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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 t... | (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 to... | 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
$$
x_{1} \geqslant 1, x_{2} \geqslant 3, x_{3} \geqslant 3, x_{4} \geqslant 0
$$... | 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 s... | 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 $5040+5040=10080$ ... | 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
$$
\begin{array}{l}
\omega^{p}=1, \omega^{-\frac{p}{2}}=-1, 2 \cos \frac{2 k \pi}{p}=\omega^{k}+\omega^{-k} . \\
\text { Hence } \prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)=\prod_{k=1}^{p-1}\left(1+\omega^{k}+\omega^{-k}\right) \\
=\prod_{k=1}^{p-1} \omega... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The sum of all real numbers $m$ that satisfy $(2-m)^{m^{2}-m-2}=1$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 2. A.
When $2-m=1$, i.e., $m=1$, it satisfies the condition.
When $2-m=-1$, i.e., $m=3$,
$$
(2-m)^{m^{2}-m-2}=(-1)^{4}=1 \text{, }
$$
it satisfies the condition.
When $2-m \neq \pm 1$, i.e., $m \neq 1$ and $m \neq 3$, by the condition, $m^{2}-m-2=0$, and $2-m \neq 0$.
Solving gives $m=-1$.
Therefore, the sum is $1+3+... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
6. Let $n$ be a positive integer, and call $n$ a "good number" if the number of prime numbers not exceeding $n$ equals the number of composite numbers not exceeding $n$. Then the sum of all good numbers is ( ).
(A) 33
(B) 34
(C) 2013
(D) 2014 | 6. B.
Since 1 is neither a prime number nor a composite number, a good number must be an odd number.
Let the number of prime numbers not exceeding $n$ be $a_{n}$, and the number of composite numbers be $b_{n}$. When $n \leqslant 15$, only consider the case where $n$ is odd (as shown in Table 1).
Table 1
\begin{tabula... | 34 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
1. Given real numbers $x, y, z$ satisfy
$$
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) \\
=... | 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,... | 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... | 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
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-... | Three, since $a, b, c$ have cyclic symmetry, without loss of generality, assume $0 < a < b < c$, then
$$
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... | 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 i... | 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 ev... | 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, ... | 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... | 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 immediatel... | 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 ... | 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[\fra... | 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... | 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 condition... | 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, ea... | 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_... | 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 gene... | 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 valu... | 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
$$
has at most two real roots. Therefore, each polynomial in $A$ appears at... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
17. Alex has 75 red cards and 75 blue cards. It is known that Alex can exchange 2 red cards for 1 silver card and 1 blue card at one stall, and can exchange 3 blue cards for 1 silver card and 1 red card at another stall. If he continues to exchange according to the above methods until he can no longer exchange any card... | 17. E.
If Alex has $a$ red cards, $b$ blue cards, and $c$ silver cards, denoted as $(a, b, c)$, then
$$
\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}
$$
Therefore, Alex has 103 silver card... | 103 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
23. In $\triangle A B C$, it is known that $A B=13, B C=14$, $C A=15$, points $D, E, F$ are on sides $B C, C A, D E$ respectively, satisfying $A D \perp B C, D E \perp A C, A F \perp B F$, the length of segment $D F$ is a reduced fraction $\frac{m}{n}\left(m, n \in \mathbf{N}_{+},(m, n)=1\right)$. Then $m+n=(\quad)$.
(... | 23. B.
As shown in Figure 5.
Let $p$ be the semi-perimeter of $\triangle ABC$. Then
$$
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.
... | 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 hold... | 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-... | 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 ... | 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, $a_{m+1}-a_{m}=2 m(m \geqslant 1)$,
$$
\be... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Given
$$
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 {. }
$$
Then $h^{\prime}(x)=\frac{x-1-\l... | 3 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Find the smallest positive integer $n$, such that the sum of the squares of all positive divisors of $n$ is $(n+3)^{2}$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note and not part of the problem statement, so it is provided in its original form as it is ... | Three, let $16 n+8, \\
\end{array}
$$
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 dis... | 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 $... | 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 p... | 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 bisec... | 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 ... | 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 th... | 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... | 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 hav... | 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 bel... | 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 ... | 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 ... | 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 ori... | 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 bou... | 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 ... | 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... | 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 $A$ appea... | 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)-(1... | 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 oth... | 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... | 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... | 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 ... | 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... | 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}
$$
Since $20... | 053 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find all odd prime numbers $p$ such that $p \mid \sum_{k=1}^{103} k^{p-1}$.
untranslated text remains the same as requested. | If $p>103$, then for $1 \leqslant k \leqslant 103$, we have
$$
\begin{array}{l}
k^{p-1} \equiv 1(\bmod p), \\
\sum_{k=1}^{103} k^{p-1} \equiv 103(\bmod p) .
\end{array}
$$
Therefore, $p \leqslant 103$.
Let $103=p q+r(0 \leqslant r < q)$, then $r=q, 103=p q+r=(p+1) r$. Since 103 is a prime number, we get $p=102, r=1$, ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Find all positive integers that are coprime with all terms of the sequence $\left\{a_{n}\right\}$ satisfying
$$
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 ... | 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.
$$
\begin{array}{l}
\text { Also, } 3^{n}+2 \time... | 4 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 5 What is the minimum degree of the highest term of a polynomial with rational coefficients that has $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots?
(2013, Joint Autonomous Admission Examination of Peking University and Other Universities) | Notice that the polynomial
$$
f(x)=\left(x^{2}-2\right)\left[(x-1)^{3}-2\right]
$$
has roots $\sqrt{2}$ and $1-\sqrt[3]{2}$, and its degree is 5. Therefore, the degree of the highest term of a rational-coefficient polynomial with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots cannot be less than 5.
If there exists a rationa... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $x$ is an integer, and satisfies the inequality system
$$
\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}
$$
When $|y-z|=6$, and $x=\f... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $x_{1}, x_{2}, \cdots, x_{15}$ take values of 1 or -1. Let
$$
S=x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{15} x_{1} \text {. }
$$
Then the smallest positive integer that $S$ can take is $\qquad$ | 4.3.
Let $y_{i}=x_{i} x_{i+1}(i=1,2, \cdots, 15)$, with the convention that $x_{16}=$ $x_{1}$. Then $y_{i}=1$ or -1.
In $y_{1}, y_{2}, \cdots, y_{15}$, let there be $a$ values that are 1 and $b$ values that are -1. Clearly, $a+b=15$.
Also, $1^{a}(-1)^{b}=y_{1} y_{2} \cdots y_{15}=\left(x_{1} x_{2} \cdots x_{15}\righ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. (25 points) As shown in Figure 1, given that $E$ is a point on side $A B$ of square $A B C D$, and the symmetric point of $A$ with respect to $D E$ is $F, \angle B F C = 90^{\circ}$. Find the value of $\frac{A B}{A E}$. | 2. As shown in Figure 4, extend $E F$ to intersect $B C$ at point $M$, connect $D M$, and let it intersect $C F$ at point $G$.
Then, Rt $\triangle D F M \cong$ Rt $\triangle D C M$. Therefore, $\angle F D M = \angle M D C$, and $F M = C M$.
Thus, $M$ is the midpoint of $B C$.
$$
\begin{array}{l}
\text { Also, } \angle ... | 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}... | 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}, ... | 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(... | 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... | 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, $M=\sum_{i=1}^{2013} x_{i}^{3}=\sum_{i=1}^{2013}\left(x_{i}^{3}-3 x_{i}\right)$
$$... | 4026 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find the remainder when $10^{10}(100$ ones$)$ is divided by 7. | \begin{array}{l}\text { Sol } 10^{10} \equiv(7+3)^{10^{10}} \equiv 3^{10} \\ \equiv(7+2)^{50 \cdots 0} \equiv 2^{5 \times 10^{9}} \equiv 2^{3 \times 106 \cdots 6+2} \\ \equiv 4(7+1)^{166 \cdots 6} \equiv 4(\bmod 7) .\end{array} | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Seven, let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\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}$.
Theref... | 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 ... | 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... | 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$ ... | -、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 o... | 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}
$$
From the given ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. $\sum_{0 \leqslant i<j \leqslant 50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}$ modulo 31 is | 8. 16 .
Notice,
$$
\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(... | 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, ... | 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 Mathemati... | 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)... | 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, $0<(5-2 \sqrt{6})^{1006}<1$,
$$
\begin{array}{l}
(5+2 \sqrt{6})^{1006}+(5-2 \sqrt{6})^{1006} \\
=2\left(C_{1006}^{0} 5^{1006}+C_{1006}^{2} 5^{1004} \times 24+\cdots+\right. \\
\left.\quad C_{1006}^{1006} 24^{503}\right) \in \mathbf... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $x=(15+\sqrt{220})^{19}+(15+\sqrt{200})^{22}$. Find the unit digit of the number $x$.
| Solve the conjugate expression
$$
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... | 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}... | 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 $-4=\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}+y_{1} y_{2}$
$$
\begin{array}{l}
=\frac{1}{16}\left(y_{1} y_{2}\right)^{2... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. In $\triangle A B C$, it is known that
$$
\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,
$$
and $b_{i} \in\left\{2,1,-\frac{1}{2}\right\}(1 \leqslant i \leqslant 8)$.... | 491 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in set $S$ is a positive integer not exceeding 100;
(2) For any two distinct elements $a, b$ in set $S$, there exists an element $c$ in $S$ such that
$$
(a, c)=(b, c)=1 \text {; }
$$
(3) For any two dist... | Each positive integer can be expressed as
$$
n=2^{k_{1}} \times 3^{k_{2}} \times 5^{k_{3}} \times 7^{k_{4}} \times 11^{k_{5}} q,
$$
where $q$ is coprime with 2, 3, 5, 7, and 11, and $k_{1}, k_{2}, \cdots, k_{5}$ are non-negative integers.
Let $A=\left\{n \leqslant 100 \mid k_{1}, k_{2}, \cdots, k_{5}\right.$ have exac... | 72 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is $\qquad$
$(2009$, National High School Mathematics League Jilin Province Preliminary) | Hint: The fractional part of $(\sqrt{2}+\sqrt{3})^{2010}$ is $1-(\sqrt{2}-\sqrt{3})^{2010}$.
Since $0.9<1-(\sqrt{2}-\sqrt{3})^{2010}<1$, the first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is 9. | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $S=\{1,2, \cdots, 98\}$. Find the smallest positive integer $n$, such that in any $n$-element subset of $S$, one can always select 10 numbers, and no matter how these 10 numbers are divided into two groups, there is always one group in which there is a number that is coprime with the other four numbers, a... | The subset of 49 even numbers in set $S$ obviously does not satisfy the coprime condition, hence $n \geqslant 50$.
To prove that any 50-element subset $T$ of set $S$ contains 10 numbers that satisfy the condition, we use the following facts.
Lemma 1 If nine elements in $T$ have a common factor greater than 1, and the... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The number of positive integers $n$ such that $n+1$ divides $n^{2012}+2012$ is $\qquad$ . | From the problem, we know
$$
\begin{array}{l}
n^{2012}+2012 \equiv(-1)^{2012}+2012 \\
=2013 \equiv 0(\bmod n+1) .
\end{array}
$$
Since $2013=3 \times 11 \times 61$, $n$ has $2^{3}-1=7$ solutions. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $P(-2,3)$ is a point on the graph of the inverse proportion function $y=\frac{k}{x}$, $Q$ is a moving point on the branch of the hyperbola in the fourth quadrant. A line is drawn through point $Q$ such that it intersects the hyperbola $y=\frac{k}{x}$ at only one point, and intersects the $x$-axis and $y$-... | 4. 48.
It is known that $k=-6, y=-\frac{6}{x}, A(-4,0), B(0,6)$.
Let the line passing through point $Q$ be $y=a x+b$.
Then $\left\{\begin{array}{l}y=a x+b, \\ y=-\frac{6}{x}\end{array}\right.$ has only one solution.
Eliminating and rearranging gives $a x^{2}+b x+6=0$.
By $\Delta=b^{2}-24 a=0 \Rightarrow b^{2}=24 a$.
T... | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) As shown in Figure 5, $P$ is a point outside circle $\odot O$, $PA$ is tangent to $\odot O$ at point $A$, and $PBC$ is a secant of $\odot O$. $AD \perp PO$ at point $D$. If $PB=4$, $CD=5$, $BC=6$, find the length of $BD$. | Connect $O A, O B, O C$. Then $O B=O C$. It is easy to see that
$$
\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 {... | 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
$$
\begin{array}{l}
\frac{0-1}{x_{n}-1}=y^{\prime}(1)=n+1 \Rightarrow x_{n}=\frac{n}{n+1} . \\
\text { Therefore, } a_{1}+a_{2}+\cdots+a_{99}=\lg \left(x_{1} x_{2} \cdots x_{99}\right) \\
=\lg \left(\frac{1}{2} \times... | -2 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $S=\{1,2, \cdots, 50\}$. Find the smallest positive integer $k$, such that in any $k$-element subset of $S$, there exist two distinct numbers $a$ and $b$ satisfying $(a+b) \mid a b$.
$(1996$, China Mathematical Olympiad) | Let the greatest common divisor of $a$ and $b$ be $d$, and $a = a_{1}d$, $b = b_{1}d$. Then $\left(a_{1}, b_{1}\right) = 1$.
Substituting into $(a+b) \mid ab$, we get
$$
\left(a_{1} + b_{1}\right) \mid a_{1} b_{1} d \Rightarrow \left(a_{1} + b_{1}\right) \mid d.
$$
Let $d = k\left(a_{1} + b_{1}\right)$, we obtain all ... | 39 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let the ellipse $C$ be:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{20}=1(a>2 \sqrt{5})
$$
with its left focus at $F$, and point $P(1,1)$. It is known that there exists a line $l$ passing through point $P$ intersecting the ellipse at points $A$ and $B$, with $M$ being the midpoint of $A B$, such that $|F M|$ i... | 11. By the median length formula, we have
$|F M|^{2}=\frac{1}{2}\left(|F A|^{2}+|F B|^{2}\right)-\frac{1}{4}|A B|^{2}$.
Also, $|F M|^{2}=|F A||F B|$, so
$|A B|^{2}=2(|F A|-|F B|)^{2}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and
$c=\sqrt{a^{2}-20}$.
Then $|F A|-|F B|=\frac{c}{a} x_{1}-\frac{c}{a} ... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Remove a $2 \times 2$ small square from any corner of a $3 \times 3$ grid, and the remaining shape is called a "corner shape". Now, place some non-overlapping corner shapes in a $9 \times 9$ grid, with the requirement that the boundaries of the corner shapes coincide with the boundaries or grid lines... | Three, first, $k_{\max }<6$. This is because, if 6 corner shapes are placed in the manner shown in Figure 4, it is impossible to place another complete corner shape on this grid.
Now, place 5 corner shapes in any manner.
Next, we will prove that it is still possible to place another complete corner shape.
Consider th... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given real numbers $a, b, c, d$ satisfy: for any real number $x$,
$a \cos x + b \cos 2x + c \cos 3x + d \cos 4x \leqslant 1$.
Find the maximum value of $a + b - c + d$ and the values of the real numbers $a, b, c, d$ at this time.
(Ninth China Southeast Mathematical Olympiad) | Let
$$
\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... | 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 th... | $\begin{array}{l}\text { 17. } 9 \\ C_{6}^{2}-C_{6}^{1}=9\end{array}$
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... | 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 $O_{1} O_{2}=$ $\qquad$ | $$
\begin{aligned}
& \angle O_{1} B O_{2}=\angle O_{1} C O_{2}=90^{\circ} \\
\Rightarrow & O_{1} 、 C 、 O_{2} 、 B \text { are concyclic } \\
\Rightarrow & \angle O_{1} C D=\angle O_{1} O_{2} B \\
\Rightarrow & \triangle O_{1} D C \backsim \triangle O_{1} B O_{2} \\
\Rightarrow & O_{1} O_{2}=\frac{O_{1} B \cdot O_{1} C}{... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. If a number, from the highest digit to the lowest digit, does not decrease at each digit, it is called a "positive number" (such as $12$, $22$, $566$, $1448$, $123456789$, etc.); if a number, from the highest digit to the lowest digit, does not increase at each digit, it is called a "negative number" (such as $21$, ... | 4.525.
According to the definition of a wavy number, we discuss in two cases:
(1) The hundreds, tens, and units digits are in the pattern "small-large-small": there are
$$
2 \times 1+3 \times 2+\cdots+9 \times 8=240 \text { (numbers). }
$$
(2) The hundreds, tens, and units digits are in the pattern "large-small-large"... | 525 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Among $m$ students, it is known that in any group of three, two of them know each other, and in any group of four, two of them do not know each other. Then the maximum value of $m$ is $\qquad$ | 3. 8 .
When $m=8$, the requirement is satisfied.
It only needs to prove: $m \leqslant 8$.
First, prove that the following two scenarios are impossible.
(1) If a student $A$ knows at least 6 people, by Ramsey's theorem, among these 6 people, there exist 3 people who either all know each other or all do not know each ot... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. If $n$ is a positive integer greater than 1, then
$$
\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}}, \ove... | 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 minimu... | 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 i... | 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 i... | 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 fo... | 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}}
$$
then the minimum value of the pos... | Let $a_{1}=\frac{1}{27}, a_{2}=a_{3}=a_{4}=3$. Then
$$
\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}
$$
Verification shows that $m=1, m=2$ do not meet the... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
$$
\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 | 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 onl... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Let non-negative real numbers $x_{1}, x_{2}, \cdots, x_{6}$ satisfy
$$
\begin{array}{c}
x_{1}+x_{2}+\cdots+x_{6}=1, x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6} \geqslant \frac{1}{540} . \\
\text { If } \max \left\{x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+x_{4} x_{5} x_{6}+\right. \\
\left.x_{5} x_{6} x_{1}+x_{6... | Let $r=x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6}$,
$$
\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}\ri... | 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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