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Example 7: There are 4 red cards, 3 blue cards, 2 yellow cards, and 1 white card. Cards of the same color are indistinguishable. Questions:
(1) How many ways are there to arrange these 10 cards in a row from left to right?
(2) How many ways are there to arrange the cards so that the first 3 cards from the left are of the same color? | (1) $\frac{10!}{4!\times 3!\times 2!\times 1!}=12600$ ways.
(2) For the left 3 cards being red, there are
$$
\frac{7!}{1!\times 3!\times 2!\times 1!}=420 \text { (ways); }
$$
For the left 3 cards being blue, there are
$$
\frac{7!}{4!\times 2!\times 1!}=105 \text { (ways). }
$$
Thus, there are $420+105=525$ ways in total. | 525 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. If numbers $a_{1}, a_{2}, a_{3}$ are taken in increasing order from the set $1, 2, \cdots, 14$, such that the following conditions are satisfied:
$$
a_{2}-a_{1} \geqslant 3 \text { and } a_{3}-a_{2} \geqslant 3 \text {. }
$$
Then the number of different ways to choose such numbers is $\qquad$ kinds. | Let $a_{1}=x_{1}, a_{2}-a_{1}=x_{2}$,
$$
a_{3}-a_{2}=x_{3}, 14-a_{3}=x_{4} \text {. }
$$
Then $x_{1}+x_{2}+x_{3}+x_{4}=14$.
Thus, the problem is transformed into finding the number of integer solutions to the equation under the conditions
$$
x_{1} \geqslant 1, x_{2} \geqslant 3, x_{3} \geqslant 3, x_{4} \geqslant 0
$$
The number of different solutions is $\mathrm{C}_{10}^{3}=120$. | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $A$ and $B$ be two sets, and $(A, B)$ is called a "pair". When $A \neq B$, $(A, B)$ and $(B, A)$ are considered different pairs. Then the number of different pairs satisfying the condition $A \cup B=\{1,2,3,4\}$ is $\qquad$ | Prompt: Following Example 3, we know there are $3^{4}=81$ pairs.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Prompt: Following Example 3, we know there are $3^{4}=81$ pairs. | 81 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5.2011 is a four-digit number whose sum of digits is 4. Then the total number of four-digit numbers whose sum of digits is 4 is $\qquad$. | In fact, the number of four-digit numbers $\overline{x_{1} x_{2} x_{3} x_{4}}$ is the number of integer solutions to the indeterminate equation
$$
x_{1}+x_{2}+x_{3}+x_{4}=4
$$
satisfying the conditions $x_{1} \geqslant 1, x_{2}, x_{3}, x_{4} \geqslant 0$. It is easy to see that there are $\mathrm{C}_{6}^{3}=20$ such solutions. | 20 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. There are 8 English letters $K, Z, A, I, G, A, K$, and $\mathrm{U}$, each written on 8 cards. Ask:
(1) How many ways are there to arrange these cards in a row?
(2) How many ways are there to arrange 7 of these cards in a row? | (1) $\frac{8!}{2!\times 2!}=10080$ ways.
(2) If the letter taken away is $K$ or $A$, then there are
$$
2 \times \frac{7!}{2!}=5040 \text { (ways); }
$$
If the letter taken away is $Z$, $I$, $G$, or $U$, then there are
$$
4 \times \frac{7!}{2!\times 2!}=5040 \text { (ways). }
$$
Therefore, there are $5040+5040=10080$ arrangements in total. | 10080 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given that $p$ is a prime number greater than 3. Find
$$
\prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)
$$
the value. | Let $\omega=\mathrm{e}^{\frac{2 \pi i}{p}}$. Then
$$
\begin{array}{l}
\omega^{p}=1, \omega^{-\frac{p}{2}}=-1, 2 \cos \frac{2 k \pi}{p}=\omega^{k}+\omega^{-k} . \\
\text { Hence } \prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)=\prod_{k=1}^{p-1}\left(1+\omega^{k}+\omega^{-k}\right) \\
=\prod_{k=1}^{p-1} \omega^{-k}\left(\omega^{2 k}+\omega^{k}+1\right) \\
=\omega^{-\left(\frac{p(p-1)}{2}\right)} \prod_{k=1}^{p-1}\left(\omega^{2 k}+\omega^{k}+1\right) \\
=(-1)^{p-1} \prod_{k=1}^{p-1} \frac{1-\omega^{3 k}}{1-\omega^{k}} .
\end{array}
$$
Since $p$ is a prime number greater than 3, $(-1)^{p-1}=1$; also, $p$ and 3 are coprime, $\{1,2, \cdots, p-1\}$ and $\{3 \times 1$, $3 \times 2, \cdots, 3 \times(p-1)\}$ both represent the reduced residue system modulo $p$.
Thus, $\prod_{k=1}^{p-1}\left(1-\omega^{3 k}\right)=\prod_{k=1}^{p-1}\left(1-\omega^{k}\right)$.
In conclusion, $\prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The sum of all real numbers $m$ that satisfy $(2-m)^{m^{2}-m-2}=1$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 2. A.
When $2-m=1$, i.e., $m=1$, it satisfies the condition.
When $2-m=-1$, i.e., $m=3$,
$$
(2-m)^{m^{2}-m-2}=(-1)^{4}=1 \text{, }
$$
it satisfies the condition.
When $2-m \neq \pm 1$, i.e., $m \neq 1$ and $m \neq 3$, by the condition, $m^{2}-m-2=0$, and $2-m \neq 0$.
Solving gives $m=-1$.
Therefore, the sum is $1+3+(-1)=3$. | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
6. Let $n$ be a positive integer, and call $n$ a "good number" if the number of prime numbers not exceeding $n$ equals the number of composite numbers not exceeding $n$. Then the sum of all good numbers is ( ).
(A) 33
(B) 34
(C) 2013
(D) 2014 | 6. B.
Since 1 is neither a prime number nor a composite number, a good number must be an odd number.
Let the number of prime numbers not exceeding $n$ be $a_{n}$, and the number of composite numbers be $b_{n}$. When $n \leqslant 15$, only consider the case where $n$ is odd (as shown in Table 1).
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline$n$ & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\
\hline$a_{n}$ & 0 & 2 & 3 & 4 & 4 & 5 & 6 & 6 \\
\hline$b_{n}$ & 0 & 0 & 1 & 2 & 4 & 5 & 6 & 8 \\
\hline
\end{tabular}
Obviously, 1, 9, 11, and 13 are good numbers.
Since $b_{15}-a_{15}=2$, when $n \geqslant 16$, based on $n=15$, every two numbers added must include one even number, which is a composite number. This means that the number of composite numbers added will not be less than the number of prime numbers added, so there must be $b_{n}-a_{n} \geqslant 2$.
Therefore, when $n \geqslant 16$, $n$ cannot be a good number.
Thus, the sum of all good numbers is
$$
1+9+11+13=34 \text {. }
$$ | 34 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
1. Given real numbers $x, y, z$ satisfy
$$
x+y=4,|z+1|=x y+2 y-9 \text {. }
$$
then $x+2 y+3 z=$ $\qquad$ | $=, 1.4$.
From $x+y=4$, we get $x=4-y$. Then
$$
\begin{array}{l}
|z+1|=x y+2 y-9 \\
=6 y-y^{2}-9=-(y-3)^{2} \\
\Rightarrow z=-1, y=3 \Rightarrow x=1 \\
\Rightarrow x+2 y+3 z=4 .
\end{array}
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. A cube is painted red on all its faces, then cut into $n^{3}(n>2)$ identical smaller cubes. If the number of smaller cubes with only one face painted red is the same as the number of smaller cubes with no faces painted red, then $n=$ $\qquad$ . | 2.8.
The total number of small cubes with only one face painted red is $6(n-2)^{2}$, and the total number of small cubes with no faces painted red is $(n-2)^{3}$. According to the problem,
$$
6(n-2)^{2}=(n-2)^{3} \Rightarrow n=8 \text {. }
$$ | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Given real numbers $a, b, c, d$ satisfy $2a^2 + 3c^2 = 2b^2 + 3d^2 = (ad - bc)^2 = 6$. Find the value of $\left(a^2 + \dot{b}^2\right)\left(c^2 + d^2\right)$. | Let $m=a^{2}+b^{2}, n=c^{2}+d^{2}$. Then
$$
\begin{array}{l}
2 m+3 n=2 a^{2}+2 b^{2}+3 c^{2}+3 d^{2}=12 . \\
\text { By }(2 m+3 n)^{2}=(2 m-3 n)^{2}+24 m n \geqslant 24 m n \\
\Rightarrow 12^{2} \geqslant 24 m n \\
\Rightarrow m n \leqslant 6 . \\
\text { Also } m n=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \\
=a^{2} c^{2}+b^{2} d^{2}+a^{2} d^{2}+b^{2} c^{2} \\
=(a c+b d)^{2}+(a d-b c)^{2} \\
\geqslant(a d-b c)^{2}=6,
\end{array}
$$
From equations (1) and (2), we get $m n=6$, that is
$$
\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=6 \text {. }
$$
[Note] Real numbers $a, b, c, d$ that satisfy the conditions exist and are not unique, for example,
$$
(a, b, c, d)=\left(\sqrt{2}, 1, \frac{\sqrt{6}}{3},-\frac{2 \sqrt{3}}{3}\right)
$$
is one of the solutions. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given that $t$ is a root of the quadratic equation
$$
x^{2}+x-1=0
$$
If positive integers $a$, $b$, and $m$ satisfy the equation
$$
(a t+m)(b t+m)=31 m
$$
find the value of $a b$. | Since $t$ is a root of the quadratic equation
$$
x^{2}+x-1=0
$$
$t$ is an irrational number, and $t^{2}=1-t$.
From the problem, we have
$$
\begin{array}{l}
a b t^{2}+m(a+b) t+m^{2}=31 m \\
\Rightarrow a b(1-t)+m(a+b) t+m^{2}=31 m \\
\Rightarrow[m(a+b)-a b] t+\left(a b+m^{2}-31 m\right)=0 .
\end{array}
$$
Since $a, b, m$ are positive integers and $t$ is an irrational number, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
m(a+b)-a b=0, \\
a b+m^{2}-31 m=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a+b=31-m \\
a b=31 m-m^{2}
\end{array}\right.
\end{array}
$$
Therefore, $a, b$ are the two integer roots of the quadratic equation in $x$
$$
x^{2}+(m-31) x+31 m-m^{2}=0
$$
The discriminant of equation (1) is
$$
\begin{array}{l}
\Delta=(m-31)^{2}-4\left(31 m-m^{2}\right) \\
=(31-m)(31-5 m) \geqslant 0 .
\end{array}
$$
Since $a, b$ are positive integers, we have
$$
a+b=31-m>0 \text {. }
$$
Thus, $0<m \leqslant \frac{31}{5}$.
Since the discriminant $\Delta$ is a perfect square, upon verification, only $m=6$ meets the requirement.
Substituting $m=6$ yields
$$
a b=31 m-m^{2}=150 .
$$ | 150 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Given $t=\sqrt{2}-1$. If positive integers $a$, $b$, and $m$ satisfy
$$
(a t+m)(b t+m)=17 m
$$
find the value of $a b$. | Given that $t=\sqrt{2}-1$, we have
$$
t^{2}=3-2 \sqrt{2} \text {. }
$$
From the problem, we know
$$
\begin{array}{l}
a b t^{2}+m(a+b) t+m^{2}=17 m \\
\Rightarrow a b(3-2 \sqrt{2})+m(a+b)(\sqrt{2}-1)+m^{2}=17 m \\
\Rightarrow \sqrt{2}[m(a+b)-2 a b]+ \\
\quad\left[3 a b-m(a+b)+m^{2}-17 m\right]=0 .
\end{array}
$$
Since $a, b, m$ are positive integers, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
m(a+b)-2 a b=0, \\
3 a b-m(a+b)+m^{2}-17 m=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a+b=2(17-m), \\
a b=17 m-m^{2} .
\end{array}\right.
\end{array}
$$
Therefore, $a, b$ are the two integer roots of the quadratic equation in $x$:
$$
x^{2}+2(m-17) x+17 m-m^{2}=0
$$
The discriminant of equation (1) is
$$
\begin{array}{l}
\Delta=4(m-17)^{2}-4\left(17 m-m^{2}\right) \\
=4(17-m)(17-2 m) \geqslant 0 .
\end{array}
$$
Since $a, b, m$ are positive integers, we have
$$
a+b=2(17-m)>0 \text {. }
$$
Thus, $0<m \leqslant \frac{17}{2}$.
Since the discriminant $\Delta$ is a perfect square, upon verification, only $m=8$ meets the requirement.
Substituting $m=8$ into the equation, we get
$$
a b=17 m-m^{2}=72 \text {. }
$$ | 72 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) If positive numbers $a, b, c$ satisfy
$$
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}=3 \text {, }
$$
find the value of the algebraic expression
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$ | Three, since $a, b, c$ have cyclic symmetry, without loss of generality, assume $0 < a < b < c$, then
$$
c-a>b>0, c-b>a>0 \text {. }
$$
Thus $\frac{b^{2}+c^{2}-a^{2}}{2 b c}=1+\frac{(c-b)^{2}-a^{2}}{2 b c}>1$,
$$
\begin{array}{l}
\frac{c^{2}+a^{2}-b^{2}}{2 c a}=1+\frac{(c-a)^{2}-b^{2}}{2 c a}>1, \\
\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{(a+b)^{2}-c^{2}}{2 a b}-13,
$$
Contradiction.
(2) If $c-1, \\
\frac{a^{2}+b^{2}-c^{2}}{2 a b}=1+\frac{(a-\dot{b})^{2}-c^{2}}{2 a b}<1 .
\end{array}
$$
From the above three equations, we get
$$
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}<3,
$$
Contradiction.
Combining (1) and (2), we know $c=a+b$.
Therefore, $\frac{b^{2}+c^{2}-a^{2}}{2 b c}=1, \frac{c^{2}+a^{2}-b^{2}}{2 c a}=1$,
$$
\begin{array}{l}
\frac{a^{2}+b^{2}-c^{2}}{2 a b}=-1 . \\
\text { Hence } \frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the function $f(x)=\frac{3+x}{1+x}$. Let
$$
\begin{array}{l}
f(1)+f(2)+f(4)+\cdots+f(1024)=m, \\
f\left(\frac{1}{2}\right)+f\left(\frac{1}{4}\right)+\cdots+f\left(\frac{1}{1024}\right)=n .
\end{array}
$$
Then $m+n=$ . $\qquad$ | Ni, 7.42.
From $f(x)=1+\frac{2}{1+x}$, we know $f\left(\frac{1}{x}\right)=1+\frac{2 x}{1+x}$.
Therefore, $f(x)+f\left(\frac{1}{x}\right)=4$.
Also, $f(1)=2$, so, $m+n=4 \times 10+2=42$. | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $\mathrm{i}$ is the imaginary unit. If
$$
z=1+\mathrm{i}+\cdots+\mathrm{i}^{2013},
$$
denote the complex conjugate of $z$ as $\bar{z}$, then $z \cdot \bar{z}=$ $\qquad$ | 8. 2 .
Let $a_{n}=\mathrm{i}^{n}$. Then $a_{n+4}=a_{n}$, and
$$
1+\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}=0 \text {. }
$$
Thus $z=1+\mathrm{i} \Rightarrow \bar{z}=1-\mathrm{i}$
$$
\Rightarrow z \cdot \bar{z}=(1+\mathrm{i})(1-\mathrm{i})=1-\mathrm{i}^{2}=2 .
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $n$ be a positive integer less than 100, and satisfies $\frac{1}{3}\left(n^{2}-1\right)+\frac{1}{5} n$ is an integer. Then the sum of all positive integers $n$ that meet the condition is $\qquad$ | 11. 635 .
Notice that,
$$
\frac{1}{3}\left(n^{2}-1\right)+\frac{1}{5} n=\frac{5 n^{2}+3 n-5}{15}
$$
is an integer, so, $15 \mid \left(5 n^{2}+3 n-5\right)$.
Thus, $5 \mid n$, and $3 \mid \left(n^{2}-1\right)$.
Therefore, $n=15 k+5$ or $15 k+10$.
Hence, the sum of all positive integers $n$ that satisfy the condition is
$$
\begin{array}{c}
\sum_{k=0}^{6}(15 k+5)+\sum_{k=0}^{5}(15 k+10) \\
=95+\sum_{k=0}^{5}(30 k+15) \\
=95+\frac{(15+165) \times 6}{2}=635 .
\end{array}
$$ | 635 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Suppose $N$ consecutive positive integers satisfy the following conditions: the sum of the digits of the 1st number is divisible by 1, the sum of the digits of the 2nd number is divisible by 2, $\cdots$. The sum of the digits of the $N$th number is divisible by $N$. Find the maximum possible value of $N$.
| Let the $N$ numbers be $a_{1}, a_{2}, \cdots, a_{N}$.
If $N \geqslant 22$, then among $a_{2}, a_{3}, \cdots, a_{21}$, these 20 numbers, at least two numbers have a units digit of 9, and among these two numbers, at least one has a tens digit that is not 9, let this number be $a_{i}$.
Thus, $i \leqslant 21$.
If $i$ is even, then $i \leqslant 20$.
Since $i \mid S\left(a_{i}\right)$, $S\left(a_{i}\right)$ is also even.
Since $S\left(a_{i+2}\right)=S\left(a_{i+1}\right)+1=S\left(a_{i}\right)-7$ is odd, we get $(i+2) \times S\left(a_{i+2}\right)$.
Therefore, $N \leqslant i+1 \leqslant 21$.
If $i$ is odd, then from $(i-1) \mid S\left(a_{i-1}\right)$, we get $S\left(a_{i-1}\right)$ is even, hence $S\left(a_{i+1}\right)$ is odd.
So, $(i+1) \times S\left(a_{i+1}\right)$.
Thus, $N \leqslant i \leqslant 21$.
In summary, $N \leqslant 21$.
Below is an example for 21.
$$
\begin{array}{l}
\text { By }[2,3, \cdots, 11]=9 \times 3080 \text {, we know that we can let } \\
a_{1}=38 \underset{3078 \uparrow}{99 \ldots 989} \text {, } \\
a_{i}=a_{1}+(i-1)(i=2,3, \cdots, 21) . \\
\end{array}
$$
Thus $2 \mid S\left(a_{2}\right)=9 \times 3080+2$,
$$
k \mid S\left(a_{k}\right)=9 \times 3081+k(k=3,4, \cdots, 11) \text {. }
$$
For $k=12,13, \cdots, 21$, $S\left(a_{k}\right)=k$.
Thus $k \mid S\left(a_{k}\right)$, but $22 \times S\left(a_{22}\right)=13$. | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Let $f(x)=x+\frac{1}{x}(x>0)$. If for any positive number $a$, there exist $m+1$ real numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ in the interval $\left[1, a+\frac{2013}{a}\right]$, such that the inequality
$$
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{m}\right)<f\left(a_{m+1}\right)
$$
holds, find the maximum value of $m$. | II. 9. Let $a=\sqrt{2013}$, then there exist $m+1$ real numbers that meet the requirements in the interval $[1,2 \sqrt{2013}]$.
Notice that $[1,2 \sqrt{2013}] \subseteq\left[1, a+\frac{2013}{a}\right]$.
Therefore, we only need to consider the existence of real numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ in $[1,2 \sqrt{2013}]$.
It is easy to see that $f(x)$ is an increasing function in the interval $[1,2 \sqrt{2013}]$.
Thus, $f(1) \leqslant f\left(a_{i}\right)(i=1,2, \cdots, m)$, $f\left(a_{m+1}\right) \leqslant f(2 \sqrt{2013})$.
Adding the first $m$ inequalities, we get
$$
\begin{array}{l}
m f(1) \leqslant f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{m}\right) \\
<f\left(a_{m+1}\right) \leqslant f(2 \sqrt{2013}),
\end{array}
$$
which implies $m<\sqrt{2013}+\frac{1}{4 \sqrt{2013}}<45$.
Therefore, $m \leqslant 44$.
When $m=44$, take $a_{1}=a_{2}=\cdots=a_{44}=1$, $a_{45}=2 \sqrt{2013}$, then the inequality in the problem holds.
Hence, the maximum value of $m$ is 44. | 44 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) In an annual super football round-robin tournament, 2013 teams each play one match against every other team. Each match awards 3 points to the winner, 0 points to the loser, and 1 point to each team in the event of a draw. After the tournament, Jia told Yi the total points of his team, and Yi immediately knew the number of wins and losses of Jia's team in the entire tournament. What is the total score of Jia's team in this tournament? | 10. Consider the case with $n$ teams.
Let the team where Jia is located win $x$ games, draw $y$ games, and lose $z$ games. Then the total score of the team is
$$
S=3 x+y \quad (x+y+z=n-1, x, y, z \geqslant 0).
$$
Consider the region $\Omega:\left\{\begin{array}{l}x+y \leqslant n-1, \\ x \geqslant 0, \\ y \geqslant 0 .\end{array}\right.$
When $S$ changes, the equation $S=3 x+y$ represents a set of parallel lines with a slope of -3. We need to determine the value of $S$ such that the line $l$ intersects the region $\Omega$ at exactly one integer point, and we call such a line $l$ a "good line."
Consider the intersection point $A(0, S)$ of the line $l$ with the $y$-axis.
When $S=0,1,2$, $l$ is a good line.
When $3 \leqslant S \leqslant n-1$, $l$ passes through at least two integer points in the region $\Omega$, so $l$ is not a good line.
When $S \geqslant n$, consider the intersection point $B\left(\frac{S}{3}, 0\right)$ of the line $l$ with the $x$-axis.
When $S=3 n-9$, $l$ passes through $B_{1}(n-3,0)$, and additionally, $l$ passes through the integer point $(n-4,3)$ in the region $\Omega$, so $l$ is not a good line.
When $S \leqslant 3 n-9$, the intersection point of $l$ with the $x$-axis lies on the segment
$O B_{1}$, and $l$ passes through at least two integer points in the region $\Omega$, so $l$ is not a good line.
When $S=3 n-4$, as shown in Figure 2, $l$ does not pass through any integer points in the region $\Omega$, so $l$ is not a good line.
When $3 n-8 \leqslant S \leqslant 3 n-5$ or $S=3 n-3$, as shown in Figure 2, $l$ passes through exactly one integer point in the region $\Omega$, so $l$ is a good line.
In summary, $S=0,1,2,3 n-8,3 n-7,3 n-6,3 n-5,3 n-3$.
Taking $n=2013$, we get
$$
S=0,1,2,6031,6032,6033,6034,6036 .
$$ | 6034 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Given a positive integer $n$. Find $\sum_{k=1}^{n}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
---
Please note that the format and line breaks have been preserved as requested. | $$
\text { Three, let } n=2^{m} a_{m}+2^{m-1} a_{m-1}+\cdots+2^{1} a_{1}+a_{0}
$$
where, $a_{m} \neq 0$.
At this point, $2^{m} \leqslant n<2^{m+1}$, so, $\left[\log _{2} n\right]=m$.
If $k \geqslant m+2$, then
$$
\frac{n}{2^{k}}-\frac{1}{2}<\frac{2^{m+1}}{2^{m+2}}-\frac{1}{2}=0 \text {, }
$$
at this time, $\left[\frac{n}{2^{k}}-\frac{1}{2}\right]=-1$.
If $k=m+1$, then
$$
\frac{n}{2^{k}}-\frac{1}{2}=\frac{n}{2^{m+1}}-\frac{1}{2} \in\left[0, \frac{1}{2}\right) \text {, }
$$
at this time, $\left[\frac{n}{2^{k}}-\frac{1}{2}\right]=0$.
If $k=m$, then
$$
\begin{array}{l}
{\left[\frac{n}{2^{k}}-\frac{1}{2}\right]=\left[\sum_{i=0}^{m} \frac{a_{m-t}}{2^{t}}-\frac{1}{2}\right]} \\
=\left[a_{m}+\frac{a_{m-1}}{2}-\frac{1}{2}\right]=a_{m-1} .
\end{array}
$$
If $k \leqslant m-1$, then,
$$
\begin{array}{l}
{\left[\frac{n}{2^{k}}-\frac{1}{2}\right]=\left[\sum_{t=0}^{m} 2^{m-k-t} a_{m-t}-\frac{1}{2}\right]} \\
=\sum_{t=k}^{m} 2^{t-k} a_{t}+a_{k-1}-1 .
\end{array}
$$
Then $\sum_{k=1}^{\left[\log _{2} n\right]}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]=\sum_{k=1}^{m}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]$
$$
\begin{array}{l}
=\sum_{k=1}^{m-1}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]+a_{m-1} \\
=a_{m-1}+\sum_{k=1}^{m-1}\left(\sum_{t=k}^{m} 2^{t-k} a_{t}+a_{k-1}-1\right) \\
=a_{m-1}-a_{m}+\sum_{t=1}^{m} \sum_{k=1}^{t} 2^{t-k} a_{t}+\sum_{k=1}^{m-1} a_{k-1}-\sum_{k=1}^{m-1} 1 \\
=a_{m}\left(2^{m}-2\right)+\sum_{t=1}^{m-1}\left(2^{t}-1\right) a_{t}+\sum_{k=1}^{m} a_{k-1}-(m-1) \\
=\sum_{t=0}^{m} 2^{t} a_{t}-m-1=n-m-1 .
\end{array}
$$
Therefore, $\sum_{k=1}^{n}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]$
$$
\begin{array}{l}
=\sum_{k=1}^{m}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]+\left[\frac{n}{2^{m+1}}-\frac{1}{2}\right]+\sum_{k=m+2}^{n}\left[\frac{n}{2^{k}}-\frac{1}{2}\right] \\
=(n-m-1)+0+\sum_{k=m+2}^{n}(-1) \\
=(n-m-1)-(n-m-1)=0 .
\end{array}
$$ | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Given $a, b, c > 0$, and $a^{2} + b^{2} + c^{2} + abc = 4$. Prove:
$$
\begin{array}{l}
\sqrt{\frac{(2-a)(2-b)}{(2+a)(2+b)}} + \sqrt{\frac{(2-b)(2-c)}{(2+b)(2+c)}} + \\
\sqrt{\frac{(2-c)(2-a)}{(2+c)(2+a)}} = 1 .
\end{array}
$$ | $$
\begin{array}{l}
16=4 a^{2}+4 b^{2}+4 c^{2}+4 a b c \text {. } \\
\text { Then } \sqrt{\frac{(2-a)(2-b)}{(2+a)(2+b)}}+\sqrt{\frac{(2-b)(2-c)}{(2+b)(2+c)}}+ \\
\sqrt{\frac{(2-c)(2-a)}{(2+c)(2+a)}} \\
=\frac{\sqrt{(4-a)^{2}\left(4-b^{2}\right)}}{(2+a)(2+b)}+ \\
\frac{\sqrt{\left(4-b^{2}\right)\left(4-c^{2}\right)}}{(2+b)(2+c)}+ \\
\frac{\sqrt{\left(4-c^{2}\right)\left(4-a^{2}\right)}}{(2+c)(2+a)} \\
=\frac{\sqrt{16-4 a^{2}-4 b^{2}+a^{2} b^{2}}}{(2+a)(2+b)}+ \\
\frac{\sqrt{16-4 b^{2}-4 c^{2}+b^{2} c^{2}}}{(2+b)(2+c)}+ \\
\frac{\sqrt{16-4 c^{2}-4 a^{2}+c^{2} a^{2}}}{(2+c)(2+a)} \\
=\frac{\sqrt{4 c^{2}+4 a b c+a^{2} b^{2}}}{(2+a)(2+b)}+ \\
\frac{\sqrt{4 a^{2}+4 a b c+b^{2} c^{2}}}{(2+b)(2+c)}+ \\
\frac{\sqrt{4 b^{2}+4 a b c+c^{2} a^{2}}}{(2+c)(2+a)} \\
=\frac{2 c+a b}{(2+a)(2+b)}+\frac{2 a+b c}{(2+b)(2+c)}+ \\
\frac{2 b+c a}{(2+c)(2+a)} \\
=\frac{(2 c+a b)(2+c)+(2 a+b c)(2+a)+(2 b+c a)(2+b)}{(2+a)(2+b)(2+c)} \\
=\frac{2\left(a^{2}+b^{2}+c^{2}\right)+2(a b+b c+c a)+4(a+b+c)+3 a b c}{(2+a)(2+b)(2+c)} \\
=\frac{a b c+2(a b+b c+c a)+4(a+b+c)+8}{(2+a)(2+b)(2+c)} \\
=\frac{(2+a)(2+b)(2+c)}{(2+a)(2+b)(2+c)}=1 . \\
\end{array}
$$ | 1 | Algebra | proof | Yes | Yes | cn_contest | false |
2. A set of 4027 points in the plane is called a "Colombian point set", where no three points are collinear, and 2013 points are red, 2014 points are blue. Draw a set of lines in the plane, which can divide the plane into several regions. If a set of lines for a Colombian point set satisfies the following two conditions, it is called a "good line set":
(1) These lines do not pass through any point in the Colombian point set;
(2) No region contains points of both colors.
Find the minimum value of $k$, such that for any Colombian point set, there exists a good line set consisting of $k$ lines. | 2. $k=2013$.
Solution 1 First, give an example to show that $k \geqslant 2013$.
Mark 2013 red points and 2013 blue points alternately on a circle, and color another point in the plane blue. This circle is divided into 4026 arcs, each with endpoints of different colors. If the requirement of the problem is to be met, each arc must intersect with some drawn line. Since each line can intersect the circle at most twice, at least $\frac{4026}{2}=2013$ lines are needed.
Next, prove that 2013 lines can meet the requirement.
Notice that for any two points of the same color $A$ and $B$, two lines can be used to separate them from other points. The method is: draw two lines parallel to $AB$ on either side of $AB$. As long as they are sufficiently close to $AB$, the strip between them will contain only the two colored points $A$ and $B$.
Let $P$ be the convex hull of all colored points, and consider the following two cases.
(1) Assume $P$ has a red vertex, denoted as $A$. Then a line can be drawn to separate point $A$ from all other colored points. This way, the remaining 2012 red points can be paired into 1006 pairs, and each pair can be separated from all other colored points using two parallel lines. Therefore, a total of 2013 lines can meet the requirement.
(2) Assume all vertices of $P$ are blue. Consider two adjacent vertices on $P$, denoted as $A$ and $B$. Then a line can be drawn to separate these two points from all other colored points. This way, the remaining 2012 blue points can be paired into 1006 pairs, and each pair can be separated from all other colored points using two lines. Therefore, a total of 2013 lines can meet the requirement.
[Note] The convex hull can be ignored, and only a line passing through two colored points $A$ and $B$ can be considered, such that all other colored points are on one side of this line. If $A$ and $B$ include a red point, then proceed as in (1); if $A$ and $B$ are both blue, then proceed as in (2).
If $m$ is odd, then there exists an $N$ such that for any $m \leqslant n \leqslant N$, $f(m, n)=m$.
For any $n>N$, $f(m, n)=m+1$.
Solution 2 Another proof that 2013 lines can meet the requirement.
First, give a more general conclusion: If there are $n$ marked points in the plane with no three points collinear, and these points are arbitrarily colored red or blue, then $\left[\frac{n}{2}\right]$ lines can meet the requirement of the problem, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Use induction on $n$.
When $n \leqslant 2$, the conclusion is obvious.
Assume $n \geqslant 3$.
Consider a line passing through two colored points $A$ and $B$ such that all other colored points are on one side of this line. For example, an edge of the convex hull of all colored points is such a line.
Temporarily remove points $A$ and $B$ from consideration. By the induction hypothesis, the remaining points can be separated using $\left[\frac{n}{2}\right]-1$ lines. Now, re-include points $A$ and $B$, and consider three cases.
(1) If points $A$ and $B$ are the same color, then a line parallel to $l$ can be drawn to separate points $A$ and $B$ from other colored points. Clearly, the $\left[\frac{n}{2}\right]$ lines thus obtained can meet the requirement.
(2) If points $A$ and $B$ are different colors, but are separated by some already drawn line, then the line parallel to $l$ also meets the requirement.
(3) If points $A$ and $B$ are different colors and are in the same region after drawing the $\left[\frac{n}{2}\right]-1$ lines. By the induction hypothesis, at least one color will have no other colored points in that region. Without loss of generality, assume the only blue point in that region is $A$. Then, a line can be drawn to separate point $A$ from all other colored points.
Thus, the induction step is completed.
[Note] Generalize the problem by replacing 2013 and 2014 with any positive integers $m$ and $n$, assuming $m \leqslant n$. Denote the solution to the corresponding problem as $f(m, n)$.
Following the idea of Solution 1, we get $m \leqslant f(m, n) \leqslant m+1$.
If $m$ is even, then $f(m, n)=m$. | 2013 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $a_{1}, a_{2}, \cdots, a_{100}$ are 100 distinct positive integers. For any positive integer $i \in\{1,2, \cdots, 100\}, d_{i}$ represents the greatest common divisor of the 99 numbers $a_{j}(j \neq i)$, and $b_{i}=a_{i}+$ $d_{i}$. Question: How many different positive integers are there at least in $b_{1}, b_{2}, \cdots, b_{100}$? | 3. Contains at least 99 different positive integers.
If we let $a_{100}=1, a_{i}=2 i(1 \leqslant i \leqslant 99)$; then $b_{1}=b_{100}=3$.
This indicates that there are at most 99 different positive integers in $b_{i}$.
Next, we prove: there are at least 99 different positive integers in $b_{i}$.
Without loss of generality, assume $a_{1}a_{j} \geqslant a_{i}+d_{i}=b_{i} .
\end{array}
$$
This indicates that $b_{i}(i \neq k)$ are all different. | 99 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $A$ be a set of ten real-coefficient quadratic polynomials. It is known that there exist $k$ consecutive positive integers $n+1$, $n+2, \cdots, n+k$, and $f_{i}(x) \in A(1 \leqslant i \leqslant k)$, such that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence. Find the maximum possible value of $k$.
| 6. Given that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence, we know there exist real numbers $a$ and $b$ such that
$$
f_{i}(n+i)=a i+b .
$$
Notice that for any quadratic polynomial $f$, the equation
$$
f(n+x)=a x+b
$$
has at most two real roots. Therefore, each polynomial in $A$ appears at most twice in $f_{1}, f_{2}, \cdots, f_{k}$.
Thus, $k \leqslant 20$.
Below is an example for $k=20$.
$$
\text { Let } P_{i}(x)=[x-(2 i-1)](x-2 i)+x \text {, }
$$
where $i=1,2, \cdots, 10$.
Then $f_{2 i-1}=f_{2 i}=P_{i}$.
Hence, $f_{i}(i)=i(1 \leqslant i \leqslant 20)$. | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
17. Alex has 75 red cards and 75 blue cards. It is known that Alex can exchange 2 red cards for 1 silver card and 1 blue card at one stall, and can exchange 3 blue cards for 1 silver card and 1 red card at another stall. If he continues to exchange according to the above methods until he can no longer exchange any cards, how many silver cards will Alex have in the end?
(A) 62
(B) 82
( C) 83
(.D) 102
(E) 103 | 17. E.
If Alex has $a$ red cards, $b$ blue cards, and $c$ silver cards, denoted as $(a, b, c)$, then
$$
\begin{array}{l}
(75,75,0) \rightarrow(1,112,37) \\
\rightarrow(38,1,74) \rightarrow(0,20,93) \\
\rightarrow(6,2,99) \rightarrow(0,5,102) \\
\rightarrow(1,2,103) .
\end{array}
$$
Therefore, Alex has 103 silver cards. | 103 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
23. In $\triangle A B C$, it is known that $A B=13, B C=14$, $C A=15$, points $D, E, F$ are on sides $B C, C A, D E$ respectively, satisfying $A D \perp B C, D E \perp A C, A F \perp B F$, the length of segment $D F$ is a reduced fraction $\frac{m}{n}\left(m, n \in \mathbf{N}_{+},(m, n)=1\right)$. Then $m+n=(\quad)$.
(A) 18
(B) 21
(C) 24
(D) 27
(E) 30 | 23. B.
As shown in Figure 5.
Let $p$ be the semi-perimeter of $\triangle ABC$. Then
$$
p=\frac{a+b+c}{2}=\frac{13+14+15}{2}=21.
$$
Thus, $S_{\triangle ABC}=\sqrt{p(p-a)(p-b)(p-c)}$
$$
\begin{array}{l}
=\sqrt{21 \times 8 \times 7 \times 6}=84 \\
\Rightarrow AD=\frac{2 S_{\triangle ABC}}{BC}=\frac{2 \times 84}{14}=12.
\end{array}
$$
Since $AD \perp BC$, we have
$$
BD=\sqrt{AB^2-AD^2}=\sqrt{13^2-12^2}=5.
$$
Given $BC=14$, then $DC=9$.
Since $DE \perp AC$, we have
$$
\begin{array}{l}
DE=\frac{AD \cdot DC}{AC}=\frac{12 \times 9}{15}=\frac{36}{5}, \\
AE=\frac{AD^2}{AC}=\frac{12^2}{15}=\frac{48}{5}.
\end{array}
$$
Since $\angle AFB=\angle ADB=90^\circ$, points $A, B, D, F$ are concyclic.
Therefore, $\angle AFE=\angle ABC$.
Thus, $\cot \angle AFE=\cot \angle ABC=\frac{5}{12}$.
In $\triangle AEF$,
$$
\begin{array}{l}
EF=AE \cot \angle AFE=\frac{48}{5} \times \frac{5}{12}=4 \\
\Rightarrow DF=DE-EF=\frac{36}{5}-4=\frac{16}{5} \\
\Rightarrow m=16, n=5 \\
\Rightarrow m+n=21.
\end{array}
$$ | 21 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $x, y, z \in \mathbf{R}_{+}$,
$$
\begin{array}{l}
S=\sqrt{x+2}+\sqrt{y+5}+\sqrt{z+10}, \\
T=\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1} .
\end{array}
$$
Then the minimum value of $S^{2}-T^{2}$ is | 4. 36 .
$$
\begin{array}{l}
S^{2}-T^{2}=(S+T)(S-T) \\
=(\sqrt{x+2}+\sqrt{x+1}+\sqrt{y+5}+ \\
\quad \sqrt{y+1}+\sqrt{z+10}+\sqrt{z+1}) \cdot \\
\quad\left(\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{4}{\sqrt{y+5}+\sqrt{y+1}}+\frac{9}{\sqrt{z+10}+\sqrt{z+1}}\right) \\
\geqslant(1+2+3)^{2}=36 .
\end{array}
$$
The equality holds if and only if $x=\frac{7}{9}, y=\frac{55}{9}, z=15$. | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. From $1,2, \cdots, 2013$, select $3 k$ different numbers to form $k$ triples $\left(a_{i}, b_{i}, c_{i}\right)(i=1,2$, $\cdots, k)$. If $a_{i}+b_{i}+c_{i}(i=1,2, \cdots, k)$ these $k$ numbers are all distinct and less than 2013, then the maximum value of $k$ is | 8. 402.
On the one hand,
$$
\begin{array}{l}
\sum_{i=1}^{k}\left(a_{i}+b_{i}+c_{i}\right) \leqslant \sum_{i=1}^{k}(2013-i) \\
=2013 k-\frac{k(k+1)}{2},
\end{array}
$$
and
$$
\begin{array}{l}
\sum_{i=1}^{k}\left(a_{i}+b_{i}+c_{i}\right) \geqslant 1+2+\cdots+3 k \\
=\frac{3 k(3 k+1)}{2} .
\end{array}
$$
Thus, $2013 k-\frac{k(k+1)}{2} \geqslant \frac{3 k(3 k+1)}{2}$.
Solving this, we get $k \leqslant 402$.
On the other hand, construct
$$
\left(a_{i}, b_{i}, c_{i}\right)=(i, 402+i, 1207-i),
$$
where, $(i=1,2, \cdots, 402)$.
The sum of the three numbers in this triplet is $1609+i$ $(i=1,2, \cdots, 402)$, which satisfies the problem's conditions. | 402 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given the sequence $\left\{a_{n}\right\}(n \in \mathbf{N})$ satisfies: $a_{1}=1$, and for any non-negative integers $m, n (m \geqslant n)$, we have
$$
a_{m+n}+a_{m-n}+m-n-1=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right) \text {. }
$$
Find the value of $\left[\frac{a_{2013}}{2012}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$). | 10. Let $m=n$, we get $a_{0}=1$.
Let $n=0$, we get $a_{2 m}=4 a_{m}+2 m-3$.
In particular, $a_{2}=3$.
Let $n=1$, we get
$$
\begin{array}{l}
a_{m+1}+a_{m-1}+m-2=\frac{1}{2}\left(a_{2 m}+a_{2}\right)=2 a_{m}+m \\
\Rightarrow a_{m+1}-a_{m}=a_{m}-a_{m-1}+2 .
\end{array}
$$
Thus, $a_{m+1}-a_{m}=2 m(m \geqslant 1)$,
$$
\begin{array}{l}
a_{m}=a_{1}+\sum_{k=1}^{m-1}\left(a_{k+1}-a_{k}\right)=1+\sum_{k=1}^{m-1} 2 k \\
=m(m-1)+1(m \geqslant 1) .
\end{array}
$$
Then $\left[\frac{a_{2013}}{2012}\right]=2013$. | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Given
$$
f(x)=\frac{1+\ln (x+1)}{x}, g(x)=\frac{k}{x+1} .
$$
Find the largest positive integer $k$, such that for any positive number $c$, there exist real numbers $a$ and $b$ satisfying $-1<a<b<c$, and
$$
f(c)=f(a)=g(b) .
$$ | 11. For positive integer $k$, it is clear that $g(x)=\frac{k}{x+1}$ is a decreasing function on the interval $(-1,+\infty)$.
Thus, for any positive number $c$,
$$
f(c)=g(b)>g(c) \text {. }
$$
When $x>0$, the inequality
$$
\begin{array}{l}
f(x)>g(x) \\
\Leftrightarrow k0) \text {. }
$$
Then $h^{\prime}(x)=\frac{x-1-\ln (x+1)}{x^{2}}$.
$$
\text { Let } \varphi(x)=x-1-\ln (x+1)(x>0) \text {. }
$$
Then $\varphi^{\prime}(x)=\frac{x}{x+1}>0$.
Hence, $\varphi(x)$ is an increasing function when $x>0$.
Also, $\varphi(2)=1-\ln 30 \text {, }
$$
Therefore, there exists a unique positive real number $x_{0}$ such that
$$
\varphi\left(x_{0}\right)=x_{0}-1-\ln \left(x_{0}+1\right)=0 \text {. }
$$
Thus, $h^{\prime}\left(x_{0}\right)=0$, and $x_{0} \in(2,3)$.
Hence, when $x \in\left(0, x_{0}\right)$, $h^{\prime}(x)0$, $h(x)$ is an increasing function.
Therefore, when $x>0$, combining with equation (2), the minimum value of $h(x)$ is $h\left(x_{0}\right)=x_{0}+1 \in(3,4)$.
Combining with equation (1), we have the positive integer
$k \leqslant 3$.
We now prove that when $k=3$, for $-10 . \\
\text { Let } \tau(x)=1-2 x+(x+1) \ln (x+1) \text {, where } \\
\text { }-1\tau(0)>0$.
Therefore, equation (4) holds.
Note that,
$g(x)=\frac{3}{x+1}(x \in(-1,+\infty))$ has a range of $(0,+\infty)$,
$f(x)=\frac{1+\ln (x+1)}{x}(x \in(0,+\infty))$ also has a range of $(0,+\infty)$,
$$
f(x)=\frac{1+\ln (x+1)}{x}(x \in(-1,0)) \text { has a range }
$$
of $\mathbf{R}$.
Combining the graphs of the functions, we know that for any positive number $c$, there exist real numbers $a 、 b$ satisfying $-1<a<b<c$, and
$$
f(c)=f(a)=g(b) \text {. }
$$
In summary, the maximum value of the positive integer $k$ is 3. | 3 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Find the smallest positive integer $n$, such that the sum of the squares of all positive divisors of $n$ is $(n+3)^{2}$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note and not part of the problem statement, so it is provided in its original form as it is not to be translated. | Three, let $16 n+8, \\
\end{array}
$$
Contradiction.
Thus, $k \leqslant 5$.
And $k=0$ clearly does not meet the problem's conditions, so $1 \leqslant k \leqslant 5$.
Let $n=\prod_{i=1}^{i} p_{i}^{\alpha_{i}}$. Then
$$
\prod_{i=1}^{t}\left(\alpha_{i}+1\right)=k+2 \in[3,7] .
$$
Thus, $1 \leqslant t \leqslant 2$.
We discuss three cases.
(1) $n=p^{\alpha}$ (where $p$ is a prime, $\alpha \in \mathbf{Z}_{+}$, and $2 \leqslant \alpha \leqslant 6$).
Then $p^{2}+p^{4}+\cdots+p^{2(\alpha-1)}=6 p^{\alpha}+8$.
Thus, $p^{2} 18 \Rightarrow p=2$.
Taking both sides modulo 8, we get $4 \equiv 0(\bmod 8)$, contradiction.
(2) $n=p q$ (where $p<q$ are primes).
Then $p^{2}+q^{2}=6 p q+8$.
Thus, $q=3 p+2 \sqrt{2 p^{2}+2}$.
Upon verification, the smallest prime $p=7$ makes $q$ a prime, at which point, $q=41, n=p q=287$.
(3) $n=p q^{2}$ (where $p, q$ are distinct primes).
Then $p^{2}+q^{2}+q^{4}+p^{2} q^{2}=6 p q^{2}+8$.
If $2 \mid n$, taking both sides of equation (1) modulo 4, we find $p=q=2$, contradiction.
Otherwise, $p, q$ are both odd, taking both sides of equation (1) modulo 8, we get $4 \equiv 6 p(\bmod 8)$, so $p=2$, contradiction.
In summary, the minimum value of $n$ is 287. | 287 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
354 Divide the sides of the equilateral $\triangle A B C$ into four equal parts, and draw lines parallel to the other two sides through each division point. The 15 points formed by the intersections of the sides of $\triangle A B C$ and these parallel lines are called lattice points. Among these 15 lattice points, if $n$ points are chosen, there will definitely be three points that can form an isosceles triangle (including equilateral triangles). Find the minimum value of $n$.
---
The problem is to find the minimum value of $n$ such that, when $n$ points are chosen from the 15 lattice points, there will always be three points that can form an isosceles triangle (including equilateral triangles). | Solve for the minimum value of $n$ being 6.
Let the three equal division points from point $A$ to $B$ on side $AB$ be $L, F, W$; the three equal division points from point $B$ to $C$ on side $BC$ be $V, D, U$; and the three equal division points from point $C$ to $A$ on side $CA$ be $N, E, M$. Denote the intersection point of lines $MV, LU, EF$ as $P$; the intersection point of lines $MV, WN, DF$ as $Q$; and the intersection point of lines $WN, LU, DE$ as $R$.
If the minimum value of $n$ is 5, and we take five grid points on side $AB$, then there do not exist three grid points that can form an isosceles triangle. Therefore, $n \geqslant 6$.
Below, we prove: If among these 15 grid points, we take $n$ points such that no three points can form an isosceles triangle, then $n \leqslant 5$.
Assume the $n$ points taken are red points, and the other points are blue points.
Since $P, Q, R$ cannot all be red points, we discuss the following three scenarios:
(1) If only two of $P, Q, R$ are red points, assume $Q, R$ are red points. Thus, $P, A, D$ are all blue points, and $U, V$ and $E, F$ are also all blue points.
Since at most one of $M, C$ can be a red point (otherwise, $M, C$ and $R$ form an isosceles triangle), at most one of $N, C$ can be a red point (otherwise, $N, C$ and $R$ form an isosceles triangle), and at most one of $M, N$ can be a red point (otherwise, $M, N$ and $Q$ form an isosceles triangle), at most one of $M, N, C$ can be a red point.
Similarly, at most one of $L, W, B$ can be a red point.
Therefore, $n \leqslant 5$.
(2) If only one of $P, Q, R$ is a red point, assume $P$ is a red point.
(i) If $D$ is a red point, then $W, N$ are both blue points, and at most one of $F, L, M, E$ can be a red point.
If $L$ (or $M$) is a red point, then $A$ is a blue point, and at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
If $F$ (or $E$) is a red point, then $A$ is a blue point, and at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
(ii) If $D$ is a blue point and $A$ is a red point, then $L, M$ and $W, N$ are all blue points, at most one of $E, F$ can be a red point; at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
(iii) If $D, A$ are both blue points, since at most one of $L, M$ can be a red point, assume $L$ is a red point, then $E, F, M$ are all blue points, at most one of $W, N$ can be a red point; at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
If $L, M$ are both blue points, when $E, F$ are both red points, $W, N$ and $B, C$ are all blue points; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
When $E, F$ have at most one red point, at most one of $W, N$ can be a red point; at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point. Therefore, $n \leqslant 5$.
(3) If $P, Q, R$ are all blue points, when each of the three-point groups $A, L, M$, $B, V, W$, $C, N, U$, $D, E, F$ have at most one red point, then there are at most four red points, otherwise, among $A, L, M$, $B, V, W$, $C, N, U$, $D, E, F$, there must be a three-point group that contains at least two red points.
(i) If $D, E, F$ contain at least two red points, assume $E, F$ are red points, then $D, A$ and $B, C$ are all blue points, at most one of $L, M$ can be a red point; at most one of $W, V$ can be a red point; at most one of $N, U$ can be a red point. Therefore, $n \leqslant 5$.
(ii) If among $A, L, M$, $B, V, W$, $C, N, U$ there is a three-point group that contains at least two red points, assume $B, V, W$ contain at least two red points.
If $V, W$ are red points, then $B, E$ and $F, D$ are all blue points. If one of $L, U$ is a red point, assume $L$ is a red point, then $C, N$ are all blue points, at most one of $A, M$ can be a red point, even if $U$ is a red point, $n \leqslant 5$; if $L, U$ are both blue points, then at most one of $M, C$ can be a red point, at most one of $A, N$ can be a red point, therefore, $n \leqslant 5$.
If $B, V$ are red points, then $W$ is a blue point.
1) If $F$ is a red point, then $D, M, N, E$ are all blue points, so at most one of $A, C$ can be a red point, at most one of $L, U$ can be a red point. Therefore, $n \leqslant 5$.
2) If $F$ is a blue point, $D$ is a red point, then $E, L$ are both blue points, at most one of $A, C$ can be a red point; at most one of $M, N, U$ can be a red point. Therefore, $n \leqslant 5$.
3) If $F, D$ are blue points, $L$ is a red point, then $U, N$ are both blue points, at most one of $A, C$ can be a red point; at most one of $M, E$ can be a red point. Therefore, $n \leqslant 5$.
4) If $F, D, L$ are blue points, $U$ is a red point, then $A$ is a blue point, at most one of $M, C$ can be a red point; at most one of $N, E$ can be a red point. Therefore, $n \leqslant 5$.
5) If $F, D, L, U$ are blue points, then at most one of $M, N$ can be a red point; at most one of $A, C$ can be a red point. Even if $E$ is a red point, $n \leqslant 5$.
Using the contrapositive, when $n \geqslant 6$, we know that there must exist | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 What is the maximum number of rational points (points with both coordinates being rational numbers) that can lie on a circle in the plane, given that the center of the circle is not a rational point. | 【Analysis】If $A, B, C$ are three rational points on a circle,
then the midpoint $D$ of $AB$ is a rational point, the slope of $AB$ is a rational number or infinite, so the equation of the perpendicular bisector of $AB$ is a linear equation with rational coefficients. Similarly, the equation of the perpendicular bisector of $BC$ is also a linear equation with rational coefficients. Therefore, the intersection of the perpendicular bisector of $AB$ and the perpendicular bisector of $BC$ is a rational point, i.e., the center of the circle is a rational point, which does not meet the condition.
Two examples of rational points: $(0,0)$ and $(1,0)$, the center of the circle is $\left(\frac{1}{2}, \sqrt{2}\right)$, which satisfies the condition.
Therefore, at most two rational points can exist on a circle such that the center of the circle is not a rational point. | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let $A, B, C$ be three non-collinear lattice points on a plane, and the side lengths of $\triangle ABC$ are all positive integers. Find the minimum value of $AB$ and the minimum perimeter. | 【Analysis】If $A B=1$, we might as well set $A(0,0), B(1,0)$. Then $|A C-B C|<A B=1$, which can only be $A C=B C$.
Thus, point $C$ lies on the perpendicular bisector of $A B$.
Therefore, the x-coordinate of point $C$ is $\frac{1}{2}$, meaning $C$ cannot be a lattice point.
Hence, $A B$ cannot be 1.
If $A B=2$, we might as well set $A(0,0), B(2,0)$.
Then $|A C-B C|<2$, which can only be $|A C-B C|=0,1$.
If $A C=B C \Rightarrow x_{c}-1 \Rightarrow A C^{2}=1+y_{c}^{2}$, thus, the difference of two perfect squares cannot be 1.
If $B C-A C=1$
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array}{l}
A C^{2}=x_{c}^{2}+y_{c}^{2}=k^{2}, \\
B C^{2}=\left(x_{C}-2\right)^{2}+y_{C}^{2}=(k+1)^{2}
\end{array}\right. \\
\Rightarrow 2 k+1=4-4 x_{C} .
\end{array}
$$
The parity of both sides of the above equation is different, so the equation does not hold, i.e., $B C-A C$ is not equal to 1.
Therefore, $A B$ cannot be 2.
If $A B=3$, we might as well set $A(0,0), B(3,0)$. Then $C(3,4)$ satisfies the conditions of the problem.
Therefore, the minimum value of $A B$ is 3.
Noting that the three vertices of $\triangle A B C$ are lattice points, using the area formula of analytic geometry, we know
$$
2 S_{\triangle A B C}=\left(\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right)
$$
The absolute value is a positive integer.
Also, the three sides of $\triangle A B C$ are all integers, with side lengths greater than or equal to 3, thus, using Heron's formula, the side lengths of 3, 3, 3; 3, 3, 4; 3, 4, 4; and 3, 3, 5 are all impossible (the area is irrational); 3, 4, 5 satisfies the conditions.
Therefore, the minimum perimeter is 12. | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 On a plane, there exist $n$ points, no three of which are collinear, and when these $n$ points are arbitrarily labeled as $A_{1}, A_{2}, \cdots, A_{n}$, the broken line $A_{1} A_{2} \cdots A_{n}$ does not intersect itself. Find the maximum value of $n$.
| 【Analysis】When $n=2,3$, it is obviously true.
When $n=4$, if the convex hull of the four points is a triangle, then it satisfies the condition.
Next, we show that when $n \geqslant 5$, it is impossible to satisfy the condition.
Consider only five points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ among these $n$ points.
If the convex hull of these five points is a pentagon $A B C D E$ or a quadrilateral $A B C D$, then the broken line $A C B D$ does not satisfy the condition.
Assume the convex hull of points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ is $\triangle P_{1} P_{2} P_{3}$, i.e., points $P_{4}, P_{5}$ are inside $\triangle P_{1} P_{2} P_{3}$. The line connecting $P_{4}, P_{5}$ does not pass through $P_{1}, P_{2}, P_{3}$, so it must intersect two sides of $\triangle P_{1} P_{2} P_{3}$. Assume without loss of generality that $P_{4} P_{5}$ intersects $P_{1} P_{3}$ and $P_{2} P_{3}$. Then, quadrilateral $P_{1} P_{2} P_{4} P_{5}$ is a convex quadrilateral, which does not satisfy the condition.
Therefore, $n \geqslant 5$ does not satisfy the condition, i.e., the maximum value of $n$ is 4.
Using the convex hull to analyze problems involving finite point sets is a common method in combinatorial geometry. | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Find the smallest $n \in \mathbf{N}_{+}$, such that for any finite point set $M$ in the plane, if any $n$ points in $M$ can be covered by two lines, then there must exist two lines that can cover the point set $M$. | The required minimum value is 6.
When $|M| \leqslant 6$, the proposition is obviously true.
If $|M|>6$, for any six points $A_{1}, A_{2}, \cdots, A_{6}$ in $M$, they can be covered by two lines $l_{1}$ and $l_{2}$. Without loss of generality, assume that $l_{1}$ contains at least three points $A_{1}, A_{2}, A_{3}$. Let $M_{1}=M \cap l_{1}$ and $M_{2}=M \backslash l_{1}$.
If $\left|M_{2}\right| \leqslant 2$, the proposition is obviously true.
Without loss of generality, assume $\left|M_{2}\right| \geqslant 3$, and take $A_{1}, A_{2}, A_{3} \in M_{1}$, $B_{1}, B_{2}, B_{3} \in M_{2}$.
Then there exist two lines $l_{1}^{\prime}$ and $l_{2}^{\prime}$ that cover these six points. Among $A_{1}, A_{2}, A_{3}$, there must be two points on one of the lines, say $A_{1}, A_{2}$ are on $l_{1}^{\prime}$, then $l_{1}=l_{1}^{\prime}$. Points $B_{1}, B_{2}, B_{3}$ are not on $l_{1}^{\prime}$, so $B_{1}, B_{2}, B_{3}$ are on $l_{2}^{\prime}$. Replacing point $B_{3}$ with another point $B_{k}$ in $M_{2}$, then $B_{1}, B_{2}, B_{k}$ are collinear. Therefore, there exist two lines that can cover the point set $M$, and the proposition is true.
When $n=5$, there exists a counterexample. Take the three vertices and the midpoints of the three sides of a regular $\triangle ABC$. Any five of these points can be covered by two lines, but these six points cannot be covered by two lines. | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. (50 points) Given that $a$, $b$, and $c$ are three distinct real numbers. If any two of the quadratic equations
$$
\begin{array}{l}
x^{2}+a x+b=0, \\
x^{2}+b x+c=0, \\
x^{2}+c x+a=0
\end{array}
$$
have exactly one common root, find the value of $a^{2}+b^{2}+$ $c^{2}$. | 2. From equations (1) and (2), we know their common root is $p=\frac{b-c}{b-a}$.
Similarly, the common roots of equations (2) and (3), and equations (1) and (3) are $q=\frac{c-a}{c-b}$ and $r=\frac{a-b}{a-c}$, respectively.
Thus, $p q r=-1$.
If any two of $p, q, r$ are equal, assume $p=q$, then the three equations have a common root $p$. Therefore,
$$
p=r \Rightarrow p=r=q=-1 \Rightarrow a=b=c,
$$
which is a contradiction.
Assume $p, q, r$ are distinct. Then the three equations have the form
$$
\begin{array}{l}
(x-p)(x-r)=0, \\
(x-p)(x-q)=0, \\
(x-q)(x-r)=0 .
\end{array}
$$
Thus, $a=-p-r=q r, b=-q-p=r p$,
$$
\begin{array}{l}
c=-r-q=p q . \\
\text { Hence }-2(p+q+r)=p q+q r+r p, \\
-1=(q+1)(p+1)(r+1) \\
=p q+q r+p+p+q+r .
\end{array}
$$
Therefore, $p+q+r=1, p q+q r+r p=-2$.
Then $a^{2}+b^{2}+c^{2}$
$$
=2\left(p^{2}+q^{2}+r^{2}+p q+q r+r p\right)=6 .
$$ | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4.12 The acrobats are numbered $1, 2, \cdots$, 12. They are to be arranged in two circles, $A$ and $B$, each with six people. In circle $B$, each acrobat stands on the shoulders of two adjacent acrobats in circle $A$. If the number of each acrobat in circle $B$ is equal to the sum of the numbers of the two acrobats below him, then such an arrangement is called a "tower". How many structurally different towers can be formed?
[Note] Towers that are the same after rotation or reflection are considered the same structure. For example, with 8 people, draw a circle and fill in the numbers of the acrobats in the bottom layer inside the circle, and the numbers of the acrobats in the top layer outside the circle. In Figure 2, the three diagrams are all towers, but the last two can be obtained from the first one by rotation or reflection, so they belong to the same structure. | 4. Let the sums of the elements in circles $A$ and $B$ be $x$ and $y$ respectively. Then $y=2x$. Therefore,
$$
3x = x + y = 1 + 2 + \cdots + 12 = 78.
$$
Solving for $x$ gives $x = 26$.
Clearly, $1, 2 \in A$ and $11, 12 \in B$.
Let $A = \{1, 2, a, b, c, d\} (a < b < c < d)$. Then $a + b + c + d = 23$, and $a \geq 3, 8 \leq d \leq 10$ (if $d \leq 7$, then $a + b + c + d \leq 4 + 5 + 6 + 7 = 22$, a contradiction).
(1) If $d = 8$, then
$$
A = \{1, 2, a, b, c, 8\}, c \leq 7, a + b + c = 15.
$$
Thus, $(a, b, c) = (3, 5, 7)$ or $(4, 5, 6)$, i.e.,
$$
A = \{1, 2, 3, 5, 7, 8\} \text{ or } \{1, 2, 4, 5, 6, 8\}.
$$
If $A = \{1, 2, 3, 5, 7, 8\}$, then
$$
B = \{4, 6, 9, 10, 11, 12\}.
$$
Since $B$ contains $4, 6, 11, 12$, in the $A$ circle, 1 must be adjacent to 3, 1 must be adjacent to 5, 5 must be adjacent to 7, and 8 must be adjacent to 3. In this case, there is only one arrangement, leading to one tower, as shown in Figure 6(a).
$$
\begin{array}{l}
\text{If } A = \{1, 2, 4, 5, 6, 8\}, \text{ then} \\
B = \{3, 7, 9, 10, 11, 12\}.
\end{array}
$$
Similarly, in the $A$ circle, 1 must be adjacent to 2, 5 must be adjacent to 6, and 4 must be adjacent to 8. In this case, there are two arrangements, leading to two towers, as shown in Figures 6(b) and (c).
(2) If $d = 9$, then
$$
A = \{1, 2, a, b, c, 9\}, c \leq 8, a + b + c = 14.
$$
Thus, $(a, b, c) = (3, 5, 6)$ or $(3, 4, 7)$, i.e.,
$$
A = \{1, 2, 3, 5, 6, 9\} \text{ or } \{1, 2, 3, 4, 7, 9\}.
$$
If $A = \{1, 2, 3, 5, 6, 9\}$, then
$$
B = \{4, 7, 8, 10, 11, 12\}.
$$
To get $4, 10, 12$ in the $B$ circle, in the $A$ circle, 1, 3, and 9 must be pairwise adjacent, which is impossible;
If $A = \{1, 2, 3, 4, 7, 9\}$, then
$$
B = \{5, 6, 8, 10, 11, 12\}.
$$
To get $6, 8, 12$ in the $B$ circle, in the $A$ circle, 2 must be adjacent to 4, 1 must be adjacent to 7, and 9 must be adjacent to 3. In this case, there are two arrangements, leading to two towers, as shown in Figure 7.
(3) If $d = 10$, then
$$
A = \{1, 2, a, b, c, 10\}, c \leq 9, a + b + c = 13.
$$
Thus, $(a, b, c) = (3, 4, 6)$, i.e.,
$$
A = \{1, 2, 3, 4, 6, 10\},
$$
$$
B = \{5, 7, 8, 9, 11, 12\}.
$$
To get $8, 9, 11, 12$ in the $B$ circle, in the $A$ circle, 6 must be adjacent to 2, 6 must be adjacent to 3, 10 must be adjacent to 1, and 10 must be adjacent to 2. In this case, there is only one arrangement, leading to one tower, as shown in Figure 8.
Therefore, there are 6 structurally distinct towers. | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $f(x)=\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{2013!}\right]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. For an integer $n$, if the equation $f(x)=n$ has a real solution, then $n$ is called a "good number". Find the number of good numbers in the set $\{1,3,5, \cdots, 2013\}$.
（Wu Genxiu, Contributed） | 5. First, give two obvious conclusions:
(1) If $m$ is a positive integer and $x$ is a real number, then
$$
\left[\frac{x}{m}\right]=\left[\frac{[x]}{m}\right] ;
$$
(2) For any integer $l$ and positive even number $m$, we have
$$
\left[\frac{2 l+1}{m}\right]=\left[\frac{2 l}{m}\right] \text {. }
$$
Returning to the original problem.
In conclusion (1), let $m=k!(k=1,2, \cdots, 2013)$, and summing up, we get
$$
f(x)=\sum_{k=1}^{2013}\left[\frac{x}{k!}\right]=\sum_{k=1}^{2013}\left[\frac{[x]}{k!}\right]=f([x]),
$$
This shows that the equation $f(x)=n$ has a real solution if and only if the equation $f(x)=n$ has an integer solution.
From now on, we only need to consider the case where $x$ is an integer.
$$
\begin{array}{l}
\text { By } f(x+1)-f(x) \\
=[x+1]-[x]+\sum_{k=2}^{2013}\left(\left[\frac{x+1}{k!}\right]-\left[\frac{x}{k!}\right]\right) \\
\geqslant 1,
\end{array}
$$
we know that $f(x)(x \in \mathbf{Z})$ is monotonically increasing.
Next, find integers $a$ and $b$ such that
$$
\begin{array}{l}
f(a-1)2013,
\end{array}
$$
so $b=1173$.
Therefore, the good numbers in $\{1,3,5, \cdots, 2013\}$ are the odd numbers in $\{f(0), f(1), \cdots, f(1173)\}$.
In equation (1), let $x=2 l(l=0,1, \cdots, 586)$, by conclusion (2) we know
$$
\begin{array}{l}
{\left[\frac{2 l+1}{k!}\right]=\left[\frac{2 l}{k!}\right](2 \leqslant k \leqslant 2013) .} \\
\text { Hence } f(2 l+1)-f(2 l) \\
=1+\sum_{k=2}^{2013}\left(\left[\frac{2 l+1}{k!}\right]-\left[\frac{2 l}{k!}\right]\right)=1,
\end{array}
$$
This shows that $f(2 l)$ and $f(2 l+1)$ have exactly one odd number.
Thus, $\{f(0), f(1), \cdots, f(1173)\}$ contains exactly $\frac{1174}{2}=587$ odd numbers, i.e., the set $\{1,3,5, \cdots, 2013\}$ contains 587 good numbers. | 587 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Remove any $2 \times 2$ small square from the corner of a $3 \times 3$ grid, and the remaining shape is called a "corner shape" (Figure 3 is an example of a corner shape). Now, place some non-overlapping corner shapes in a $10 \times 10$ grid (Figure 4). The boundaries of the corner shapes must coincide with the boundaries or grid lines of the grid. Find the maximum value of the positive integer $k$, such that no matter how $k$ corner shapes are placed, it is always possible to place one more complete corner shape in the grid. | 7. First, $k_{\max }$
$<8$. This is because, if eight corner shapes are placed in the manner shown in Figure 9, it is impossible to place another corner shape in the grid.
Next, we prove that after placing seven corner shapes arbitrarily, it is still possible to place another complete corner shape.
Cover the 5th and 6th rows and the 5th and 6th columns of the $10 \times 10$ grid, leaving four $4 \times 4$ smaller grids. After placing seven corner shapes, since each corner shape cannot intersect with two of the aforementioned $4 \times 4$ smaller grids, according to the pigeonhole principle, there must exist a $4 \times 4$ smaller grid $S$ such that at most one corner shape intersects with $S$. Since a corner shape can be contained within a $3 \times 3$ smaller grid, the part of the smaller grid $S$ occupied by the corner shape must be contained within a $3 \times 3$ smaller grid at one of its corners.
As shown in Figure 10, a new corner shape can be placed in the remaining part of the smaller grid $S$ after removing a $3 \times 3$ smaller grid from one of its corners. Therefore, $k=7$ satisfies the condition.
In conclusion, $k_{\max }=7$. | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $A$ be a set of ten real-coefficient polynomials of degree five. It is known that there exist $k$ consecutive positive integers $n+1$, $n+2, \cdots, n+k$, and $f_{i}(x) \in A(1 \leqslant i \leqslant k)$, such that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence. Find the maximum possible value of $k$.
| 6. Given that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence, we know there exist real numbers $a$ and $b$ such that
$$
f_{i}(n+i)=a i+b .
$$
Notice that, for any fifth-degree polynomial $f$, the equation
$$
f(n+x)=a x+b
$$
has at most five real roots. Therefore, each polynomial in $A$ appears at most five times in $f_{1}, f_{2}, \cdots, f_{k}$.
Thus, $k \leqslant 50$.
Below is an example for $k=50$.
$$
\begin{array}{l}
\text { Let } P_{k}(x) \\
=[x-(5 k-4)][x-(5 k-3)] \cdots(x-5 k)+x,
\end{array}
$$
where $k=1,2, \cdots, 10$.
Then $f_{5 k}=f_{5 k-1}=f_{5 k-2}=f_{5 k-3}=f_{5 k-4}=P_{k}$, where $1 \leqslant k \leqslant 10$.
Thus, $f_{k}(k)=k(1 \leqslant k \leqslant 50)$. | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 4, in the Cartesian coordinate system, $O$ is the origin, the diagonals of $\square A B O C$ intersect at point $M$, and the hyperbola $y=\frac{k}{x}(x<0)$ passes through points $B$ and $M$. If the area of $\square A B O C$ is 24, then $k=$ . $\qquad$ | 2, 1. -8.
Let $M\left(\frac{k}{y}, y\right)$. Then $B\left(\frac{k}{2 y}, 2 y\right), C\left(\frac{3 k}{2 y}, 0\right)$.
From $S_{\text {OABOC }}=4 S_{\triangle O C M}=2\left|x_{c} y_{M}\right|$
$\Rightarrow 24=2\left|\frac{3 k}{2}\right| \Rightarrow|k|=8 \Rightarrow k=-8$. | -8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $m$, $n$, and $p$ are real numbers. If $x-1$ and $x+4$ are both factors of the polynomial $x^{3}+m x^{2}+n x+p$, then
$$
2 m-2 n-p+86=
$$
$\qquad$. | 2. 100 .
From the divisibility property of polynomials, we know
$$
\left\{\begin{array} { l }
{ 1 + m + n + p = 0 , } \\
{ - 6 4 + 1 6 m - 4 n + p = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
p=12-4 m, \\
n=3 m-13 .
\end{array}\right.\right.
$$
Therefore, $2 m-2 n-p+86$
$$
\begin{array}{l}
=2 m-2(3 m-13)-(12-4 m)+86 \\
=100 .
\end{array}
$$ | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
During the Teachers' Day, 200 teachers at a school sent text messages to greet each other, with each teacher sending exactly 1 text message to another teacher. Now, from them, the maximum number of $k$ teachers can be selected to attend an award ceremony, such that none of them has sent a text message to any of the others. Try to find the minimum value of $k$. | Let the 200 teachers be denoted as $A_{1}, A_{2}, \cdots, A_{200}$, and let $G=\left\{A_{1}, A_{2}, \cdots, A_{200}\right\}$. If $A_{i}$ sends a message to $A_{j}$, then we mark an arrow from $A_{i}$ to $A_{j}$ as
$A_{i} \rightarrow A_{j}(1 \leqslant i \neq j \leqslant 200)$.
First, construct an instance for $k=67$.
In $G$, let
$$
A_{1} \rightarrow A_{2} \rightarrow A_{1}, A_{3 i} \rightarrow A_{3 i+1} \rightarrow A_{3 i+2} \rightarrow A_{3 i} \text {, }
$$
resulting in 1 pair of two people and 66 triangles $\triangle A_{3 i} A_{3 i+1} A_{3 i+2}$. From the pair, we can select one person, and from each triangle, we can select one person $A_{3 i}(i=1,2, \cdots, 66)$, hence
$$
A=\left\{A_{1}, A_{3}, A_{6}, \cdots, A_{195}, A_{198}\right\}
$$
satisfies the requirement.
Next, prove that for the largest $k$ that satisfies the requirement, $k \geqslant 67$.
Assume $A=\left\{A_{1}, A_{2}, \cdots, A_{k}\right\}$ is a set of $k$ representatives that satisfy the requirement, and let
$$
B=G \backslash A=\left\{A_{k+1}, A_{k+2}, \cdots, A_{200}\right\} .
$$
Next, calculate the number of arrows in $G$.
On one hand, by the problem statement, $G$ has exactly 200 arrows.
On the other hand, estimate the number of arrows in $G$.
Note that $G=A \cup B$, and consider two cases.
(1) The number of arrows within $A$ is 0.
(2) There are at least $200-k$ arrows between $A$ and $B$ (since each person in $B$ must have at least one arrow to someone in $A$). Otherwise, if there exists a person $A_{j}(k+1 \leqslant j \leqslant 200)$ in $B$ who has no arrow to anyone in $A$, then $A^{\prime}=A \cup\left\{A_{j}\right\}$ also satisfies the requirement, but $\left|A^{\prime}\right|>k$, contradicting the maximality of $k$.
(3) There are at least $200-2 k$ arrows among the people in $B$.
In fact, partition $B$ into $m$ non-empty subsets $B_{1}$, $B_{2}, \cdots, B_{m}$, such that any two people in each $B_{i}$ can be connected by a sequence of arrows. By the maximality of $k$, we have $m \leqslant k$. By the definition of $B_{i}$, each $B_{i}$ has at least $\left|B_{i}\right|-1$ $(i=1,2, \cdots, m)$ arrows. Therefore, $B$ has at least
$$
\sum_{i=1}^{m}\left(\left|B_{i}\right|-1\right)=(200-k)-m \geqslant 200-2 k
$$
arrows.
Thus, $G$ has at least
$$
(200-k)+(200-2 k)=400-3 k
$$
arrows. Therefore,
$$
400-3 k \leqslant 200 \Rightarrow 3 k \geqslant 200 \Rightarrow k \geqslant 67 \text {. }
$$
In summary, the minimum value of $k$ is 67. | 67 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) There are 2014 points distributed on a circle, which are arbitrarily colored red and yellow. If starting from a certain point and moving around the circle in any direction to any position, the number of red points (including the starting point) is always greater than the number of yellow points, then the point is called a "good point." To ensure that there is at least one good point on the circle, find the maximum number of yellow points on the circle.
保留源文本的换行和格式，直接输出翻译结果如下：
```
Three. (50 points) There are 2014 points distributed on a circle, which are arbitrarily colored red and yellow. If starting from a certain point and moving around the circle in any direction to any position, the number of red points (including the starting point) is always greater than the number of yellow points, then the point is called a "good point." To ensure that there is at least one good point on the circle, find the maximum number of yellow points on the circle.
``` | Three, the advantages must be red points.
First, consider the simple case.
When there are $1, 2, 3, 4, 5, 6, 7$ points on the circumference, if there is at least one advantage point on the circumference, then the maximum number of yellow points on the circumference are $0, 0, 0, 1, 1, 1, 2$. From this, we can derive a general conclusion:
When there are $3n+1$ points on the circumference, and they are arbitrarily colored red and yellow. To ensure that there is at least one advantage point on the circumference, the number of yellow points must not exceed $n$.
Next, we will prove this using mathematical induction.
When $n=1$, there are four points on the circumference (one yellow point and three red points). Among the three consecutive red points, take the middle one, and it is easy to see that it is an advantage point. Therefore, the proposition holds for $n=1$.
Assume that the proposition holds for $n=k$, i.e., when there are $3k+1$ points on the circumference, colored red and yellow, to ensure that there is at least one advantage point on the circumference, the number of yellow points must not exceed $k$.
When $n=k+1$, take any one of the $k+1$ yellow points, denoted as $A$, and take the nearest red points on either side of $A$, denoted as $B$ and $C$. Remove these three points from the circumference, leaving $3k+1$ points, which satisfy the conditions of the proposition for $n=k$.
By the induction hypothesis, there is at least one advantage point on the circumference, denoted as $P$. Now, put $A, B, C$ back to their original positions.
We need to prove that point $P$ is still an advantage point.
Since $P$ is a red point, $P$ must be outside the arc $\overparen{BAC}$. Therefore, from point $P$ to any position on the arc $\overparen{BA}(\overparen{BC})$, the difference between the number of red points and yellow points is at least 1, and it is at least 0 when reaching point $A$. Hence, $P$ is still an advantage point.
Thus, the proposition holds for $n=k+1$.
On the other hand, if the number of yellow points among the $3n+1$ points is $n+1$, divide the circumference into $n+1$ segments, and place $2n$ red points into each segment, with at most two points in each segment, then no red point can be an advantage point.
Therefore, the maximum number of yellow points on the circumference is $n$.
Since $2014 = 3 \times 671 + 1$, the maximum number of yellow points on the circumference is 671. | 671 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given 361, find the smallest positive integer $n$, such that 16 numbers can be selected from $1,2, \cdots, n$ and filled into a $4 \times 4$ grid so that the product of the numbers in each row and each column is equal. | The smallest value of the positive integer $n$ is 27.
First, when $n=27$, a valid arrangement is shown in Figure 3.
Next, we prove that the smallest value of $n$ is 27.
Consider the prime factorization of the product $P$ of each row or column.
(1) If $P$ has only one prime factor, then $n \geqslant 2^{15} > 27$.
(2) If $P$ has exactly two prime factors, to minimize the largest number, these two prime factors should be as small as possible, taken as 2 and 3. In this case, at most 16 numbers can be selected from $1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, \cdots$, clearly, $n > 27$.
(3) If $P$ has exactly three prime factors, to minimize the largest number, these three prime factors should be as small as possible, taken as 2, 3, and 5. In this case, at most 16 numbers can be selected from $1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, \cdots$, and only 25 has two 5s, which cannot be selected, clearly, $n \geqslant 27$.
(4) If $P$ has more than three prime factors, then $n \geqslant 4 \times 7 = 28$.
In summary, the smallest value of the positive integer $n$ is 27.
(Song Hongjun, Fuyang High School, Zhejiang, 311400) | 27 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Find the last three digits of $2013^{2013^{2013}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Notice,
$$
\begin{array}{l}
2013 \equiv 3(\bmod 5), 2013^{2} \equiv-1(\bmod 5), \\
2013^{4} \equiv 1(\bmod 5), 2013 \equiv 13(\bmod 25), \\
2013^{2} \equiv-6(\bmod 25), 2013^{4} \equiv 11(\bmod 25), \\
2013^{8} \equiv-4(\bmod 25), 2013^{12} \equiv 6(\bmod 25), \\
2013^{20} \equiv 1(\bmod 25) .
\end{array}
$$
Since $2013 \equiv 1(\bmod 4), 2013^{20} \equiv 1(\bmod 4)$, therefore, $2013^{20} \equiv 1(\bmod 100)$.
$$
\begin{array}{l}
\text { Hence } 2013^{2013} \equiv 2013^{20 \times 100+13} \equiv 2013^{13} \equiv 13^{13} \\
\equiv(10+3)^{13} \equiv 10 \mathrm{C}_{13}^{12} \cdot 3^{12}+3^{13} \\
\equiv 3^{12} \times 133 \equiv(80+1)^{3} \times 33 \equiv 241 \times 33 \\
\equiv 41 \times 33 \equiv 53(\bmod 100) .
\end{array}
$$
$$
\begin{array}{l}
\text { By } 2013^{20} \equiv 13^{20} \equiv(10+3)^{20} \\
\equiv \mathrm{C}_{20}^{19} \cdot 10 \times 3^{19}+3^{20} \equiv 3^{19} \times 203 \\
\equiv\left(3^{5}\right)^{3} \times 3^{4} \times 78 \\
\equiv-7^{3} \times 81 \times 78 \equiv 51(\bmod 125),
\end{array}
$$
we get $2013^{40} \equiv(1+50)^{2} \equiv 101(\bmod 125)$,
$$
2013^{80} \equiv(1+100)^{2} \equiv-49(\bmod 125) \text {. }
$$
Hence $2013^{100} \equiv 1(\bmod 125)$.
$$
\text { By } 2013 \equiv 5(\bmod 8), 2013^{2} \equiv 1(\bmod 8) \text {, }
$$
we get $2013^{100} \equiv(\bmod 1000)$.
From $2013^{2013}=100 k+53$, we have
$$
\begin{array}{l}
2013^{2013^{2013}} \equiv 2013^{100 k+53} \equiv 2013^{53} \\
\equiv(2000+13)^{53} \equiv 13^{53} \equiv(10+3)^{53} \\
\equiv C_{53}^{2} \cdot 10^{2} \times 3^{51}+C_{58}^{1} \cdot 10 \times 3^{52}+3^{53} \\
\equiv 3^{51} \times 399(\bmod 1000) \\
\text { Also } 3^{7} \equiv 187(\bmod 1000), \\
3^{10} \equiv 187 \times 27 \equiv 49(\bmod 1000), \\
3^{17} \equiv 187 \times 49 \equiv 163(\bmod 1000), \\
3^{51} \equiv 163^{3} \equiv 747(\bmod 1000),
\end{array}
$$
Therefore, $3^{51} \times 399 \equiv 53(\bmod 1000)$.
In summary, the last three digits are 053. | 053 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find all odd prime numbers $p$ such that $p \mid \sum_{k=1}^{103} k^{p-1}$.
untranslated text remains the same as requested. | If $p>103$, then for $1 \leqslant k \leqslant 103$, we have
$$
\begin{array}{l}
k^{p-1} \equiv 1(\bmod p), \\
\sum_{k=1}^{103} k^{p-1} \equiv 103(\bmod p) .
\end{array}
$$
Therefore, $p \leqslant 103$.
Let $103=p q+r(0 \leqslant r < q)$, then $r=q, 103=p q+r=(p+1) r$. Since 103 is a prime number, we get $p=102, r=1$, which is a contradiction.
If $p \leqslant q$, then
$$
103=p q+r \geqslant p^{2} \Rightarrow p=3,5,7 \text {. }
$$
When $p=3$, $q=34 \equiv 1=r(\bmod 3)$;
When $p=5$, $q=20 \equiv 2(\bmod 3), 2 \neq r=3$;
When $p=7$, $q=14 \equiv 2(\bmod 3), 2 \neq r=5$. Therefore, $p=3$ is the solution. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Find all positive integers that are coprime with all terms of the sequence $\left\{a_{n}\right\}$ satisfying
$$
a_{n}=2^{n}+3^{n}+6^{n}-1\left(n \in Z_{+}\right)
$$ | Solution: Clearly, $\left(1, a_{n}\right)=1$.
Let $m(m>1)$ be a positive integer that is coprime with all terms in $\left\{a_{n}\right\}$, and let $p$ be a prime factor of $m$.
If $p>3$, then by Fermat's Little Theorem,
$$
\begin{array}{l}
2^{p-1} \equiv 1(\bmod p), 3^{p-1} \equiv 1(\bmod p), \\
6^{p-1} \equiv 1(\bmod p) .
\end{array}
$$
Let $2^{p-1}=m p+1,3^{p-1}=n p+1$,
$$
6^{p-1}=t p+1\left(m, n, t \in \mathbf{Z}_{+}\right) \text {. }
$$
Then $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1$
$$
\begin{array}{l}
=\frac{m p+1}{2}+\frac{n p+1}{3}+\frac{t p+1}{6}-1 \\
=\frac{3 m p+2 n p+t p}{6}=p \cdot \frac{3 m+2 n+t}{6} .
\end{array}
$$
Since $a_{p-2}$ is an integer, $(p, 6)=1$, thus,
$$
6|(3 m+2 n+t), p| a_{p-2} \text {. }
$$
This contradicts $\left(m, a_{p-2}\right)=1$.
If $p=2$ or 3, then $a_{2}=48=2^{4} \times 3 \Rightarrow p \mid a_{2}$, which is also a contradiction.
Therefore, the only positive integer that is coprime with all terms in the sequence $\left\{a_{n}\right\}$ is 1. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Prove: For any positive integer $n, 3^{n}+2 \times 17^{n}$ is not a multiple of 5, and find the smallest positive integer $n$, such that
$$
11 \mid\left(3^{n}+2 \times 17^{n}\right) .
$$ | $$
\begin{array}{l}
3^{2 k}+2 \times 17^{2 k} \equiv(-1)^{k}+2 \times 2^{2 k} \\
\equiv 3(-1)^{k}(\bmod 5) \\
3^{2 k+1}+2 \times 17^{2 k+1} \equiv 3(-1)^{k}+4(-1)^{k} \\
\equiv 2(-1)^{k}(\bmod 5)
\end{array}
$$
Therefore, $3^{n}+2 \times 17^{n}$ is not a multiple of 5.
$$
\begin{array}{l}
\text { Also, } 3^{n}+2 \times 17^{n} \equiv 3^{n}+2 \times 6^{n} \\
\equiv 3^{n}\left(1+2^{n+1}\right) \equiv 0(\bmod 11) \\
\Leftrightarrow 111\left(1+2^{n+1}\right), \\
\text { so } n_{\min }=4 .
\end{array}
$$ | 4 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 5 What is the minimum degree of the highest term of a polynomial with rational coefficients that has $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots?
(2013, Joint Autonomous Admission Examination of Peking University and Other Universities) | Notice that the polynomial
$$
f(x)=\left(x^{2}-2\right)\left[(x-1)^{3}-2\right]
$$
has roots $\sqrt{2}$ and $1-\sqrt[3]{2}$, and its degree is 5. Therefore, the degree of the highest term of a rational-coefficient polynomial with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots cannot be less than 5.
If there exists a rational-coefficient polynomial of degree no more than 4
$$
g(x)=a x^{4}+b x^{3}+c x^{2}+d x+e
$$
with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots, where $a, b, c, d, e$ are not all zero, then we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
g(\sqrt{2})=(4 a+2 c+e)+(2 b+d) \sqrt{2}=0, \\
g(1-\sqrt[3]{2}) \\
=-(7 a+b-c-d-e)-(2 a+3 b+2 c+d) \sqrt[3]{2}+ \\
(6 a+3 b+c) \sqrt[3]{4}=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
4 a+2 c+e=0, \\
2 b+d=0, \\
7 a+b-c-d-e=0, \\
2 a+3 b+2 c+d=0, \\
6 a+3 b+c=0 .
\end{array}\right.
\end{array}
$$
(1) + (3) gives
$$
11 a+b+c-d=0.
$$
(6) + (2), (6) + (4) respectively give
$$
\begin{array}{l}
11 a+3 b+c=0, \\
13 a+4 b+3 c=0 .
\end{array}
$$
(7) - (5) gives $a=0$.
Substituting into equations (7), (8) gives $b=c=0$,
Substituting into equations (1), (2) gives $d=e=0$.
Thus, $a=b=c=d=e=0$, which contradicts the assumption that $a, b, c, d, e$ are not all zero.
In conclusion, the minimum degree of the highest term of a rational-coefficient polynomial with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots is 5. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $x$ is an integer, and satisfies the inequality system
$$
\left\{\begin{array}{l}
x-1>0, \\
2 x-1<4,
\end{array}\right.
$$
then $x=$ $\qquad$ | $$
=, 1.2 \text {. }
$$
From the given, we know that $1<x<\frac{5}{2}$. Therefore, the integer $x=2$. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Let real numbers $x, y, z$ satisfy
$$
x^{2}+y^{2}+z^{2}-x y-y z-z x=27 \text {. }
$$
Then the maximum value of $|y-z|$ is $\qquad$ | 3. 6 .
The original equation is equivalent to a quadratic equation in $x$
$$
\begin{array}{l}
x^{2}-(y+z) x+y^{2}+z^{2}-y z-27=0 . \\
\text { And } \Delta=(y+z)^{2}-4\left(y^{2}+z^{2}-y z-27\right) \geqslant 0 \\
\Rightarrow(y-z)^{2} \leqslant 36 \Rightarrow|y-z| \leqslant 6 .
\end{array}
$$
When $|y-z|=6$, and $x=\frac{y+z}{2}$, the equality holds.
Therefore, the maximum value of $|y-z|$ is 6 . | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $x_{1}, x_{2}, \cdots, x_{15}$ take values of 1 or -1. Let
$$
S=x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{15} x_{1} \text {. }
$$
Then the smallest positive integer that $S$ can take is $\qquad$ | 4.3.
Let $y_{i}=x_{i} x_{i+1}(i=1,2, \cdots, 15)$, with the convention that $x_{16}=$ $x_{1}$. Then $y_{i}=1$ or -1.
In $y_{1}, y_{2}, \cdots, y_{15}$, let there be $a$ values that are 1 and $b$ values that are -1. Clearly, $a+b=15$.
Also, $1^{a}(-1)^{b}=y_{1} y_{2} \cdots y_{15}=\left(x_{1} x_{2} \cdots x_{15}\right)^{2}$ $=1$, so $b$ must be even.
Since $S$ is a positive integer, $S=a-b=15-2 b \geqslant 3$, at this point, $b=6$.
On the other hand, in $x_{1}, x_{2}, \cdots, x_{15}$,
$$
x_{2}=x_{5}=x_{8}=-1 \text {, }
$$
the rest are all 1, then $S=3$.
Therefore, the smallest positive integer that $S$ can take is 3. | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. (25 points) As shown in Figure 1, given that $E$ is a point on side $A B$ of square $A B C D$, and the symmetric point of $A$ with respect to $D E$ is $F, \angle B F C = 90^{\circ}$. Find the value of $\frac{A B}{A E}$. | 2. As shown in Figure 4, extend $E F$ to intersect $B C$ at point $M$, connect $D M$, and let it intersect $C F$ at point $G$.
Then, Rt $\triangle D F M \cong$ Rt $\triangle D C M$. Therefore, $\angle F D M = \angle M D C$, and $F M = C M$.
Thus, $M$ is the midpoint of $B C$.
$$
\begin{array}{l}
\text { Also, } \angle E D M = \angle E D F + \angle F D M \\
= \frac{1}{2} \angle A D F + \frac{1}{2} \angle F D C = 45^{\circ} .
\end{array}
$$
Rotate $\triangle M D C$ around point $D$ by $90^{\circ}$ counterclockwise to get $\triangle H D A$.
Then, $\triangle M D E \cong \triangle H D E$.
Therefore, $E M = H E = A E + M C$.
Let the side length of the square be $1$, and $A E = x$.
By $E B^{2} + B M^{2} = E M^{2}$, we have
$$
\begin{array}{l}
(1-x)^{2} + \left(\frac{1}{2}\right)^{2} = \left(x + \frac{1}{2}\right)^{2} \\
\Rightarrow x = \frac{1}{3} \Rightarrow \frac{A B}{A E} = 3 .
\end{array}
$$ | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. (25 points) On a circle, there are $n$ different positive integers $a_{1}$, $a_{2}, \cdots, a_{n}$ placed in a clockwise direction. If for any number $b$ among the ten positive integers $1, 2, \cdots, 10$, there exists a positive integer $i$ such that $a_{i}=b$ or $a_{i}+a_{i+1}=b$, with the convention that $a_{n+1}=a_{1}$, find the minimum value of the positive integer $n$. | 3. From the conditions, we know that the $2n$ numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{1}+a_{2}, a_{2}+a_{3}, \cdots, a_{n}+a_{1}$ should include the ten positive integers $1,2, \cdots, 10$. Therefore, $2 n \geqslant 10 \Rightarrow n \geqslant 5$.
When $n=5$, the ten numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{1}+a_{2}, a_{2}+a_{3}, \cdots, a_{n}+a_{1}$ are all distinct and take values from $1 \sim 10$.
$$
\begin{array}{l}
\text { Hence } a_{1}+a_{2}+\cdots+a_{n}+\left(a_{1}+a_{2}\right)+ \\
\left(a_{2}+a_{3}\right)+\cdots+\left(a_{n}+a_{1}\right) \\
=1+2+\cdots+10 \\
\Rightarrow 3\left(a_{1}+a_{2}+\cdots+a_{n}\right)=55,
\end{array}
$$
which is a contradiction.
Thus, when $n=5$, there do not exist five positive integers $a_{1}, a_{2}, \cdots, a_{5}$ that satisfy the conditions.
When $n=6$, construct $a_{1}=1, a_{2}=10, a_{3}=2, a_{4}=6, a_{5}=3, a_{6}=4$, and verify that these six numbers satisfy the conditions.
In conclusion, the minimum value of the positive integer $n$ is 6. | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Find the number of polynomials $f(x)=a x^{3}+b x$ that satisfy the following two conditions:
(1) $a, b \in\{1,2, \cdots, 2013\}$;
(2) The difference between any two numbers in $f(1), f(2), \cdots, f(2013)$ is not a multiple of 2013. (Wang Bin) | 4. It is known that the prime factorization of $2013=3 \times 11 \times 61$.
Let $p_{1}=3, p_{2}=11, p_{3}=61$.
For $a, b \in\{1,2, \cdots, 2013\}$, let
$a \equiv a_{i}\left(\bmod p_{i}\right), b \equiv b_{i}\left(\bmod p_{i}\right)$,
where $i=1,2,3$.
By the Chinese Remainder Theorem, we know that $(a, b)$ and $\left(a_{1}{ }^{\prime}, a_{2}, a_{3}\right.$, $\left.b_{1}, b_{2}, b_{3}\right)$ are in one-to-one correspondence.
Let $f_{i}(x)=a_{i} x^{3}+b_{i} x(i=1,2,3)$.
A polynomial is called a "good polynomial modulo $n$" if $f(0), f(1), \cdots, f(n-1)$ have distinct remainders modulo $n$.
If $f(x)=a x^{3}+b x$ is not a good polynomial modulo 2013, then there exist $x_{1} \neq x_{2}(\bmod 2013)$ such that
$f\left(x_{1}\right) \equiv f\left(x_{2}\right)(\bmod 2013)$.
Assume $x_{1} \neq x_{2}\left(\bmod p_{i}\right)$, and let $u_{1}, u_{2}$ be the remainders of $x_{1}$ and $x_{2}$ modulo $p_{i}$. Then
$u_{1} \equiv u_{2}\left(\bmod p_{i}\right)$,
and $f_{i}\left(u_{1}\right) \equiv f_{i}\left(u_{2}\right)\left(\bmod p_{i}\right)$.
Thus, $f_{i}(x)$ is not a good polynomial modulo $p_{i}$.
If $f(x)=a x^{3}+b x$ is a good polynomial modulo 2013, then each $f_{i}(x)$ is a good polynomial modulo $p_{i}$.
In fact, for any different
$r_{1}, r_{2} \in\left\{0,1, \cdots, p_{i}-1\right\}$,
there exist $x_{1}, x_{2} \in\{1,2, \cdots, 2013\}$ such that
$$
x_{1} \equiv x_{2}\left(\bmod \frac{2013}{p_{i}}\right)
$$
and $x_{1} \equiv r_{1}\left(\bmod p_{i}\right), x_{2} \equiv r_{2}\left(\bmod p_{i}\right)$.
Since $f\left(x_{1}\right) \neq f\left(x_{2}\right)(\bmod 2013)$ and
$f\left(x_{1}\right) \equiv f\left(x_{2}\right)\left(\bmod \frac{2013}{p_{i}}\right)$,
it follows that $f\left(r_{1}\right) \neq f\left(r_{2}\right)\left(\bmod p_{i}\right)$, and the conclusion holds.
Therefore, the problem reduces to finding the number of good polynomials $f_{i}(x)$ modulo $p_{i}$.
For $p_{1}=3$, by Fermat's Little Theorem, we have
$f_{1}(x) \equiv a_{1} x+b_{1} x \equiv\left(a_{1}+b_{1}\right) x(\bmod 3)$
is a good polynomial if and only if $a_{1}+b_{1}$ is not a multiple of 3, and there are 6 such $f_{1}(x)$.
For $i=2,3$, if $f_{i}(x)$ is a good polynomial modulo $p_{i}$,
then for any $u, v \not\equiv 0\left(\bmod p_{i}\right)$,
$f_{i}(u+v) \not\equiv f_{i}(u-v)\left(\bmod p_{i}\right)$
$\Rightarrow p_{i} \nmid\left[f_{i}(u+v)-f_{i}(u-v)\right]$
$\Rightarrow p_{i} \nmid 2 v\left[a_{i}\left(3 u^{2}+v^{2}\right)+b_{i}\right]$.
If $a_{i} \neq 0$, the sets
$A=\left\{3 a_{i} u^{2} \mid u=0,1, \cdots, \frac{p_{i}-1}{2}\right\}$
and $B=\left\{-b_{i}-a_{i} v^{2} \mid v=1,2, \cdots, \frac{p_{i}-1}{2}\right\}$,
have non-overlapping remainders modulo $p_{i}$.
Clearly, the elements of sets $A$ and $B$ are distinct modulo $p_{i}$, and $|A|+|B|=p_{i}$, so the $p_{i}$ elements of $A \cup B$ have remainders that are a permutation of $0,1, \cdots, p_{i}-1$, and thus their sum is a multiple of $p_{i}$, i.e.,
$$
\sum_{u=0}^{\frac{p_{i}-1}{2}} 3 a_{i} u^{2}+\sum_{v=1}^{\frac{p_{i}-1}{2}}\left(-b_{i}-a_{i} v^{2}\right) \equiv 0\left(\bmod p_{i}\right) .
$$
Notice that,
$$
\begin{array}{l}
1^{2}+2^{2}+\cdots+\left(\frac{p_{i}-1}{2}\right)^{2} \\
=\frac{1}{6} \times \frac{p_{i}-1}{2} \times \frac{p_{i}+1}{2} p_{i}
\end{array}
$$
is a multiple of $p_{i}$, so $-\frac{p_{i}-1}{2} b_{i}$ is also a multiple of $p_{i}$. Thus, $b_{i}$ is a multiple of $p_{i}$, meaning that at least one of $a_{i}$ or $b_{i}$ is 0, and they cannot both be 0.
If $a_{i}=0, b_{i} \neq 0$, then $f_{i}(x)=b_{i} x$ is clearly a good polynomial.
Thus, there are $p_{i}-1$ such good polynomials.
If $a_{i} \neq 0, b_{i}=0$, then $f_{i}(x)=a_{i} x^{3}$.
For $p_{2}=11$, by Fermat's Little Theorem, we have
$\left(x^{3}\right)^{?}=x^{21} \equiv x(\bmod 11)$.
Thus, for $x_{1} \equiv x_{2}(\bmod 11), x_{1}^{3} \equiv x_{2}^{3}(\bmod 11)$, $f_{2}(x)=a_{2} x^{3}$ is a good polynomial.
Thus, there are 10 such good polynomials.
For $p_{3}=61$, since
$4^{3}=64 \equiv 125=5^{3}(\bmod 61)$,
$f_{3}(x)=a_{3} x^{3}$ is not a good polynomial.
In summary, the number of polynomials $f(x)$ is
$6 \times(10+10) \times 60=7200$. | 7200 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Find the largest positive integer $n(n \geqslant 3)$, such that there exists a convex $n$-gon, where the tangent values of all its interior angles are integers.
(Proposed by the Problem Committee) | On the one hand, since each interior angle of a regular octagon is $135^{\circ}$, and its tangent value is -1, $n=8$ satisfies the condition.
On the other hand, if $n \geqslant 9$, let the exterior angles of the $n$-sided polygon be $\angle A_{1}, \angle A_{2}, \cdots, \angle A_{n}\left(0<\angle A_{1} \leqslant \angle A_{2}\right.$ $\left.\leqslant \cdots \leqslant \angle A_{n}\right)$. Then
$$
\begin{array}{l}
\angle A_{1}+\angle A_{2}+\cdots+\angle A_{n}=2 \pi \\
\Rightarrow 0<\angle A_{1} \leqslant \frac{2 \pi}{n}<\frac{\pi}{4} \Rightarrow 0<\tan A_{1}<1 .
\end{array}
$$
Thus, the tangent value of the corresponding interior angle of $\angle A_{1}$ is not an integer. Therefore, when $n \geqslant 9$, there is no polygon that satisfies the condition. In summary, the maximum value of $n$ is 8. | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Let $x_{k} \in[-2,2](k=1,2, \cdots, 2013)$,
and
$x_{1}+x_{2}+\cdots+x_{2013}=0$. Try to find
$$
M=x_{1}^{3}+x_{2}^{3}+\cdots+x_{2013}^{3}
$$
the maximum value.
(Liu Kangning) | Given $x_{i} \in[-2,2](i=1,2, \cdots, 2013)$, we know that $x_{i}^{3}-3 x_{i}=\left(x_{i}-2\right)\left(x_{i}+1\right)^{2}+2 \leqslant 2$.
The equality holds if and only if $x_{i}=2$ or -1.
Noting that, $\sum_{i=1}^{2013} x_{i}=0$.
Thus, $M=\sum_{i=1}^{2013} x_{i}^{3}=\sum_{i=1}^{2013}\left(x_{i}^{3}-3 x_{i}\right)$
$$
\leqslant \sum_{i=1}^{2013} 2=4026 \text {. }
$$
When $x_{1}, x_{2}, \cdots, x_{2013}$ include 671 values of 2 and 1342 values of -1, the equality holds.
Therefore, the maximum value of $M$ is 4026. | 4026 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find the remainder when $10^{10}(100$ ones$)$ is divided by 7. | \begin{array}{l}\text { Sol } 10^{10} \equiv(7+3)^{10^{10}} \equiv 3^{10} \\ \equiv(7+2)^{50 \cdots 0} \equiv 2^{5 \times 10^{9}} \equiv 2^{3 \times 106 \cdots 6+2} \\ \equiv 4(7+1)^{166 \cdots 6} \equiv 4(\bmod 7) .\end{array} | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Seven, let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=1, \\
a_{n+1}=\left(1+\frac{k}{n}\right) a_{n}+1(n=1,2, \cdots) .
\end{array}
$$
Find all positive integers $k$ such that every term in the sequence $\left\{a_{n}\right\}$ is an integer.
(Zhang Lei) | When $k=1$, $a_{2}=3, a_{3}=\frac{11}{2}$, which does not satisfy the condition.
When $k=2$, by the given condition we have
$$
\frac{a_{n+1}}{(n+1)(n+2)}=\frac{a_{n}}{n(n+1)}+\frac{1}{(n+1)(n+2)} \text {. }
$$
Thus, $\frac{a_{n}}{n(n+1)}=\frac{a_{1}}{1 \times 2}+\sum_{i=2}^{n} \frac{1}{i(i+1)}=1-\frac{1}{n+1}$.
Therefore, $a_{n}=n^{2}$ is a positive integer, satisfying the condition.
When $k \geqslant 3$, by the given condition we have
$$
\begin{array}{l}
\frac{a_{n+1}}{(n+1)(n+2) \cdots(n+k)} \\
=\frac{a_{n}}{n(n+1) \cdots(n+k-1)}+\frac{1}{(n+1)(n+2) \cdots(n+k)} .
\end{array}
$$
Thus, $\frac{a_{n}}{n(n+1) \cdots(n+k-1)}$
$$
\begin{array}{l}
=\frac{a_{1}}{1 \times 2 \times \cdots \times k}+\sum_{i=2}^{n} \frac{1}{i(i+1) \cdots(i+k-1)} \\
=\frac{1}{k-1}\left[\frac{1}{1 \times 2 \times \cdots \times(k-1)}-\frac{1}{(n+1)(n+2) \cdots(n+k-1)}\right] \\
\Rightarrow a_{n}=\frac{1}{k-1} \cdot \frac{n(n+1) \cdots(n+k-1)}{1 \times 2 \times \cdots \times(k-1)}-\frac{n}{k-1} \\
=\frac{(n+k-1)(n+k-2)}{(k-1)^{2}} \mathrm{C}_{n+k-3}^{k-2}-\frac{n}{k-1} .
\end{array}
$$
When $n+k-2=(k-1)^{2}, n=(k-1)^{2}-k+2$, by $\frac{(n+k-1)(n+k-2)}{(k-1)^{2}} \mathrm{C}_{n+k-3}^{k-2} \in \mathbf{N}, \frac{n}{k-1}$ $\notin \mathbf{N}$, we know
$$
a_{n}=\frac{(n+k-1)(n+k-2)}{(k-1)^{2}} \mathrm{C}_{n+k-3}^{k-2}-\frac{n}{k-1} \notin \mathbf{N} \text {. }
$$
In summary, the value of the positive integer $k$ that satisfies the condition is 2. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Write 2013 different real numbers on 2013 cards. Place the cards face down on the table. Two players, A and B, play the following game: in each round, A can arbitrarily select ten cards, and B will tell A one of the ten numbers written on these cards (B does not tell A which card the number is written on). Find the maximum positive integer $t$, such that A can certainly determine the numbers on $t$ cards after a finite number of rounds. | 4. The maximum value of $t$ is $1986=2013-27$.
Let $A_{1}, A_{2}, \cdots, A_{2013}$ be these 2013 cards.
First, note that player B has a strategy to prevent player A from determining the number written on any of the cards $A_{1}, A_{2}, \cdots, A_{27}$.
B divides $T=\{1,2, \cdots, 27\}$ into nine groups
$T_{i}=\{3 i-2,3 i-1,3 i\}(1 \leqslant i \leqslant 9)$.
For any ten-element subset $B$ of $A=\{1,2, \cdots, 2013\}$, consider the following two cases.
(1) If $B \backslash T \neq \varnothing$, let
$i_{0}=\min \{i \mid i \in B \backslash T\}$,
then B tells A the number on $A_{i_{0}}$.
(2) If $B \backslash T=\varnothing$, let
$i_{0}=\min \left\{i\left|T_{i} \cap B\right| \geqslant 2\right\}$,
then consider the following two sub-cases.
(i) If $\left|T_{i_{0}} \cap B\right|=3$, then B tells A the number on $A_{3 i_{0}}$;
(ii) If $\left|T_{i_{0}} \cap B\right|=2$, then B has two options to tell A.
For $T_{i_{0}} \cap B$
$=\{3 i-2,3 i-1\},\{3 i-1,3 i\},\{3 i, 3 i-2\}$.
【Option 1】B tells A the numbers on $A_{3 i-2}, A_{3 i-1}, A_{3 i}$ respectively.
【Option 2】B tells A the numbers on $A_{3 i-1}, A_{3 i}, A_{3 i-2}$ respectively.
Since A does not know which option B uses, A cannot determine the number on any of the first 27 cards (in fact, if the number on $A_{3 i-2}$ is rewritten to $A_{3 i-1}$, the number on $A_{3 i-1}$ is rewritten to $A_{3 i}$, and the number on $A_{3 i}$ is rewritten to $A_{3 i-2}$, then for any ten cards specified by A, B's response for the rewritten Option 2 is exactly the same as for the original Option 1).
Next, we show that for any 28 cards, A has a way to determine the number on one of them.
To do this, we first prove a lemma in graph theory.
Lemma $n(n \geqslant 2)$ is a positive integer. In a graph with at least $3 n-2$ vertices and at most $3 n-2$ edges, there must exist $n$ vertices such that no two of them are connected by an edge.
Proof By induction on $n$.
Assume the graph has exactly $3 n-2$ vertices and $3 n-2$ edges.
When $n=2$, if the number of edges is less than $\mathrm{C}_{4}^{2}=6$, then there are two vertices not connected by an edge.
Assume $n>2$. Let $d_{1}, d_{2}, \cdots, d_{3 n-2}$ be the degrees of the vertices. Then
$d_{1}+d_{2}+\cdots+d_{3 n-2}=2(3 n-2)$.
This indicates that either all $d_{i}=2$ or there exist $i, j$ such that
$d_{i}<2<d_{j}$.
In the first case, remove any vertex (e.g., the $i$-th) and its two adjacent vertices; in the second case, remove the $i$-th and $j$-th vertices and the vertex adjacent to the $i$-th vertex. This removes at most three vertices and at least three edges. Apply the induction hypothesis to the remaining graph to get $n-1$ vertices not connected by any edge, and add the $i$-th vertex to get $n$ vertices not connected by any edge.
The lemma is proved.
Let $A_{1}, A_{2}, \cdots, A_{28}$ be 28 cards.
$M$ denotes the set of all ten-element subsets of $\left\{A_{1}, A_{2}, \cdots, A_{28}\right\}$.
Define a function $f$ on $M$: for any $B \in M, f(B)$ represents the number B tells A for $B$. Let $\left\{c_{1}, c_{2}, \cdots, c_{k}\right\}$ $(k \leqslant 28)$ be the range of $f$.
Let $S_{i}=\bigcap_{B \in f^{-1}\left(c_{i}\right)} B$.
If for some $i, S_{i}=\left\{A_{m}\right\}$ is a singleton, then the number on $A_{m}$ is $c_{i}$.
If for any $i(1 \leqslant i \leqslant k), S_{i}$ contains at least two elements, select two elements from each $S_{i}$ and connect them with an edge, forming a graph with 28 vertices and no more than 28 edges. According to the lemma $(n=10)$, there exist 10 vertices in the graph that are not connected by any edge.
Let the set of these 10 vertices be $B \in M$, $f(B)=c_{i}$. Then $S_{i} \subset B$. $S_{i}$ contains at least two vertices connected by an edge, i.e., there are two vertices in $B$ connected by an edge, which is a contradiction. | 1986 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $S$ be a set of $n(n \geqslant 5)$ points in the plane. If any four points chosen from $S$ have at least one point connected to the other three, then which of the following conclusions is correct? $\qquad$
(1) There is no point in $S$ that is connected to all other points;
(2) There is at least one point in $S$ that is connected to all other points;
(3) There are at most two points in $S$ that are not connected to all other points;
(4) There are at most two points in $S$ that are connected to all other points. | -、1. (2).
In the point set $S$, all points are connected to each other, which clearly satisfies the problem. Therefore, conclusions (1) and (4) are incorrect.
Suppose $A$, $B$, and $C$ are three points in the point set $S$ that are not connected to each other, but the remaining $n-3$ points are all connected to each other. This also clearly satisfies the problem. Hence, conclusion (3) is also incorrect.
If all points in the point set $S$ are connected to each other, then conclusion (2) is obviously true; otherwise, assume points $P$ and $Q$ are not connected. Take any two other points in the point set $S$, say $R$ and $T$. Among the four points $P$, $Q$, $R$, and $T$, there must be one point that is connected to the other three. This point cannot be $P$ or $Q$, so it must be one of $R$ or $T$ (let's say $R$). By the arbitrariness of points $R$ and $T$, we know that there is at least one point in the point set $S$ that is connected to all other points. Therefore, conclusion (2) is correct. | 2 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
6. Given that the three vertices of $\triangle A B C$ are all on the parabola $y^{2}=2 p x(p>0)$, and the centroid of $\triangle A B C$ is exactly the focus of the parabola. If the equation of the line on which side $B C$ lies is $4 x+y$ $-20=0$, then $p=$ $\qquad$ . | 6. $p=8$.
Let $A\left(\frac{y_{1}^{2}}{2 p}, y_{1}\right), B\left(\frac{y_{2}^{2}}{2 p}, y_{2}\right), C\left(\frac{y_{3}^{2}}{2 p}, y_{3}\right)$.
$$
\begin{array}{l}
\text { From }\left\{\begin{array}{l}
y^{2}=2 p x, \\
4 x+y-20=0
\end{array}\right. \\
\Rightarrow 2 y^{2}+p y-20 p=0 .
\end{array}
$$
From the given information,
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{y_{1}^{2}}{2 p}+\frac{y_{2}^{2}}{2 p}+\frac{y_{3}^{2}}{2 p}=\frac{3 p}{2}, \\
y_{1}+y_{2}+y_{3}=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
y_{2}+y_{3}=-\frac{p}{2}, \\
y_{2} y_{3}=-10 p, \\
y_{1}+y_{2}+y_{3}=0, \\
y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=3 p^{2} .
\end{array}\right.
\end{array}
$$
From equations (1) and (3), we get $y_{1}=\frac{p}{2}$.
Substituting into equation (4) and solving with (1) and (2) yields $p=8$. | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. $\sum_{0 \leqslant i<j \leqslant 50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}$ modulo 31 is | 8. 16 .
Notice,
$$
\begin{array}{l}
\sum_{0 \leqslant i<j \leqslant 50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}=\frac{1}{2} \sum_{0 \leqslant i \neq j \leqslant 50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j} \\
=\frac{1}{2}\left[\sum_{i=0}^{50} \sum_{j=0}^{50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}-\sum_{j=0}^{50}\left(\mathrm{C}_{50}^{i}\right)^{2}\right] \\
=\frac{1}{2}\left[\left(\sum_{i=0}^{50} \mathrm{C}_{50}^{i}\right)\left(\sum_{j=0}^{50} \mathrm{C}_{50}^{j}\right)-\sum_{i=0}^{50}\left(\mathrm{C}_{50}^{i}\right)^{2}\right] \\
=\frac{1}{2}\left(2^{100}-\mathrm{C}_{100}^{50}\right)=2^{99}-\frac{1}{2} \mathrm{C}_{100}^{50} . \\
\text { Also, } 2^{99}=2^{4} \times\left(2^{5}\right)^{19} \equiv 16(\bmod 31), \text { and } \\
\frac{1}{2} \mathrm{C}_{100}^{50}=\frac{99 \times 98 \times \cdots \times 51}{49 \times 48 \times \cdots \times 2 \times 1}
\end{array}
$$
is a multiple of 31, hence the remainder when $\sum_{0 \leqslant i<j \leqslant 50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}$ is divided by 31 is 16. | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given real numbers
$$
x_{i} \in[-6,10](i=1,2, \cdots, 10), \sum_{i=1}^{10} x_{i}=50 \text {. }
$$
Find the maximum value of $\sum_{i=1}^{10} x_{i}^{2}$, and the conditions that should be satisfied when the maximum value is achieved. | 10. Let $a_{i}=x_{i}+6$. Then $a_{i} \in[0,16]$, and
$$
\begin{array}{l}
\sum_{i=1}^{10} a_{i}=110, \\
\sum_{i=1}^{10} a_{i}^{2}=\sum_{i=1}^{10} x_{i}^{2}+12 \sum_{i=1}^{10} x_{i}+360 \\
=\sum_{i=1}^{10} x_{i}^{2}+960 .
\end{array}
$$
When there are at least two numbers in $a_{i}$ that are not both 0 and not both 16, let these two numbers be $p$ and $q$.
$$
\begin{array}{l}
\text { (1) When } p+q \geqslant 16 \text {, since } \\
16^{2}+(p+q-16)^{2}-\left(p^{2}+q^{2}\right) \\
=2 \times 16^{2}+2(q-16) p-32 q \\
>2 \times 16^{2}+2(q-16) \times 16-32 q=0,
\end{array}
$$
thus, without changing the other numbers, replacing $p$ with 16 and $p+q-16$ with $q$ keeps the sum unchanged but increases the sum of squares.
(2) When $p+q<16$,
$$
\begin{array}{l}
0^{2}+(p+q)^{2}-\left(p^{2}+q^{2}\right) \\
=2pq>0,
\end{array}
$$
thus, replacing $p$ with 0 and $p+q$ with $q$ increases the sum of squares.
Given $110=16 \times 6+14$, when six of the $a_{i}$ are 16, three are 0, and one is 14, the maximum value of $\sum_{i=1}^{10} x_{i}^{2}$ is
$$
6 \times 10^{2}+3 \times(-6)^{2}+8^{2}=772 .
$$ | 772 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given $\tan \alpha+\tan \beta+\tan \gamma=\frac{17}{6}$, $\cot \alpha+\cot \beta+\cot \gamma=-\frac{4}{5}$, $\cot \alpha \cdot \cot \beta+\cot \beta \cdot \cot \gamma+\cot \gamma \cdot \cot \alpha=-\frac{17}{5}$.
Then $\tan (\alpha+\beta+\gamma)=$ $\qquad$ [5]
(2012, Xin Zhi Cup Shanghai High School Mathematics Competition) | Let $\tan \alpha=x, \tan \beta=y, \tan \gamma=z$.
Then $x+y+z=\frac{17}{6}$,
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{4}{5}$,
$\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=-\frac{17}{5}$.
(1) $\div$ (3) gives
$x y z=-\frac{5}{6}$.
(2) $\times$ (4) gives $x y+y z+z x=\frac{2}{3}$.
Therefore, $\tan (\alpha+\beta+\gamma)=\frac{x+y+z-x y z}{1-(x y+y z+z x)}=11$. | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Try to find the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$.
[2] | Notice that,
$$
(\sqrt{2}+\sqrt{3})^{2012}=(5+2 \sqrt{6})^{1006} \text {. }
$$
Clearly, $0<(5-2 \sqrt{6})^{1006}<1$,
$$
\begin{array}{l}
(5+2 \sqrt{6})^{1006}+(5-2 \sqrt{6})^{1006} \\
=2\left(C_{1006}^{0} 5^{1006}+C_{1006}^{2} 5^{1004} \times 24+\cdots+\right. \\
\left.\quad C_{1006}^{1006} 24^{503}\right) \in \mathbf{Z}_{+} .
\end{array}
$$
Therefore, the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$ is
$$
2\left(\mathrm{C}_{1006}^{0} 5^{1006}+\mathrm{C}_{1006}^{2} 5^{1004} \times 24+\cdots+\mathrm{C}_{1006}^{100} 24^{503}\right)-1 \text {. }
$$
And $2\left(\mathrm{C}_{1006}^{0} 5^{1006}+\mathrm{C}_{1006}^{2} 5^{1004} \times 24+\cdots+\right.$
$$
\begin{array}{l}
\left.C_{1006}^{1006} 24^{503}\right)-1 \\
\equiv 2 \times 24^{503}-1 \equiv 2 \times(25-1)^{503}-1 \\
\equiv 2 \times(-1)^{503}-1 \equiv 7(\bmod 10),
\end{array}
$$
That is, the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$ is 7. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $x=(15+\sqrt{220})^{19}+(15+\sqrt{200})^{22}$. Find the unit digit of the number $x$.
| Solve the conjugate expression
$$
y=(15-\sqrt{220})^{19}+(15-\sqrt{220})^{82} \text {. }
$$
Then $x+y$
$$
\begin{aligned}
= & (15+\sqrt{220})^{19}+(15-\sqrt{220})^{19}+ \\
& (15+\sqrt{220})^{82}+(15-\sqrt{220})^{82} .
\end{aligned}
$$
By the binomial theorem, for any $n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
(15+\sqrt{220})^{n}+(15-\sqrt{220})^{n} \\
=2\left(\mathrm{C}_{n}^{0} 15^{n}+\mathrm{C}_{n}^{2} 15^{n-2} \times 220+\cdots\right) \in \mathbf{Z}_{+},
\end{array}
$$
and the last digit is zero.
Therefore, $x+y$ is a positive integer with the last digit being zero. Also, $0<15-\sqrt{220}=\frac{5}{15+\sqrt{220}}<\frac{5}{25}=\frac{1}{5}$, and $(15-\sqrt{220})^{82}<(15-\sqrt{220})^{19}$, so $0<y<2(15-\sqrt{220})^{19}<2 \times 0.2^{19}<0.4$. Therefore, the last digit of $x$ is 9. | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Positive integers $a_{1}, a_{2}, \cdots, a_{2006}$ (allowing repetition). What is the minimum number of distinct numbers in $a_{2}, \cdots, a_{2006}$? ${ }^{[1]}$
(2006, China Mathematical Olympiad) | This problem is solved using combinatorial geometry methods.
Let the positive integers $a_{1}, a_{2}, \cdots, a_{n}$ be such that $\frac{a_{i}}{a_{i+1}} (1 \leqslant i \leqslant n-1)$ are all distinct. Then the minimum number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{n}$ is $k=\left\lceil\frac{1+\sqrt{4 n-7}}{2}\right\rceil$, where $\lceil x\rceil$ denotes the smallest integer not less than the real number $x$.
Since the number of distinct ratios of $k$ distinct positive integers is at most $k(k-1)+1$, we have
$$
k(k-1)+1 \geqslant n-1 \Rightarrow k \geqslant\left\lceil\frac{1+\sqrt{4 n-7}}{2}\right\rceil .
$$
Let $p_{1}, p_{2}, \cdots, p_{k}$ be $k$ distinct prime numbers, and construct a directed graph with these primes as vertices. Connect each pair of distinct vertices $p_{i}$ and $p_{j}$ with two directed edges $p_{i} \rightarrow p_{j}$ and $p_{j} \rightarrow p_{i}$, and add a loop at vertex $p_{1}$. Since this graph is connected and each vertex has equal in-degree and out-degree, it has an Eulerian circuit. Traversing this circuit (which has $k(k-1)+1$ edges) and recording the vertices visited, we obtain a sequence of $k(k-1)+2$ terms that satisfies the given condition (because the prime equality $\frac{p_{i}}{p_{j}}=\frac{p_{k}}{p_{m}}$ holds if and only if $(i, j)=(k, m)$ or $(i, k)=(j, m)$).
Therefore, the minimum number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{2006}$ is
$$
\lceil\sqrt{2004.25}+0.5\rceil=46
$$ | 46 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
A=\{2,0,1,3\}, \\
B=\left\{x \mid -x \in A, 2-x^{2} \notin A\right\} .
\end{array}
$$
Then the sum of all elements in set $B$ is | ,$- 1 .-5$.
It is easy to know that $B \subseteq\{-2,0,-1,-3\}$.
When $x=-2,-3$, $2-x^{2}=-2,-7 \notin A$; when $x=0,-1$, $2-x^{2}=2,1 \in A$.
Therefore, the set $B=\{-2,-3\}$.
Thus, the sum of all elements in set $B$ is -5. | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In the Cartesian coordinate system $x O y$, it is known that points $A$ and $B$ lie on the parabola $y^{2}=4 x$, and satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B}=-4, F$ is the focus of the parabola. Then $S_{\triangle O F A} \cdot S_{\triangle O F B}=$ $\qquad$ . | 2. 2 .
From the problem, we know the point $F(1,0)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Thus, $x_{1}=\frac{y_{1}^{2}}{4}, x_{2}=\frac{y_{2}^{2}}{4}$.
Then $-4=\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}+y_{1} y_{2}$
$$
\begin{array}{l}
=\frac{1}{16}\left(y_{1} y_{2}\right)^{2}+y_{1} y_{2} \\
\Rightarrow \frac{1}{16}\left(y_{1} y_{2}+8\right)^{2}=0 \\
\Rightarrow y_{1} y_{2}=-8 .
\end{array}
$$
Therefore, $S_{\triangle O F A} \cdot S_{\triangle O F B}$
$$
\begin{array}{l}
=\left(\frac{1}{2}|O F|\left|y_{1}\right|\right)\left(\frac{1}{2}|O F|\left|y_{2}\right|\right) \\
=\frac{1}{4}|O F|^{2}\left|y_{1} y_{2}\right|=2 .
\end{array}
$$ | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. In $\triangle A B C$, it is known that
$$
\sin A=10 \sin B \cdot \sin C, \cos A=10 \cos B \cdot \cos C \text {. }
$$
Then $\tan A=$ $\qquad$ | 3. 11 .
From $\sin A-\cos A$
$$
\begin{array}{l}
=10(\sin B \cdot \sin C-\cos B \cdot \cos C) \\
=-10 \cos (B+C)=10 \cos A \\
\Rightarrow \sin A=11 \cos A \\
\Rightarrow \tan A=11 .
\end{array}
$$ | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given a sequence $\left\{a_{n}\right\}$ with nine terms, where $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in \left\{2,1,-\frac{1}{2}\right\}$. Then the number of such sequences is | 8.491.
Let $b_{i}=\frac{a_{i+1}}{a_{i}}(1 \leqslant i \leqslant 8)$. Then for each sequence $\left\{a_{n}\right\}$ that meets the conditions, it satisfies
$$
\prod_{i=1}^{8} b_{i}=\prod_{i=1}^{8} \frac{a_{i+1}}{a_{i}}=\frac{a_{9}}{a_{1}}=1,
$$
and $b_{i} \in\left\{2,1,-\frac{1}{2}\right\}(1 \leqslant i \leqslant 8)$.
Conversely, a sequence $\left\{b_{n}\right\}$ of eight terms that meets the above conditions can uniquely determine a sequence $\left\{a_{n}\right\}$ of nine terms that meets the problem's conditions.
Let the number of sequences $\left\{b_{n}\right\}$ that meet the conditions be $N$. Clearly, $b_{i}(1 \leqslant i \leqslant 8)$ contains $2 k$ terms of $-\frac{1}{2}$; thus, there are $2 k$ terms of $2$ and $8-4 k$ terms of $1$. When $k$ is given, the number of ways to choose $\left\{b_{n}\right\}$ is $\mathrm{C}_{8}^{2 k} \mathrm{C}_{8-2 k}^{2 k}$. It is evident that the possible values of $k$ are only $0, 1, 2$, hence
$$
\begin{array}{l}
N=1+\mathrm{C}_{8}^{2} \mathrm{C}_{6}^{2}+\mathrm{C}_{8}^{4} \mathrm{C}_{4}^{4} \\
=1+28 \times 15+70 \times 1=491 .
\end{array}
$$
Therefore, by the principle of correspondence, the number of sequences $\left\{a_{n}\right\}$ that meet the conditions is 491. | 491 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in set $S$ is a positive integer not exceeding 100;
(2) For any two distinct elements $a, b$ in set $S$, there exists an element $c$ in $S$ such that
$$
(a, c)=(b, c)=1 \text {; }
$$
(3) For any two distinct elements $a, b$ in set $S$, there exists an element $d$ in $S$ different from $a, b$ such that
$$
(a, d)>1,(b, d)>1 \text {. }
$$ | Each positive integer can be expressed as
$$
n=2^{k_{1}} \times 3^{k_{2}} \times 5^{k_{3}} \times 7^{k_{4}} \times 11^{k_{5}} q,
$$
where $q$ is coprime with 2, 3, 5, 7, and 11, and $k_{1}, k_{2}, \cdots, k_{5}$ are non-negative integers.
Let $A=\left\{n \leqslant 100 \mid k_{1}, k_{2}, \cdots, k_{5}\right.$ have exactly one or two non-zero values\}.
By definition, set $A$ clearly satisfies condition (1).
For any $a, b \in S$, since $a$ and $b$ are each divisible by at most
$$
P=\{2,3,5,7,11\}
$$
two integers, there exists $p \in P \subseteq S$ that is coprime with both $a$ and $b$. Taking $c=p$, we see that set $A$ satisfies condition (2).
Furthermore, if $a$ and $b$ have a common factor $u \in P$, then $p u \in A, p u \neq a, b$.
Taking $d=p u$, we see that set $A$ satisfies condition (3).
If $a$ and $b$ do not have a common factor in $P$, and they have factors $v, w \in P$, then $v w \in A, v w \neq a, b$. Taking $d=v w$ also shows that set $A$ satisfies condition (3).
Therefore, set $A$ satisfies the given conditions.
By the principle of inclusion-exclusion, we have
$$
\begin{array}{l}
|A|=\sum_{i \in P}\left[\frac{100}{i}\right]-\sum_{\substack{i, j \in P \\
i < j}}\left[\frac{100}{i j}\right]+\sum_{\substack{i, j, k \in P \\
i < j < k}}\left[\frac{100}{i j k}\right]-\sum_{\substack{i, j, k, l \in P \\
i < j < k < l}}\left[\frac{100}{i j k l}\right]+\left[\frac{100}{2 \times 3 \times 5 \times 7 \times 11}\right] \\
=50+33+20+14+9-(16+10+7+5+4+3+2)+(3+2+1+1+1)-(0+0+0+0+0)+0 \\
=126-47+8-0+0 \\
=87.
\end{array}
$$
Since $2 \times 3 \times 5 \times 7 \times 11 > 100$, there does not exist such a $d$. There are 21 prime numbers greater than 10 and less than 100. Let
$T=\{1,2, \cdots, 100\} \backslash\{1$ and prime numbers greater than 10 $\}$.
We will prove that at least seven numbers in set $T$ do not belong to
set $S$, thus, $|S| \leqslant 100-1-20-7=72$.
(i) If set $S$ does not contain any prime number greater than 10, then the smallest prime factor of each number in set $S$ is one of $2,3,5,7$. By condition (2), if $a b$ is divisible by $2 \times 3 \times 5 \times 7$, then $a$ and $b$ cannot both belong to set $S$. Therefore, at least one number in each of the following seven pairs does not belong to set $S$:
$$
\begin{array}{l}
(3,2 \times 5 \times 7),(5,2 \times 3 \times 7), \\
(7,2 \times 3 \times 5),(2 \times 3,5 \times 7), \\
(2 \times 5,3 \times 7),(2 \times 7,3 \times 5), \\
\left(2^{2} \times 7,3^{2} \times 5\right) .
\end{array}
$$
(ii) If set $S$ contains a prime number $p$ greater than 10, then the smallest prime factor of each number in set $S$ is one of $2, 3, 5, 7, p$. By conditions (2) and (3), we know:
1) When $7 p \in S$,
$2 \times 3 \times 5, 2^{2} \times 3 \times 5, 2 \times 3^{2} \times 5 \notin S$;
while $7 p \notin S$, $7, 7^{2}, 7 \times 11, 7 \times 13 \notin S$.
2) When $5 p \in S$, $2 \times 3 \times 7, 2^{2} \times 3 \times 7 \notin S$;
while $5 p \notin S$, $5, 5^{2} \notin S$.
3) $3 p$ and $2 \times 5 \times 7$ cannot both belong to set $S$.
4) $2 \times 3 \times p$ and $5 \times 7$ cannot both belong to set $S$.
5) When $5 p, 7 p \notin S$, $5 \times 7 \notin S$.
When $p=11,13$, by 1) to 4), we know that at least $3+2+1+1=7$ numbers in set $T$ do not belong to set $S$;
When $p=17,19$, by 1) to 3), we know that at least $4+2+1=7$ numbers in set $T$ do not belong to set $S$;
When $p \geqslant 23$, by 1), 2), and 5), we know that at least $4+2+1=7$ numbers in set $T$ do not belong to set $S$.
In summary, the maximum number of elements in a set $S$ that satisfies the given conditions is 72. | 72 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is $\qquad$
$(2009$, National High School Mathematics League Jilin Province Preliminary) | Hint: The fractional part of $(\sqrt{2}+\sqrt{3})^{2010}$ is $1-(\sqrt{2}-\sqrt{3})^{2010}$.
Since $0.9<1-(\sqrt{2}-\sqrt{3})^{2010}<1$, the first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is 9. | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $S=\{1,2, \cdots, 98\}$. Find the smallest positive integer $n$, such that in any $n$-element subset of $S$, one can always select 10 numbers, and no matter how these 10 numbers are divided into two groups, there is always one group in which there is a number that is coprime with the other four numbers, and in the other group, there is always a number that is not coprime with the other four numbers. | The subset of 49 even numbers in set $S$ obviously does not satisfy the coprime condition, hence $n \geqslant 50$.
To prove that any 50-element subset $T$ of set $S$ contains 10 numbers that satisfy the condition, we use the following facts.
Lemma 1 If nine elements in $T$ have a common factor greater than 1, and there is another element $x$ that is coprime with all of them, then these 10 numbers satisfy the condition.
Lemma 2 If the number of even elements in $T$ is $E(T)$, and there exists an odd element $x$ in $T$ such that the number of even elements in $T$ that are not coprime with $x$ is $f(x) \leqslant E(T)-9$, then $T$ contains 10 numbers that satisfy the condition.
Let $P_{1}=\{1,53,59,61,67,71,73,79,83,89,97\}$ (1 and the primes in set $S$ greater than 50), $\left|P_{1}\right|=11$, $P_{2}=\{13,17,19,23,29,31,37,41,43,47\}$ (primes greater than 11 and less than 50), $\left|P_{2}\right|=10$, $P_{3}=\{5,7,11\}$, $\left|P_{3}\right|=3$.
Since set $S$ contains only 49 even numbers, $T$ must contain odd numbers.
(1) If $T$ contains $x \in P_{1}$, $x$ is coprime with all other numbers in set $S$. Since there are 33 numbers in set $S$ that are neither divisible by 2 nor by 3, $T$ must contain at least $50-33=17$ numbers that are divisible by 2 or 3, among which there must be nine numbers that have a common factor of 2 or 3. $x$ is coprime with these nine numbers.
By Lemma 1, $T$ contains 10 numbers that satisfy the condition.
(2) If $T$ does not contain elements from $P_{1}$ but contains $x \in P_{2}$, $f(x) \leqslant\left[\frac{98}{2 \times 13}\right]=3$. Since $T$ contains no more than $49-11=38$ odd numbers, hence
$$
E(T) \geqslant 50-38=12 \geqslant f(x)+9.
$$
By Lemma 2, $T$ contains 10 numbers that satisfy the condition.
(3) If $T$ does not contain elements from $P_{1} \cup P_{2}$ but contains $x \in P_{3}$, $f(x) \leqslant\left[\frac{98}{2 \times 5}\right]=9$. In this case, $T$ contains no more than $49-11-10=28$ odd numbers, hence
$$
E(T) \geqslant 50-28=22>f(x)+9.
$$
$T$ also contains 10 numbers that satisfy the condition.
(4) If $T$ does not contain elements from $P_{1} \cup P_{2} \cup P_{3}$, $T$ contains no more than $49-11-10-3=25$ odd numbers, then
$$
E(T) \geqslant 50-25=25.
$$
Notice that, the odd numbers in $S \backslash\left(P_{1} \cup P_{2} \cup P_{3}\right)$ are divided into
$$
\begin{array}{c}
Q_{1}=\left\{3,3^{2}, 3^{3}, 3^{4}, 5^{2}, 5 \times 7,5 \times 11,5 \times 13,\right. \\
\left.5 \times 17,5 \times 19,7^{2}, 7 \times 11,7 \times 13\right\}
\end{array}
$$
and
$$
\begin{aligned}
Q_{2}= & \left\{3 \times 5,3^{2} \times 5,3 \times 5^{2}, 3 \times 7,3^{2} \times 7,\right. \\
& 3 \times 11,3 \times 13,3 \times 17,3 \times 19,3 \times 23, \\
& 3 \times 29,3 \times 31\} .
\end{aligned}
$$
If $T$ contains $x \in Q_{1}$, then either
$$
f(x) \leqslant f(3) \leqslant\left[\frac{98}{2 \times 3}\right]=16,
$$
or
$$
f(x) \leqslant\left[\frac{98}{2 \times 5}\right]+\left[\frac{98}{2 \times 7}\right]=16,
$$
both are no greater than $E(T)-9$.
If $T$ does not contain elements from $Q_{1}$ but contains $x \in Q_{2}$, then $f(x) \leqslant\left[\frac{98}{2 \times 3}\right]+\left[\frac{98}{2 \times 5}\right]=25$.
Hence $E(T) \geqslant 25+13=38>f(x)+9$.
In summary, any 50-element subset $T$ of set $S$ contains 10 numbers that satisfy the condition.
Therefore, the minimum value of $n$ is 50. | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The number of positive integers $n$ such that $n+1$ divides $n^{2012}+2012$ is $\qquad$ . | From the problem, we know
$$
\begin{array}{l}
n^{2012}+2012 \equiv(-1)^{2012}+2012 \\
=2013 \equiv 0(\bmod n+1) .
\end{array}
$$
Since $2013=3 \times 11 \times 61$, $n$ has $2^{3}-1=7$ solutions. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $P(-2,3)$ is a point on the graph of the inverse proportion function $y=\frac{k}{x}$, $Q$ is a moving point on the branch of the hyperbola in the fourth quadrant. A line is drawn through point $Q$ such that it intersects the hyperbola $y=\frac{k}{x}$ at only one point, and intersects the $x$-axis and $y$-axis at points $C$ and $D$, respectively. Another line $y=\frac{3}{2} x+6$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Then the minimum value of the area of quadrilateral $A B C D$ is $\qquad$. | 4. 48.
It is known that $k=-6, y=-\frac{6}{x}, A(-4,0), B(0,6)$.
Let the line passing through point $Q$ be $y=a x+b$.
Then $\left\{\begin{array}{l}y=a x+b, \\ y=-\frac{6}{x}\end{array}\right.$ has only one solution.
Eliminating and rearranging gives $a x^{2}+b x+6=0$.
By $\Delta=b^{2}-24 a=0 \Rightarrow b^{2}=24 a$.
Then $O C \cdot O D=(-b)\left(-\frac{b}{a}\right)=24=O A \cdot O B$.
Let $O C=m$. Then
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=3 m+\frac{48}{m}+24 \\
=3\left(\sqrt{m}-\frac{4}{\sqrt{m}}\right)^{2}+48 .
\end{array}
$$
Obviously, when $\sqrt{m}=\frac{4}{\sqrt{m}} \Rightarrow m=4$, the minimum value 48 is obtained. | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) As shown in Figure 5, $P$ is a point outside circle $\odot O$, $PA$ is tangent to $\odot O$ at point $A$, and $PBC$ is a secant of $\odot O$. $AD \perp PO$ at point $D$. If $PB=4$, $CD=5$, $BC=6$, find the length of $BD$. | Connect $O A, O B, O C$. Then $O B=O C$. It is easy to see that
$$
\begin{array}{l}
P A \perp O A, \\
P A^{2}=P B \cdot P C=P D \cdot P O \\
\Rightarrow \frac{P O}{P C}=\frac{P B}{P D} .
\end{array}
$$
Since $\angle B P D$ is a common angle, we have
$$
\begin{array}{l}
\triangle P O B \sim \triangle P C D . \\
\text { Then } \frac{P C}{C D}=\frac{P O}{O B}=\frac{P O}{O C} .
\end{array}
$$
At the same time, $\triangle P O C \sim \triangle P B D \Rightarrow \frac{P O}{O C}=\frac{P B}{B D}$.
$$
\text { Hence } \frac{P C}{C D}=\frac{P B}{B D} \Rightarrow B D=\frac{P B \cdot C D}{P C}=\frac{4 \times 5}{4+6}=2 \text {. }
$$ | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. For each positive integer $n$, let the tangent line to the curve $y=x^{n+1}$ at the point $(1,1)$ intersect the $x$-axis at a point with abscissa $x_{n}$. Let $a_{n}=\lg x_{n}$. Then
$$
a_{1}+a_{2}+\cdots+a_{99}=
$$
$\qquad$ | 3. -2 .
Given $y=x^{n+1}$, we know $y^{\prime}=(n+1) x^{n}$.
Then according to the condition, we have
$$
\begin{array}{l}
\frac{0-1}{x_{n}-1}=y^{\prime}(1)=n+1 \Rightarrow x_{n}=\frac{n}{n+1} . \\
\text { Therefore, } a_{1}+a_{2}+\cdots+a_{99}=\lg \left(x_{1} x_{2} \cdots x_{99}\right) \\
=\lg \left(\frac{1}{2} \times \frac{2}{3} \times \cdots \times \frac{99}{100}\right)=-2 .
\end{array}
$$ | -2 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $S=\{1,2, \cdots, 50\}$. Find the smallest positive integer $k$, such that in any $k$-element subset of $S$, there exist two distinct numbers $a$ and $b$ satisfying $(a+b) \mid a b$.
$(1996$, China Mathematical Olympiad) | Let the greatest common divisor of $a$ and $b$ be $d$, and $a = a_{1}d$, $b = b_{1}d$. Then $\left(a_{1}, b_{1}\right) = 1$.
Substituting into $(a+b) \mid ab$, we get
$$
\left(a_{1} + b_{1}\right) \mid a_{1} b_{1} d \Rightarrow \left(a_{1} + b_{1}\right) \mid d.
$$
Let $d = k\left(a_{1} + b_{1}\right)$, we obtain all pairs satisfying the condition:
$$
a = k a_{1}\left(a_{1} + b_{1}\right), \quad b = k b_{1}\left(a_{1} + b_{1}\right).
$$
Assume $a_{1} < b_{1} \leqslant 6$.
List all 23 pairs $(a, b)$ in the set $S$ as follows:
$$
\begin{array}{l}
(3,6), (6,12), (9,18), (12,24), (15,30), \\
(18,36), (21,42), (24,48), (4,12), (8,24), \\
(12,36), (16,48), (10,15), (20,30), (30,45), \\
(5,20), (10,40), (21,28), (6,30), (14,35), \\
(24,40), (36,45), (7,42).
\end{array}
$$
Now construct a graph with $S$ as the vertex set, such that the graph has 26 isolated points:
$$
\begin{array}{l}
1, 2, 11, 13, 17, 19, 22, 23, 25, 26, 27, 29, \\
31, 32, 33, 34, 37, 38, 39, 41, 43, 44, 46, \\
47, 49, 50,
\end{array}
$$
The remaining 24 points (23 edges) are divided into 3 connected components, as shown in Figure 3.
From Figure 3, we can select 38 points such that no two points are adjacent: 26 isolated points and 12 points:
$$
14, 7, 21, 5, 4, 9, 8, 16, 3, 40, 15, 45.
$$
Therefore, $k \geqslant 39$.
On the other hand, in any 39-element subset, at least 13 points are not isolated. They are distributed among 12 adjacent pairs:
$$
\begin{array}{l}
(14,35), (7,42), (21,28), (5,20), \\
(9,18), (4,12), (8,24), (16,48), \\
(3,6), (40,10), (15,30), (45,36),
\end{array}
$$
At least one pair of these points will belong to the subset, and they satisfy the divisibility condition.
Therefore, the minimum value of $k$ is 39. | 39 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let the ellipse $C$ be:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{20}=1(a>2 \sqrt{5})
$$
with its left focus at $F$, and point $P(1,1)$. It is known that there exists a line $l$ passing through point $P$ intersecting the ellipse at points $A$ and $B$, with $M$ being the midpoint of $A B$, such that $|F M|$ is the geometric mean of $|F A|$ and $|F B|$. Find the smallest positive integer value of $a$, and the equation of $l$ at this value of $a$. | 11. By the median length formula, we have
$|F M|^{2}=\frac{1}{2}\left(|F A|^{2}+|F B|^{2}\right)-\frac{1}{4}|A B|^{2}$.
Also, $|F M|^{2}=|F A||F B|$, so
$|A B|^{2}=2(|F A|-|F B|)^{2}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and
$c=\sqrt{a^{2}-20}$.
Then $|F A|-|F B|=\frac{c}{a} x_{1}-\frac{c}{a} x_{2}$.
Thus, $\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=\frac{2 c^{2}}{a^{2}}\left(x_{1}-x_{2}\right)^{2}$, which means
$\left(y_{1}-y_{2}\right)^{2}=\frac{2 c^{2}-a^{2}}{a^{2}}\left(x_{1}-x_{2}\right)^{2}$
$=\frac{a^{2}-40}{a^{2}}\left(x_{1}-x_{2}\right)^{2}$.
Therefore, $a^{2} \geqslant 40$.
When $a \in \mathbf{N}_{+}$, $a \geqslant 7$.
When $a=7$, $\left(y_{1}-y_{2}\right)^{2}=\frac{9}{49}\left(x_{1}-x_{2}\right)^{2}$.
Thus, the slope of line $l$ is $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}= \pm \frac{3}{7}$.
Since point $P(1,1)$ is clearly inside the ellipse, both lines $y-1= \pm \frac{3}{7}(x-1)$ satisfy the condition.
In summary, the smallest positive integer value of $a$ is 7, and at this time, the equations of line $l$ are
$3 x-7 y+4=0$ or $3 x+7 y-10=0$. | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Remove a $2 \times 2$ small square from any corner of a $3 \times 3$ grid, and the remaining shape is called a "corner shape". Now, place some non-overlapping corner shapes in a $9 \times 9$ grid, with the requirement that the boundaries of the corner shapes coincide with the boundaries or grid lines of the grid. Find the maximum positive integer $k$ such that, no matter how $k$ corner shapes are placed, it is always possible to place one more complete corner shape in the grid. | Three, first, $k_{\max }<6$. This is because, if 6 corner shapes are placed in the manner shown in Figure 4, it is impossible to place another complete corner shape on this grid.
Now, place 5 corner shapes in any manner.
Next, we will prove that it is still possible to place another complete corner shape.
Consider the five shaded areas $A, B, C, D, E$ in Figure 5. If there exists a region that does not intersect with any of the 5 placed corner shapes, then it is clearly possible to place a complete corner shape in that region.
Assume that all five regions contain a part of a corner shape. Note that a corner shape cannot intersect with two of the regions $A, B, C, D, E$ simultaneously. Therefore, each of the five regions $A, B, C, D, E$ corresponds to a unique corner shape that intersects with it, which we denote as $f(A), f(B), f(C), f(D), f(E)$.
According to Figure 5, the four $4 \times 4$ sub-grids in the corners of the $9 \times 9$ grid are labeled as $a, b, c, d$.
Note that the corner shape $f(E)$ cannot intersect with all four regions $a, b, c, d$ simultaneously. Without loss of generality, assume $f(E)$ does not intersect with $a$. Considering that the corner shapes $f(B), f(C), f(D)$ also do not intersect with $a$, the only corner shape that intersects with $a$ is $f(A)$. However, the part of region $a$ occupied by the corner shape $f(A)$ must be contained within a $3 \times 3$ sub-grid in one of the corners of $a$. Without loss of generality, assume it is contained in the shaded area shown in Figure 6, then it is possible to place a new corner shape in the remaining part of $a$ as shown in Figure 6.
The above demonstrates that $k=5$ satisfies the problem's requirements.
In conclusion, $k_{\max }=5$. | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given real numbers $a, b, c, d$ satisfy: for any real number $x$,
$a \cos x + b \cos 2x + c \cos 3x + d \cos 4x \leqslant 1$.
Find the maximum value of $a + b - c + d$ and the values of the real numbers $a, b, c, d$ at this time.
(Ninth China Southeast Mathematical Olympiad) | Let
$$
\begin{array}{l}
f(x)=a \cos x+b \cos 2 x+c \cos 3 x+d \cos 4 x . \\
\text { By } f(0)=a+b+c+d, \\
f(\pi)=-a+b-c+d, \\
f\left(\frac{\pi}{3}\right)=\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2},
\end{array}
$$
then \(a+b-c+d\)
$$
=f(0)+\frac{2}{3} f(\pi)+\frac{4}{3} f\left(\frac{\pi}{3}\right) \leqslant 3 .
$$
Equality holds if and only if \(f(0)=f(\pi)=f\left(\frac{\pi}{3}\right)=1\), i.e., \(a=1, b+d=1, c=-1\).
At this point, let \(t=\cos x(-1 \leqslant t \leqslant 1)\). Then
$$
\begin{array}{l}
f(x)-1 \\
=\cos x+b \cos 2 x-\cos 3 x+d \cos 4 x-1 \\
=t+(1-d)\left(2 t^{2}-1\right)-\left(4 t^{3}-3 t\right)+ \\
d\left(8 t^{4}-8 t^{2}+1\right)-1 \\
=2(t-1)(t+1)(2 t-1)[2 d t-(1-d)] \\
\leqslant 0 \\
\end{array}
$$
for any real number \(t \in[-1,1]\).
Thus, \(d>0\), and \(\frac{2 d}{2}=\frac{1-d}{1}\), i.e., \(d=\frac{1}{2}\).
Therefore, the maximum value of \(a+b-c+d\) is 3, and at this time,
$$
(a, b, c, d)=\left(1, \frac{1}{2},-1, \frac{1}{2}\right) \text {. }
$$ | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
17. A moving point moves on the integer points in the first quadrant of the Cartesian coordinate system (including the integer points on the $x$-axis and $y$-axis of the first quadrant), with the movement rules being $(m, n) \rightarrow(m+1, n+1)$ or $(m, n) \rightarrow$ $(m+1, n-1)$. If the moving point starts from the origin and reaches the point $(6,2)$ after 6 steps, then there are $\qquad$ different movement paths. | $\begin{array}{l}\text { 17. } 9 \\ C_{6}^{2}-C_{6}^{1}=9\end{array}$
The translation is as follows:
$\begin{array}{l}\text { 17. } 9 \\ C_{6}^{2}-C_{6}^{1}=9\end{array}$
Note: The original text is already in a mathematical format, which is universal and does not require translation. However, if you intended to have the problem statement or any surrounding text translated, please provide that context. | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure $4, \triangle A B C$ has an incircle $\odot O_{1}$ that touches side $B C$ at point $D, \odot O_{2}$ is the excircle of $\triangle A B C$ inside $\angle A$. If $O_{1} B=6, O_{1} C=3, O_{1} D=2$, then $O_{1} O_{2}=$ $\qquad$ | $$
\begin{aligned}
& \angle O_{1} B O_{2}=\angle O_{1} C O_{2}=90^{\circ} \\
\Rightarrow & O_{1} 、 C 、 O_{2} 、 B \text { are concyclic } \\
\Rightarrow & \angle O_{1} C D=\angle O_{1} O_{2} B \\
\Rightarrow & \triangle O_{1} D C \backsim \triangle O_{1} B O_{2} \\
\Rightarrow & O_{1} O_{2}=\frac{O_{1} B \cdot O_{1} C}{O_{1} D}=\frac{3 \times 6}{2}=9 .
\end{aligned}
$$ | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. If a number, from the highest digit to the lowest digit, does not decrease at each digit, it is called a "positive number" (such as $12$, $22$, $566$, $1448$, $123456789$, etc.); if a number, from the highest digit to the lowest digit, does not increase at each digit, it is called a "negative number" (such as $21$, $22$, $665$, $8441$, etc.). A number that is neither a positive number nor a negative number is called a "swinging number" (such as $253$, $3473$, $5887$, etc.). Then among the three-digit numbers, there are $\qquad$ swinging numbers. | 4.525.
According to the definition of a wavy number, we discuss in two cases:
(1) The hundreds, tens, and units digits are in the pattern "small-large-small": there are
$$
2 \times 1+3 \times 2+\cdots+9 \times 8=240 \text { (numbers). }
$$
(2) The hundreds, tens, and units digits are in the pattern "large-small-large": there are
$$
1^{2}+2^{2}+\cdots+9^{2}=285 \text { (numbers). }
$$
In summary, there are $240+285=525$ wavy numbers. | 525 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Among $m$ students, it is known that in any group of three, two of them know each other, and in any group of four, two of them do not know each other. Then the maximum value of $m$ is $\qquad$ | 3. 8 .
When $m=8$, the requirement is satisfied.
It only needs to prove: $m \leqslant 8$.
First, prove that the following two scenarios are impossible.
(1) If a student $A$ knows at least 6 people, by Ramsey's theorem, among these 6 people, there exist 3 people who either all know each other or all do not know each other. If it is the former, then $A$ and these 3 people form a group of 4 people who all know each other, which contradicts the given condition; if it is the latter, this contradicts the given condition that among any 3 people, there are 2 who know each other.
(2) If a student $A$ knows at most $m-5$ people, then at least 4 people do not know $A$, thus, these 4 people all know each other, which contradicts the given condition.
Secondly, when $m \geqslant 10$, either (1) or (2) must occur, which is impossible.
When $m=9$, to make (1) and (2) not occur, each student must know exactly 5 other people, thus, the number of friend pairs (pairs of people who know each other) generated by these 9 people is $\frac{9 \times 5}{2} \notin Z$, which is a contradiction.
In summary, the maximum value of $m$ is 8. | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. If $n$ is a positive integer greater than 1, then
$$
\begin{array}{l}
\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\cos \frac{6 \pi}{n}+\cdots+\cos \frac{2 n \pi}{n} \\
=
\end{array}
$$ | 5. 0 .
$$
\sum_{k=1}^{n} \cos \frac{2 k \pi}{n}=\operatorname{Re} \sum_{k=1}^{n} \mathrm{e}^{\frac{2 k \pi i}{n}}=0 .
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Let the 10 complex roots of the equation $x^{10}+(13 x-1)^{10}=0$ be $x_{1}, x_{2}, \cdots, x_{10}$. Then
$$
\frac{1}{x_{1} \overline{x_{1}}}+\frac{1}{x_{2} \overline{x_{2}}}+\cdots+\frac{1}{x_{5} \overline{x_{5}}}=
$$
$\qquad$ | 7. 850 .
Let $\varepsilon=\cos \frac{\pi}{10}+\mathrm{i} \sin \frac{\pi}{10}$. Then $\varepsilon^{10}=-1$.
Given that the 10 complex roots of the equation $x^{10}+(13 x-1)^{10}=0$ are $x_{1}, x_{2}, \cdots, x_{10}$, we can assume them to be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, $\overline{x_{1}}, \overline{x_{2}}, \overline{x_{3}}, \overline{x_{4}}, \overline{x_{5}}$.
From $(13 x-1)^{10}=-x^{10}$, we know
$$
13 x_{k}-1=x_{k} \varepsilon^{2 k-1}(k=1,2, \cdots, 5) \text {. }
$$
Thus, $\frac{1}{x_{k}}=13-\varepsilon^{2 k-1}$.
$$
\begin{array}{l}
\text { Hence } \frac{1}{x_{1} \overline{x_{1}}}+\frac{1}{x_{2} \overline{x_{2}}}+\cdots+\frac{1}{x_{5} \overline{x_{5}}} \\
=\sum_{k=1}^{5}\left(13-\varepsilon^{2 k-1}\right)\left(13-\bar{\varepsilon}^{2 k-1}\right) \\
=\sum_{k=1}^{5}\left[170-13\left(\varepsilon^{2 k-1}+\bar{\varepsilon}^{2 k-1}\right)\right] \\
=850-13 \sum_{k=1}^{5}\left(\varepsilon^{2 k-1}+\bar{\varepsilon}^{2 k-1}\right)=850 .
\end{array}
$$ | 850 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) 11 interest classes, several students participate (can participate repeatedly), and each interest class has the same number of students (full, unknown number). It is known that any nine interest classes include all students, while any eight interest classes do not include all students. Find the minimum total number of students. | Let the set of students in 11 interest classes be $A_{1}$, $A_{2}, \cdots, A_{11}$.
By the problem, we know $\left|A_{1}\right|=\left|A_{2}\right|=\cdots=\left|A_{11}\right|=x$.
Let $T=A_{1} \cup A_{2} \cup \cdots \cup A_{11}$.
By the problem, we know the union of any nine sets is $T$, and the union of any eight sets is a proper subset of $T$.
Construct a table, if student $a_{i} \in A_{j}$, then fill the cell in the $i$-th row and $j$-th column with 1, otherwise fill it with 0.
By the condition, we know that the union of any eight sets is not $T$, i.e., for any eight columns, there must be a row where the intersecting cells are all 0, which we call a "zero row".
Furthermore, by the condition that the union of any nine sets is $T$, any two zero rows cannot be the same row.
Thus, we form a one-to-one mapping from every 8 columns to one row.
Therefore, $\mathrm{C}_{11}^{8} \leqslant|T|$.
On the other hand, each row has 3 zeros, and the columns of zeros in any two rows are not all the same, with a total of $\mathrm{C}_{11}^{3}$ ways to fill. And $\mathrm{C}_{11}^{3}=\mathrm{C}_{11}^{8}$, so each row is filled in exactly one way. At this point, the conditions of the problem are exactly satisfied.
In summary, the minimum number of students is $\mathrm{C}_{11}^{8}=165$. | 165 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 A philanthropist recruits members for his club in the following way: each member can introduce two others to join, where these two are not introduced by anyone else; and each new member can also introduce two others to join. For a member $A$, the members introduced by $A$ and the members introduced by those introduced by $A$ are collectively referred to as $A$'s "downlines." If each of the two members $B_{1}, B_{2}$ introduced by a member $A$ has at least 200 downlines, then member $A$ will receive a dinner voucher from the philanthropist as a reward at the end of the year.
Determine: If a total of 2012 people have become members, then at most how many people can enjoy the philanthropist's dinner at the end of the year? | Solve using reverse thinking.
For this, let $200=k$, and denote the minimum number of members when exactly $r$ people receive coupons as $f(r)$. Members who join without being introduced (or without a superior) are called "bosses", and the members they introduce are called "followers". Each boss along with all their followers and subordinates form a "gang"; each gang forms a tree diagram.
To minimize the number of members, in any gang, a boss who cannot win a prize can actually be removed. Thus, some of the followers under their control may become bosses, forming new gangs along with the subtrees they manage.
When $r=1$, the minimum number of members is $2k+3$ (at this time, only $A$ wins, and $A, B_1, B_2$ are all members, and $B_1, B_2$ each have at least $k$ subordinates), i.e., $f(1)=2k+3$.
When $r=2$, the winners are $A$ and one of $B_1, B_2$ (let it be $B_1$), and according to the above discussion, the branch containing $B_1$ (including $B_1$) has at least $2k+3$ people. Since $A$ is a winner, $B_2$ must have at least $k$ subordinates. Therefore, the number of members is at least
$$
1+(2k+3)+(k+1)=3k+5,
$$
i.e., $f(2)=3k+5$.
When $r=3$, there are two scenarios for the winners.
(1) $A, B_1, B_2$;
(2) $A, B_1$, and one of $B_1$'s subordinates (let it be $C_1$).
For (1), according to the above discussion, the branch containing $B_1$ (including $B_1$) has at least $2k+3$ people; the branch containing $B_2$ (including $B_2$) also has at least $2k+3$ people. Therefore, including $A$, the number of members is at least $4k+7$.
For (2), $B_1$'s branch has two winners, and this branch has at least $3k+5$ people; since $A$ is a winner, $B_2$ must have at least $k$ subordinates. Therefore, the number of members is at least
$$
1+(3k+5)+(k+1)=4k+7.
$$
Thus, both scenarios result in $f(3)=4k+7$.
$$
\text{By } f(2)-f(1)=k+2, f(3)-f(2)=k+2,
$$
we can guess that for each positive integer $n$,
$$
\begin{array}{l}
f(r+1)-f(r)=k+2. \\
\text{Thus, } f(r)=r(k+2)+(k+1).
\end{array}
$$
It is sufficient to prove equation (1) for the case of only one gang.
In fact, if there are $s$ gangs, each gang has $n_1, n_2, \cdots, n_s$ people, and the number of winners are $r_1, r_2, \cdots, r_s$, then
$$
\begin{array}{l}
\text{Total number of members } n=\sum_{i=1}^{1} n_{i}, \\
\text{Total number of winners } r=\sum_{i=1}^{1} r_{i}. \\
\text{Thus, } n=\sum_{i=1}^{s} n_{i} \geqslant \sum_{i=1}^{\prime} f\left(r_{i}\right) \\
=\sum_{i=1}^{s}\left[r_{i}(k+2)+(k+1)\right] \\
=(k+2) \sum_{i=1}^{s} r_{i}+s(k+1) \\
=r(k+2)+s(k+1) \\
\geqslant r(k+2)+(k+1).
\end{array}
$$
Next, we prove equation (1) using mathematical induction.
When $r=1,2,3$, it has already been established.
Assume equation (1) holds for fewer than $r$ winners.
When there are $r$ winners and the number of members in this gang has reached the minimum, the boss $A$ must be a winner. Let the two direct followers of boss $A$ be $B, C$. If the number of people in $B$'s sub-gang (including $B$ and all their followers) is $n_B$, and the number of winners among them is $r_B$, similarly define $n_C, r_c$.
According to the induction hypothesis, if $r_B>0$ (i.e., $B$'s sub-gang has winners), then the number of members in this sub-gang is
$$
n_B \geqslant r_B(k+2)+(k+1).
$$
If $r_B=0$, since $A$ is a winner, $B$ must have introduced at least $k$ people, including $B$ themselves, $B$'s sub-gang has at least $k+1$ members. Thus,
$$
n_B \geqslant k+1=r_B(k+2)+(k+1).
$$
Therefore, in any case,
$$
n_B \geqslant r_B(k+2)+(k+1).
$$
Similarly, $n_C \geqslant r_c(k+2)+(k+1)$.
Since the total number of members in $A$'s gang is $n=n_B+n_C+1$, and the total number of winners is $r=r_B+r_C+1$, we have
$$
\begin{array}{l}
n=n_B+n_C+1 \\
\geqslant\left(r_B+r_C\right)(k+2)+2(k+1)+1 \\
=r(k+2)+(k+1), \\
f(r)=r(k+2)+(k+1).
\end{array}
$$
Thus, $f(r)=r(k+2)+(k+1)$.
According to equation (1), when there are $r$ winners, the number of members $n$ satisfies $n \geqslant f(r)=r(k+2)+(k+1)$.
When $n=2012, k=200$, the above equation becomes
$$
202 r \leqslant 1811 \Rightarrow r \leqslant\left[\frac{1811}{202}\right]=8.
$$
Therefore, at most 8 people can receive the philanthropist's banquet at the end of the year. | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 If the positive integer $m$ makes it true that for any set of positive numbers $a_{1}, a_{2}, a_{3}, a_{4}$ satisfying $a_{1} a_{2} a_{3} a_{4}=1$, we have
$$
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}
$$
then the minimum value of the positive integer $m$ is $\qquad$ [2] | Let $a_{1}=\frac{1}{27}, a_{2}=a_{3}=a_{4}=3$. Then
$$
\begin{array}{l}
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m}=\left(\frac{1}{27}\right)^{m}+3 \times 3^{m}, \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}=27+3 \times \frac{1}{3}=28 .
\end{array}
$$
Verification shows that $m=1, m=2$ do not meet the requirements.
Hence $m \geqslant 3$.
$$
\begin{array}{l}
\text { By } \frac{a_{1}^{3}+a_{2}^{3}+a_{3}^{3}}{3} \geqslant a_{1} a_{2} a_{3}, \frac{a_{1}^{3}+a_{2}^{3}+a_{4}^{3}}{3} \geqslant a_{1} a_{2} a_{4}, \\
\frac{a_{1}^{3}+a_{3}^{3}+a_{4}^{3}}{3} \geqslant a_{1} a_{3} a_{4}, \frac{a_{2}^{3}+a_{3}^{3}+a_{4}^{3}}{3} \geqslant a_{2} a_{3} a_{4}, \\
a_{1} a_{2} a_{3} a_{4}=1,
\end{array}
$$
we get
$$
\begin{array}{l}
a_{1}^{3}+a_{2}^{3}+a_{3}^{3}+a_{4}^{3} \\
\geqslant a_{1} a_{2} a_{3}+a_{1} a_{2} a_{4}+a_{1} a_{3} a_{4}+a_{2} a_{3} a_{4} \\
=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}} .
\end{array}
$$
Therefore, $m=3$ meets the requirements.
Thus, the smallest positive integer $m$ is 3. | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { Example } 6 \text { Let } u=1+\frac{x^{3}}{3!}+\frac{x^{6}}{6!}+\cdots, \\
v=\frac{x}{1!}+\frac{x^{4}}{4!}+\frac{x^{7}}{7!}+\cdots, w=\frac{x^{2}}{2!}+\frac{x^{5}}{5!}+\frac{x^{8}}{8!}+\cdots
\end{array}
$$
Prove: $u^{3}+v^{3}+w^{3}-3 u v w=1$. | Let $\lambda=\mathrm{e}^{\frac{2 \pi}{3}}$ be a unit cube root.
Then $1+\lambda+\lambda^{2}=0$.
Thus $u^{3}+v^{3}+w^{3}-3 u v w$
$$
=(u+v+w)\left(u+\lambda v+\lambda^{2} w\right)\left(u+\lambda^{2} v+\lambda w\right)
$$
(By Corollary 3)
$$
=\mathrm{e}^{x} \mathrm{e}^{\lambda x} \mathrm{e}^{\lambda^{2} x}=\mathrm{e}^{0}=1 .
$$ | 1 | Algebra | proof | Yes | Yes | cn_contest | false |
2. Given $x, y, z \in \mathbf{R}_{+}$, satisfying $x^{2}+y^{2}+z^{2}=1$. Then $\min \left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)=$ $\qquad$ . | Notice,
$$
\begin{array}{l}
\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)\left(x^{2}+y^{2}+z^{2}\right) \\
\geqslant 3 \sqrt[3]{\frac{1}{x^{2}} \cdot \frac{1}{y^{2}} \cdot \frac{1}{z^{2}}} \times 3 \sqrt[3]{x^{2} y^{2} z^{2}}=9,
\end{array}
$$
when and only when $x=y=z=\frac{\sqrt{3}}{3}$,
$$
\min \left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)=9 .
$$ | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Let non-negative real numbers $x_{1}, x_{2}, \cdots, x_{6}$ satisfy
$$
\begin{array}{c}
x_{1}+x_{2}+\cdots+x_{6}=1, x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6} \geqslant \frac{1}{540} . \\
\text { If } \max \left\{x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+x_{4} x_{5} x_{6}+\right. \\
\left.x_{5} x_{6} x_{1}+x_{6} x_{1} x_{2}\right\}=\frac{p}{q},(p, q)=11 \text {, find } p+q .{ }^{[3]}
\end{array}
$$ | Let $r=x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6}$,
$$
\begin{aligned}
s= & x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+ \\
& x_{4} x_{5} x_{6}+x_{5} x_{6} x_{1}+x_{6} x_{1} x_{2} .
\end{aligned}
$$
By the AM-GM inequality,
$$
\begin{array}{l}
r+s=\left(x_{1}+x_{4}\right)\left(x_{2}+x_{5}\right)\left(x_{3}+x_{6}\right) \\
\leqslant\left[\frac{\left(x_{1}+x_{4}\right)+\left(x_{2}+x_{5}\right)+\left(x_{3}+x_{6}\right)}{3}\right]^{3}=\frac{1}{27},
\end{array}
$$
with equality if and only if $x_{1}+x_{4}=x_{2}+x_{5}=x_{3}+x_{6}=\frac{1}{3}$.
Therefore, $s \leqslant \frac{1}{27}-\frac{1}{540}=\frac{19}{540}$.
Let $x_{1}=x_{3}=\frac{3}{10}, x_{5}=\frac{1}{60}, x_{2}=\frac{1}{3}-x_{5}=\frac{19}{60}$,
$$
x_{4}=\frac{1}{3}-x_{1}=\frac{1}{30}, x_{6}=\frac{1}{3}-x_{3}=\frac{1}{30} \text {. }
$$
Then $r=\frac{1}{540}, s=\frac{19}{540}$.
Thus, $p+q=559$. | 559 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Calculate:
$$
\frac{2013^{2}+2011}{2011^{2}-2013} \times \frac{4020^{2}-8040}{2011 \times 2014-4}=
$$
$\qquad$ | $=1.4$.
Let $a=2$ 011. Then
$$
\begin{array}{l}
\text { Original expression }=\frac{(a+2)^{2}+a}{a^{2}-a-2} \cdot \frac{(2 a-2)^{2}-4(a-1)}{a(a+3)-4} \\
=\frac{(a+1)(a+4)}{(a-2)(a+1)} \cdot \frac{4(a-1)(a-2)}{(a+4)(a-1)}=4 .
\end{array}
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. A sequence of numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ is constructed according to the following rule: $a_{1}=7, a_{k}=a_{k-1}^{2}$'s digit sum $+1(k=2,3, \cdots)$. For example, $a_{2}=14, a_{3}=17$, and so on.
Then $a_{2013}=$ $\qquad$ . | 2.8.
From the problem, we know
$$
\begin{array}{l}
a_{1}=7, a_{2}=14, a_{3}=17, a_{4}=0, \\
a_{5}=5, a_{6}=8, a_{7}=11, a_{8}=5 .
\end{array}
$$
Thus, $a_{8}=a_{5}$, meaning from $a_{5}$ onwards, the sequence repeats with a period of 3.
Therefore, $a_{2013}=a_{6+3 \times 669}=8$. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
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