. | In the natural numbers from $1 \sim 10000$, there are 100 perfect squares.
$$
\begin{array}{l}
\text { Because } 22^{3}=10648>10000, \\
21^{3}=9261<10000,
\end{array}
$$
Therefore, there are 21 perfect cubes.
Next, consider the number of natural numbers from $1 \sim 10000$ that are both perfect squares and perfect cubes (i.e., perfect sixth powers). Since $4^{6}=4096,5^{6}=15625$, there are 4 perfect sixth powers.
By the principle of inclusion-exclusion, the number of natural numbers from $1 \sim 10000$ that are neither perfect squares nor perfect cubes is
$$
10000-100-21+4=9883 .
$$ | 9883 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the sequence of natural numbers from $1 \sim 8$ be $a_{1}, a_{2}$, $\cdots, a_{8}$. Then
$$
\begin{array}{l}
\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\left|a_{3}-a_{4}\right|+\left|a_{4}-a_{5}\right|^{\prime}+ \\
\left|a_{5}-a_{6}\right|+\left|a_{6}-a_{7}\right|+\left|a_{7}-a_{8}\right|+\left|a_{8}-a_{1}\right|^{\prime}
\end{array}
$$
The maximum value is $\qquad$ | 5. 32 .
From the problem, we have
$$
\begin{aligned}
S= & \left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\left|a_{3}-a_{4}\right|+ \\
& \left|a_{4}-a_{5}\right|+\left|a_{5}-a_{6}\right|+\left|a_{6}-a_{7}\right|+ \\
& \left|a_{7}-a_{8}\right|+\left|a_{8}-a_{1}\right| .
\end{aligned}
$$
Removing the absolute value signs in any term of this sum, we get one positive and one negative natural number. To achieve the maximum possible value of the sum, we need to take the numbers from 1 to 4 as negative and the numbers from 5 to 8 as positive. Thus,
$$
\begin{array}{l}
S=2[(8+7+6+5)-(4+3+2+1)]=32 \text {. } \\
\text { For example, }|8-4|+|4-7|+|7-1|+|1-5|+ \\
|5-2|+|2-6|+|6-3|+|3-8| \\
=32 .
\end{array}
$$ | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Five, (15 points) A school assigns numbers to the contestants participating in a math competition, with the smallest number being 0001 and the largest number being 2014. No matter which contestant steps forward to calculate the average of the numbers of all other contestants in the school, the average is always an integer. How many contestants can the school have at most?
Will the above text be translated into English, preserving the original text's line breaks and format? Please output the translation directly. | Let the school have a total of $n$ participants, whose admission numbers are
$$
1=x_{1}<x_{2}<\cdots<x_{n-1}<x_{n}=2014 .
$$
According to the problem, we have
$$
S_{k}=\frac{x_{1}+x_{2}+\cdots+x_{n}-x_{k}}{n-1}(k=1,2, \cdots, n) \in \mathbf{Z}_{+} .
$$
For any $i, j(1 \leqslant i<j \leqslant n)$, we have
$$
S_{i}-S_{j}=\frac{x_{j}-x_{i}}{n-1} \in \mathbf{Z}_{+} \text {. }
$$
Thus, $x_{j}-x_{i} \geqslant n-1$.
$$
\begin{array}{l}
\text { Hence } x_{n}-x_{1} \\
=\left(x_{n}-x_{n-1}\right)+\left(x_{n-1}-x_{n-2}\right)+\cdots+\left(x_{2}-x_{1}\right) \\
\geqslant(n-1)^{2} \\
\Rightarrow(n-1)^{2} \leqslant x_{n}-x_{1}=2013 \Rightarrow n \leqslant 45 .
\end{array}
$$
Since $\frac{2014-1}{n-1}$ is an integer, it follows that $n-1$ is a divisor of 2013.
Noting that $2013=3 \times 11 \times 61$, the largest divisor not exceeding 45 is 33. Therefore, the maximum value of $n$ is 34, meaning the maximum number of participants is 34.
Such 34 participants' numbers can be realized. For example,
$$
x_{i}=33 i-32(i=1,2, \cdots, 33), x_{34}=2014 \text {. }
$$
Therefore, the maximum number of participants in the competition is 34. | 34 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1}=a_{n}+a_{n-1}(n \geqslant 2) \text {. }
$$
If $a_{7}=8$, then $a_{1}+a_{2}+\cdots+a_{10}=$ $\qquad$ | 3.88.
From the problem, we know that $a_{7}=8 a_{2}+5 a_{1}$.
Therefore, $a_{1}+a_{2}+\cdots+a_{10}$
$$
=88 a_{2}+55 a_{1}=11 a_{7}=88 .
$$ | 88 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Arrange the numbers in the set $\left\{2^{x}+2^{y}+2^{z} \mid x 、 y 、 z \in \mathbf{N}, x<y<z\right\}$ in ascending order. The 100th number is $\qquad$ (answer with a number).
| 5.577.
Notice that the number of combinations $(x, y, z)$ such that $0 \leqslant x<y<z \leqslant n$ is $\mathrm{C}_{n+1}^{3}$.
Since $\mathrm{C}_{9}^{3}=84<100<120=\mathrm{C}_{10}^{3}$, the 100th number must satisfy $z=9$.
Also notice that the number of combinations $(x, y)$ such that $0 \leqslant x<y \leqslant m$ is $\mathrm{C}_{m+1}^{2}$.
Since $\mathrm{C}_{9}^{3}+\mathrm{C}_{6}^{2}=99$, the 100th number must satisfy $y=6, x=0$, which means the 100th number is
$$
2^{0}+2^{6}+2^{9}=577 .
$$ | 577 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the function $f(x)$ satisfies
$$
f(x)=\left\{\begin{array}{ll}
x-3, & x \geqslant 1000 ; \\
f(f(x+5)), & x<1000 .
\end{array}\right.
$$
Then $f(84)=$ . $\qquad$ | 6.997.
Let $f^{(n)}(x)=\underbrace{f(f(\cdots f(x)))}_{n \uparrow}$. Then
$$
\begin{aligned}
& f(84)=f(f(89))=\cdots=f^{(184)}(999) \\
= & f^{(185)}(1004)=f^{(184)}(1001)=f^{(183)}(998) \\
= & f^{(184)}(1003)=f^{(183)}(1000)=f^{(182)}(997) \\
= & f^{(183)}(1002)=f^{(182)}(999)=f^{(183)}(1004) \\
= & f^{(182)}(1001)=f^{(181)}(998)=f^{(182)}(1003) \\
= & f^{(181)}(1000)=\cdots=f(1000)=997 .
\end{aligned}
$$
Therefore, $f(84)=997$. | 997 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $N>1$ be a positive integer, and $m$ denote the largest divisor of $N$ that is less than $N$. If $N+m$ is a power of 10, find $N$.
| 5. $N=75$.
Let $N=m p$. Then $p$ is the smallest prime factor of $N$.
By the problem, we know $m(p+1)=10^{k}$.
Since $10^{k}$ is not a multiple of 3, therefore, $p>2$.
Hence, $N$ and $m$ are both odd.
Thus, $m=5^{*}$.
If $s=0, N=p=10^{k}-1$ is a multiple of 9, which is a contradiction.
Then $s \geqslant 1,5 \mid N$.
Therefore, $p \leqslant 5$.
If $p=3$, then $4 \times 5^{s}=10^{k}$.
So $k=2, m=25, N=75$.
If $p=5$, we get $6110^{k}$, which is a contradiction. | 75 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. In the figure shown in Figure 3, on both sides of square $P$, there are $a$ and $b$ squares to the left and right, and $c$ and $d$ squares above and below, where $a$, $b$, $c$, and $d$ are positive integers, satisfying
$$
(a-b)(c-d)=0 \text {. }
$$
The shape formed by these squares is called a "cross star".
There is a grid table consisting of 2014 squares, forming a $38 \times 53$ grid. Find the number of cross stars in this grid table.
(Tao Pingsheng, provided) | 8. For a cross, the cell $P$ referred to in the problem is called the "center block" of the cross.
When $a=b$, the cross is called "standing"; when $c=d$, it is called "lying" (some crosses are both standing and lying).
If the union of a row and a column of a rectangle $R$ is exactly a cross $S$, then $R$ is called the "source rectangle" of $S$. In the source rectangle $R$, the cross $S$ is uniquely determined by the position of the center block, and the crosses corresponding to different source rectangles must be different. Moreover, by the definition of the cross, the number of rows and columns of the source rectangle is at least 3; a source rectangle with an even number of rows does not contain a lying cross; a source rectangle with an even number of columns does not contain a standing cross.
Now consider the $m \times n$ grid $P_{m, n}$. The number of source rectangles of size $(2k+1) \times l$ is $(m-2k)(n-l+1)$, where $k, l$ satisfy
$$
3 \leqslant 2k+1 \leqslant m, 3 \leqslant l \leqslant n.
$$
For each $(2k+1) \times l$ source rectangle, the center block of a lying cross must be in the $(k+1)$-th row and in the 2nd to $(l-1)$-th columns, so this source rectangle corresponds to $l-2$ lying crosses.
$$
\begin{array}{l}
\text { Hence } A=\sum_{k=1}^{\left[\frac{m-1}{2}\right]} \sum_{l=3}^{n}(l-2)(m-2k)(n-l+1) \\
=\left[\sum_{k=1}^{\left[\frac{m-1}{2}\right]}(m-2k)\right]\left[\sum_{i=1}^{n-2} l(n-l-1)\right], \\
\text { where, } \sum_{k=1}^{\left[\frac{m-1}{2}\right]}(m-2k) \\
=\left[\frac{m-1}{2}\right] \frac{(m-2)+\left(m-2\left[\frac{m-1}{2}\right]\right)}{2} \\
\quad=\left[\frac{m-1}{2}\right]\left(m-\left[\frac{m+1}{2}\right]\right), \\
\sum_{i=1}^{n-2} l(n-l-1)=\left(\sum_{l=1}^{n-2} l\right)(n-1)-\sum_{l=1}^{n-2} l^{2} \\
=\frac{(n-2)(n-1)^{2}}{2}-\frac{(n-2)(n-1)(2n-3)}{6} \\
=\frac{n(n-1)(n-2)}{6} .
\end{array}
$$
Thus, $A=\left[\frac{m-1}{2}\right]\left(n-\left[\frac{m+1}{2}\right]\right) \frac{n(n-1)(n-2)}{6}$.
Similarly, the number of source rectangles of size $k \times (2l+1)$ in the grid $P_{m, n}$ is $(m-k+1)(n-2l)$, and each $k \times (2l+1)$ source rectangle corresponds to $k-2$ standing crosses. Therefore,
$$
\begin{aligned}
B & =\sum_{k=3}^{m}\left[\sum_{k=1}^{\left.n-\frac{n-1}{2}\right]}(k-2)(m-k+1)(n-2l)\right. \\
& =\left[\frac{n-1}{2}\right]\left(n-\left[\frac{n+1}{2}\right]\right) \frac{m(m-1)(m-2)}{6} .
\end{aligned}
$$
Each $(2k+1) \times (2l+1)$ source rectangle corresponds to exactly one cross that is both standing and lying (the center block is at the center of the source rectangle), and the grid $P_{m, n}$ contains $(m-2k)(n-2l)$ such source rectangles, so similarly we get
$$
\begin{aligned}
C & \left.=\sum_{k=1}^{\left[\frac{m-1}{2}\right]}\right]\left[\frac{n-1}{2}\right] \\
& =\left[\frac{m-1}{2}\right]\left(m-\left[\frac{m+1}{2}\right]\right)\left[\frac{n-1}{2}\right]\left(n-\left[\frac{n+1}{2}\right]\right), \text { (3) }
\end{aligned}
$$
which is the number of crosses that are counted twice in $A+B$.
In particular, for the case $m=38, n=53$, we have
$$
\begin{array}{l}
{\left[\frac{m-1}{2}\right]\left(m-\left[\frac{m+1}{2}\right]\right)=18 \times 19=342,} \\
\frac{m(m-1)(m-2)}{6}=8436, \\
{\left[\frac{n-1}{2}\right]\left(n-\left[\frac{n+1}{2}\right]\right)=26 \times 26=676,} \\
\frac{n(n-1)(n-2)}{6}=23426 .
\end{array}
$$
Substituting into equations (1), (2), and (3), we get
$$
\begin{array}{l}
A=342 \times 23426=8011692, \\
B=676 \times 8436=5702736, \\
C=342 \times 676=231192 .
\end{array}
$$
Thus, the number of crosses in the grid $P_{38,53}$ is
$$
A+B-C=13483236
$$ | 13483236 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $1 \leqslant x, y, z \leqslant 6$.
The number of cases where the product of the positive integers $x, y, z$ is divisible by 10 is
$\qquad$ kinds. | 8. 72 .
(1) The number of ways to choose $x, y, z$ is $6^{3}$;
(2) The number of ways to choose $x, y, z$ without taking $2, 4, 6$ is $3^{3}$; (3) The number of ways to choose $x, y, z$ without taking 5 is $5^{3}$;
(4) The number of ways to choose $x, y, z$ without taking $2, 4, 5, 6$ is $2^{3}$. Therefore, the number of ways for the product of $x, y, z$ to be divisible by 10 is $6^{3}-3^{3}-5^{3}+2^{3}=72$. | 72 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given the sequence $\left\{a_{n}\right\}_{n \geqslant 0}$ satisfies $a_{0}=0$, $a_{1}=1$, and for all positive integers $n$,
$$
a_{n+1}=2 a_{n}+2013 a_{n-1} \text {. }
$$
Find the smallest positive integer $n$ such that $2014 \mid a_{n}$. | 10. Below are $a_{n}$ modulo 2014.
Then $a_{n+1} \equiv 2 a_{n}-a_{n-1} \Rightarrow a_{n+1}-a_{n} \equiv a_{n}-a_{n-1}$.
Therefore, the sequence $\left\{a_{n}\right\}$ has the characteristics of an arithmetic sequence modulo 2014.
Since $a_{0}=0, a_{1}=1$, we have $a_{n} \equiv n$.
Thus, the smallest positive integer $n$ such that $2014 \mid a_{n}$ is 2014. | 2014 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $f(x)$ be a function defined on $\mathbf{R}$, for any $x \in \mathbf{R}$, we have
$$
f(x+3) \leqslant f(x)+3, f(x+2) \geqslant f(x)+2 .
$$
Let $g(x)=f(x)-x$. If $f(4)=2014$, then
$$
f(2014)=
$$
$\qquad$ | 6.4024.
Let $g(x)=f(x)-x$, then we have
$$
\begin{array}{l}
g(x+2)=f(x+2)-x-2, \\
g(x+3)=f(x+3)-x-3 .
\end{array}
$$
Also, from $f(x+3) \leqslant f(x)+3$,
$$
f(x+2) \geqslant f(x)+2 \text {, }
$$
we get
$$
\begin{array}{l}
g(x+2) \geqslant f(x)+2-x-2=f(x)-x, \\
g(x+3) \leqslant f(x)+3-x-3=f(x)-x .
\end{array}
$$
From equation (1), we have
$$
\begin{array}{l}
g(x+4) \geqslant f(x+2)-x-2 \\
\geqslant f(x)+2-x-2=f(x)-x .
\end{array}
$$
Thus, $g(x+6) \geqslant f(x+2)-x-2 \geqslant f(x)-x$.
From equation (2), we have
$$
g(x+6) \leqslant f(x+3)-x-3 \leqslant f(x)-x \text {. }
$$
Therefore, $g(x+6)=f(x)-x=g(x)$.
Hence, $g(x)$ is a periodic function with a period of 6.
Notice that,
$$
\begin{array}{l}
g(2014)=g(335 \times 6+4)=g(4) \\
=f(4)-4=2014-4=2010 .
\end{array}
$$
Thus, $f(2014)=g(2014)+2014=4024$. | 4024 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. If non-negative integers $m, n$ add up with exactly one carry (in decimal), then the ordered pair $(m, n)$ is called "good". The number of all good ordered pairs whose sum is 2014 is $\qquad$ . | 7. 195 .
If the carry is in the units place, then the combination of units and tens is $5+9$, $6+8$, $7+7$, $8+6$, $9+5$, a total of 5 kinds; the hundreds place can only be $0+0$, a total of 1 kind; the thousands place is $0+2$, $1+1$, $2+0$, a total of 3 kinds. In this case, there are $5 \times 1 \times 3=15$ pairs.
(1) If the carry is in the tens place, this would cause the hundreds place to also carry, which does not meet the requirement;
(2) If the carry is in the hundreds place, then the units place is $0+4$, $1+3$, $2+2$, $3+1$, $4+0$, a total of 5 kinds; the tens place is $0+1$, $1+0$, a total of 2 kinds; the combination of hundreds and thousands place is $1+19$, $2+18$, ..., $9+11$, $11+9$, $12+8$, ..., $19+1$, a total of 18 kinds. In this case, there are $5 \times 2 \times 18=180$ pairs.
The thousands place cannot carry.
Therefore, the number of ordered pairs is $180+15=195$. | 195 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) Prove: There exists a set $S$ consisting of 2014 positive integers, with the following property: if a subset $A$ of $S$ satisfies that for any $a, a' \in A, a \neq a'$, we have $a + a' \notin S$, then $|A| \leq 152$.
---
The translation maintains the original text's formatting and line breaks. | For $1<k<2014$, let
$2014=k q+r(0 \leqslant r<k)$.
For $i=1,2, \cdots, k$, let
$S_{i}=\left\{2^{i-1} m \mid q \leqslant m \leqslant 2 q-1\right\}$.
Then $\left|S_{i}\right|=q$, and for any
$1 \leqslant i<j \leqslant k, q \leqslant m_{1}, m_{2} \leqslant 2 q-1$,
we have $2^{i-1} m_{1}=2^{j-1} m_{2} \Leftrightarrow 2^{j-i} m_{2}=m_{1}$.
From $2 m_{2} \leqslant 2^{j-i} m_{2}=m_{1} \leqslant 2 q-1$, we get $m_{2}<q$, which is a contradiction.
Thus, $S_{i} \cap S_{j}=\varnothing$.
Let $S_{0}$ be a set composed of $r$ positive integers, such that for any $1 \leqslant i \leqslant k$, we have $S_{0} \cap S_{i}=\varnothing$.
Let $S=\bigcup_{i=0}^{k} S_{i}$. Then $|S|=k q+r=2014$.
Thus, the set $S$ has a subset $A$, satisfying for any $a, a^{\prime} \in A$, $a \neq a^{\prime}$, we have
$a+a^{\prime} \notin S$ (for example, $S_{1}=\{q, q+1, \cdots, 2 q-1\}$).
Therefore, for any subset $A$ and $1 \leqslant i \leqslant k-1$, we have $\left|S_{i} \cap A\right| \leqslant 2$.
If there exists $i(1 \leqslant i \leqslant k-1)$ such that $\left|S_{i} \cap A\right| \geqslant 3$, let $2^{i-1} m_{1}, 2^{i-1} m_{2}, 2^{i-1} m_{3}$ be three different numbers in $S_{i} \cap A$, where $q \leqslant m_{1}, m_{2}, m_{3} \leqslant 2 q-1$. Then among $m_{1}, m_{2}, m_{3}$, there are two with the same parity (let's say $m_{1}, m_{2}$), so,
$m_{1}+m_{2}=2 m\left(m \in \mathbf{Z}_{+}\right)$, and $q \leqslant m \leqslant 2 q-1$.
Thus, $2^{i-1} m_{1}+2^{i-1} m_{2}=2^{i} m \in S_{i+1}$.
But $S_{i+1} \subseteq S$, which means the sum of the numbers $2^{i-1} m_{1}, 2^{i-1} m_{2}$ in subset $A$ is in set $S$, a contradiction.
Therefore, $\left|S_{i} \cap A\right| \leqslant 2$ for $1 \leqslant i \leqslant k-1$.
From this and $r \leqslant k-1$, we get
$$
\begin{array}{l}
|A| \leqslant\left|S_{0}\right|+\left|S_{k}\right|+2(k-1) \\
=r+q+2(k-1) \\
\leqslant 3 k+\left[\frac{2014}{k}\right]-3
\end{array}
$$
$$
\leqslant 3 k+\frac{2014}{k}-3=3\left(k+\frac{\frac{2014}{3}}{k}\right)-3,
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Taking $k=\left[\sqrt{\frac{2014}{3}}\right]+1=26$, then
$$
|A| \leqslant 3 \times 26+\frac{2014}{26}-3 \text {. }
$$
Thus, $|A| \leqslant 3 \times 26+\left[\frac{2014}{26}\right]-3=152$.
(Wang Xin, Fuzhou No.1 High School, 350001, China) | 152 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 2 Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfying $f(0) \neq 0$, and for any $x, y \in \mathbf{R}$ we have
$$
f\left((x-y)^{2}\right)=f^{2}(x)-2 x f(y)+y^{2} .
$$
Then $f(2012)=$ $\qquad$ | Let $x=y=0$.
Then $f(0)=f^{2}(0) \Rightarrow f(0)=1$ or 0 (discard 0).
Let $y=x$.
$$
\begin{array}{l}
\text { Then } f(0)=f^{2}(x)-2 x f(x)+x^{2}=(f(x)-x)^{2} \\
\Rightarrow f(x)=x \pm 1 .
\end{array}
$$
If there exists $x_{0}$ such that $f\left(x_{0}\right)=x_{0}-1$, let
$$
\begin{array}{l}
x=x_{0}, y=0 \text {. } \\
\text { Then } f\left(x_{0}^{2}\right)=f^{2}\left(x_{0}\right)-2 x_{0} \\
=\left(x_{0}-1\right)^{2}-2 x_{0}=x_{0}^{2}-4 x_{0}+1 \text {. }
\end{array}
$$
Now let $x_{0}=0, y=x_{0}$.
$$
\begin{array}{l}
\text { Then } f\left(x_{0}^{2}\right)=f^{2}(0)+x_{0}^{2}=1+x_{0}^{2} \\
\Rightarrow x_{0}^{2}-4 x_{0}+1=1+x_{0}^{2} \\
\Rightarrow x_{0}=0 \Rightarrow f(0)=-1 .
\end{array}
$$
This contradicts $f(0)=1$,
Therefore, for any $x \in \mathbf{R}$, we have
$$
f(x)=x+1 \text {. }
$$
Thus, $f(2012)=2013$. | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
f(x)=x\left(\sqrt{36-x^{2}}+\sqrt{64-x^{2}}\right)
$$
Find the maximum value of the function. | Algebraic solution Using the Cauchy-Schwarz inequality, we get
$$
\begin{aligned}
f(x) & =x \sqrt{36-x^{2}}+x \sqrt{64-x^{2}} \\
& \leqslant \sqrt{\left(x^{2}+36-x^{2}\right)\left(64-x^{2}+x^{2}\right)}=48 .
\end{aligned}
$$
Geometric solution Construct $\triangle A B C, A D \perp B C$, and let $A B=6$, $A C=8, A D=x$. Then
$$
\begin{array}{l}
2 S_{\triangle A B C}=2 S_{\triangle A D B}+2 S_{\triangle A D C} \\
=x \sqrt{36-x^{2}}+x \sqrt{64-x^{2}} \\
=6 \times 8 \sin \angle B A C .
\end{array}
$$
Obviously, when
$$
\angle B A C=90^{\circ} \Leftrightarrow A B \perp A C \Leftrightarrow B C=10, A D=\frac{24}{5}
$$
$S_{\triangle A B C}$ achieves its maximum value of 48. | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 The function $f$ is defined on the set of ordered pairs of positive integers, and satisfies
$$
\begin{array}{c}
f(x, x)=x, f(x, y)=f(y, x), \\
(x+y) f(x, y)=y f(x, x+y) .
\end{array}
$$
Calculate $f(14,52)$. | Since $f(x, x+y)=\frac{x+y}{y} f(x, y)$, we have,
$$
\begin{aligned}
& f(14,52)=f(14,14+38)=\frac{52}{38} f(14,38) \\
= & \frac{26}{19} f(14,14+24)=\frac{13}{6} f(14,24) \\
= & \frac{13}{6} f(14,14+10)=\frac{26}{5} f(14,10) \\
= & \frac{26}{5} f(10,14)=\frac{26}{5} f(10,10+4) \\
= & \frac{91}{5} f(10,4)=\frac{91}{5} f(4,10)=\frac{91}{3} f(4,6) \\
= & 91 f(4,2)=91 f(2,4)=182 f(2,2) \\
= & 182 \times 2=364 .
\end{aligned}
$$ | 364 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If positive numbers $a, b$ satisfy
$$
2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b) \text {, }
$$
then $\frac{1}{a}+\frac{1}{b}=$ $\qquad$ . | $-, 1.108$
Let $2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)=k$. Then
$$
\begin{array}{l}
a=2^{k-2}, b=3^{k-3}, a+b=6^{k} . \\
\text { Therefore } \frac{1}{a}+\frac{1}{b}=\frac{a+b}{a b}=\frac{6^{k}}{2^{k-2} \times 3^{k-3}} \\
=2^{2} \times 3^{3}=108 .
\end{array}
$$ | 108 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example: Let $f(x)$ be a function defined on $\mathbf{R}$, for any $x, y \in \mathbf{R}$, we have
$$
f(x+3) \leqslant f(x)+3, f(x+2) \geqslant f(x)+2 .
$$
Let $g(x)=f(x)-x$.
(1) Prove: $g(x)$ is a periodic function;
(2) If $f(998)=1002$, find the value of $f(2000)$. | (1) Proof: From $g(x)=f(x)-x$, we get
$$
\begin{array}{l}
g(x+2)=f(x+2)-x-2, \\
g(x+3)=f(x+3)-x-3 .
\end{array}
$$
Substituting into the inequality in the problem, we get
$$
\begin{array}{l}
g(x+2) \geqslant f(x)+2-x-2=f(x)-x, \\
g(x+3) \leqslant f(x)+3-x-3=f(x)-x .
\end{array}
$$
From equation (1), we get
$$
\begin{array}{l}
g(x+4) \geqslant f(x+2)-x-2 \\
\geqslant f(x)+2-x-2=f(x)-x, \\
g(x+6) \geqslant f(x+2)-x-2 \geqslant f(x)-x .
\end{array}
$$
From equation (2), we get
$$
g(x+6) \leqslant f(x+3)-x-3 \leqslant f(x)-x \text {. }
$$
From equations (3) and (4), we know
$$
g(x+6)=f(x)-x=g(x) \text {. }
$$
Therefore, $g(x)$ is a periodic function (6 is one of its periods).
(2) Solution: Note that, $2000-998=1002$ is a multiple of 6, thus,
$$
\begin{array}{l}
g(2000)=g(998) \\
\Rightarrow f(2000)-2000=f(998)-998 \\
\Rightarrow f(2000)=f(998)+1002 \\
=1002+1002=2004 .
\end{array}
$$ | 2004 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{\pi}{6}, a_{n+1}=\arctan \left(\sec a_{n}\right)\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Find the positive integer $m$ such that
$$
\sin a_{1} \cdot \sin a_{2} \cdots \cdot \sin a_{m}=\frac{1}{100} .
$$ | 10. From the problem, we know that for any positive integer $n$,
$$
a_{n+1} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),
$$
and $\tan a_{n+1}=\sec a_{n}$.
Since $\sec a_{n}>0$, then $a_{n+1} \in\left(0, \frac{\pi}{2}\right)$.
From equation (1), we get $\tan ^{2} a_{n+1}=\sec ^{2} a_{n}=1+\tan ^{2} a_{n}$.
Thus, $\tan ^{2} a_{n}=n-1+\tan ^{2} a_{1}$
$$
\begin{array}{l}
=n-1+\frac{1}{3}=\frac{3 n-2}{3} \\
\Rightarrow \tan a_{n}=\sqrt{\frac{3 n-2}{3}} .
\end{array}
$$
Therefore, $\sin a_{1} \cdot \sin a_{2} \cdots \cdot \sin a_{m}$
$=\frac{\tan a_{1}}{\sec a_{1}} \cdot \frac{\tan a_{2}}{\sec a_{2}} \cdots \cdots \cdot \frac{\tan a_{m}}{\sec a_{m}}$
$=\frac{\tan a_{1}}{\tan a_{2}} \cdot \frac{\tan a_{2}}{\tan a_{3}} \cdots \cdot \frac{\tan a_{m}}{\tan a_{m+1}}$ (using equation (1))
$=\frac{\tan a_{1}}{\tan a_{m+1}}=\sqrt{\frac{1}{3 m+1}}$.
From $\sqrt{\frac{1}{3 m+1}}=\frac{1}{100} \Rightarrow m=3333$. | 3333 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Let $f(n)$ be a function defined on $\mathbf{N}$ taking non-negative integer values, and for all $m, n \in \mathbf{N}$ we have
$$
f(m+n)-f(m)-f(n)=0 \text{ or } 1 \text{, }
$$
and $f(2)=0, f(3)>0, f(6000)=2000$.
Find $f(5961)$. | Solve: From $0=f(2) \geqslant 2 f(1) \Rightarrow f(1)=0$;
From $f(3)-f(2)-f(1)=0$ or 1
$$
\Rightarrow 0 \leqslant f(3) \leqslant 1 \text {. }
$$
But $f(3)>0$, hence $f(3)=1$.
By the problem statement, we know
$$
\begin{array}{l}
f(3 n+3)=f(3 n)+3+0 \text { or } 1 \\
\Rightarrow f(3(n+1)) \geqslant f(3 n)+1 .
\end{array}
$$
In the above, let $n=1,2, \cdots, k$, respectively, we get
$$
\begin{array}{l}
f(3 \times 2) \geqslant f(3 \times 1)+1, \\
f(3 \times 3) \geqslant f(3 \times 2)+1, \\
\cdots \cdots \\
f(3 k) \geqslant f(3(k-1))+1 .
\end{array}
$$
Adding all the equations, we get
$$
f(3 k) \geqslant f(3)+(k-1)=k \text {. }
$$
Thus, for all natural numbers $k$,
$$
f(3 k) \geqslant k \text {. }
$$
When $k2000 .
\end{array}
$$
This contradicts the problem statement.
Therefore, $f(3 k)=k$.
Since $1987<2000$, we have
$$
f(5961)=f(3 \times 1987)=1987 .
$$ | 1987 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The function $f(x)(x \neq 1)$ defined on $\mathbf{R}$ satisfies $f(x)+2 f\left(\frac{x+2002}{x-1}\right)=4015-x$. Then $f(2004)=(\quad)$. | Let $x=2, x=2004$, we get
$$
f(2004)=2005 .
$$ | 2005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. People numbered $1,2, \cdots, 2015$ are arranged in a line, and a position-swapping game is played among them, with the rule that each swap can only occur between adjacent individuals. Now, the person numbered 100 and the person numbered 1000 are to swap positions, with the minimum number of swaps required being $\qquad$ times. | $$
-, 1.1799 .
$$
Using the formula, the minimum number of swaps required is
$$
(1000-100) \times 2-1=1799
$$
times.
| 1799 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a b c=-1, a+b+c=4, \\
\frac{a}{a^{2}-3 a-1}+\frac{b}{b^{2}-3 b-1}+\frac{c}{c^{2}-3 c-1}=1 .
\end{array}
$$
Find the value of $a^{2}+b^{2}+c^{2}$. | From the conditions given in the problem, we have
$$
\frac{1}{a}=-b c, \quad a=4-b-c \text {. }
$$
Notice that,
$$
\begin{array}{l}
\frac{a}{a^{2}-3 a-1}=\frac{1}{a-3-\frac{1}{a}}=\frac{1}{b c-b-c+1} \\
=\frac{1}{(b-1)(c-1)} .
\end{array}
$$
Similarly, $\frac{b}{b^{2}-3 b-1}=\frac{1}{(c-1)(a-1)}$,
$\frac{c}{c^{2}-3 c-1}=\frac{1}{(a-1)(b-1)}$.
Therefore, $\frac{a}{a^{2}-3 a-1}+\frac{b}{b^{2}-3 b-1}+\frac{c}{c^{2}-3 c-1}$
$=\frac{1}{(b-1)(c-1)}+\frac{1}{(c-1)(a-1)}+\frac{1}{(b-1)(c-1)}$
$=\frac{a+b+c-3}{(a-1)(b-1)(c-1)}$
$=\frac{1}{(a-1)(b-1)(c-1)}=1$
$\Rightarrow (a-1)(b-1)(c-1)$
$=a b c-(a b+b c+c a)+(a+b+c)-1$
$=1$
$\Rightarrow a b+b c+c a=1$
$\Rightarrow a^{2}+b^{2}+c^{2}$
$=(a+b+c)^{2}-2(a b+b c+c a)=14$. | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Find the maximum value of the positive integer $r$ that satisfies the following condition: for any five 500-element subsets of the set $\{1,2, \cdots, 1000\}$, there exist two subsets that have at least $r$ elements in common. ${ }^{\text {[2] }}$
(2013, Romanian National Team Selection Exam) | 【Analysis】Similarly, map the five subsets of 500 elements each to five vectors in a 1000-dimensional linear space. Since the requirement is the number of elements rather than their parity, we can consider the Euclidean space.
Let $v_{1}, v_{2}, v_{3}, v_{4}, v_{5}$ be the five vectors after transformation.
Notice,
$$
\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}=\sum_{i=1}^{5}\left|v_{i}\right|^{2}+2 \sum_{1<i<j \in s} v_{i} \cdot v_{j} \text {, (1) }
$$
and $\left|v_{i}\right|^{2}=500$.
Therefore, to minimize the maximum value of $v_{i} \cdot v_{j}$, we need $\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}$ to be as small as possible, and the values of $v_{i} \cdot v_{j}$ to be as evenly distributed as possible.
First, find the minimum value of $\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}$.
Let $v_{1}+v_{2}+\cdots+v_{5}=v=\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{1000}\right)$, where $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{1000} \in \mathbf{N}$, and their sum is 2500.
Since $\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}=\sum_{i=1}^{1000} \alpha_{i}^{2}$, the minimum value of the right-hand side is conjectured to be when 500 of $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{1000}$ are 2 and the other 500 are 3.
This can be proven using the adjustment method, or simply by
$$
\left(\alpha_{i}-2\right)\left(\alpha_{i}-3\right) \geqslant 0 \Rightarrow \alpha_{i}^{2} \geqslant 5 \alpha_{i}-6 \text {. }
$$
Summing over $i=1,2, \cdots, 1000$ gives
$$
\sum_{i=1}^{1000} \alpha_{i}^{2} \geqslant 5 \times 2500-6000=6500 \text {. }
$$
Combining equation (1) and $\left|v_{i}\right|^{2}=500$, we get
$$
\sum_{1 \leqslant i<j \leqslant 5} v_{i} \cdot v_{j} \geqslant 2000 \Rightarrow \max _{1 \leqslant i<j \leqslant 5} v_{i} \cdot v_{j} \geqslant 200 \text {. }
$$
Next, consider the construction part. We only need to construct five vectors in a 10-dimensional space, each with five components being 1 and the other five components being 0, and the inner product of any two vectors is 2. Then, by replicating each of these five vectors 100 times, we obtain the required vectors in the 1000-dimensional space.
Using the cyclic construction method (the first five components cycle, and the last five components cycle), it is not difficult to construct the five vectors as:
$$
\begin{array}{l}
\boldsymbol{w}_{1}=(1,1,1,0,0,1,0,1,0,0), \\
w_{2}=(0,1,1,1,0,0,1,0,1,0), \\
w_{3}=(0,0,1,1,1,0,0,1,0,1), \\
w_{4}=(1,0,0,1,1,1,0,0,1,0), \\
w_{5}=(1,1,0,0,1,0,1,0,0,1) .
\end{array}
$$
In summary, the maximum value of $r$ is 200. | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Function
$$
y=\tan 2013 x-\tan 2014 x+\tan 2015 x
$$
The number of zeros of the function in the interval $[0, \pi]$ is $\qquad$. | 2. 2014 .
$$
\begin{aligned}
y & =\tan 2013 x-\tan 2014 x+\tan 2015 x \\
& =\frac{\sin (2013 x+2015 x)}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{\sin 4028 x}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{2 \sin 2014 x \cdot \cos 2014 x}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{\sin 2014 x \cdot(2 \cos 2014 x-\cos 2013 x \cdot \cos 2015 x)}{\cos 2013 x \cdot \cos 2014 x \cdot \cos 2015 x} \\
& =\frac{\sin 2014 x \cdot(1+\cos 4028 x-\cos 2013 x \cdot \cos 2015 x)}{\cos 2013 x \cdot \cos 2014 x \cdot \cos 2015 x} \\
& =\frac{\sin 2014 x \cdot(1-\sin 2013 x \cdot \sin 2015 x)}{\cos 2013 x \cdot \cos 2014 x \cdot \cos 2015 x} .
\end{aligned}
$$
From the domain of $y$, we know that
$$
\sin 2013 x \cdot \sin 2015 x \neq 1 \text{.}
$$
Therefore, the zeros of $y$ are
$$
x=\frac{k \pi}{2014}(k=0,1, \cdots, 2014) \text{.}
$$
However, when $k=1007$, the function is undefined.
Thus, $y$ has 2014 zeros in the interval $[0, \pi]$. | 2014 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $P_{1}$ and $P_{2}$ be two points on a plane, $P_{2 k+1}\left(k \in \mathbf{Z}_{+}\right)$ be the symmetric point of $P_{2 k}$ with respect to $P_{1}$, and $P_{2 k+2}$ be the symmetric point of $P_{2 k+1}$ with respect to $P_{2}$. If $\left|P_{1} P_{2}\right|=1$, then $\left|P_{2013} P_{2014}\right|=$ $\qquad$ . | 4.4024.
From the problem, we know
$$
\begin{array}{l}
\left\{\begin{array}{l}
P_{2 k+1}=2 P_{1}-P_{2 k}, \\
P_{2 k+2}=2 P_{2}-P_{2 k+1}
\end{array}\right. \\
\Rightarrow P_{2 k+2}=2\left(P_{2}-P_{1}\right)+P_{2 k} \\
\Rightarrow\left\{\begin{array}{l}
P_{2 k+2}=2 k\left(P_{2}-P_{1}\right)+P_{2}, \\
P_{2 k+1}=2 k\left(P_{1}-P_{2}\right)+P_{2}
\end{array}\right. \\
\Rightarrow \overrightarrow{P_{2 k+1} P_{2 k+2}}=4 k \overrightarrow{P_{1} P_{2}} \text {. } \\
\end{array}
$$
In particular, $\left|P_{2013} P_{2014}\right|=4024$. | 4024 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=0, a_{2}=1$, and for all $n \geqslant 3, a_{n}$ is the smallest positive integer greater than $a_{n-1}$ such that there is no subsequence of $a_{1}, a_{2}, \cdots, a_{n}$ that forms an arithmetic sequence. Find $a_{2014}$. | 4. First, prove a lemma using mathematical induction.
Lemma A non-negative integer appears in the sequence if and only if its ternary expansion contains only 0 and 1.
Proof It is obvious that the proposition holds for 0.
Assume the proposition holds for all non-negative integers less than \( N \), and consider \( N \).
If the ternary expansion of \( N \) contains the digit 2, replace all occurrences of 2 with 0 to get the number \( N_{0} \); replace all occurrences of 2 with 1 to get the number \( N_{1} \). Thus, the ternary expansions of \( N_{0} \) and \( N_{1} \) do not contain the digit 2.
By the induction hypothesis, the numbers \( N_{0} \) and \( N_{1} \) are in the sequence.
Since \( N_{0} \), \( N_{1} \), and \( N \) form an arithmetic sequence, \( N \) does not appear in the sequence. If the ternary expansion of \( N \) does not contain the digit 2, then we need to prove that \( N \) must appear in the sequence. If not, then there exist terms \( N_{0} \) and \( N_{1} \) in the sequence such that \( N_{0} \), \( N_{1} \), and \( N \) form an arithmetic sequence. Let the common difference be \( d \), and \( 3^{k} \| d \). Then the ternary expansions of these three numbers have the same lowest \( k-1 \) digits, and the \( k \)-th digit is different for each pair, so one of the numbers must have a 2 in the \( k \)-th digit, which contradicts the assumption.
Returning to the original problem.
We only need to find the 2014th non-negative integer whose ternary expansion does not contain the digit 2.
Notice that under this restriction, the ternary carry-over method is equivalent to binary, so we just need to write 2013 (note \( a_{1}=0 \)) in binary \((11111011101)_{2}\) and then convert it to ternary \( a_{2014}=88327 \). | 88327 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Prove: In the prime factorization of the product of any 10 consecutive three-digit numbers, there are at most 23 distinct prime factors. | First, in the prime factorization of each three-digit number, at most two prime factors greater than 10 can appear; otherwise, their product would exceed 1000, which is impossible.
Second, in any sequence of 10 consecutive three-digit numbers, there is one that is a multiple of 10, and in its prime factorization, at most one prime factor greater than 10 can appear. Therefore, at most 19 prime factors greater than 10 can appear. Including $2, 3, 5, 7$, at most 23 distinct prime numbers can appear. | 23 | Number Theory | proof | Yes | Yes | cn_contest | false |
3. In the 100th year of Besmiki's tenure as the President of the Currency Authority, he decided to issue new gold coins. In this year, he put into circulation an unlimited number of gold coins with a face value of $2^{100}-1$ yuan. In the following year, he put into circulation an unlimited number of gold coins with a face value of $2^{101}-1$ yuan. This continued every year until the day when the face value of a newly issued gold coin equaled the sum of the face values of some gold coins issued in previous years, at which point he was dismissed. When did this situation occur in Besmiki's tenure as President? | 3. It happens in the 200th year of Besmiki's presidency.
Assume that the described scenario occurs in the $k$-th year of Besmiki's presidency. Then,
$$
2^{k}-1=a_{1}+a_{2}+\cdots+a_{n}=N-n,
$$
where $N$ is the sum of some powers of 2, all of which are divisible by $2^{100}$.
Since $2^{k}$ is also divisible by $2^{100}$, it follows that $n-1$ is also divisible by $2^{100}$.
Clearly, $n>1$.
Thus, $n \geqslant 2^{100}+1$.
This implies
$$
2^{k}-1 \geqslant\left(2^{100}-1\right)\left(2^{100}+1\right) \geqslant 2^{200}-1 \text {. }
$$
This indicates that $k \geqslant 200$.
Therefore, Besmiki cannot be dismissed from the presidency before the 200th year.
However, from $\left(2^{100}-1\right)\left(2^{100}+1\right)=2^{200}-1$, we know that Besmiki is dismissed in the 200th year. | 200 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let the decimal representation of the positive integer $N$ consist only of the digits 1 and 2. By deleting digits from $N$, one can obtain all 10000 different positive integers formed by 9999 digits 1 and 1 digit 2. Find the minimum possible number of digits in $N$. | 7. The minimum possible number of digits in $N$ is 10198.
For example, $N=\underbrace{1 \cdots 1}_{99 \uparrow} \underbrace{1 \cdots}_{100 \uparrow} 12 \underbrace{1 \cdots}_{100 \uparrow} \underbrace{2}_{98 \uparrow} \underbrace{1 \cdots 1}_{99 \uparrow}$.
For a number formed by 9999 digits 1 and 1 digit 2, if there are $100 m+n (0 \leqslant m, n \leqslant 99)$ digits 1 before the digit 2, then delete the other 2, leaving the $(m+1)$-th digit 2 from the left, and then delete the $99-n$ digits 1 before it and the $n$ digits 1 after it.
The following explains: The number of digits in $N$ cannot be less than 10198.
Obviously, in $N$, there do not need to be two digits 2 next to each other, otherwise one of them can be deleted.
Assume there are $k$ digits 2 in $N$, with $a_{0}$ digits 1 before the first 2, $a_{1}$ digits 1 between the first 2 and the second 2, and so on, with $a_{k}$ digits 1 after the $k$-th 2.
Let $s=a_{0}+a_{1}+\cdots+a_{k}$.
To ensure that the number obtained has only one 1 before the 2, at least $a_{0}-1$ digits 1 need to be deleted, thus, $s-\left(a_{0}-1\right)$ should be no less than 9999, i.e., $s-a_{0} \geqslant 9998$.
To ensure that the number obtained has $a_{0}+1$ digits 1 before the 2, the first 2 must be deleted and at least $a_{1}-1$ digits 1 must be deleted, thus, $s-a_{1} \geqslant 9998$.
To ensure that the number obtained has $a_{0}+a_{1}+1$ digits 1 before the 2, the first two 2s must be deleted and at least $a_{2}-1$ digits 1 must be deleted, thus, $s-a_{2} \geqslant 9998$.
By similar reasoning, for $i=0,1, \cdots, k-1$, we have $s-a_{i} \geqslant 9998$.
And to ensure that the 2 appears in the last position in the number obtained, we need $s-a_{k} \geqslant 9999$.
Adding these inequalities, we get
$$
\begin{array}{l}
(k+1) s-s \geqslant 9998(k+1)+1 \\
\Rightarrow k s>9998(k+1) \\
\Rightarrow s>9998+\frac{9998}{k} .
\end{array}
$$
Since there are also $k$ digits 2 in $N$, the number of digits in $N$ is more than
$$
9998+\frac{9998}{k}+k \geqslant 9998+2 \sqrt{9998}>10197 .
$$ | 10198 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. The integer $n$ that satisfies $\left(1+\frac{1}{n}\right)^{n+1}=\left(1+\frac{1}{2014}\right)^{2014}$ is $=$. | 6. -2015 .
Notice that for any $x \in(-1,+\infty)$ we have
$$
\frac{x}{1+x} \leqslant \ln (1+x) \leqslant x \text {. }
$$
Then for $f(x)=\left(1+\frac{1}{x}\right)^{x+1}(x>0)$ and
$$
g(x)=\left(1+\frac{1}{x}\right)^{x}(x>0)
$$
the derivatives are respectively
$$
\begin{array}{l}
f^{\prime}(x)=\left(1+\frac{1}{x}\right)^{x+1}\left[\ln \left(1+\frac{1}{x}\right)-\frac{1}{x}\right]0 .
\end{array}
$$
Thus, $f(x)$ is decreasing on the interval $(0,+\infty)$, $g(x)$ is increasing on the interval $(0,+\infty)$, and for any $x \in(0,+\infty)$ we have $f(x)>\mathrm{e}>g(x)$.
Therefore, for any $m 、 n \in \mathbf{Z}_{+}$ we have
$$
\left(1+\frac{1}{n}\right)^{n+1}>\mathrm{e}>\left(1+\frac{1}{m}\right)^{m} \text {. }
$$
Thus, the integer $n$ that satisfies $\left(1+\frac{1}{n}\right)^{n+1}=\left(1+\frac{1}{2014}\right)^{2014}$ must be negative.
Let $n=-k\left(k \in \mathbf{Z}_{+}\right)$, substituting into the given equation we get
$$
\left(1+\frac{1}{2014}\right)^{2014}=\left(1-\frac{1}{k}\right)^{-k+1}=\left(1+\frac{1}{k-1}\right)^{k-1} \text {. }
$$
Hence $k-1=2014, n=-k=-2015$. | -2015 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. If $x, y, z > 0$ satisfy
$$
\left\{\begin{array}{l}
\frac{2}{5} \leqslant z \leqslant \min \{x, y\}, \\
x z \geqslant \frac{4}{15}, \\
y z \geqslant \frac{1}{5},
\end{array}\right.
$$ | 7.13.
From the problem, we have
$$
\frac{1}{\sqrt{x}} \leqslant \frac{\sqrt{15 z}}{2}, \frac{1}{z} \leqslant \frac{5}{2}, \frac{1}{\sqrt{y}} \leqslant \sqrt{5 z} \text {. }
$$
Then
$$
f=\frac{2}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}}+\frac{1}{z}\left(1-\frac{z}{x}\right)+
$$
$$
\begin{aligned}
& 2\left[\frac{2}{\sqrt{y}} \cdot \frac{1}{\sqrt{y}}+\frac{1}{z}\left(1-\frac{z}{y}\right)\right] \\
\leqslant & \frac{2}{\sqrt{x}} \cdot \frac{\sqrt{15 z}}{2}+\frac{5}{2}\left(1-\frac{z}{x}\right)+ \\
& 2\left[\frac{2}{\sqrt{y}} \cdot \sqrt{5 z}+\frac{5}{2}\left(1-\frac{z}{y}\right)\right] \\
= & \frac{5}{2}+\sqrt{15} \cdot \sqrt{\frac{z}{x}}-\frac{5}{2} \cdot \frac{z}{x}+ \\
& 2\left[\frac{5}{2}+2 \sqrt{5} \cdot \sqrt{\frac{z}{y}}-\frac{5}{2} \cdot \frac{z}{y}\right] \\
= & 13-\frac{5}{2}\left(\sqrt{\frac{z}{x}}-\sqrt{\frac{3}{5}}\right)^{2}-5\left(\sqrt{\frac{z}{y}}-\frac{2}{\sqrt{5}}\right)^{2}
\end{aligned}
$$
$$
\leqslant 13 \text {. }
$$
When $(x, y, z)=\left(\frac{2}{3}, \frac{1}{2}, \frac{2}{5}\right)$, the equality holds. Therefore, $f_{\max }=13$. | 13 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. Rearrange the six-element array $(1,2,3,4,5,6)$ to
$$
\begin{array}{l}
A=\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right), \\
\text { and } \quad B=\left(b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\right) .
\end{array}
$$
Then the minimum value of $P=\sum_{i=1}^{6} i a_{i} b_{i}$ is | 8. 162 .
From the geometric mean
$$
G=\sqrt[6]{\prod_{i=1}^{6} i a_{i} b_{i}}=\sqrt[6]{(6!)^{3}}=12 \sqrt{5} \in(26,27),
$$
we know that there exists at least one term not less than 27, and at least one term not greater than 25.
Let $i_{1} a_{i_{1}} b_{i_{1}} \leqslant 25, i_{2} a_{i_{2}} b_{i_{2}} \geqslant 27$.
$$
\begin{array}{l}
\text { Then } P \geqslant\left(\sqrt{i_{2} a_{i_{2}} b_{i_{2}}}-\sqrt{i_{1} a_{i_{1}} b_{i_{1}}}\right)^{2}+ \\
2\left(\sqrt{i_{1} a_{i_{1}} b_{i_{1}} i_{2} a_{i_{2}} b_{i_{2}}}+\sqrt{i_{3} a_{i_{3}} b_{i_{3}} i_{4} a_{i_{4}} b_{i_{4}}}+\right. \\
\left.\sqrt{i_{5} a_{i 5} b_{i 5} i_{6} a_{i_{6}} b_{i_{6}}}\right) \\
\geqslant(3 \sqrt{3}-5)^{2}+6 \sqrt[6]{(6!)^{3}} \\
=0.19^{2}+72 \sqrt{5}>161 \text {. } \\
\end{array}
$$
Thus, $P \geqslant 162$, and the equality holds when
$$
A=(4,6,3,2,5,1) \text { and } B=(6,2,3,4,1,5) \text {. }
$$
Therefore, $P_{\min }=162$. | 162 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. The largest positive integer $n$ for which the inequality $\frac{9}{17}<\frac{n}{n+k}<\frac{8}{15}$ holds for a unique integer $k$ is $\qquad$ | 2. 144 .
From the problem, we know that $\frac{7}{8}<\frac{k}{n}<\frac{8}{9}$.
By the uniqueness of $k$, we have
$$
\begin{array}{l}
\frac{k-1}{n} \leqslant \frac{7}{8}, \text { and } \frac{k+1}{n} \geqslant \frac{8}{9} . \\
\text { Therefore, } \frac{2}{n}=\frac{k+1}{n}-\frac{k-1}{n} \geqslant \frac{8}{9}-\frac{7}{8}=\frac{1}{72} \\
\Rightarrow n \leqslant 144 .
\end{array}
$$
When $n=144$, from $\frac{7}{8}<\frac{k}{n}<\frac{8}{9}$, we get
$$
126<k<128 \text {. }
$$
Thus, $k$ can take the unique integer value 127. Therefore, the maximum value of the positive integer $n$ that satisfies the condition is 144. | 144 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Given positive integers $a, b, c$ satisfy
$$
1<a<b<c, a+b+c=111, b^{2}=a c \text {. }
$$
then $b=$ $\qquad$ | 4. 36.
Let $(a, c)=d, a=a_{1} d, c=c_{1} d, a_{1}, c_{1}$ be positive integers, and $\left(a_{1}, c_{1}\right)=1, a_{1}<c_{1}$.
Then $b^{2}=a c=d^{2} a_{1} c_{1} \Rightarrow d^{2}\left|b^{2} \Rightarrow d\right| b$.
Let $b=b_{1} d\left(b_{1} \in \mathbf{Z}_{+}\right)$. Then $b_{1}^{2}=a_{1} c_{1}$.
Since $\left(a_{1}, c_{1}\right)=1$, $a_{1}, c_{1}$ are both perfect squares.
Let $a_{1}=m^{2}, c_{1}=n^{2}$. Then
$$
b_{1}=m n\left(m, n \in \mathbf{Z}_{+} \text {, and }(m, n)=1, m<n\right) \text {. }
$$
Also, $a+b+c=111$, so
$$
\begin{array}{l}
d\left(a_{1}+b_{1}+c_{1}\right)=111 \\
\Rightarrow d\left(m^{2}+n^{2}+m n\right)=111 .
\end{array}
$$
Notice that,
$$
m^{2}+n^{2}+m n \geqslant 1^{2}+2^{2}+1 \times 2=7 \text {. }
$$
Therefore, $d=1$ or 3.
If $d=1$, then $m^{2}+n^{2}+m n=111$.
By trial, only $m=1, n=10$ satisfy the equation, but in this case, $a=1$ (discard).
If $d=3$, then $m^{2}+n^{2}+m n=37$.
By trial, only $m=3, n=4$ satisfy the equation, in this case, $a=27, b=36, c=48$, which meets the requirements.
Therefore, $b=36$. | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Let $n$ be an integer. If there exist integers $x, y, z$ satisfying
$$
n=x^{3}+y^{3}+z^{3}-3 x y z \text {, }
$$
then $n$ is said to have property $P$.
(1) Determine whether $1, 2, 3$ have property $P$;
(2) Among the 2014 consecutive integers $1, 2, \cdots, 2014$, how many do not have property $P$? | (1) Let $x=1, y=z=0$, we get
$$
1=1^{3}+0^{3}+0^{3}-3 \times 1 \times 0 \times 0 \text {. }
$$
Thus, 1 has property $P$.
Let $x=y=1, z=0$, we get
$$
2=1^{3}+1^{3}+0^{3}-3 \times 1 \times 1 \times 0 \text {. }
$$
Thus, 2 has property $P$.
If 3 has property $P$, then there exist integers $x, y, z$ such that
$$
\begin{array}{l}
3=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx) \\
\Rightarrow 3\left|(x+y+z)^{3} \Rightarrow 3\right|(x+y+z) \\
\Rightarrow 9 \mid\left[(x+y+z)^{3}-3(x+y+z)(xy+yz+zx)\right] \\
\Rightarrow 9 \mid 3,
\end{array}
$$
which is impossible.
Therefore, 3 does not have property $P$.
(2) Let $f(x, y, z)=x^{3}+y^{3}+z^{3}-3xyz$. Then
$$
\begin{array}{l}
f(x, y, z)=(x+y)^{3}+z^{3}-3xy(x+y)-3xyz \\
=(x+y+z)^{3}-3z(x+y)(x+y+z)- \\
3xy(x+y+z) \\
=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx) \\
= \frac{1}{2}(x+y+z)\left(x^{2}+y^{2}+z^{2}-xy-yz-zx\right) \\
= \frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right],
\end{array}
$$
i.e.,
$$
\begin{array}{l}
f(x, y, z) \\
=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] .
\end{array}
$$
Assume without loss of generality that $x \geqslant y \geqslant z$.
If $x-y=1, y-z=0, x-z=1$, i.e.,
$$
x=z+1, y=z \text {, }
$$
then $f(x, y, z)=3z+1$;
If $x-y=0, y-z=1, x-z=1$, i.e.,
$$
x=y=z+1 \text {, }
$$
then $f(x, y, z)=3z+2$;
If $x-y=1, y-z=1, x-z=2$, i.e., $x=z+2, y=z+1$,
then $f(x, y, z)=9(z+1)$.
Thus, numbers of the form $3k+1$ or $3k+2$ or $9k$ $(k \in \mathbf{Z})$ all have property $P$.
Note that,
$$
\begin{array}{l}
f(x, y, z) \\
=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx) .
\end{array}
$$
If $3 \mid f(x, y, z)$, then
$3\left|(x+y+z)^{3} \Rightarrow 3\right|(x+y+z)$
$\Rightarrow 9 \mid f(x, y, z)$.
In summary, integers $n$ do not have property $P$ if and only if $n=9k+3$ or $n=9k+6$ $(k \in \mathbf{Z})$.
Since $2014=9 \times 223+7$, in the 2014 consecutive integers from 1 to 2014, the number of integers that do not have property $P$ is $224 \times 2=448$. | 448 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If $n$ is a positive integer, then
$$
\sum_{n=1}^{2014}\left(\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]\right)=
$$
$\qquad$ | 7. 2027091.
Let $f(n)=\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]$. Then
$$
\begin{array}{l}
f(0)=0, f(1)=0, f(2)=1, \\
f(3)=2, f(4)=3, f(5)=3 .
\end{array}
$$
For a positive integer $k$, we have
$$
\begin{array}{l}
f(6 k)=\left[\frac{6 k}{2}\right]+\left[\frac{6 k}{3}\right]+\left[\frac{6 k}{6}\right] \\
=3 k+2 k+k=6 k \text {, } \\
f(6 k+1) \\
=\left[\frac{6 k+1}{2}\right]+\left[\frac{6 k+1}{3}\right]+\left[\frac{6 k+1}{6}\right] \\
=3 k+2 k+k=6 k \text {, } \\
f(6 k+2) \\
=\left[\frac{6 k+2}{2}\right]+\left[\frac{6 k+2}{3}\right]+\left[\frac{6 k+2}{6}\right] \\
=(3 k+1)+2 k+k=6 k+1 \text {, } \\
f(6 k+3) \\
=\left[\frac{6 k+3}{2}\right]+\left[\frac{6 k+3}{3}\right]+\left[\frac{6 k+3}{6}\right] \\
=(3 k+1)+(2 k+1)+k=6 k+2 \text {, } \\
f(6 k+4) \\
=\left[\frac{6 k+4}{2}\right]+\left[\frac{6 k+4}{3}\right]+\left[\frac{6 k+4}{6}\right] \\
=(3 k+2)+(2 k+1)+k=6 k+3 \text {, } \\
f(6 k+5) \\
=\left[\frac{6 k+5}{2}\right]+\left[\frac{6 k+5}{3}\right]+\left[\frac{6 k+5}{6}\right] \\
=(3 k+2)+(2 k+1)+k=6 k+3 \text {. } \\
\text { Hence } \sum_{n=1}^{2014}\left(\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]\right) \\
=\sum_{n=1}^{2014} f(n)=\sum_{n=0}^{2014} f(n) \\
=\sum_{k=0}^{335} \sum_{i=0}^{5} f(6 k+i)-f(2015) \\
=\sum_{k=0}^{335}(36 k+9)-f(6 \times 335+5) \\
=\frac{(9+12069) \times 336}{2}-(6 \times 335+3) \\
=2027091 . \\
\end{array}
$$ | 2027091 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. As shown in Figure 1, under the rules of Chinese chess, the "pawn" at point $A$ can reach point $B$ through a certain path (before crossing the river, the pawn can only move to the adjacent intersection directly in front of it each step; after crossing the river, it can move to the adjacent intersections in front, to the left, or to the right each step, but it cannot move backward. The "river" refers to the area between the 5th and 6th horizontal lines on the chessboard in Figure 1). During the pawn's movement, if each intersection on the chessboard is not visited more than once by the pawn, such a path is called a "non-repeating path." Therefore, the number of different non-repeating paths is $\qquad$. | 7. 6561.
Assume the chessboard has 10 horizontal lines from bottom to top, sequentially labeled as the 1st, 2nd, ..., 10th rows, and 9 vertical lines from left to right, sequentially labeled as the 1st, 2nd, ..., 9th columns. For example, point $A$ is located at the 4th row and 5th column.
Note that, during the movement of the pawn from point $A$ to $B$, it cannot move downward. Therefore, the step from the $i$-th row to the $(i+1)$-th row is unique. If the starting and ending points of this step are in the $j$-th column, we denote $a_{i}=j(4 \leqslant i \leqslant 9,1 \leqslant j \leqslant 9)$.
According to the rules, it is easy to see that
$$
a_{4}=a_{5}=5, a_{6} 、 a_{7} 、 a_{8} 、 a_{9} \in\{1,2, \cdots, 9\} \text {. }
$$
There are $1^{2} \times 9^{4}=6561$ such ordered arrays $\left(a_{4}, a_{5}, \cdots, a_{9}\right)$, and each array corresponds one-to-one to a unique non-repeating path from point $A$ to $B$. This is because, after the pawn reaches the $i$-th row ($i=4,5, \cdots, 9$), there is exactly one way to move to the $a_{i}$-th column in the $i$-th row (otherwise, the path would have a repeat), and then move forward to the $(i+1)$-th row. Therefore, the total number of different non-repeating paths is 6561. | 6561 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $n+2$ real numbers
$$
a_{1}, a_{2}, \cdots, a_{n}, 16, a_{n+2} \text {, }
$$
where the average of the first $n$ numbers is 8, the average of the first $n+1$ numbers is 9, and the average of these $n+2$ numbers is 10. Then the value of $a_{n+2}$ is $\qquad$ | 2. 18 .
From the condition, $\frac{8 n+16}{n+1}=9 \Rightarrow n=7$.
Also, $\frac{8 \times 7+16+a_{n+2}}{7+1+1}=10 \Rightarrow a_{n+2}=18$. | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. If $a \in A$, and $a-1 \notin A, a+1 \notin A$, then $a$ is called an isolated element of set $A$. Therefore, the number of four-element subsets of set $M=\{1,2, \cdots, 9\}$ without isolated elements is $\qquad$ . | 9. 21 .
Consider the smallest element $i$ and the largest element $j$ in a set that satisfies the condition.
Let this set be $A$. Then $i+1 \in A, j-1 \in A$ (otherwise, $i$ or $j$ would be an isolated element).
Thus, $A=\{i, i+1, j-1, j\}$.
And $2 \leqslant i+1 < j-1 \leqslant 8$, so the number of ways to choose $i+1, j-1$ is $\mathrm{C}_{7}^{2}=21$. Therefore, the number of four-element subsets of the set $M=\{1,2, \cdots, 9\}$ without isolated elements is 21. | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Divide the sequence of positive integers $1,2, \cdots$ from left to right into segments such that the first segment has $1 \times 2$ numbers, the second segment has $2 \times 3$ numbers, $\cdots$, the $n$-th segment has $n \times(n+1)$ numbers, $\cdots$. Then 2014 is in the $\qquad$ segment. | $-, 1.18$.
$$
\begin{array}{l}
\text { Let } S_{n}=1 \times 2+2 \times 3+\cdots+n(n+1) \\
=\left(1^{2}+2^{2}+\cdots+n^{2}\right)+(1+2+\cdots+n) \\
=\frac{n(n+1)(n+2)}{3} .
\end{array}
$$
If 2014 is in the $(n+1)$-th segment, since there are $S_{n}$ numbers before this segment, then $S_{n}<2014 \leqslant S_{n+1}$.
Since $S_{17}=1938, S_{18}=2280$, and
$$
S_{17}<2014 \leqslant S_{18},
$$
Therefore, 2014 is in the 18th segment. | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. If the fraction $\frac{p}{q}\left(p, q \in \mathbf{Z}_{+}\right)$ is converted to a decimal as
$$
\frac{p}{q}=0.198 \cdots,
$$
then when $q$ takes the minimum value, $p+q=$ . $\qquad$ | 6. 121 .
Given $\frac{p}{q}=0.198 \cdots5 p$.
Let $q=5 p+m\left(m \in \mathbf{Z}_{+}\right)$. Then
$$
\begin{array}{l}
\frac{p}{5 p+m}=0.198 \cdots \\
\Rightarrow 0.198(5 p+m)<p<0.199(5 p+m) \\
\Rightarrow 19.8 m<p<39.8 m .
\end{array}
$$
When $m=1$, $20 \leqslant p \leqslant 39$, taking $p=20, m=1$, $q$ is minimized to 101.
Upon verification, $\frac{20}{101}=0.19801980 \cdots$ meets the requirement. Therefore, when $q$ is minimized, $p+q=121$. | 121 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. $A=\left[\frac{8}{9}\right]+\left[\frac{8^{2}}{9}\right]+\cdots+\left[\frac{8^{0114}}{9}\right]$ when divided by 63 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 9. 56 .
Notice that, for any positive integer $k$, $\frac{8^{2 k-1}}{9}$ and $\frac{8^{2 k}}{9}$ are not integers, and
$$
\frac{8^{2 k-1}}{9}+\frac{8^{2 k}}{9}=8^{2 k-1} \text {. }
$$
Therefore, for any positive integer $k$, we have
$$
\begin{array}{l}
{\left[\frac{8^{2 k-1}}{9}\right]+\left[\frac{8^{2 k}}{9}\right]=\frac{8^{2 k-1}}{9}+\frac{8^{2 k}}{9}-1} \\
=8^{2 k-1}-1 \equiv 7(\bmod 63) . \\
\text { Hence } A=\left[\frac{8}{9}\right]+\left[\frac{8^{2}}{9}\right]+\cdots+\left[\frac{8^{2014}}{9}\right] \\
\equiv 1007 \times 7 \equiv 56(\bmod 63) .
\end{array}
$$ | 56 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For any positive integer $n$, the function $f(n)$ is the sum of the digits (i.e., the digital sum) of $n^{2}+3 n+1$ in decimal notation. Question: Does there exist an integer $n$ such that
$$
f(n)=2013 \text { or } 2014 \text { or } 2015 \text { ? }
$$ | When $3 \mid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 1(\bmod 3) ;
$$
When $3 \nmid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 2(\bmod 3) \text {. }
$$
Thus, $3 \nmid f(n)$.
Since 312013, there does not exist an integer $n$ such that
$$
f(n)=2013 \text {. }
$$
If there exists an integer $n$ such that $f(n)=2014$, then by $2014 \equiv 1(\bmod 3) \Rightarrow 3 \mid n$. Hence $n^{2}+3 n+1 \equiv 1 \equiv f(n)(\bmod 9)$. But $f(n) \equiv 7(\bmod 9)$, which is a contradiction.
Therefore, there does not exist an integer $n$ such that $f(n)=2014$.
When $k>2, n=10^{k}-3$,
$$
\begin{array}{l}
n^{2}+3 n+1=\left(10^{k}-3\right)^{2}+3\left(10^{k}-3\right)+1 \\
=\underset{k-1 \uparrow}{99 \cdots 9} 7 \underset{k-1 \uparrow}{00 \cdots 01} .
\end{array}
$$
Since $2015=9 \times 223+8$, when $k=224$, i.e., $n=10^{224}-3$, $f(n)=2015$.
(Song Hongjun, Zhejiang Fuyang High School, 311400) | 2015 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A. The minimum value of the algebraic expression $\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}+9}$ is | $=、 6$. A. 13.
This problem can be transformed into finding the minimum value of the sum of distances from a point $(x, 0)$ on the $x$-axis to the points $(0,2)$ and $(12,3)$ in a Cartesian coordinate system.
Since the symmetric point of $(0,2)$ with respect to the $x$-axis is $(0,-2)$, the length of the line segment connecting $(0,-2)$ and $(12,3)$ is the minimum value of the expression $\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}+9}$.
In fact, when $x=\frac{24}{5}$, the expression reaches its minimum value
$$
\sqrt{(12-0)^{2}+(3+2)^{2}}=13 \text {. }
$$ | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Let real numbers $x_{1}, x_{2}, \cdots, x_{2014}$ satisfy
$$
\left|x_{1}\right|=99,\left|x_{n}\right|=\left|x_{n-1}+1\right| \text {, }
$$
where, $n=2,3, \cdots, 2014$. Find the minimum value of $x_{1}+x_{2}+\cdots+x_{2014}$. | 11. From the given, we have
$$
x_{n}^{2}=x_{n-1}^{2}+2 x_{n}+1(n=2,3, \cdots, 2014) \text {. }
$$
Adding the above 2013 equations, we get
$$
\begin{array}{l}
x_{2014}^{2}=x_{1}^{2}+2\left(x_{1}+x_{2}+\cdots+x_{2013}\right)+2013 \\
\Rightarrow 2\left(x_{1}+x_{2}+\cdots+x_{2014}\right) \\
\quad=x_{2014}^{2}+2 x_{2014}-2013-x_{1}^{2} \\
\quad=\left(x_{2014}+1\right)^{2}-2014-99^{2} .
\end{array}
$$
Since $x_{1}$ is odd, from the given conditions, we know that $x_{2}$ is even, $x_{3}$ is odd, $\cdots, x_{2014}$ is even.
Thus, $\left(x_{2014}+1\right)^{2} \geqslant 1$.
Therefore, $x_{1}+x_{2}+\cdots+x_{2014}$
$$
\geqslant \frac{1}{2}\left(1-2014-99^{2}\right)=-5907 \text {. }
$$
When $x_{1}=-99, x_{2}=-98, \cdots, x_{99}=-1, x_{100}$
$$
=0, x_{101}=-1, x_{102}=0, x_{103}=-1, \cdots, x_{2013}=
$$
$-1, x_{2014}=0$, the equality holds.
Hence, the minimum value of $x_{1}+x_{2}+\cdots+x_{2014}$ is -5907. | -5907 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Consider each permutation of $1,2, \cdots, 8$ as an eight-digit number. Then the number of eight-digit numbers that are multiples of 11 is $\qquad$ | 8. 4608 .
For each such eight-digit number $\overline{a_{1} a_{2} \cdots a_{8}}$, let
$$
A=\left\{a_{1}, a_{3}, a_{5}, a_{7}\right\}, B=\left\{a_{2}, a_{4}, a_{6}, a_{8}\right\} .
$$
Let $S(A)$ and $S(B)$ denote their digit sums, and assume $S(A) \geqslant S(B)$.
Then $S(A)+S(B)=36$.
Thus, $S(A)$ and $S(B)$ have the same parity, and $S(A)+S(B)$ and $S(A)-S(B)$ have the same parity.
Therefore, $S(A)-S(B)$ is an even number and a multiple of 11. If $S(A)-S(B)=22$, then $S(B)=7$, which is impossible (since the sum of the smallest four numbers is at least 10).
Thus, $S(A)-S(B)=0$, i.e.,
$S(A)=S(B)=18$.
Hence, $S(A)=S(B)$ has only four cases:
1278, 3456; 1368, 2457;
1467, 2358; 2367, 1458.
In each group, the four numbers can be fully permuted, resulting in $4! \times 4!$ eight-digit numbers.
Considering the two scenarios for the four numbers in odd or even positions, the total number of eight-digit numbers that satisfy the condition is
$$
4 \times 2 \times 4! \times 4! = 4608 \text{ (numbers). }
$$ | 4608 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given natural numbers $a, b, c$ whose sum is $S$, satisfying $a+b=1014, c-b=497, a>b$. Then the maximum value of $S$ is ( ).
(A) 2014
(B) 2015
(C) 2016
(D) 2017 | -1. D.
From the given, we have $a \geqslant b+1$.
Then $1014=a+b \geqslant 2 b+1$
$$
\begin{array}{l}
\Rightarrow b \leqslant 506.5 \Rightarrow b \leqslant 506 . \\
\text { Also } S=(a+b)+(c-b)+b \\
=1014+497+b=1511+b \\
\leqslant 1511+506=2017,
\end{array}
$$
Therefore, the maximum value of $S$ is 2017. | 2017 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given a set of data consisting of seven positive integers, the only mode is 6, and the median is 4. Then the minimum value of the sum of these seven positive integers is $\qquad$ | 2, 1.26.
Arrange these seven positive integers in ascending order, it is clear that the fourth number is 4. If 6 appears twice, then the other four numbers are $1, 2, 3, 5$, their sum is smaller, being 27; if 6 appears three times, the other three numbers are $1, 1, 2$, their sum is 26.
Therefore, the minimum sum of the seven positive integers is 26. | 26 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Given the sequence $\left\{a_{n}\right\}$ satisfies
$a_{1}=1, a_{2 n}=\left\{\begin{array}{ll}a_{n}, & n \text { is even; } \\ 2 a_{n}, & n \text { is odd, }\end{array}\right.$
$a_{2 n+1}=\left\{\begin{array}{ll}2 a_{n}+1, & n \text { is even; } \\ a_{n}, & n \text { is odd. }\end{array}\right.$
Find the number of positive integers $n$ in the range $[1,2014]$ that satisfy $a_{n}=a_{2014}$. | Solve: From the given, we have
$$
\begin{array}{l}
a_{2014}=2 a_{1007}=2 a_{503}=2 a_{251}=2 a_{125} \\
=2\left(2 a_{62}+1\right)=2\left(4 a_{31}+1\right) \\
=2\left(4 a_{15}+1\right)=2\left(4 a_{7}+1\right) \\
=2\left(4 a_{3}+1\right)=2\left(4 a_{1}+1\right)=10 .
\end{array}
$$
In fact, $a_{n}$ can be obtained through the following process.
Write $n$ in binary, then replace consecutive "1"s with a single "1" and consecutive "0"s with a single "0". Denote this sequence as $\left\{b_{n}\right\}$.
For example, $2014=(11111011110)_{2}$,
$b_{2014}=(1010)_{2}=10$.
We will prove by mathematical induction that:
$$
a_{n}=b_{n}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Clearly, $a_{1}=1=b_{1}, a_{2}=2=b_{2}, a_{3}=1=b_{3}$. Assume that $a_{t}=b_{t}$ holds for any positive integer $t<n$.
(1) When $n=2 k$, if $k$ is even, then $a_{2 k}=$ $a_{k}$, and the last two digits of $n=2 k$ in binary are 00, so $b_{2 k}$ $=b_{k}$, thus, $a_{2 k}=b_{2 k}$; if $k$ is odd, then $a_{2 k}=2 a_{k}$, and the last two digits of $n=2 k$ in binary are 10, so $b_{2 k}=$ $2 b_{k}$, thus, $a_{2 k}=b_{2 k}$.
(2) When $n=2 k+1$, if $k$ is even, then $a_{2 k+1}=2 a_{k}+1$, and the last two digits of $n=2 k$ in binary are 01, so $b_{2 k+1}=2 b_{k}+1$, thus, $a_{2 k+1}=b_{2 k+1}$; if $k$ is odd, then $a_{2 k+1}=a_{k}$, and the last two digits of $n=2 k$ in binary are 11, so $b_{2 k+1}=b_{k}$, thus, $a_{2 k+1}=$ $b_{2 k+1}$.
Therefore, the original problem is equivalent to
$a_{n}=b_{n}=b_{2014}=(1010)_{2}$
$=10$ the number of positive integers $n$
$\Leftrightarrow$ the number of $n$ such that when written in binary, it becomes $(1010)_{2}$ according to the rules.
Since $n<2048=2^{11}$, i.e., $n$ in binary has at most 11 digits, and numbers larger than
$$
2014=(11111011110)_{2}
$$
do not meet the requirements.
This can be understood as selecting 4 positions out of 11 to fill (1010), with 0s before the first 1 and the remaining spaces filled with the digit before the space.
The $n$ that do not meet the requirements can be understood as selecting 3 positions out of the last 5 to fill (010), with 1s before the first 0 and the remaining spaces filled with the digit before the space.
Therefore, there are $\mathrm{C}_{11}^{4}-\mathrm{C}_{5}^{3}=320$ positive integers $n$ that meet the requirements. | 320 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that Figure 1 is the graph of an even function $f(x)$, and Figure 2 is the graph of an odd function $g(x)$.
Let the number of real roots of the equations $f(f(x))=0, f(g(x))=0$,
$$
g(g(x))=0, g(f(x))=0
$$
be $a, b, c, d$ respectively. Then
$$
a+b+c+d=
$$
$\qquad$ | 3. 30 .
From the graph, we know that the range of the function $y=f(x)$ is $[-1,1]$, and the range of $y=g(x)$ is $[-2,2]$.
Notice that, the roots of the equation $f(x)=0$ are $0, x_{1}, x_{2}$, with $\left|x_{1}\right|=\left|x_{2}\right| \in(1,2)$;
The roots of the equation $g(x)=0$ are $0, x_{3}, x_{4}$, with $\left|x_{3}\right|=\left|x_{4}\right| \in(0,1)$.
Therefore, the roots of the equation $f(f(x))=0$ are the roots of the equations $f(x)=0$ or $f(x)=x_{1}$ or $f(x)=x_{2}$.
Since the range of the function $y=f(x)$ is $[-1,1]$, and $\left|x_{1}\right|=\left|x_{2}\right| \in(1,2)$, the equations $f(x)=x_{1}$ and $f(x)=x_{2}$ have no roots, while the equation $f(x)=0$ has three roots. Thus, the number of roots of the equation $f(f(x))=0$ is 3.
The roots of the equation $f(g(x))=0$ are the roots of the equations $g(x)=0$ or $g(x)=x_{1}$ or $g(x)=x_{2}$.
Since the range of the function $y=g(x)$ is $[-2,2]$, and $\left|x_{1}\right|=\left|x_{2}\right| \in(1,2)$, the equations $g(x)=x_{1}$ and $g(x)=x_{2}$ each have three roots, and the equation $g(x)=0$ also has three roots. Thus, the number of roots of the equation $f(g(x))=0$ is 9.
Similarly, the equations $g(g(x))=0$ and $g(f(x))=0$ each have 9 roots.
Therefore, $a+b+c+d=3+9+9+9=30$. | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the set $P=\{1,2, \cdots, 2014\}, A \cong P$. If any two numbers in set $A$ have a difference that is not a multiple of 99, and the sum of any two numbers is also not a multiple of 99, then the set $A$ can contain at most $\qquad$ elements. | 5.50.
Let the set
$$
B_{i}=\{99 \times 1+i, 99 \times 2+i, \cdots, 99 \times 20+i\} \text {, }
$$
where, $i=0,1, \cdots, 34$;
$$
B_{j}=\{99 \times 1+j, 99 \times 2+j, \cdots, 99 \times 19+j\},
$$
where, $j=35,36, \cdots, 98$.
Take any $a, b \in A$.
Since the difference between any two numbers in set $A$ is not a multiple of 99, $a$ and $b$ do not belong to the same set $B_{i}$ or $B_{j}$.
Furthermore, since the sum of any two numbers in set $A$ is also not a multiple of 99, $a$ and $b$ are not a pair in the set
$$
M=\left\{\left(B_{1}, B_{98}\right),\left(B_{2}, B_{97}\right), \cdots,\left(B_{49}, B_{50}\right)\right\}
$$
Therefore, the set $A$ can have at most 50 elements, such as taking one element from each of the sets $B_{0}, B_{1}, \cdots, B_{49}$. | 50 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $0<x<\frac{\pi}{2}, \sin x-\cos x=\frac{\pi}{4}$. If $\tan x+\frac{1}{\tan x}$ can be expressed in the form $\frac{a}{b-\pi^{c}}$ ($a$, $b$, $c$ are positive integers), then $a+b+c=$ $\qquad$ . | 8. 50 .
Squaring both sides of $\sin x-\cos x=\frac{\pi}{4}$ and rearranging yields $\sin x \cdot \cos x=\frac{16-\pi^{2}}{32}$.
Therefore, $\tan x+\frac{1}{\tan x}=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$
$$
=\frac{1}{\sin x \cdot \cos x}=\frac{32}{16-\pi^{2}} \text {. }
$$
Thus, $a=32, b=16, c=2$.
Hence, $a+b+c=50$. | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Let the positive integer $n$ satisfy $31 \mid\left(5^{n}+n\right)$. Then the minimum value of $n$ is $\qquad$ . | 9. 30 .
Given $5^{3} \equiv 1(\bmod 31)$, when $n=3 k\left(k \in \mathbf{Z}_{+}\right)$, $5^{n}+n \equiv 1+n \equiv 0(\bmod 31)$, at this time, the minimum value of $n$ is $n_{\text {min }}=30$;
When $n=3 k+1\left(k \in \mathbf{Z}_{+}\right)$,
$$
5^{n}+n \equiv 5+n \equiv 0(\bmod 31),
$$
at this time, the minimum value of $n$ is $n_{\text {min }}=88$;
When $n=3 k+2\left(k \in \mathbf{Z}_{+}\right)$,
$$
5^{n}+n \equiv 25+n \equiv 0(\bmod 31),
$$
at this time, the minimum value of $n$ is $n_{\text {min }}=68$.
Therefore, the minimum value of $n$ that satisfies the condition is $n_{\min }=30$. | 30 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Given
$$
S_{n}=|n-1|+2|n-2|+\cdots+10|n-10| \text {, }
$$
where, $n \in \mathbf{Z}_{+}$. Then the minimum value of $S_{n}$ is $\qquad$ | 10. 112 .
From the problem, we know
$$
\begin{array}{l}
S_{n+1}-S_{n} \\
=|n|+2|n-1|+\cdots+10|n-9|- \\
{[|n-1|+2|n-2|+\cdots+10|n-10|] } \\
=|n|+|n-1|+\cdots+|n-9|-10|n-10| .
\end{array}
$$
When $n \geqslant 10$, $S_{n+1}-S_{n}>0$, thus, $S_{n}$ is monotonically increasing; $\square$
When $n=0$, $S_{1}-S_{0}>0$;
When $n=6$, $S_{7}-S_{6}<0$.
Therefore, $S_{n}$ is monotonically decreasing in the interval $[1,7]$ and monotonically increasing in the interval $[7,+\infty)$.
Hence, the minimum value of $S_{n}$ is $S_{7}=112$. | 112 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $15 \mid \overline{\text { xaxax }}$, then the sum of all five-digit numbers $\overline{x a x a x}$ that satisfy the requirement is ( ).
(A) 50505
(B) 59595
(C) 110100
(D) 220200 | - 1. D.
Notice that, $15| \overline{\text { xaxax }} \Leftrightarrow\left\{\begin{array}{l}5 \mid \overline{\text { xaxax }}, \\ 3 \mid \overline{\text { xaxax }} .\end{array}\right.$
From $51 \overline{\text { xaxax }} \Rightarrow x=5$;
From $31 \overline{x a x a x} \Rightarrow 31(x+a+x+a+x)$
$$
\Rightarrow 3|2 a \Rightarrow 3| a \Rightarrow a=0,3,6,9 \text {. }
$$
Thus, the five-digit numbers that meet the requirements are
$$
50505,53535,56565,59595 \text {, }
$$
their sum is 220200. | 220200 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
1. Let $a_{1}, a_{2}, \cdots, a_{2015}$ be a sequence of numbers taking values from $-1, 0, 1$, satisfying
$$
\sum_{i=1}^{2015} a_{i}=5 \text {, and } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040,
$$
where $\sum_{i=1}^{n} a_{i}$ denotes the sum of $a_{1}, a_{2}, \cdots, a_{n}$.
Then the number of 1's in this sequence is $\qquad$ | $=, 1.510$.
Let the number of -1's be $x$, and the number of 0's be $y$.
Then, from $\sum_{i=1}^{2015} a_{i}=5$, we know the number of 1's is $x+5$. Combining this with
$$
\begin{array}{c}
\text { the equation } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040, \text { we get } \\
\left\{\begin{array}{l}
x+(x+5)+y=2015, \\
4(x+5)+y=3040 .
\end{array}\right.
\end{array}
$$
Solving these equations, we get $x=505$.
Therefore, the number of 1's is 510. | 510 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If the integer $n$
satisfies
$$
(n+1)^{2} \mid\left(n^{2015}+1\right) \text {, }
$$
then the minimum value of $n$ is $\qquad$ | 3. -2016 .
From $n^{2015}+1=(n+1) \sum_{i=0}^{2014}(-1)^{i} n^{2014-i}$, we know
$$
\begin{array}{l}
(n+1)^{2} \mid\left(n^{2015}+1\right) \\
\left.\Leftrightarrow(n+1)\right|_{i=0} ^{2014}(-1)^{i} n^{2014-i} \\
\Leftrightarrow(n+1) \mid \sum_{i=0}^{2014}(-1)^{i}(-1)^{2014-i} \\
\Leftrightarrow(n+1) \mid \sum_{i=0}^{2014}(-1)^{2014} \\
\Leftrightarrow(n+1) \mid 2015 .
\end{array}
$$
Therefore, the minimum value of $n$ is -2016. | -2016 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. To color the eight vertices of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with four different colors, such that the two endpoints of the same edge have different colors, there are a total of coloring methods. | 8. 2652 .
First, color the four points $A, B, C, D$ above, with 84 coloring methods. Then consider the four points below, using the principle of inclusion-exclusion, the total number of methods is
$$
\begin{array}{l}
84\left\{84-C_{4}^{1}[3 \times(3+2 \times 2)+\right. \\
(3+2 \times 2)]+2\left(\frac{36}{84} \times 9+\frac{48}{84} \times 4\right)- \\
\left.C_{4}^{3}\left(\frac{36}{84} \times 3+\frac{48}{84} \times 2\right)+C_{4}^{4}\right\} \\
=2652 .
\end{array}
$$ | 2652 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? ${ }^{[5]}$
$(2012$, Girls' Mathematical Olympiad) | 【Analysis】Perform prime factorization $2012=2^{2} \times 503$.
First consider 503 I $\mathrm{C}_{2012}^{k}$.
By Corollary 3, we know that when and only when $k$ and $2012-k$ do not produce a carry when added in base 503, $\left(503, \mathrm{C}_{2012}^{k}\right)=1$.
Writing 2012 in base 503 gives $2012=(40)_{503}$.
Let $k=\left(a_{1} a_{2}\right)_{503}$.
No carry must have $a_{2}=0, a_{1} \in\{0,1,2,3,4\}$.
Converting to decimal, we know that when and only when
$k \in\{0,503,1006,1509,2012\}$
then $503 \nmid \mathrm{C}_{2012}^{k}$.
Next, consider the opposite.
Determine $k$ such that $v_{2}\left(\mathrm{C}_{2012}^{k}\right) \leqslant 1$.
There are two cases.
(1) $v_{2}\left(\mathrm{C}_{2012}^{k}\right)=0$.
By Corollary 3, we know that $k$ and $2012-k$ do not produce a carry when added in binary, and
$2012=(11111011100)_{2}$,
From this, it is easy to see that there are $2^{8}=256$ such $k$.
(2) $v_{2}\left(\mathrm{C}_{2012}^{k}\right)=1$.
By Corollary 2, we know that $k$ and $2012-k$ produce exactly one carry when added in binary.
First, if this carry occurs at a position where 2012's binary representation is 1, then the position to the right of this 1 must also produce a carry during the addition, resulting in at least two carries, which is a contradiction.
Thus, this single carry must occur at a position where 2012's binary representation is 0.
The following discussion is in binary.
(i) If the carry occurs at the first position from the right, note that $(11111011100)_{2}$ has a 0 at the second position from the right, so a carry must also occur at the second position, which is a contradiction.
(ii) If the carry occurs at the sixth position from the right, note that $(11111011100)_{2}$ has a 1 at the seventh position from the right, and combining this with the fact that only one carry occurs, we deduce that the seventh position of $k$ must be 0.
Thus, there are $2^{7}=128$ such $k$.
(iii) If the carry occurs at the second position from the right, similarly, there are $2^{7}=128$ such $k$.
Combining (1) and (2), there are 512 $k$ such that $v_{2}\left(\mathrm{C}_{2012}^{k}\right) \leqslant 1$.
It is easy to verify that $\mathrm{C}_{2012}^{0}=\mathrm{C}_{2012}^{2012} \equiv 1(\bmod 2)$,
$\mathrm{C}_{2012}^{503}=\mathrm{C}_{2012}^{1599} \equiv 0(\bmod 4)$,
$\mathrm{C}_{2012}^{1006} \equiv 0(\bmod 4)$.
Thus, the number of non-negative integers $k$ in the set that satisfy $2012 । \mathrm{C}_{2012}^{k}$ is
$2013-5-512+2=1498$
That is, there are 1498 elements $k$ in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012. | 1498 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a_{1}, a_{2}, \cdots, a_{2014}$ be a permutation of the positive integers $1,2, \cdots$, 2014. Denote
$$
S_{k}=a_{1}+a_{2}+\cdots+a_{k}(k=1,2, \cdots, 2014) \text {. }
$$
Then the maximum number of odd numbers in $S_{1}, S_{2}, \cdots, S_{2014}$ is $\qquad$ | 6.1511.
If $a_{i}(2 \leqslant i \leqslant 2014)$ is odd, then $S_{i}$ and $S_{i-1}$ have different parities. From $a_{2}$ to $a_{2014}$, there are at least $1007-1=1006$ odd numbers. Therefore, $S_{1}, S_{2}, \cdots, S_{2014}$ must change parity at least 1006 times.
Thus, $S_{1}, S_{2}, \cdots, S_{2014}$ must have at least 503 even numbers, meaning there are at most $2014-503=1511$ odd numbers.
Take $a_{1}=1, a_{2}=3, a_{3}=5, \cdots, a_{1007}=2013$, $a_{1008}=2, a_{1009}=4, \cdots, a_{2014}=2014$.
Then $S_{1}, S_{3}, \cdots, S_{1007}$ and $S_{1008} \sim S_{2014}$ are all odd, totaling $\frac{1008}{2}+1007=1511$ odd numbers.
In conclusion, the maximum number of odd numbers in $S_{1} \sim S_{2014}$ is 1511. | 1511 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Print 90000 five-digit numbers
$$
10000,10001, \cdots, 99999
$$
on cards, with one five-digit number on each card. Some cards (such as 19806, which reads 90861 when flipped) have numbers that can be read in two different ways, causing confusion. The number of cards that will not cause confusion is $\qquad$ cards. | 8. 88060 .
Among the ten digits $0 \sim 9$, the digits that can still represent numbers when inverted are $0, 1, 6, 8, 9$.
Since the first digit cannot be 0 and the last digit cannot be 0, the number of such five-digit numbers that can be read when inverted is $4 \times 5 \times 5 \times 5 \times 4=2000$.
Among these 2000 five-digit numbers, we need to exclude the five-digit numbers that do not cause confusion when read backwards (such as 10801, 60809).
Now, divide the digits of the five-digit number into three groups.
【First Group】The first and last digits form a group with four possibilities: $(1,1),(8,8),(6,9),(9,6)$;
【Second Group】The second and fourth digits form a group with five possibilities:
$(0,0),(1,1),(8,8),(6,9),(9,6)$;
【Third Group】The third digit can be $0, 1, 8$ three possibilities.
Thus, the number of five-digit numbers that read the same forwards and backwards is $4 \times 5 \times 3 = 60$.
Therefore, the number of five-digit numbers that do not cause confusion is $90000-(2000-60)=88060$ (numbers). | 88060 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $A$ be a set composed of any 100 distinct positive integers, and let
$$
B=\left\{\left.\frac{a}{b} \right\rvert\, a 、 b \in A \text { and } a \neq b\right\},
$$
$f(A)$ denotes the number of elements in set $B$. Then the sum of the maximum and minimum values of $f(A)$ is $\qquad$ . | 11. 10098.
From the problem, when the elements in set $B$ are pairwise coprime, $f(A)$ reaches its maximum value $\mathrm{A}_{100}^{2}=9900$; when the elements in set $B$ form a geometric sequence with a common ratio not equal to 1, $f(A)$ reaches its minimum value of $99 \times 2=198$.
Therefore, the sum of the maximum and minimum values is 10098. | 10098 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Let
$$
\begin{array}{l}
A=\{1,2, \cdots, 2014\}, \\
B_{i}=\left\{x_{i}, y_{i}\right\}(i=1,2, \cdots, t)
\end{array}
$$
be $t$ pairwise disjoint binary subsets of $A$, and satisfy the conditions
$$
\begin{array}{l}
x_{i}+y_{i} \leqslant 2014(i=1,2, \cdots, t), \\
x_{i}+y_{i} \neq x_{j}+y_{j}(1 \leqslant i<j \leqslant t) .
\end{array}
$$
Then the maximum value of $t$ is | 8. 805 .
On the one hand, for any $1 \leqslant i<j \leqslant t$, we have
$$
\left\{x_{i}, y_{i}\right\} \cap\left\{x_{j}, y_{j}\right\}=\varnothing \text {. }
$$
Thus, $x_{1}, x_{2}, \cdots, x_{t}, y_{1}, y_{2}, \cdots, y_{t}$ are $2 t$ distinct integers. Therefore,
$$
\sum_{i=1}^{t}\left(x_{i}+y_{i}\right) \geqslant \sum_{i=1}^{2 t} i=t(2 t+1) .
$$
On the other hand, for any $1 \leqslant i<j \leqslant t$, we have
$$
\begin{array}{l}
x_{i}+y_{i} \neq x_{j}+y_{j}, \text { and } x_{i}+y_{i} \leqslant 2014, \\
\text { then } \sum_{i=1}^{t}\left(x_{i}+y_{i}\right) \\
\leqslant \sum_{i=1}^{t}(2015-i)=\frac{t(4029-t)}{2} .
\end{array}
$$
From equations (1) and (2), we get $t \leqslant 805$.
Below are two examples showing that $t=805$ meets the requirements.
$$
\begin{array}{l}
\text { (1) }\{k, 1208+k\}(k=1,2, \cdots, 402), \\
\{k, 403+k\}(k=403,404, \cdots, 805) ; \\
\text { (2) }\{k, 1611-2 k\}(k=1,2, \cdots, 402), \\
\{k, 2016-k\}(k=403,404, \cdots, 805) .
\end{array}
$$ | 805 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the set $A=\{1,2,3\}, f$ and $g$ are functions from set $A$ to $A$. Then the number of function pairs $(f, g)$ whose image sets intersect at the empty set is $\qquad$ . | 4. 42 .
When the image set of function $f$ is 1 element, if the image set of function $f$ is $\{1\}$, at this time the image set of function $g$ is a subset of $\{2,3\}$, there are $2^{3}=8$ kinds, so there are $3 \times 8=24$ pairs of functions $(f, g)$ that meet the requirements.
When the image set of function $f$ is 2 elements, if the image set of function $f$ is $\{1,2\}$, at this time $f(1) 、 f(2) 、 f(3)$ have two numbers as 1 and one number as 2, there are $2 \times 3=6$ kinds, so there are $3 \times 6=18$ pairs of functions $(f, g)$ that meet the requirements.
Therefore, the number of function pairs $(f, g)$ is 42 . | 42 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $f(x)=\left(x^{2}+3 x+2\right)^{\cos \pi x}$. Then the sum of all $n$ that satisfy the equation
$$
\left|\sum_{k=1}^{n} \log _{10} f(k)\right|=1
$$
is | 5. 21 .
It is known that for integer $x$, we have
$$
f(x)=[(x+1)(x+2)]^{(-1)^{x}} \text {. }
$$
Thus, when $n$ is odd,
$$
\sum_{k=1}^{n} \log _{10} f(k)=-\log _{10} 2-\log _{10}(n+2) \text {; }
$$
When $n$ is even,
$$
\sum_{k=1}^{n} \log _{10} f(k)=-\log _{10} 2+\log _{10}(n+2) \text {. }
$$
Therefore, the $n$ that satisfies the condition is 3 or 18, and their sum is 21. | 21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given real numbers $a_{0}, a_{1}, \cdots, a_{2015}$, $b_{0}, b_{1}, \cdots, b_{2011}$ satisfy
$$
\begin{array}{l}
a_{n}=\frac{1}{65} \sqrt{2 n+2}+a_{n-1}, \\
b_{n}=\frac{1}{1009} \sqrt{2 n+2}-b_{n-1},
\end{array}
$$
where $n=1,2, \cdots, 2015$.
If $a_{0}=b_{2015}$, and $b_{0}=a_{2015}$, find the value of the following expression
$$
\sum_{k=1}^{2015}\left(a_{k} b_{k-1}-a_{k-1} b_{k}\right) .
$$ | 10. Notice that, for any $k=1,2 \cdots, 2015$, we have
$$
\begin{array}{l}
a_{k}-a_{k-1}=\frac{1}{65} \sqrt{2 k+2}, \\
b_{k}+b_{k-1}=\frac{1}{1009} \sqrt{2 k+2} .
\end{array}
$$
Multiplying the two equations, we get
$$
\begin{array}{l}
a_{k} b_{k}-a_{k-1} b_{k-1}+a_{k} b_{k-1}-a_{k-1} b_{k} \\
=\frac{2 k+2}{65 \times 1009} .
\end{array}
$$
Summing up, we have
$$
\begin{array}{l}
a_{2015} b_{2015}-a_{0} b_{0}+\sum_{k=1}^{2015}\left(a_{k} b_{k-1}-a_{k-1} b_{k}\right) \\
=\frac{2015(4+4032)}{2 \times 65 \times 1009}=62 .
\end{array}
$$
Therefore, the required value is 62. | 62 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. A positive integer that can be expressed as the difference of squares of two positive integers is called a "wise number". For example, $9=5^{2}-4^{2}, 9$ is a wise number.
(1) Try to determine which numbers among the positive integers are wise numbers, and explain the reason;
(2) In the sequence of wise numbers arranged in ascending order, find the 2015th wise number. | 14. (1) It is easy to know that the positive integer 1 cannot be expressed as the difference of squares of two positive integers, i.e., 1 is not a wise number.
For odd numbers greater than 1, we have
$$
2 k+1=(k+1)^{2}-k^{2}(k=1,2, \cdots),
$$
which means that all odd numbers greater than 1 are wise numbers.
When $k=2,3, \cdots$, we have
$4 k=(k+1)^{2}-(k-1)^{2}$.
When $k=1$, if $4=a^{2}-b^{2}$, then
$$
4=(a+b)(a-b) \text {, }
$$
we know that there do not exist positive integers $a, b$ that satisfy this equation, i.e., 4 is not a wise number.
Therefore, numbers greater than 4 and divisible by 4 are all wise numbers.
For positive integers of the form $4 k+2(k=0$, $1, \cdots)$, assume $4 k+2$ is a wise number. Then
$$
4 k+2=x^{2}-y^{2}=(x+y)(x-y) \text {, }
$$
where $x, y$ are positive integers.
When $x, y$ have the same parity, $4 \mid(x+y)(x-y)$, but $4 \nmid(4 k+2)$, which is a contradiction;
When $x, y$ have different parities, $(x+y)(x-y)$ is odd, but $4 k+2$ is even, which is a contradiction.
Therefore, numbers of the form $4 k+2(k=1,2, \cdots)$ are not wise numbers.
Hence, only numbers of the form $2 k+1$ and $4(k+1)(k=1,2, \cdots)$ are wise numbers.
(2) Starting from 1, the first four consecutive positive integers form a group, and by (1), only 3 is a wise number in the first group, and in each subsequent group of four numbers, three are wise numbers.
Notice that, $2016=3 \times 672$, and $4 \times 672=$ 2688, and since the first group has only 1 wise number, 2688 is the 2014th wise number.
Therefore, 2689 is the 2015th wise number. | 2689 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
16. The basketball league has eight teams. Each season, each team plays two games (home and away) against each of the other teams in the league, and each team also plays four games against opponents outside the league. Therefore, in one season, the eight teams in the league play a total of ( ) games.
(A) 60
(B) 88
(C) 96
(D) 144
(E) 160 | 16. B.
There are $\mathrm{C}_{8}^{2}=28$ pairs that can be formed from the eight teams in the league.
Since each pair of teams plays both a home and an away match, the eight teams in the league play a total of $28 \times 2=56$ matches.
Since each team plays four matches against opponents outside the league, the eight teams play a total of $4 \times 8=32$ matches against teams outside the league.
Therefore, in one season, these eight teams play a total of $56+32=88$ matches. | 88 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
14. Let the angle between vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ be $\frac{\pi}{3}$, the angle between vectors $\boldsymbol{c}-\boldsymbol{a}$ and $\boldsymbol{c}-\boldsymbol{b}$ be $\frac{2 \pi}{3}$, $|\boldsymbol{a}-\boldsymbol{b}|=5$, and $|\boldsymbol{c}-\boldsymbol{a}|=2 \sqrt{3}$. Then the maximum value of $\boldsymbol{a} \cdot \boldsymbol{c}$ is | 14. 24 .
Let $\overrightarrow{O A}=a, \overrightarrow{O B}=b, \overrightarrow{O C}=c$. Then
$$
|\overrightarrow{A C}|=|c-a|=2 \sqrt{3},|\overrightarrow{A B}|=|a-b|=5 \text {. }
$$
Also, $\angle A O B=\frac{\pi}{3}, \angle A C B=\frac{2 \pi}{3}$, at this time, $O, A, C, B$ are concyclic.
By the Law of Sines, we get
$$
\sin \angle A B C=\frac{3}{5} \Rightarrow \cos \angle A B C=\frac{4}{5} \text {. }
$$
In $\triangle A C O$, since $\angle A O C=\angle A B C$, by the Law of Cosines, we have
$$
\begin{array}{l}
A C^{2}=|\boldsymbol{a}|^{2}+|\boldsymbol{c}|^{2}-2|\boldsymbol{a}||\boldsymbol{c}| \cos \angle A O C \\
\Rightarrow 12 \geqslant 2|\boldsymbol{a}||\boldsymbol{c}|-\frac{8}{5}|\boldsymbol{a}||\boldsymbol{c}| \\
\Rightarrow|\boldsymbol{a}||\boldsymbol{c}| \leqslant 30 \\
\Rightarrow \boldsymbol{a} \cdot \boldsymbol{c}=|\boldsymbol{a}||\boldsymbol{c}| \cos \angle A O C \leqslant 24 . \\
\text { When } \angle A C O=\frac{\pi}{4}+\frac{1}{2} \arctan \frac{4}{3} \text {, the above inequalities hold with equality. }
\end{array}
$$
When $\angle A C O=\frac{\pi}{4}+\frac{1}{2} \arctan \frac{4}{3}$, the above inequalities hold with equality.
Therefore, the maximum value of $a \cdot c$ is 24. | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Allocate 24 volunteer slots to 3 schools. Then the number of allocation methods where each school gets at least one slot and the number of slots for each school is different is $\qquad$ kinds. ${ }^{[2]}$ | Let the quotas allocated to schools A, B, and C be $x$, $y$, and $z$ respectively.
First, without considering that $x, y, z$ are pairwise distinct.
From $x+y+z=24$, we get a total number of combinations $\mathrm{C}_{23}^{2}$.
Next, we separate out the number of positive integer solutions $(x, y, z)$ that are not pairwise distinct.
From $2a+b=24$, we know $b=2c$.
Therefore, $a+c=12$, which has 11 solutions.
Among these, one solution satisfies $a=b=8$, corresponding to one triplet $(8,8,8)$;
Additionally, 10 solutions $(a, a, b) (a \neq b)$ correspond to $10 \mathrm{C}_{3}^{1}$ positive integer triplets $(x, y, z)$.
Thus, the number of pairwise distinct positive integer triplets $(x, y, z)$ is
$$
\mathrm{C}_{23}^{2}-1-10 \mathrm{C}_{3}^{1}=222 .
$$ | 222 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. The sum of the ages of three people, A, B, and C, represented by $x, y, z$ is 120, and $x, y, z \in (20,60)$. Then the number of ordered triples $(x, y, z)$ is $\qquad$ | 8. 1141 .
Notice a basic conclusion:
The number of positive integer solutions $(a, b, c)$ to the indeterminate equation $a+b+c=n$ is $\mathrm{C}_{n-1}^{2}$.
Thus, the number of solutions to the indeterminate equation
$$
(x-20)+(y-20)+(z-20)=60
$$
satisfying $x, y, z>20$ is $\mathrm{C}_{59}^{2}$.
Among these $\mathrm{C}_{59}^{2}$ solutions, the number of solutions where $x \geqslant 60$ and do not meet the requirements is
$$
(x-59)+(y-20)+(z-20)=21
$$
satisfying $x \geqslant 60, y, z>20$, the number of such solutions is $\mathrm{C}_{20}^{2}$.
Similarly, the number of solutions where $y \geqslant 60$ is $\mathrm{C}_{20}^{2}$, and the number of solutions where $z \geqslant 60$ is $\mathrm{C}_{20}^{2}$.
Since no two individuals can simultaneously reach 60 years old, the number of ordered triples $(x, y, z)$ is $\mathrm{C}_{59}^{2}-3 \mathrm{C}_{20}^{2}=1141$. | 1141 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The sequence $\left\{a_{n}\right\}$ has 9 terms, where $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in\left\{2,1,-\frac{1}{2}\right\}$. Find the number of such sequences. | Prompt: Categorized Count. Answer: 491. | 491 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If numbers $1,2, \cdots, 14$ are taken in ascending order as $a_{1}, a_{2}, a_{3}$, such that $a_{2}-a_{1} \geqslant 3$, and $a_{3}-a_{2} \geqslant 3$, find the number of different ways to choose them. | From the given information, we have
$$
\begin{aligned}
a_{1} & \leqslant a_{2}-3 \leqslant a_{3}-6 \\
& \Rightarrow 1 \leqslant a_{1}<a_{2}-2<a_{3}-4 \leqslant 10 .
\end{aligned}
$$
Substitute $\left(a_{1}, a_{2}-2, a_{3}-4\right)=(x, y, z)$.
Then the number of tuples $\left(a_{1}, a_{2}, a_{3}\right)$ equals the number of ways to choose three different elements from $\{1,2, \cdots, 10\}$, which is $\mathrm{C}_{10}^{3}=120$. | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 For any non-empty subset $X$ of the set $M=\{1,2, \cdots, 1000\}$, let $\alpha_{X}$ denote the sum of the maximum and minimum numbers in $X$. Find the arithmetic mean of all such $\alpha_{X}$. | 【Analysis】According to the problem, the required average is
$$
f=\frac{\sum_{\varnothing \neq X \subseteq M} \alpha_{X}}{2^{1000}-1} \text {. }
$$
The key to solving this is to calculate the sum in the numerator
$$
N=\sum_{\varnothing \neq X \subseteq M} \alpha_{X} \text {. }
$$
To compute such an "unordered sum", one must first choose a "sequence". For this, one can consider the contribution of each element to the total sum, then accumulate, often referred to as the "contribution method"; or consider pairing two subsets so that their "sum" is easy to calculate.
Solution 1 (Contribution Method) Take any $i \in M$, then $i$ contributes to the sum (1) in two ways: $i$ contributes as the maximum number the number of times the subset $\{i\} \cup P(P \subseteq\{1,2, \cdots, i-1\})$ appears, which is $2^{i-1}$; $i$ contributes as the minimum number the number of times the subset $\{i\} \cup P(P \subseteq\{i+1, i+2, \cdots, 1000\})$ appears, which is $2^{1000-i}$.
Thus, the total contribution of $i$ to the numerator is $i\left(2^{i-1}+2^{1000-i}\right)$, and the cumulative "ordered" sum for (1) is
$$
N=\sum_{i=1}^{1000} i\left(2^{i-1}+2^{1000-i}\right) \text {. }
$$
Using the method of summing "arithmetic-geometric series" (or Abel's summation formula) we get
$$
N=1001\left(2^{1000}-1\right) \Rightarrow f=1001 \text {. }
$$
Solution 2 (Pairing Method) Take any $\varnothing \neq X \subseteq M$, define $X^{\prime}=\{1001-x \mid x \in X\}$.
If $X=X^{\prime}$, then
$\min X=\min X^{\prime}=1001-\max X$
$\Rightarrow \alpha_{X}=\max X+\min X=1001$.
If $X \neq X^{\prime}$, then
$\min X^{\prime}=1001-\max X$ and $\max X^{\prime}=1001-\min X$
$\Rightarrow \alpha_{X^{\prime}}=1001 \times 2-\alpha_{X}$
$\Rightarrow \alpha_{X}+\alpha_{X^{\prime}}=1001 \times 2$.
Let $k$ be the number of non-empty subsets among the $2^{1000}-1$ non-empty subsets that satisfy $X^{\prime}=X$. Then
$$
\begin{array}{l}
N=\sum_{\varnothing \neq X \subseteq M} \alpha_{X} \\
=1001 k+\frac{2^{1000}-1-k}{2} \times 2002 \\
=1001\left(2^{1000}-1\right) . \\
\text { Hence } f=\frac{N}{2^{1000}-1}=1001 .
\end{array}
$$ | 1001 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Divide a circle with a circumference of 24 into 24 equal segments, and select eight points from the 24 points, such that the arc length between any two points is not equal to 3 and 8. Find the number of different ways to select such a group of eight points. | 【Analysis】First analyze the essential structure of the data: Label 24 points in a clockwise direction as $1,2, \cdots, 24$. Then arrange them into the following $3 \times 8$ number table:
$$
\left(\begin{array}{cccccccc}
1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
17 & 20 & 23 & 2 & 5 & 8 & 11 & 14
\end{array}\right) .
$$
In the number table, the arc length between two points represented by adjacent numbers in the same column is 8, and the arc length between two points represented by adjacent numbers in the same row is 3 (the first column and the eighth column are also adjacent, as are the first row and the third row). Therefore, any two adjacent numbers in the table cannot be taken simultaneously, meaning exactly one number is taken from each column, and the numbers taken from adjacent columns cannot be in the same row.
Generalize and construct a recursive relationship.
From a $3 \times n(n \geqslant 3)$ number table, take one number from each column such that the numbers taken from adjacent columns are not in the same row (the first column and the $n$-th column are considered adjacent), and let the number of ways to do this be $a_{n}$.
Then $a_{3}=6$.
When $n \geqslant 4$, we have
$$
a_{n}+a_{n-1}=3 \times 2^{n-1} \Rightarrow a_{n}=2^{n}+2(-1)^{n} \text {. }
$$
Recursive calculation yields $a_{8}=2^{8}+2(-1)^{8}=258$. | 258 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Fill two $a$s and two $b$s into the 16 cells shown in Figure 3, with at most one letter per cell. If the same letters must not be in the same row or column, find the number of different ways to fill the cells. | Prompt: Categorized Count. Answer: 3960. | 3960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If a positive integer has eight positive divisors, and the sum of these eight positive divisors is 3240, then this positive integer is called a "good number". For example, 2006 is a good number, because the sum of its positive divisors $1, 2, 17, 34, 59, 118, 1003, 2006$ is 3240. Find the smallest good number.
(28th Brazilian Mathematical Olympiad) | Let $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, where $\alpha_{i} \in \mathbf{Z}_{+}, p_{i} (i=1,2, \cdots, k)$ are prime numbers, and $p_{1}<p_{2}<\cdots<p_{k}$.
From $\tau(n)=\prod_{i=1}^{k}\left(1+\alpha_{i}\right)=8=2^{3}$, we know $k=1,2,3$.
(1) If $k=1$, then
$\alpha_{1}=7, n=p^{7}$ ($p$ is a prime number), $\sum_{i=0}^{7} p^{i}=3240$. By $\sum_{i=0}^{7} 2^{i}=255<3240<\sum_{i=0}^{7} 3^{i}=3280$, we know such a prime number $p$ does not exist.
(2) If $k=2$, then let $n=p^{3} q$ ($p, q$ are prime numbers).
Thus $\left(1+p+p^{2}+p^{3}\right)(1+q)$
$$
=3240=2^{3} \times 3^{4} \times 5.
$$
Since $1+q \geqslant 3$, we have
$$
1+p+p^{2}+p^{3} \leqslant 1080 \Rightarrow p \leqslant 7.
$$
(i) If $p=2$, then $1+q=216 \Rightarrow q=215$;
(ii) If $p=3$, then $1+q=81 \Rightarrow q=80$;
(iii) If $p=5$, then $1+q=\frac{270}{13} \notin \mathbf{Z}$;
(iv) If $p=7$, then $1+q=\frac{41}{5} \notin \mathbf{Z}$.
All these cases contradict the fact that $q$ is a prime number.
(3) If $k=3$, then let
$n=p q r (p, q, r$ are prime numbers, $p<q<r)$.
Thus $(1+p)(1+q)(1+r)=3240$.
(i) If $p=2$, then
$$
(1+q)(1+r)=1080=2^{3} \times 3^{3} \times 5 \text{ (it is easy to see that }
$$
$1+q, 1+r$ are both even numbers
$$
\begin{array}{l}
\Rightarrow(1+q, 1+r) \\
=(4,270),(6,180),(10,108),(12,90), \\
(18,60),(20,54),(30,36) \\
\Rightarrow(q, r)=(3,269),(5,179),(11,89), \\
(17,59),(19,53) \\
\Rightarrow n=2 \times 3 \times 269=1614 \text{ or } \\
n= 2 \times 5 \times 179=1790 \text{ or } \\
n= 2 \times 11 \times 89=1958 \text{ or } \\
n=2 \times 17 \times 59=2006 \text{ or } \\
n=2 \times 19 \times 53=2014.
\end{array}
$$
(ii) If $p \geqslant 3$, then $1+p, 1+q, 1+r$ are all even numbers.
$$
\text{Let } 1+p=2 p_{1}, 1+q=2 q_{1}, 1+r=2 r_{1}.
$$
Then $p_{1} q_{1} r_{1}=405=3^{4} \times 5 (2 \leqslant p_{1}<q_{1}<r_{1})$.
Thus $\left(p_{1}, q_{1}, r_{1}\right)=(3,5,27),(3,9,15)$
$$
\Rightarrow(p, q, r)=(5,9,53),(5,17,29).
$$
Since $q$ is a prime number, we get
$$
\begin{array}{l}
(p, q, r)=(5,17,29) \\
\Rightarrow n=5 \times 17 \times 29=2465.
\end{array}
$$
In summary, $n_{\min }=2 \times 3 \times 269=1614$. | 1614 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Question 3 Let positive real numbers $a, b, c$ satisfy $ab + bc + ca = 48$. Try to find the minimum value of $f = \left(a^{2} + 5\right)\left(b^{2} + 5\right)\left(c^{2} + 5\right)$. | Given $A=48, k=5$, hence
$$
f \geqslant 5(48-5)^{2}=9245 \text {. }
$$
When $a=5,\{b, c\}=\left\{\frac{43+\sqrt{97}}{12}, \frac{43-\sqrt{97}}{12}\right\}$,
the equality holds.
Therefore, $f_{\min }=9245$. | 9245 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 1, in the right $\triangle A B C$, it is known that $\angle A C B$ $=90^{\circ}, A C=21, B C=$ 28, and a square $A B D E$ is constructed outward with $A B$ as one side. The angle bisector of $\angle A C B$ intersects $D E$ at point $F$. Then the length of line segment $D F$ is $\qquad$ | 8. 15 .
As shown in Figure 6, let $CF$ intersect $AB$ and $AD$ at points $G$ and $O$ respectively.
Then $\angle BCO = 45^{\circ}$
$= \angle OAB$.
Thus, $O, A, C, B$
are concyclic.
Hence $\angle ABO$
$$
= \angle OCA = 45^{\circ}.
$$
Therefore, $OA = OB$.
Thus, $O$ is the center of the square $ABDE$.
By symmetry, $DF = AG$.
Hence $\frac{AG}{GB} = \frac{AC}{BC} = \frac{21}{28} = \frac{3}{4}$
$\Rightarrow AG = \frac{3}{7} AB = \frac{3}{7} \times 35 = 15$.
Therefore, $DF = 15$. | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. Given that $a$ and $b$ are real numbers, the system of inequalities about $x$
$$
\left\{\begin{array}{l}
20 x+a>0, \\
15 x-b \leqslant 0
\end{array}\right.
$$
has only the integer solutions $2, 3, 4$. Then the maximum value of $ab$ is . $\qquad$ | 10. -1200 .
From the conditions, we know that $-\frac{a}{20}<x \leqslant \frac{b}{15}$.
The integer solutions of the inequality system are only $2, 3, 4$, so $1 \leqslant-\frac{a}{20}<2,4 \leqslant \frac{b}{15}<5$, which means $-40<a \leqslant-20,60 \leqslant b<75$.
Therefore, $-3000<a b \leqslant-1200$.
Thus, when $a=-20, b=60$, $a b$ has the maximum value -1200. | -1200 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
16. (25 points) Given $A \subseteq\{1,2, \cdots, 2014\}$, let real numbers $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3} 、 x_{1} 、 x_{2} 、 x_{3}$ satisfy
(i) $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3} \in\{-1,0,1\}$ and not all are 0;
(ii) $x_{1}, x_{2} 、 x_{3} \in A$;
(iii) If $x_{i}=x_{j}$, then $\lambda_{i} \lambda_{j} \neq-1(1 \leqslant i 、 j \leqslant 3)$.
If all numbers of the form $x_{1} x_{2} x_{3}$ and $\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3}$ are not multiples of 2014, then the set $A$ is called a "good set". Find the maximum number of elements in a good set $A$. | 16. (1) Construct a good set $A$ with 503 elements.
Let $A=\{1,3,5, \cdots, 1005\}$.
If $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3}$ are all non-zero, then
$$
\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3} \equiv x_{1}+x_{2}+x_{3} \equiv 1(\bmod 2) \text {. }
$$
Thus, $\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3}$ is odd and cannot be a multiple of 2014.
If $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3}$ include 0, assume $\lambda_{3}=0$, then by condition (i), at least one of $\lambda_{1} 、 \lambda_{2}$ is non-zero.
By condition (iii), $\lambda_{1} x_{1}+\lambda_{2} x_{2} \neq 0$.
Notice that,
$$
\begin{array}{l}
\left|\lambda_{1} x_{1}+\lambda_{2} x_{2}\right| \leqslant\left|\lambda_{1} x_{1}\right|+\left|\lambda_{2} x_{2}\right| \\
\leqslant\left|x_{1}\right|+\left|x_{2}\right| \leqslant 2 \times 10051007$, then changing $\lambda_{i}$ to $-\lambda_{i}$ and $x_{i}$ to $2014-x_{i}$ results in a number that is congruent to $\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3}$ modulo 2014.
We now discuss different cases for $r$.
1) If $d \leqslant r<2 d$, then at most $d-1$ numbers in $d+1, d+2, \cdots, d+r$ belong to the set $S$.
$$
\begin{array}{l}
\text { Hence }|S| \leqslant 1+(d-1)+d q=d q+d \\
\leqslant d q+\frac{r}{2}+\frac{d}{2}=503 .
\end{array}
$$
2) If $0 \leqslant r \leqslant d-1$, then
$$
\begin{array}{l}
|S| \leqslant 1+r+d q \leqslant d q+\frac{r}{2}+\frac{d}{2}+\frac{1}{2}=503.5 \\
\Rightarrow|S| \leqslant 503 .
\end{array}
$$
Therefore, any good set $S$ must satisfy $|S| \leqslant 503$. From (1) and (2), we know that the maximum number of elements in a good set $A$ is 503. | 503 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. If positive integers $m, n$ satisfy $\frac{(m+n)!}{n!}=5040$, then the value of $m!n$ is $\qquad$ . | 9. 144 .
$$
\begin{array}{l}
\text { Given } \frac{(m+n)!}{n!} \\
=(m+n)(m+n-1) \cdots(n+1), \\
5040=10 \times 9 \times 8 \times 7,
\end{array}
$$
we know $\left\{\begin{array}{l}m+n=10, \\ n+1=7\end{array} \Rightarrow\left\{\begin{array}{l}m=4, \\ n=6 .\end{array}\right.\right.$
Therefore, $m!n=144$. | 144 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. Let the monotonic increasing sequence $\left\{a_{n}\right\}$ consist of positive integers, and $a_{7}=120, a_{n+2}=a_{n}+a_{n+1}\left(n \in \mathbf{Z}_{+}\right)$. Then $a_{8}=$ . $\qquad$ | 10. 194 .
From $a_{n+2}=a_{n}+a_{n+1}$, we get $a_{7}=5 a_{1}+8 a_{2}=120, a_{8}=8 a_{1}+13 a_{2}$. Since $(5,8)=1$, and $a_{1}, a_{2}$ are both positive integers, it follows that $8\left|a_{1}, 5\right| a_{2}$.
Let $a_{1}=8 k, a_{2}=5 m\left(k, m \in \mathbf{Z}_{+}\right)$.
Then $k+m=3$.
Also, $a_{1}<a_{2}$, so, $k=1, m=2$.
Thus, $a_{1}=8, a_{2}=10$.
Therefore, $a_{8}=8 a_{1}+13 a_{2}=194$. | 194 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $n$ be the smallest positive integer satisfying the following conditions:
(1) $n$ is a multiple of 75;
(2) $n$ has exactly 75 positive divisors (including 1 and itself).
Find $\frac{n}{75}$.
(Eighth American Mathematical Invitational) | ```
Given: $n=75 k=3 \times 5^{2} k$.
\[
\begin{array}{l}
\text { By } 75=3 \times 5 \times 5 \\
=(2+1)(4+1)(4+1),
\end{array}
\]
we know the number of prime factors is at most three.
By discussing the number of prime factors, we can solve to get
\[
\left(\frac{n}{75}\right)_{\min }=432 \text {. }
\]
``` | 432 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Choose three different angles from $1^{\circ}, 2^{\circ}, \cdots, 179^{\circ}$ to form the three interior angles of a triangle. There are $\qquad$ different ways to do this. | 3.2611.
Notice that, the equation $x+y+z=180$ has $\mathrm{C}_{179}^{2}=$ 15931 sets of positive integer solutions, among which, the solution where $x=y=z$ is 1 set, and the solutions where exactly two of $x, y, z$ are equal are $88 \times 3=$ 264 sets.
Therefore, the number of selection methods is $\frac{15931-1-264}{\mathrm{~A}_{3}^{3}}=2611$ kinds. | 2611 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $f_{0}(x)=|x|-2015$,
$$
f_{n}(x)=\left|f_{n-1}(x)\right|-1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Then the number of zeros of the function $y=f_{2015}(x)$ is
$\qquad$ | 8.4031 .
From the graph, it is easy to see that the function $y=f_{1}(x)$ has 4 zeros, the function $y=f_{2}(x)$ has 6 zeros, $\cdots \cdots$ and so on, the function $y=f_{2014}(x)$ has 4030 zeros. However, the intersection point of the function $y=f_{2014}(x)$ with the y-axis is $(0,1)$, therefore, the function $y=f_{2015}(x)$ has a total of 4031 zeros. | 4031 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. There are infinitely many cards, each with a real number written on it. For each real number $x$, there is exactly one card with the number $x$ written on it. Two players each select a set of 100 cards, denoted as $A$ and $B$, such that the sets are disjoint. Formulate a rule to determine which of the two players wins, satisfying the following conditions:
(1) The winner depends only on the relative order of these 200 cards: if these 200 cards are placed face down in increasing order, and the audience is informed which card belongs to which player, but not what number is written on each card, the audience can still determine who will win;
(2) If the elements of the two sets are written in increasing order as
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}, B=\left\{b_{1}, b_{2}, \cdots, b_{100}\right\},
$$
where, for all $i \in\{1,2, \cdots, 100\}$, $a_{i}>b_{i}$, then $A$ defeats $B$;
(3) If three players each select a set of 100 cards, denoted as $A$, $B$, and $C$, and $A$ defeats $B$, and $B$ defeats $C$, then $A$ defeats $C$.
Question: How many such rules are there?
[Note] Two different rules mean that there exist two sets $A$ and $B$ such that in one rule, $A$ defeats $B$, and in the other rule, $B$ defeats $A$. | 6. There are 100 ways.
To prove the more general case, where each person selects $n$ cards, there are $n$ ways that satisfy the conditions. Let $A>B$ or $B>A$ if and only if $a_{k}>b_{k}$.
Such rules satisfy all three conditions, and different $k$ correspond to different rules, hence there are at least $n$ different rules.
Next, we prove that there are no other ways to define such rules.
Assume a rule satisfies the conditions, let $k \in\{1,2$, $\cdots, n\}$ be the smallest positive integer satisfying the following property:
$$
\begin{array}{l}
A_{k}=\{1,2, \cdots, k, n+k+1, n+k+2, \cdots, 2 n\} \\
B_{k-1}$.
Thus, $U>W$ (trivial when $k=1$).
The elements in the set $V \cup W$ arranged in increasing order are the same as the elements in the set $A_{k} \cup B_{k}$ arranged in increasing order.
By the choice of $k$, $A_{k}<B_{k}$.
Thus, $V<W$.
Therefore, $X<V<W<U<Y$.
By condition (3), $X<Y$.
Hence, the conclusion holds. | 100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Given odd prime numbers $x, y, z$ satisfying
$$
x \mid \left(y^{5}+1\right), y \mid \left(z^{5}+1\right), z \mid \left(x^{5}+1\right) \text {. }
$$
Find the minimum value of the product $x y z$. (Cheng Chuanping, problem contributor) | 7. Let's assume $x$ is the minimum of $x, y, z$.
(1) If $x=3$, then
$$
3^{5}+1=244=2^{2} \times 61 \Rightarrow z=61.
$$
Since $3 \mid (y^{5}+1)$, we have $y \equiv -1 \pmod{3}$.
Clearly, $5 \nmid (61^{5}+1)$.
After calculation, we find $11 \mid (61^{5}+1)$.
Thus, $y_{\text{min}}=11$.
Therefore, $(x y z)_{\min}=3 \times 11 \times 61=2013$.
This indicates that the smallest set of three consecutive odd primes greater than 2013 is $11 \times 13 \times 17=2431$.
Hence, we only need to consider the cases where $x=5$ or $x=7$.
(2) If $x=5$, note that,
$$
5^{5}+1=3126=2 \times 3 \times 521.
$$
Since $5 \times 521 > 2013$, this does not meet the requirement.
(3) If $x=7$, we estimate the value of $7^{5}+1$.
Since $7 \mid (y^{5}+1)$, we have $y \equiv -1 \pmod{7}$.
If $y=13$, then $13 \mid (z^{5}+1)$.
Thus, $z \equiv -1 \pmod{13}$.
Hence, $z_{\text{min}}=103$, and $(x y z)_{\min} > 2013$.
In summary, the minimum value of $x y z$ is 2013. | 2013 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. A coin collector has 100 coins that look the same. The collector knows that 30 of them are genuine, and 70 are fake, and that all genuine coins weigh the same, while all fake coins weigh different and are heavier than the genuine ones. The collector has a balance scale that can be used to compare the weight of two groups of coins of the same number. To ensure finding at least one genuine coin, how many weighings are needed at minimum? | 8. First, it is clear that 70 weighings can certainly find at least one genuine coin.
In fact, each time, one coin is placed on each side of the balance. If they weigh the same, both are genuine; if they do not, the heavier one must be a counterfeit. Thus, each weighing either finds two genuine coins or one counterfeit.
After 70 weighings, either at least two genuine coins have been found, or all 70 counterfeits have been identified. This indicates that all genuine coins can be found.
Next, we prove that when the weight of the genuine coin is \(2^{100}\), and the weight of the counterfeit coins is
\[
m_{i}=2^{100}+2^{i} \quad (1 \leqslant i \leqslant 70),
\]
69 weighings cannot guarantee finding one genuine coin.
It is easy to see that during the weighing (let's say \(k\) coins on each side), the side with the heaviest counterfeit among the \(2k\) coins will be heavier.
Assume the collector follows his plan and is informed of the results of his weighings and the weights assigned to some counterfeit coins.
The first time he weighs, designate any one of the heavier side as weight \(m_{70}\).
Generally, when he weighs, if neither side has any counterfeit coins with already assigned weights, assign the heaviest unassigned weight to any one of the counterfeit coins on the heavier side; if there are counterfeit coins with previously assigned weights on the balance, do not assign any weight to any counterfeit coin this round.
After 69 rounds, the weight \(m_{1}\) has not been assigned, and the counterfeit coin with weight \(m_{1}\) is indistinguishable from 30 genuine coins.
In summary, finding one genuine coin requires at least 70 weighings.
(Li Weiguo provided) | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
1. The function $f(x)$ defined on $\mathbf{R}$ satisfies $f\left(\frac{2 a+b}{3}\right)=\frac{2 f(a)+f(b)}{3}(a, b \in \mathbf{R})$,
and $f(1)=1, f(4)=7$.
Then $f(2015)=$ | $$
-, 1.4029 .
$$
From the function $f(x)$ being concave or convex in reverse on $\mathbf{R}$, we know its graph is a straight line, $f(x)=a x+b$.
Given $f(1)=1$ and $f(4)=7$, we have
$$
\begin{array}{l}
a=2, b=-1 \Rightarrow f(x)=2 x-1 \\
\Rightarrow f(2015)=4029 .
\end{array}
$$ | 4029 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Fill the numbers $1,2, \cdots, 36$ in a $6 \times 6$ grid, with each cell containing one number, such that the numbers in each row are in increasing order from left to right. Then the minimum value of the sum of the six numbers in the third column is $\qquad$ | 4. 63.
Let the six numbers filled in the third column, arranged in ascending order, be $A, B, C, D, E, F$.
Since the row where $A$ is located needs to fill in two numbers smaller than $A$ before it, then $A \geqslant 3$; since the row where $B$ is located needs to fill in two numbers smaller than $B$, and $A$ and the two numbers before $A$ are all smaller than $B$, then $B \geqslant 6$.
Similarly, $C \geqslant 9, D \geqslant 12, E \geqslant 15, F \geqslant 18$.
Therefore, the sum of the six numbers filled in the third column is
$$
\begin{array}{l}
A+B+C+D+E+F \\
\geqslant 3+6+9+12+15+18=63 .
\end{array}
$$
Table 1 is one way to fill the numbers so that the sum of the six numbers in the third column reaches the minimum value (the filling method for the last three columns is not unique).
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline 1 & 2 & 3 & 19 & 20 & 21 \\
\hline 4 & 5 & 6 & 25 & 27 & 29 \\
\hline 7 & 8 & 9 & 22 & 23 & 24 \\
\hline 10 & 11 & 12 & 26 & 28 & 30 \\
\hline 13 & 14 & 15 & 31 & 34 & 35 \\
\hline 16 & 17 & 18 & 32 & 33 & 36 \\
\hline
\end{tabular} | 63 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
In the plane rectangular coordinate system $x O y$, the set of points
$$
\begin{aligned}
K= & \{(x, y) \mid(|x|+|3 y|-6) . \\
& (|3 x|+|y|-6) \leqslant 0\}
\end{aligned}
$$
corresponds to a plane region whose area is $\qquad$ | 6. 24 .
Let $K_{1}=\{(x, y)|| x|+| 3 y |-6 \leqslant 0\}$.
First, consider the part of the point set $K_{1}$ in the first quadrant, at this time, $x+3 y \leqslant 6$. Therefore, these points correspond to $\triangle O C D$ and its interior in Figure 2.
By symmetry, the region corresponding to the point set $K_{1}$ is the rhombus $A B C D$ and its interior centered at the origin $O$ in Figure 2.
Similarly, let
$$
K_{2}=\{(x, y)|| 3 x|+| y |-6 \leqslant 0\} \text {. }
$$
Then the region corresponding to the point set $K_{2}$ is the rhombus $E F G H$ and its interior centered at $O$ in Figure 2.
By the definition of the point set $K$, the plane region corresponding to $K$ is the part covered by exactly one of the point sets $K_{1}$ and $K_{2}$.
Therefore, what is required in this problem is the area $S$ of the shaded region in Figure 2.
By $l_{C D}: x+3 y=6, l_{G H}: 3 x+y=6$, we know that the intersection point of the two lines is $P\left(\frac{3}{2}, \frac{3}{2}\right)$.
By symmetry,
$$
S=8 S_{\triangle C P G}=8 \times \frac{1}{2} \times 4 \times \frac{3}{2}=24 .
$$ | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. For a four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b 、 c 、 d \leqslant$ $9)$, if $a>b, b<c, c>d$, then $\overline{a b c d}$ is called a $P$ class number; if $a<c, c<d$, 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 $Q$ class numbers, respectively. Then the value of $N(P)-N(Q)$ is $\qquad$ | 8. 285.
Let the sets of all P-type numbers and Q-type numbers be denoted as $A$ and $B$, respectively. Let the set of all P-type numbers whose unit digit is zero be denoted as $A_{0}$, and the set of all P-type numbers whose unit digit is not zero be denoted as $A_{1}$.
For any four-digit number $\overline{a b c d} \in A_{1}$, map it to the four-digit number $\overline{d c b a}$.
Note that, $a > b$, and $bd \geqslant 1$.
Then, $\overline{d c b a} \in B$.
Conversely, each $\overline{d c b a} \in B$ uniquely corresponds to an element $\overline{a b c d} \in A_{1}$.
Therefore, a one-to-one correspondence is established between $A_{1}$ and $B$.
Hence, $N(P) - N(Q) = |A| - |B|$
$= |A_{0}| + |A_{1}| - |B| = |A_{0}|$.
Next, we calculate $|A_{0}|$.
For any four-digit number $\overline{a b c 0} \in A_{0}$, $b$ can take values $0, 1, \cdots, 9$. For each $b$, given $b < a \leqslant 9$ and $b < c \leqslant 9$, $a$ and $c$ each have $9 - b$ possible values.
Thus, $|A_{0}| = \sum_{b=0}^{9} (9 - b)^{2} = \sum_{k=1}^{9} k^{2} = 285$.
Therefore, $N(P) - N(Q) = 285$. | 285 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ be an arithmetic sequence, and
$$
\sum_{i=1}^{n}\left|a_{i}+j\right|=2028(j=0,1,2,3) \text {. }
$$
Then the maximum value of the number of terms $n$ is | 6. 52.
Since the equation $|x|=|x+1|=|x+2|$ has no solution, we have $n \geqslant 2$ and the common difference is not 0.
Assume the general term of the sequence is $a-k d(1 \leqslant k \leqslant n, d>0)$.
Construct the function $f(x)=\sum_{k=1}^{n}|x-k d|$.
The given condition is equivalent to $f(x)=2028$ having at least four distinct roots $a, a+1, a+2, a+3$, which means the graph of $y=f(x)$ intersects the line $l: y=2028$ at least four times.
Notice that the graph of $y=f(x)$ is a downward convex broken line with $n+1$ segments symmetric about the line $y=\frac{(n+1) d}{2}$. It intersects the line $l$ at least four times if and only if there is a horizontal segment of the broken line on the line $l$, which happens if and only if $n=2 m$ and $a, a+1, a+2, a+3 \in[m d,(m+1) d]$, $f(m d)=2028$, i.e., $d \geqslant 3$ and $m^{2} d=2028$.
This gives $m^{2} \leqslant \frac{2028}{3}=676 \Rightarrow m \leqslant 26$.
Clearly, when $m=26$, taking $d=3, a=78$ satisfies the conditions of the problem.
Therefore, the maximum value of $n$ is 52. | 52 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example: $10 f(x)$ is a continuous function defined on the interval $[0,2015]$, and $f(0)=f(2015)$. Find the minimum number of real number pairs $(x, y)$ that satisfy the following conditions:
(1) $f(x)=f(y)$;
(2) $x-y \in \mathbf{Z}_{+}$.
$(2015$, Peking University Mathematics Summer Camp) | 【Analysis】Consider a continuous function $f(x)$ on the interval $[0, n]$ that satisfies $f(0)=f(n)$. Let the minimum number of real number pairs $(x, y)$ that satisfy (1) and (2) be $F(n)$.
Obviously, $F(1)=1, F(2)=2$.
We will prove: $F(n)=n\left(n \in \mathbf{Z}_{+}\right)$.
Assume that for $n \leqslant k-1$, equation (1) holds.
We need to prove that $F(k)=k$.
First, we prove that $F(k) \geqslant k$.
In fact, take any continuous function $f(x)$ on the interval $[0, k]$ that satisfies $f(0)=f(k)$.
Without loss of generality, assume $f(0)=f(k)=0$.
If there exists $t \in\{1,2, \cdots, k-1\}$ such that $f(t)=0$, then consider the two intervals $[0, t]$ and $[t, k]$.
By the induction hypothesis, it is easy to see that
$$
\begin{array}{l}
F(k) \geqslant F(t)+F(k-t)+1 \\
=t+(k-t)+1=k+1 .
\end{array}
$$
Now assume that $f(1), f(2), \cdots, f(k-1)$ are all non-zero. Without loss of generality, assume $f(1)>0$ (if $f(1)<0$, the proof is similar).
Then
$f(k-1)-f(0)=f(k-1)>0$,
$f(k)-f(1)=-f(1)<0$.
Hence, there exists $t_0 \in (1, k-1)$ such that $f(t_0)=0$.
Since $f(k-1)-f(t_0-1)>0$, there exists $j \in (k-1, k)$ such that $f(j)=f(j+k-t_0)$.
By the induction hypothesis, the number of real number pairs $(x, y)$ that satisfy (1) and (2) in the interval $\left[j, j+k-t_0\right]$ is at least $k-t_0$.
In the interval $\left[0, t_0\right]$,
$f(0)=0, f(1)>0, f(2)>0$,
$\cdots, f(t_0-1)>0, f(t_0)<0, f(t_0)-f(t_0-m)<0$.
Then there exist $x, y \in \left[0, t_0\right]$ such that
$f(x)=f(y)$, and $x-y=m$.
Additionally, $(x, y)=(2015,0)$ is also a pair that satisfies (1) and (2).
Therefore, $F(k) \geqslant (k-t_0) + (t_0-1) + 1 = k$.
In summary, $F(k) \geqslant k$.
Next, we provide an example to show that $F(k)=k$.
In fact, any function $f(x)$ that first increases and then decreases satisfies $F(k)=k$.
This is because, for any positive integer $m \leqslant k$, the number of pairs $(x, y)$ that satisfy $f(x)=f(y)$ and $x-y=m$ is exactly one.
Thus, $F(k)=k$, i.e., for any $n \in \mathbf{Z}_{+}$, we have
$F(n)=n$.
Therefore, $F(2015)=2015$. | 2015 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 In $\triangle ABC$ with a fixed perimeter, it is known that $AB=6$, and when vertex $C$ is at a fixed point $P$, $\cos C$ has a minimum value of $\frac{7}{25}$.
(1) Establish an appropriate coordinate system and find the equation of the locus of vertex $C$;
(2) Draw a line through point $A$ intersecting the curve from (1) at points $M$ and $N$, and find the minimum value of $|\overrightarrow{B M}||\overrightarrow{B N}|$. | (1) Establish a Cartesian coordinate system with the line $AB$ as the $x$-axis and the perpendicular bisector of segment $AB$ as the $y$-axis.
Let $|CA| + |CB| = 2a (a > 3)$ be a constant. Then the locus of point $C$ is an ellipse with foci at $A$ and $B$.
Thus, the focal distance $2c = |AB| = 6$.
Notice,
$$
\begin{array}{l}
\cos C = \frac{|CA|^2 + |CB|^2 - |AB|^2}{2|CA||CB|} \\
= \frac{(|CA| + |CB|)^2 - 2|CA||CB| - |AB|^2}{2|CA||CB|} \\
= \frac{2a^2 - 18}{|CA||CB|} - 1.
\end{array}
$$
Also, $|CA||CB| \leq \left(\frac{|CA| + |CB|}{2}\right)^2 = a^2$, so $\cos C \geq 1 - \frac{18}{a^2}$.
From the problem, $1 - \frac{18}{a^2} = \frac{7}{25} \Rightarrow a^2 = 25$.
At this point, $|PA| = |PB|$, and $P(0, \pm 4)$.
Therefore, the equation of the locus of point $C$ is
$$
\frac{x^2}{25} + \frac{y^2}{16} = 1 (y \neq 0).
$$
(2) Establish a polar coordinate system with $A$ as the pole and $Ax$ as the polar axis. Then the polar equation of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1 (y \neq 0)$ is
$$
\rho = \frac{16}{5 - 3 \cos \theta} (\theta \neq 0 \text{ and } \theta \neq \pi).
$$
Let the inclination angle of line $MN$ be $\theta$. Then
$$
|AM| = \frac{16}{5 - 3 \cos \theta}, |BM| = \frac{16}{5 + 3 \cos \theta}.
$$
By the definition of the ellipse,
$$
\begin{array}{l}
|\overrightarrow{BM}||\overrightarrow{BN}| = (10 - |AM|)(10 - |BM|) \\
= 100 - 10\left(\frac{16}{5 - 3 \cos \theta} + \frac{16}{5 + 3 \cos \theta}\right) + \\
\frac{16}{5 - 3 \cos \theta} \cdot \frac{16}{5 + 3 \cos \theta} \\
= 100 - \frac{16 \times 84}{25 - 9 \cos^2 \theta}.
\end{array}
$$
If there are no restrictions, when $\cos^2 \theta = 1$, $|\overrightarrow{BM}||\overrightarrow{BN}|$ achieves its minimum value of 16.
However, $0 < \theta < \pi$, so such $M$ and $N$ do not exist, meaning the set of minimum values of $|\overrightarrow{BM}||\overrightarrow{BN}|$ is empty. | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. How many different right-angled triangles with integer side lengths have an area that is 999 times their perimeter (considering congruent triangles as the same)? (Provided by Lin Chang)
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. Let the three sides of a right-angled triangle be \(a, b, c\) (with \(c\) being the hypotenuse).
From the Pythagorean triple formula, we have
\[
a = k \cdot 2uv, \quad b = k(u^2 - v^2), \quad c = k(u^2 + v^2),
\]
where the greatest common divisor of the three sides \(k\) is a positive integer, \(u\) and \(v\) are coprime, \(u > v\), and \(u\) and \(v\) are of opposite parity.
\[
\begin{array}{l}
\text{Then } \frac{1}{2} ab = 999(a + b + c) \\
\Leftrightarrow k^2 uv(u^2 - v^2) = 999 \times 2u(u + v) \\
\Leftrightarrow k v(u - v) = 1998 = 2 \times 3^3 \times 37.
\end{array}
\]
Notice that \(u - v\) is odd, so the factor 2 can only be assigned to \(k\) or \(v\), giving two ways; \(v\) and \(u - v\) are coprime, and the odd prime factor \(p^\alpha\) can be assigned to \(k\), \(v\), or \(u - v\) in the following ways: \((\alpha, 0, 0)\) or \((i, \alpha - i, 0)\), \((i, 0, \alpha - i) (1 \leqslant i \leqslant \alpha)\), giving \(2\alpha + 1\) ways.
By the multiplication principle, the number of ways to assign the prime factors is
\[
2(2 \times 3 + 1)(2 \times 1 + 1) = 42.
\]
Thus, there are 42 such triangles.
[Note] Generally, if the multiple is \(m = 2^\alpha p_1^{\beta_1} p_2^{\beta_2} \cdots p_n^{\beta_n}\) \((p_1, p_2, \cdots, p_n\) are distinct odd primes), then the number of such triangles is
\[
(\alpha + 2)(2\beta_1 + 1)(2\beta_2 + 1) \cdots (2\beta_n + 1).
\] | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The function $f(x)$ defined on $\mathbf{R}$, for any real number $x$, satisfies
$$
\begin{array}{l}
f(x+3) \leqslant f(x)+3, \\
f(x+2) \geqslant f(x)+2,
\end{array}
$$
and $f(1)=2$. Let $a_{n}=f(n)\left(n \in \mathbf{Z}_{+}\right)$, then
$$
f(2015)=
$$
$\qquad$ | 5.2016.
Notice,
$$
\begin{array}{l}
f(x)+3 \geqslant f(x+3) \\
=f(x+1+2) \geqslant f(x+1)+2 \\
\Rightarrow f(x)+1 \geqslant f(x+1) . \\
\text { Also } f(x)+4 \leqslant f(x+2)+2 \leqslant f(x+4) \\
=f(x+1+3) \leqslant f(x+1)+3 \\
\Rightarrow f(x+1) \geqslant f(x)+1 .
\end{array}
$$
Therefore, $f(x+1)=f(x)+1$.
Thus, $f(2015)=2016$. | 2016 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=n^{4}+6 n^{3}+11 n^{2}+6 n \text {. }
$$
Then the sum of the first 12 terms $S_{12}=$ $\qquad$ | 7. 104832 .
Notice that,
$$
\begin{array}{l}
a_{n}=n^{4}+6 n^{3}+11 n^{2}+6 n \\
=n(n+1)(n+2)(n+3) . \\
\text { Let } f(n)=\frac{1}{5} n(n+1)(n+2)(n+3)(n+4) .
\end{array}
$$
Then $a_{n}=f(n)-f(n-1), S_{n}=f(n)$.
Therefore, $S_{12}=f(12)=104832$. | 104832 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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