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Given $a, b, x, y$ are non-negative real numbers, and $a+b=27$. Try to find the maximum value of $\lambda$, such that the inequality
$$
\left(a x^{2}+b y^{2}+4 x y\right)^{3} \geqslant \lambda\left(a x^{2} y+b x y^{2}\right)^{2}
$$
always holds, and find the conditions for equality. | Let $a=0, b=27, x=27, y=2$, then the original inequality is $\lambda \leqslant 4$.
We will now prove that $\lambda=4$ makes the given inequality always true.
It is only necessary to prove under the original conditions that
$$
\left(a x^{2}+b y^{2}+4 x y\right)^{3} \geqslant 4(a x+b y)^{2} x^{2} y^{2}.
$$
When $x$ or $... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 12 A student, in order to plot the graph of the function $y=$ $a x^{2}+b x+c(a \neq 0)$, took 7 values of the independent variable: $x_{1}<x_{2}<\cdots<x_{7}$, and $x_{2}-x_{1}=x_{3}-x_{2}=\cdots$ $=x_{7}-x_{6}$, and calculated the corresponding $y$ values, listing them in Table 1.
But due to carelessness, one... | Explanation: Let $x_{2}-x_{1}=x_{3}-x_{2}=\cdots=x_{7}-x_{6}$ $=d$, and the function value corresponding to $x_{i}$ is $y_{i}$. Then
$$
\begin{array}{l}
\Delta_{k}=y_{k+1}-y_{k} \\
=\left(a x_{k+1}^{2}+b x_{k+1}+c\right)-\left(a x_{k}^{2}+b x_{k}+c\right) \\
=a\left[\left(x_{k}+d\right)^{2}-x_{k}^{2}\right]+b\left[\lef... | 551 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $\sqrt{x^{2}+32}-\sqrt{65-x^{2}}=5$. Then $3 \sqrt{x^{2}+32}+2 \sqrt{65-x^{2}}=$ $\qquad$ | 3.35.
Observing the experiment, we know that $x^{2}=49$, so the original expression $=3 \times 9+2 \times 4=35$. | 35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. There are two positive integers $a$ and $b$, the sum of their squares is 585, and the sum of their greatest common divisor and least common multiple is 87. Then $a+b=$ $\qquad$ . | 4.33.
Let the required positive integers be $a<b$, and let $(a, b)=d$, then we have $a=d x, b=d y$, and $(x, y)=1$.
Thus, $[a, b]=d x y$. According to the problem, we have $\left\{\begin{array}{l}d^{2} x^{2}+d^{2} y^{2}=585, \\ d+d x y=87,\end{array}\right.$ which simplifies to $\left\{\begin{array}{l}x^{2}+y^{2}=\fra... | 33 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (17 points) The quadratic trinomial $x^{2}-x-2 n$ can be factored into the product of two linear factors with integer coefficients.
(1) If $1 \leqslant n \leqslant 30$, and $n$ is an integer, how many such $n$ are there?
(2) When $n \leqslant 2005$, find the largest integer $n$.
| (1) Notice that
$$
x^{2}-x-2 n=\left(x-\frac{1+\sqrt{1+8 n}}{2}\right)\left(x-\frac{1-\sqrt{1+8 n}}{2}\right),
$$
then we should have
$$
1+8 n=9,25,49,81,121,169,225,289, \cdots \text {. }
$$
The corresponding solutions for $n$ are $1,3,6,10,15,21,28,36$ (discard the last one).
Therefore, when $1 \leqslant n \leqslan... | 1953 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four. (17 points) As shown in Figure 4, given the line
$$
l: y=k x+2-4 k \text{ ( } k \text{ is a real number). }
$$
- (1) Prove that regardless of the value of $k$, the line $l$ always passes through a fixed point $M$, and find the coordinates of point $M$;
(2) If the line $l$ intersects the positive $x$-axis and $y$-... | (1) Let $k=1$, we get $y=x-2$; let $k=2$, we get $y=2x-6$. Solving these simultaneously, we get $x=4, y=2$. Therefore, the fixed point is $M(4,2)$.
Substituting the coordinates of point $M(4,2)$ into the equation of line $l$, we get $2=2$, which is an identity independent of $k$. Hence, regardless of the value of $k$,... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. On the coordinate plane, a point $(x, y)$ where both the x-coordinate and y-coordinate are integers is called an integer point. If the area enclosed by the quadratic function $y=-x^{2}+8 x-\frac{39}{4}$ and the x-axis is colored red, then the number of integer points inside this red region and on its boundary is ( )... | 5.C.
$y=-\left(x-\frac{3}{2}\right)\left(x-\frac{13}{2}\right)$ intersects the $x$-axis at two points $M\left(\frac{3}{2}, 0\right)$ and $N\left(\frac{13}{2}, 0\right)$. Between $x=\frac{3}{2}$ and $x=\frac{13}{2}$, there are 5 integers: $2, 3, 4, 5, 6$.
Rewrite the function as
$$
y=-(x-4)^{2}+\frac{25}{4}.
$$
When $x... | 25 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. Given $a, b$ are positive integers, $a=b-2005$. If the equation $x^{2}-a x+$ $b=0$ has positive integer solutions, then the minimum value of $a$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2.95.
Let the two roots of the equation be $x_{1}$ and $x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=a, \\
x_{1} x_{2}=b .
\end{array}\right.
$$
Since $x_{1}$ and $x_{2}$ include one positive integer, the other must also be a positive integer. Without loss of generality, let $x_{1} \leqslant x_{2}$. From equat... | 95 | Number Theory | proof | Yes | Yes | cn_contest | false |
3. A person adds the page numbers of a book in the order $1,2,3, \cdots$, with one page number being added an extra time, resulting in an incorrect total sum of 2005. Then the page number that was added extra is $\qquad$ . | 3.52 pages.
Let the total number of pages in the book be $n$, and the page number that was added extra be $x(1 \leqslant x \leqslant n)$. We have $\frac{n(n+1)}{2}+x=2005$.
And $\frac{n(n+1)}{2}+1 \leqslant 2005 \leqslant \frac{n(n+1)}{2}+n$, which means $n^{2}+n+2 \leqslant 4010 \leqslant n(n+3)$.
Since $\sqrt{4010} ... | 52 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $n$ be a natural number. If 2005 can be written as the sum of $n$ positive odd composite numbers, then $n$ is called a "good number". The number of such good numbers is $\qquad$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result di... | 4.111.
Let $a_{1}, a_{2}, \cdots, a_{n}$ be odd composite numbers, and $a_{1}+a_{2}+\cdots+a_{n}=$ 2005, then $n$ is odd. Since 9 is the smallest odd composite number, and 2005 $<2007=9 \times 223$, hence $n<223$. Therefore, $n \leqslant 221$.
$$
2005=1980+25=\underbrace{9+9+\cdots}_{20 \uparrow}+9+25 \text {, }
$$
... | 111 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Six, let the arithmetic mean of all elements in the set $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be denoted as $P(A)$ $\left(P(A)=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)$. If $B$ is a non-empty subset of $A$ and $P(B)=P(A)$, then $B$ is called a "balanced subset" of $A$. Try to find the number of all "balanced ... | $$
\begin{array}{l}
\text { Six, since } P(M)=5 \text {, let } \\
M^{\prime}=\{x-5 \mid x \in M\} \\
=\{-4,-3,-2,-1,0,1,2,3,4\},
\end{array}
$$
then $P\left(M^{\prime}\right)=0$. According to this translation relationship, the balanced subsets of $M$ and $M^{\prime}$ can be one-to-one correspondence. Let $f(k)$ denote... | 51 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
(1) Discuss the number of roots of the equation with respect to $x$
$$
|x+1|+|x+2|+|x+3|=a
$$
(2) Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence, and
$$
\begin{array}{l}
\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right| \\
=\left|a_{1}+1\right|+\left|a_{2}+1\right|+\cdots+\left|a_{n}+1\right... | (1) According to the graph of the function $y=|x+1|+|x+2|+|x+3|$ (as shown in Figure 3), we can see that:
When $a2$, the equation has two solutions.
(2) Since the equation $|x|=|x+1|=|x-2|$ has no solution, it follows that $n \geqslant 2$ and the common difference is not 0. Let's assume the terms of the sequence are $... | 26 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. A university has 10001 students, some of whom join and form several clubs (a student can belong to different clubs), and some clubs join together to form several associations (a club can belong to different associations). It is known that there are $k$ associations in total. Assume the following conditions are met:
... | 1. Replace 10001 with $n$, and use two methods to calculate the number of ordered triples $(a, R, S)$, where $a, R, S$ represent a student, a club, and a society, respectively, and satisfy $a \in R, R \in S$. We call such triples "acceptable".
Fix a student $a$ and a society $S$. By condition (2), there is a unique cl... | 5000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. If $x=\frac{13}{4+\sqrt{3}}$, then $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+3}{x^{3}-7 x^{2}+5 x+15}=$ $\qquad$ | $$
\text { II.1. }-5 \text {. }
$$
Since $x=\frac{13}{4+\sqrt{3}}=4-\sqrt{3}$, then $x^{2}-8 x+13=0$.
$$
\begin{array}{l}
\text { Hence } x^{4}-6 x^{3}-2 x^{2}+18 x+3 \\
=\left(x^{2}+2 x+1\right)\left(x^{2}-8 x+13\right)-10=-10, \\
x^{3}-7 x^{2}+5 x+15 \\
=(x+1)\left(x^{2}-8 x+13\right)+2=2 .
\end{array}
$$
Therefore... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. A natural number minus 69 is a perfect square, and this natural number plus 20 is still a perfect square. Then this natural number is $\qquad$ . | 2.2005 .
Let this natural number be $x$. According to the problem, we have
$$
\left\{\begin{array}{l}
x-69=m^{2}, \\
x+20=n^{2},
\end{array}\right.
$$
where $m, n$ are both natural numbers.
Subtracting the two equations gives $n^{2}-m^{2}=89$, which is $(n-m)(n+m)=89$.
Since $n>m$, and 89 is a prime number, we have
$... | 2005 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given real numbers $x_{1}, x_{2}, y_{1}, y_{2}$ satisfy
$$
\begin{array}{l}
x_{1}^{2}+25 x_{2}^{2}=10, \\
x_{2} y_{1}-x_{1} y_{2}=25, \\
x_{1} y_{1}+25 x_{2} y_{2}=9 \sqrt{55} .
\end{array}
$$
Then $y_{1}^{2}+25 y_{2}^{2}=$ $\qquad$ | 3.2008 .
Notice that
$$
\begin{array}{l}
\left(x_{1}^{2}+a x_{2}^{2}\right)\left(y_{1}^{2}+a y_{2}^{2}\right) \\
=\left(x_{1} y_{1}+a x_{2} y_{2}\right)^{2}+a\left(x_{2} y_{1}-x_{1} y_{2}\right)^{2},
\end{array}
$$
where $a$ is any real number.
For this problem, take $a=25$, and it is easy to find that $y_{1}^{2}+25 ... | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given that the sum of 10 natural numbers is 1001. What is the maximum possible value of their greatest common divisor?
untranslated text:
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 part of the text is a note about the translation instruction and is not part of the content to be translated. Here is ... | Solution: Let the greatest common divisor be $d$, and the 10 numbers be $b_{1} d, b_{2} d, \cdots, b_{10} d$. According to the problem, we have
$$
\left(b_{1}+b_{2}+\cdots+b_{10}\right) d=1001 .
$$
Since $b_{1}+b_{2}+\cdots+b_{10} \geqslant 10$, it follows that $d \leqslant$ 100 and is a divisor of 1001.
Given $1001=... | 91 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Find the smallest natural number $n$ such that $\frac{n-13}{5 n+6}$ is a non-zero reducible fraction.
(6th IMO) | Solution: Let the common divisor of the numerator and denominator be $d(d>1)$, and let
$$
\begin{array}{l}
n-13=k d, \\
5 n+6=l d .
\end{array}
$$
(2) $-5 \times$ (1) gives
$$
(l-5 k) d=71 \text{. }
$$
Since 71 is a prime number and $d>1$, we have
$$
d=71 \text{ and } l-5 k=1 \text{. }
$$
Thus, $n=13+k \times 71$.
Wh... | 84 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given that $x, y$ are integers, and
$$
15 x^{2} y^{2}=35 x^{2}-3 y^{2}+412 \text {. }
$$
then $15 x^{2} y^{2}=$ $\qquad$ . | Solution: Transform the given equation into
$$
\left(5 x^{2}+1\right)\left(3 y^{2}-7\right)=405=3^{4} \times 5 \text {. }
$$
Since $5 x^{2}+1$ is not a multiple of 5, and $3 y^{2}-7$ is not a multiple of 3, it can only be
$$
5 x^{2}+1=3^{4}, 3 y^{2}-7=5 \text {. }
$$
Therefore, $5 x^{2}=80,3 y^{2}=12$.
Thus, $15 x^{2... | 960 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3: There are $n$ people, and it is known that any two of them make at most one phone call. The total number of phone calls made among any $n-2$ people is equal, and it is $3^{k}$ times, where $k$ is a positive integer. Find all possible values of $n$.
(2000, National High School Mathematics Competition)
Analysi... | Let $n$ people be denoted as $A_{1}, A_{2}, \cdots, A_{n}$. Let the number of calls made by $A_{i}$ be $m_{i}$, and the number of calls between $A_{i}$ and $A_{j}$ be $\lambda_{i j}(1 \leqslant i, j \leqslant n)$, where $\lambda_{i j}=0$ or 1.
Clearly, $n \geqslant 5$. Therefore,
$$
\begin{array}{l}
\left|m_{i}-m_{j}\r... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 The polynomial $x^{4}+m x^{3}+n x-16$ contains the factors $x-1$ and $x-2$. Then $m n=$ $\qquad$
(1989, Sichuan Province Junior High School Mathematics League) | Solution: Let $f(x)=x^{4}+m x^{3}+n x-16$. By the Factor Theorem, we have
$$
\left\{\begin{array} { l }
{ f ( 1 ) = 0 , } \\
{ f ( 2 ) = 0 , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
m+n=15, \\
8 m+2 n=0 .
\end{array}\right.\right.
$$
Solving, we get $m=-5, n=20$.
Therefore, $m n=-100$. | -100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given $a 、 b 、 c$ are real numbers, and
$$
a^{2}+b^{2}+c^{2}+2 a b=1, a b\left(a^{2}+b^{2}+c^{2}\right)=\frac{1}{8} \text {, }
$$
The roots of the quadratic equation $(a+b) x^{2}-(2 a+c) x-(a+b)=0$ are $\alpha 、 \beta$. Find the value of $2 \alpha^{3}+\beta^{-5}-\beta^{-1}$. | Solution: From the given, we have
$$
\left\{\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)+2 a b=1, \\
2 a b\left(a^{2}+b^{2}+c^{2}\right)=\frac{1}{4} .
\end{array}\right.
$$
Thus, \(a^{2}+b^{2}+c^{2}\) and \(2 a b\) are the roots of the equation \(t^{2}-t+\frac{1}{4}=0\).
Since the roots of the equation \(t^{2}-t+\f... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 When $m=$ $\qquad$, the polynomial
$$
12 x^{2}-10 x y+2 y^{2}+11 x-5 y+m
$$
can be factored into the product of two linear factors.
(1992, Zhengzhou City Junior High School Mathematics Competition) | Solution: First, factorize the quadratic term, we have
$$
12 x^{2}-10 x y+2 y^{2}=(3 x-y)(4 x-2 y) \text {. }
$$
Therefore, we can set
$$
\begin{array}{l}
12 x^{2}-10 x y+2 y^{2}+11 x-5 y+m \\
=(3 x-y+a)(4 x-2 y+b) \\
=(3 x-y)(4 x-2 y)+(4 a+3 b) x- \\
(2 a+b) y+a b .
\end{array}
$$
Comparing the coefficients of the ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The axial cross-section of a wine glass is part of a parabola, whose equation is $x^{2}=2 y(0 \leqslant y<15)$. If a glass ball with a radius of 3 is placed inside the cup, then the distance from the highest point of the ball to the bottom of the cup is $\qquad$ | 5.8 .
As shown in the figure, let the center of the sphere be at $(0, b)$.
Then we have
$$
x^{2}+(y-6)^{2}=9 \text {. }
$$
Therefore, the system of equations
$$
\left\{\begin{array}{l}
x^{2}=2 y \\
x^{2}+(y-b)^{2}=9
\end{array}\right.
$$
has two solutions. For $y$,
$$
y^{2}+2(1-b) y+b^{2}-9=0 \text {. }
$$
By $\Delt... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Fill the numbers $1,2, \cdots, 8$ into the 8 squares surrounding the four sides of a $3 \times 3$ chessboard, so that the sum of the absolute values of the differences between adjacent numbers in these 8 squares is maximized. Find this maximum value. | Solution: Let the sum of the absolute values of the differences between adjacent numbers in the 8 squares be $M$. Note that rotating the 8 numbers in the outer ring of the $3 \times 3$ chessboard does not change the value of $M$. As shown in Figure 6, let the numbers in each square be denoted as $a_{1}, a_{2}, \cdots, ... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Fill the numbers $1,2, \cdots, 9$ into a $3 \times 3$ chessboard, such that the sum of the absolute values of the differences between adjacent (sharing a common edge) cells is maximized. Find this maximum value. | Solution: Let the sum of the absolute values of the differences between adjacent cells be $M$. Note that $M$ can be divided into two parts:
(1) The sum of the absolute values of the differences between adjacent numbers in the 8 surrounding cells, denoted as $M_{1}$;
(2) The sum of the 4 absolute values of the differenc... | 58 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Question 4 If real numbers $a, b, c$ satisfy
$$
\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1 \text{, }
$$
find the value of $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}$.
(1999, Changsha Junior High School Mathematics Competition) | Solution: Construct the identity
$$
\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1 \text {. }
$$
Subtract the above identity from the given equation to get
$$
\begin{array}{l}
\left(\frac{a}{b+c}-\frac{a}{a+b+c}\right)+\left(\frac{b}{a+c}-\frac{b}{a+b+c}\right)+ \\
\left(\frac{c}{a+b}-\frac{c}{a+b+c}\right)=0 .
\en... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Find the smallest positive integer $n$ such that there exists a positive integer $k$ for which $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$ holds. | Solution: For the inequality $\frac{8}{15}1$, i.e., $m>8$.
Therefore, $n>56$.
When $n=7 m+p\left(m \in \mathbf{N}_{+}, p \in \mathbf{N}\right.$, and $1 \leqslant p$ $\leqslant 6$), we have
$$
\begin{array}{l}
6 m+\frac{6}{7} p0$, i.e., $m>p$.
Since $1 \leqslant p \leqslant 6$, to find the smallest $m$, when $p=1$, $m_... | 15 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 1, the diagonals $A C$ and $B D$ of quadrilateral $A B C D$ intersect at point $O$. $S_{\triangle A O B}=4, S_{\triangle C O D}=9$. Then the minimum value of $S_{\text {quadrilateral } A B C D}$ is ( ).
(A) 22
(B) 25
(C) 28
(D) 32 | 4. B.
As shown in Figure 1, let $S_{\triangle M O D}=x, S_{\triangle B O C}=y$. Then $S_{\text {quadrilateral } A B C D}=4+9+x+y \geqslant 13+2 \sqrt{x y}$. From $\frac{x}{9}=\frac{4}{y}$, we have $x y=36$. Therefore, $S_{\text {quadrilateral } A B C D} \geqslant 13+2 \sqrt{x y}=13+12=25$. Hence, the minimum value of ... | 25 | Geometry | MCQ | Yes | Yes | cn_contest | false |
1. As shown in Figure 2, two diameters $A C$ and $B D$ of the large circle intersect perpendicularly at point $O$. Four semicircles are drawn outside the large circle with $A B$, $B C$, $C D$, and $D A$ as diameters, respectively. The total area of the four "crescent" shaded regions in the figure is $2 \mathrm{~cm}^{2}... | $=.1 .1$.
By the Pythagorean theorem, we know $A D^{2}+C D^{2}=A C^{2}$, so the area of the upper half of the large circle is equal to the sum of the areas of the two semicircles with diameters $A D$ and $C D$. Similarly, the area of the lower half of the large circle is equal to the sum of the areas of the two semicir... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $a^{2}+b c=14, b^{2}-2 b c=-6$. Then $3 a^{2}+4 b^{2}-5 b c=$ $\qquad$ | 3.18 .
$$
\begin{array}{l}
3 a^{2}+4 b^{2}-5 b c=3\left(a^{2}+b c\right)+4\left(b^{2}-2 b c\right) \\
=3 \times 14+4 \times(-6)=18
\end{array}
$$ | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. There are 6 natural numbers that have the same remainder when divided by 12, and their product is 971425. Then the minimum value of the sum of these 6 natural numbers is $\qquad$ . | 5. 150.
Since 971425 leaves a remainder of 1 when divided by 12, and
$$
971425=5 \times 5 \times 7 \times 7 \times 13 \times 61,
$$
among the prime factors, there are two each that leave a remainder of 5, 7, and 1 when divided by 12. Since the product of two numbers that leave a remainder of 5 (or 7) when divided by ... | 150 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (15 points) Given non-zero real numbers $a, b, c$ satisfy $a+b+c=0$. Prove:
(1) $a^{3}+b^{3}+c^{3}=3 a b c$;
(2) $\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9$. | Three, (1) From $a+b+c=0$, we get $a+b=-c$. Therefore, $(a+b)^{3}=-c^{3}$.
Thus, $a^{3}+3 a^{2} b+3 a b^{2}+b^{3}=-c^{3}$.
Hence $a^{3}+b^{3}+c^{3}=-3 a b(a+b)=-3 a b(-c)=3 a b c$.
$$
\begin{array}{l}
\text { (2) }\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot \frac{c}{a-b} \\
=1+\left(\frac{b-c}{a}+\frac... | 9 | Algebra | proof | Yes | Yes | cn_contest | false |
12. If the sum of the digits of a natural number $a$ equals 7, then $a$ is called a "lucky number". Arrange all lucky numbers in ascending order as $a_{1}, a_{2}, a_{3}, \cdots$, if $a_{n}=$ 2005, then $a_{5 n}=$ $\qquad$ | 12.52000 .
It is known that the number of integer solutions to the equation $x_{1}+x_{2}+\cdots+x_{k}=m$ that satisfy $x_{1} \geqslant 1$, $x_{i} \geqslant 0(i \geqslant 2)$ is $\mathrm{C}_{m+k-2}^{m-1}$. Taking $m=7$, we know that the number of $k$-digit lucky numbers is $P(k)=\mathrm{C}_{k+5}^{6}$.
Since 2005 is th... | 52000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $n(n \geqslant 3)$ be a positive integer. If there are $n$ lattice points $P_{1}, P_{2}, \cdots, P_{n}$ in the plane such that: when $\left|P_{i} P_{j}\right|$ is a rational number, there exists $P_{k}$ such that $\left|P_{i} P_{k}\right|$ and $\left|P_{j} P_{k}\right|$ are both irrational; when $\left|P_{i} P_{... | 6. We assert that the smallest good number is 5, and 2005 is a good number.
In a triplet $\left(P_{i}, P_{j}, P_{k}\right)$, if $\left|P_{i} P_{j}\right|$ is a rational number (or irrational number), and $\left|P_{i} P_{k}\right|, \left|P_{j} P_{k}\right|$ are irrational numbers (or rational numbers), then $\left(P_{i... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Among the seven-digit numbers formed by the digits $0,1,2,3,4,5,6$, the number of permutations that do not contain "246" and "15" is ( ).
(A) 3606
(B) 3624
(C) 3642
(D) 4362 | 6.C.
Since the total number of permutations of the 7 digits $0,1, \cdots, 6$ into a seven-digit number with the first digit not being zero is $\mathrm{A}_{7}^{7}-\mathrm{A}_{6}^{6}=4320$, but among these, the number where “246” are together is $\mathrm{A}_{4}^{4} \mathrm{C}_{4}^{1}=96$, and the number where “15” are t... | 3642 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
2. The number of non-negative integer solutions to the equation $x_{1}+x_{2}+\cdots+x_{99}+2 x_{100}=3$ is $\qquad$ . | 2. 166749 .
Classify by $x_{100}$:
(1) When $x_{100}=1$, there are 99 non-negative integer solutions;
(2) When $x_{100}=0$, there are
$$
\mathrm{C}_{99}^{1}+\mathrm{A}_{99}^{2}+\mathrm{C}_{99}^{3}=166 \text { 650 (solutions). }
$$
Combining (1) and (2), the total number of non-negative integer solutions is
$$
99+1666... | 166749 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Given positive integers $a, b, c, d$ satisfying $b<a<d<c$, and the sums of each pair are $26, 27, 41, 101, 115, 116$. Find the value of $(100a + b) - (100d - c)$. | Three, from $b<a<d<c$, we can get
$$
\begin{array}{l}
a+b<b+d<b+c<a+c<d+c, \\
b+d<a+d<a+c .
\end{array}
$$
Therefore, $a+d$ and $b+c$ are both between $b+d$ and $a+c$.
It is easy to know that $a+b=26, b+d=27, a+c=115, d+c=116$.
If $a+d=101$, then $b+c=41$.
Solving this gives $b=-24$ (discard).
Thus, $a+d=41, b+c=101$.... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. In the Cartesian coordinate system $x O y$, it is known that points $A(6,4)、B(4,0)、C(0,3)$, and the line $l: y=k x$ bisects the area of $\triangle A B C$, $k=\frac{-b+\sqrt{c}}{a}(a、b、c$ are all positive integers, and $c$ has no square factors). Then $a+b+c=$ $\qquad$ . | 6.584.
As shown in Figure 5, take the midpoint $M\left(2, \frac{3}{2}\right)$ of side $BC$. Let the line $l$ intersect sides $BC$ and $CA$ at points $P$ and $Q$, respectively.
Since $AM$ and $PQ$ both bisect the area of $\triangle ABC$, we have $S_{\triangle PQC} = S_{\triangle MMC}$. Subtracting the common area $S_{... | 584 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
For each positive integer $n$, define the function
$$
f(n)=\left\{\begin{array}{cl}
0, & \text { when } n \text { is a perfect square; } \\
{\left[\frac{1}{\{\sqrt{n}\}}\right],} & \text { when } n \text { is not a perfect square. }
\end{array}\right.
$$
where $[x]$ denotes the greatest integer not exceeding $x$, and ... | Solution: If $k$ is not a perfect square, then there exists $a \in \mathbf{N}$, such that $a^{2}<k<(a+1)^{2}$, then $a<\sqrt{k}<a+1$, i.e.,
$$
\{\sqrt{k}\}=\sqrt{k}-a \text {. }
$$
Therefore, $\left[\frac{1}{\{\sqrt{k}\}}\right]=\left[\frac{1}{\sqrt{k}-a}\right]=\left[\frac{\sqrt{k}+a}{k-a^{2}}\right]$.
$$
\begin{alig... | 768 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$6 \rightarrow$ In a regular $n$-sided polygon, the difference between any two adjacent interior angles is $18^{\circ}$, find the maximum value of $n$, | Hedgehog, this problem can be solved in three steps:
(1) Prove that $n$ is even
(2) Prove that $n<40$, and $n \leqslant 38$;
(3) Prove that $n=38$, i.e., the maximum value of $n$ is 38. | 38 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. The six faces of a unit cube are painted in 6 different colors, and a different number of roosters are drawn on each face. The colors and the number of roosters on each face correspond as shown in Table 1:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Color on the face & Red & Yellow & Blue & Green & Purple & Gree... | 5.17.
Since the 4 aforementioned unit cubes are identical, it can be observed that the colors of the 4 adjacent faces to the red face are blue, yellow, cyan, and purple, so the color of the face opposite the red face is green. The colors of the 4 adjacent faces to the yellow face are cyan, blue, red, and green, thus t... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
9. In the arithmetic sequences $3,10,17, \cdots, 2005$ and $3,8, 13, \cdots, 2003$, the number of terms that have the same value is $\qquad$. | 9.58.
Subtract 3 from each term of the two sequences, transforming them into $0,7,14, \cdots$, 2002 and $0,5,10, \cdots, 2000$. The first sequence represents multiples of 7 not exceeding 2002, and the second sequence can be seen as multiples of 5 within the same range. Therefore, the common terms are multiples of 35. ... | 58 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. If $2^{6}+2^{9}+2^{n}$ is a perfect square, then the positive integer $n=$ $\qquad$ . | 11. 10 .
$$
2^{6}+2^{9}=2^{6}\left(1+2^{3}\right)=2^{6} \times 3^{2}=24^{2} \text {. }
$$
Let $24^{2}+2^{n}=a^{2}$, then
$$
(a+24)(a-24)=2^{n} \text {. }
$$
Thus, $a+24=2^{r}, a-24=2^{t}$,
$$
2^{r}-2^{t}=48=2^{4} \times 3,2^{t}\left(2^{r-t}-1\right)=2^{4} \times 3 \text {. }
$$
Then $t=4, r-t=2$.
So $r=6, n=t+r=10$. | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
12. Using weights labeled $1 \mathrm{~g}, 2 \mathrm{~g}, 3 \mathrm{~g}, 15 \mathrm{~g}, 40 \mathrm{~g}$, each one of them, to weigh objects on a balance scale without graduations. If weights can be placed on both ends of the balance, then the maximum number of different gram weights (positive integer weights) that can ... | 12.55.
Using $1 \mathrm{~g}, 2 \mathrm{~g}, 3 \mathrm{~g}$ these three weights, all integer grams in the interval $A=$ $[1,6]$ can be measured; after adding a $15 \mathrm{~g}$ weight, the measurement range is expanded to the interval $B=[15-6,15+6]=[9,21]$. After adding a $40 \mathrm{~g}$ weight, the measurement range... | 55 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. Find the smallest positive integer $n$, such that among any $n$ consecutive positive integers, there is at least one number whose sum of digits is a multiple of 7. | 15. First, we can point out 12 consecutive positive integers, for example,
$$
994,995, \cdots, 999,1000,1001, \cdots, 1005 \text {, }
$$
where the sum of the digits of any number is not a multiple of 7. Therefore, $n \geqslant 13$.
Next, we prove that in any 13 consecutive positive integers, there must be one number ... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7.3. On an island, there live 100 people, some of whom always lie, while the rest always tell the truth. Each resident of the island worships one of three gods: the Sun God, the Moon God, and the Earth God. Each resident was asked three questions:
(1) Do you worship the Sun God?
(2) Do you worship the Moon God?
(3) Do ... | 7.3. People who always tell the truth are called "honest people", and those who always lie are called "liars". Each honest person will only answer "yes" to one question, while each liar will answer "yes" to two questions. Let the number of honest people be $x$, and the number of liars be $y$. Therefore, $x+2y=130$. Sin... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
7.3. In the guard of a grand duke, there are 1000 warriors. Any two warriors are either friends, enemies, or do not know each other. The warriors are all honest and only speak to their friends. However, the current situation makes every warrior unhappy, because for each warrior, any two of his friends are enemies, and ... | 7.3. We point out that any samurai has no more than 2 friends. In fact, if a certain samurai has at least 3 friends $A, B, C$, then according to the problem, $A, B, C$ must be enemies with each other. This leads to a contradiction: since $B$ and $C$ are both enemies of $A$, they should be friends, not enemies. Therefor... | 200 | Combinatorics | proof | Yes | Yes | cn_contest | false |
8.3. Let $p(n, k)$ denote the number of divisors of the positive integer $n$ that are not less than $k$. Try to find
$$
\begin{array}{l}
p(1001,1)+p(1002,2)+\cdots+ \\
p(2000,1000) .
\end{array}
$$ | 8.3. If we write down all the divisors of each positive integer $1000+k$ that are not less than $k$ $(k=1,2, \cdots, 1000)$, then the sum we are looking for is the sum of the number of all these divisors.
Below we prove that each positive integer $n$ from 1 to 2000 is written exactly once. If $n>1000$, then it is only... | 2000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $x=\frac{3-\sqrt{5}}{2}$. Then $x^{3}-3 x^{2}+3 x+$ $\frac{6}{x^{2}+1}=$ . $\qquad$ | $$
=, 1.6 \text {. }
$$
Given $x=\frac{3-\sqrt{5}}{2}$, we know that
$$
\begin{array}{l}
x^{2}-3 x+1=0, x+\frac{1}{x}=3 . \\
\text { Therefore, } x^{3}-3 x^{2}+3 x+\frac{6}{x^{2}+1} \\
=x\left(x^{2}-3 x\right)+3 x+\frac{6}{x^{2}+1}=2 x+\frac{6}{3 x} \\
=2 x+\frac{2}{x}=2\left(x+\frac{1}{x}\right)=6 .
\end{array}
$$ | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a$ and $b$ are real numbers, and $a \geqslant 1$. If the equation $x^{2}-2 b x-\left(a-2 b^{2}\right)=0$ has real solutions, and satisfies $2 a^{2}-a b^{2}-5 a+b^{2}+4=0$, then $a^{2}+b^{2}=$ | 3.6 .
From the equation $x^{2}-2 b x-\left(a-2 b^{2}\right)=0$ having real solutions, we get $\Delta=4 b^{2}+4\left(a-2 b^{2}\right) \geqslant 0$, which means $a \geqslant b^{2}$.
From $2 a^{2}-a b^{2}-5 a+b^{2}+4=0$, we can derive $2 a^{2}-5 a+4=a b^{2}-b^{2}=b^{2}(a-1) \leqslant a(a-1)$. Therefore, $2 a^{2}-5 a+4 \l... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 A mall installs an escalator from the first floor to the second floor, which moves upward at a uniform speed. A boy and a girl start walking up the escalator at the same time (the escalator is also moving). If both of their movements are considered uniform, and the boy walks twice as many steps per minute as ... | Explanation: (1) Let the girl's speed be $x$ steps/min, the escalator's speed be $y$ steps/min, the boy's speed be $2x$ steps/min, and the stairs have $s$ steps. Then
$$
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{s-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y} .
\end{array}\right.
$$
(1)
$$
\frac{3}{4}=\frac{s-27}{s-18} \te... | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 A pedestrian and a cyclist are traveling south simultaneously on a road parallel to a railway. The pedestrian's speed is $3.6 \mathrm{~km} / \mathrm{h}$, and the cyclist's speed is $10.8 \mathrm{~km} / \mathrm{h}$. If a train comes from behind and takes $22 \mathrm{~s}$ to pass the pedestrian and $26 \mathrm{... | Explanation: First, convert the two speed units to: pedestrian $1 \mathrm{~m} / \mathrm{s}$, cyclist $3 \mathrm{~m} / \mathrm{s}$. Let the length of the train be $l \mathrm{~m}$, and the speed of the train be $v \mathrm{~m} / \mathrm{s}$. According to the problem, we have
$$
\left\{\begin{array}{l}
l=(v-1) \times 22, \... | 286 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the function $f(n)=\frac{20^{n}+3^{n}}{n!}, n \in \mathbf{N}$. Then, the value of $n$ that maximizes $f(n)$ is, $n=$ | 3.19.
$$
\begin{array}{l}
G(n)=f(n+1)-f(n) \\
=\frac{20^{n+1}+3^{n+1}}{(n+1)!}-\frac{20^{n}+3^{n}}{n!} \\
=\frac{(19-n) \times 20^{n}-(n-2) \times 3^{n}}{(n+1)!} \\
=\frac{3^{n}}{(n+1)!}\left[(19-n) \times\left(\frac{20}{3}\right)^{n}-(n-2)\right] .
\end{array}
$$
It is easy to see that when $n \geqslant 19$, $G(n) < ... | 19 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The maximum value of the algebraic expression $a \sqrt{2-b^{2}}+b \sqrt{2-a^{2}}$ is $\qquad$ . | 5.2.
It is known that $|a| \leqslant \sqrt{2},|b| \leqslant \sqrt{2}$. Let
$$
a=\sqrt{2} \sin \alpha, b=\sqrt{2} \sin \beta, \alpha, \beta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text {. }
$$
Then $a \sqrt{2-b^{2}}+b \sqrt{2-a^{2}}$
$$
\begin{array}{l}
=2(\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \alp... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Six. (15 points) A square is divided into 4 squares, the number of edges in the division diagram is 12. If a square is divided into 2005 convex polygons, try to find the maximum number of edges in the division diagram.
| Six, according to Euler's theorem, the number of vertices $a$, the number of faces $b$, and the number of edges $e$ of a simple polyhedron have the following relationship: $a+b-e=2$.
From Euler's theorem, it is easy to see that if a convex polygon is divided into $n$ convex polygons, then the number of vertices $a$, t... | 6016 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a regular $n$-sided polygon inscribed in a circle. If the line connecting a vertex to the center of the circle is exactly on the perpendicular bisector of a certain side, then $n \equiv$ $\qquad$ $(\bmod 2)$ | 2.1.
As shown in Figure 8, if the line connecting vertex $A_{1}$ and the center $O$ perpendicularly bisects side $A_{i} A_{j}$, then $\triangle A_{1} A_{i} A_{j}$ is an isosceles triangle. Since equal chords subtend equal arcs, the number of vertices in $\overparen{A_{1} A_{i}}$ is equal, and their sum is even. Adding... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given a sequence of numbers $a_{1}, a_{2}, \cdots, a_{2006}$, where $a_{1}$ $=1$, and the sum of each pair of consecutive terms is 3. Then $a_{1}-a_{2}+$ $a_{3}-a_{4}+\cdots+a_{2003}-a_{2004}+a_{2005}=$ $\qquad$ | 3. -1001 .
Given $a_{1}+a_{2}=a_{2}+a_{3}=\cdots=a_{2004}+a_{2000}=3$, and $a_{1}=1$, we can sequentially deduce that
$$
\begin{array}{l}
a_{1}=a_{3}=\cdots=a_{2005}=1, a_{2}=a_{4}=\cdots=a_{2004}=2 . \\
\text { Then } a_{1}-a_{2}+a_{3}-a_{4}+\cdots-a_{2004}+a_{2000} \\
=\left(a_{1}-a_{2}\right)+\left(a_{3}-a_{4}\righ... | -1001 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Find the smallest positive integer $k$, such that there exists a positive integer $n$, satisfying $10^{n}=29 k+2$.
---
The translation maintains the original text's format and line breaks. | $$
\begin{array}{l}
\text { Given } \frac{1}{29}=0.0334482758620689655172413793 \text { i, } \\
\frac{2}{29}=0.0068965517241379310344827586 \text { 2e, }
\end{array}
$$
we know that when $n=11$, $k$ takes the minimum value 3448275862.
Explanation: By appending the 28-digit repeating cycle of $\frac{1}{29}$ to 34482758... | 3448275862 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $A=\{1,2,3,4,5\}$. Then the number of mappings $f: A \rightarrow A$ that satisfy the condition $f(f(x))$ $=f(x)$ is $\qquad$ (answer with a number) | 3.196 .
It can be generalized to the case where $A$ has $n$ elements. From the condition, if $a \in A$ and $a$ is in the range of $f$, then it must be that $f(a)=a$. Therefore, we can classify $f$ based on the number of elements in its range. The number of $f$ with $k\left(k \in \mathbf{N}_{+}\right)$ elements in its ... | 196 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. In the geometric sequence $\left\{a_{n}\right\}$, $a_{1}=\frac{1}{8}$, the geometric mean of the first $n$ terms is 8. If the geometric mean of the remaining terms after removing one term from the first $n$ terms is $4 \sqrt{2}$, then the removed term is the $\qquad$th term. | 6.13.
Let the common ratio be $q$, the product of the first $n$ terms be $M$, and the $k$-th term be removed, then $M=8^{n}$, and $\frac{M}{a_{k}}=(4 \sqrt{2})^{n-1}$.
Thus, $a_{k}=(\sqrt{2})^{n+5}$, which means $\frac{1}{8} q^{k-1}=(\sqrt{2})^{n+5}$.
Therefore, $q^{k-1}=(\sqrt{2})^{n+11}$.
Also, $M=8^{n}$, i.e., $\le... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) If a positive integer $n$ has the sum of its digits in base 3 divisible by 3, then $n$ is called a "proper number." Find the sum of all proper numbers in $S=\{1,2, \cdots, 2005\}$. | For $m \in \mathbf{N}$, in ternary (base 3) representation, there are $3^{m+1}$ non-negative integers with at most $m+1$ digits. Let the number of these integers whose digit sums are congruent to 0, 1, and 2 modulo 3 be $a_{m}$, $b_{m}$, and $c_{m}$, respectively. When $m \geqslant 1$, by classifying the numbers whose ... | 671007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 A positive integer $n$ cannot be divisible by $2$ or $3$, and there do not exist non-negative integers $a$, $b$ such that $\left|2^{a}-3^{b}\right|=n$. Find the minimum value of $n$.
(2003, National Training Team Problem) | Explanation: When $n=1$, $\left|2^{1}-3^{1}\right|=1$;
When $n=5$, $12^{2}-3^{2} \mid=5$;
When $n=7$, $\left|2^{1}-3^{2}\right|=7$;
When $n=11$, $\left|2^{4}-3^{3}\right|=11$;
When $n=13$, $\left|2^{4}-3^{1}\right|=13$;
When $n=17$, $12^{6}-3^{4} \mid=17$;
When $n=19$, $12^{3}-3^{3} \mid=19$;
When $n=23$, $12^{5}-3^{2}... | 35 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Find the smallest positive integer $a$, such that there exists a positive odd integer $n$, satisfying $2001 \mid \left(55^{n}+a \times 32^{n}\right)$.
（14th Irish Mathematical Olympiad） | (Given $2001=87 \times 23$, we get $87 \mid\left(55^{n}+a \times\right.$ $\left.32^{n}\right)$, and $231\left(55^{n}+a \times 32^{n}\right)$. Then $0 \equiv 55^{n}+a \times 32^{n} \equiv$ $(-32)^{n}+a \times 32^{n}=32^{n}(a-1)(\bmod 87)$. Hence $a-1=$ $0(\bmod 87)$. Similarly, we get $a+1 \equiv 0(\bmod 23)$. Let $a=87... | 436 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 As shown in Figure 1, quadrilateral $ABCD$ is a rectangle. Two people, A and B, start from points $A$ and $B$ respectively at the same time, and move counterclockwise along the rectangle. In which circle does B possibly catch up with A for the first time? In which circle does B certainly catch up with A at th... | Explanation: Let $A D=B C=a \mathrm{~m}, A B=C D=b . \mathrm{m}$, and suppose the first time Yi catches up with Jia is after
$$
\frac{2 a+b}{74-65}=\frac{2 a+b}{9}
$$
minutes. The distance Yi has run when he first catches up with Jia is
$$
\frac{2 a+b}{9} \times 74(\mathrm{~m}) \text {. }
$$
At this point, the number... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: Team A and Team B each send out 7 members to participate in a Go team tournament according to a pre-arranged order. Both sides start with the No. 1 member competing, the loser is eliminated; the winner then competes with the No. 2 member of the losing side, ... until all members of one side are eliminated, a... | Solution: Since the losing side is completely eliminated, the total number of eliminations is 7. Let's assume side A wins, and let the $i$-th member of side A eliminate $x_{i}$ members of side B. Then the problem is equivalent to finding the number of non-negative integer solutions to the equation
$$
x_{1}+x_{2}+\cdots... | 3432 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 From the numbers $1,2, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, and $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways of selection are there that meet the conditions? | Solution: Notice that
$$
a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\left(14-a_{3}\right)=14 \text {, }
$$
where $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3,14-a_{3}$ $\geqslant 0$, i.e.,
$$
\begin{array}{l}
a_{1}-1+\left(a_{2}-a_{1}-3\right)+\left(a_{3}-a_{2}-3\right)+ \\
\left(1... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4: Arrange 5 white stones and 10 black stones in a horizontal row, such that the right neighbor of each white stone must be a black stone. How many arrangements are there?
(1996, Japan Mathematical Olympiad Preliminary) | Let the number of black stones between the $i$-th white stone and the $(i+1)$-th white stone be $x_{i+1}(i=1,2,3,4)$, with $x_{1}$ black stones at the far left and $x_{6}$ black stones at the far right, then
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=10,
$$
where $x_{1} \geqslant 0, x_{k} \geqslant 1, k=2,3,4,5,6$.
$$
\be... | 252 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 In coin tossing, if Z represents heads and F represents tails, then the sequence of coin tosses is represented by a string composed of Z and F. We can count the number of occurrences of heads followed by tails (ZF), heads followed by heads (ZZ)...... For example, the sequence ZZFFZZZZFZZFFFF is the result of ... | Solution: The sequences that meet the requirements have the following two possible forms:
(1) Starting with F: $F \cdots F Z \cdots Z F \cdots F Z \cdots Z$;
(2) Starting with Z: $\mathrm{Z} \cdots \mathrm{ZF} \cdots \mathrm{FZ} \cdots \mathrm{ZF} \cdots \mathrm{F}$.
Since the sequence is required to have exactly 3 $\m... | 560 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 In a live military exercise, the Red side has set up 20 posts along a straight line. To test 5 new types of weapons, it is planned to equip 5 posts with these new weapons, with the requirement that the first and last posts are not equipped with new weapons, and among every 5 adjacent posts, at least one post ... | Let the 20 positions be ordered as $1,2, \cdots, 20$, and let the sequence number of the $k$-th new weapon be $a_{k}$,
$$
k=1,2,3,4,5 \text {. }
$$
$$
\begin{array}{l}
\text { Let } x_{1}=a_{1}, x_{2}=a_{2}-a_{1}, x_{3}=a_{3}-a_{2}, \\
x_{4}=a_{4}-a_{3}, x_{5}=a_{5}-a_{4}, x_{6}=20-a_{5} .
\end{array}
$$
Then we have
... | 69600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 In a family photo album, there are 10 photos. Each photo has 3 men, the person standing on the left is the son of the person in the middle, and the person on the right is the middle person's brother. It is known that the 10 people in the middle of the 10 photos are all different. How many different people are... | Solution: As shown in Figure 1, there are 16 people, where the horizontal lines represent brothers, and the vertical and diagonal lines represent father-son relationships. The 10 photos taken by them are as follows:
$$
\begin{array}{l}
\quad\{3,1,2\},\{16, \\
2,1\},\{5,3,4\},\{15, \\
4,3\},\{7,5,6\},\{9, \\
6,5\},\{11,... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Question 4 Let the function $f(x)=\frac{1}{k-x}$, and denote
$$
\underbrace{f(f(\cdots)}_{\text {j times }}
$$
as $f_{j}(x)$. Given that $f_{n}(x)=x$, and
$$
f_{i}(x) \neq f_{j}(x)(i \neq j, 1 \leqslant i, j \leqslant n) \text {. }
$$
Prove: $\prod_{i=1}^{n} f_{i}(x)=-1$. | Proof: Consider the recursive sequence $g_{j}=k g_{j-1}-g_{j-2}$, $g_{0}=0, g_{1}=1$. We have
$$
\begin{array}{l}
f_{1}(x)=\frac{1}{k-x}=\frac{x g_{0}-g_{1}}{x g_{1}-g_{2}}, \\
f_{2}(x)=\frac{x g_{1}-g_{2}}{x g_{2}-g_{3}}, \\
\cdots \cdots \\
f_{n}(x)=\frac{x g_{n-1}-g_{n}}{x g_{n}-g_{n+1}} .
\end{array}
$$
Multiplyin... | -1 | Algebra | proof | Yes | Yes | cn_contest | false |
4. For a positive integer $n$, if there exist positive integers $a$ and $b$ such that $n=a+b+a b$, then $n$ is called a "good number". For example, $3=1+1+1 \times 1$, so 3 is a good number. Then, among the 20 positive integers from $1 \sim 20$, the number of good numbers is $\qquad$ . | 4.12.
$n+1=a+b+a b+1=(a+1)(b+1)$ is a composite number, so the required $n$ is the number obtained by subtracting 1 from the composite numbers between 2 and 21. The composite numbers between 2 and 21 are $4,6,8,9,10,12,14,15,16,18,20,21$, a total of 12. Therefore, the required $n$ has 12 values: $3,5,7,8,9,11,13,14,15,... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Place five different files $A, B, C, D, E$ into a row of seven drawers numbered $1, 2, 3, 4, 5, 6, 7$, with each drawer containing at most one file. If files $A$ and $B$ must be placed in adjacent drawers, and files $C$ and $D$ must also be placed in adjacent drawers, then the number of different ways to place the f... | 5.C.
Place files $A$ and $B$ in adjacent drawers, denoted as “$AB$”, place files $C$ and $D$ in adjacent drawers, denoted as “$CD$”, and place file $E$ in a drawer, denoted as “$E$”. Thus, “$AB$”, “$CD$”, “$E$”, and the two empty drawers can be considered as five elements, and the number of all permutations of these f... | 240 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
6. Let the set $M=\left\{a \left\lvert\, a=\frac{x+y}{t}\right., 2^{x}+2^{y}=\right.$ $2^{t}$, where $x, y, t, a$ are all integers $\}$. Then the sum of all elements in the set $M$ is ( ).
(A) 1
(B) 4
(C) 7
(D) 8 | 6. D.
Assume $x \leqslant y$, then $2^{x}=2^{x}+2^{y} \leqslant 2^{y}+2^{y}=2^{y+1}$.
Thus, $t \leqslant y+1$.
From $2^{x}>0$, we get $2^{t}=2^{x}+2^{y}>2^{y}$. Therefore, $t>y$.
So, $y<t \leqslant y+1$.
Given that $x, y, t$ are all integers, then $t=y+1$. Hence, $2^{y+1}=$ $2^{x}+2^{y}$, which means $2^{x}=2^{y}$. Th... | 8 | Algebra | MCQ | Yes | Yes | cn_contest | false |
7. Given a fixed point $A(4, \sqrt{7})$. If a moving point $P$ is on the parabola $y^{2}=4 x$, and the projection of point $P$ on the $y$-axis is point $M$, then the maximum value of $|P A|-|P M|$ is $\qquad$. | $$
\begin{array}{l}
|P M|=|P N|-|M N|=|P F|-1 \downarrow \\
\text { Then }|P A|-|P M|=|P A|-(|P F|-1) \\
=(|P A|-|P F|)+1 \leqslant|A F|+1=4+1=5 .
\end{array}
$$ | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n}+a_{n+1}=1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
If $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, then the value of $S_{2003}-2 S_{2004}+S_{2000}$ is $\qquad$ | 9.3.
According to the problem, when $n$ is even, we have $a_{1}+a_{2}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n}{2}$ pairs, so $S_{n}=\frac{n}{2}$. Therefore, $S_{2004}=1002$.
When $n$ is odd, we have $a_{2}+a_{3}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n-1}{2}$ pairs, so
$$... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. As shown in Figure 2, the side length of rhombus $A B C D$ is 1, and $\angle A B C=120^{\circ}$. If
$E$ is any point on the extension of $B C$,
$A E$
intersects $C D$ at point $F$, then
the angle between vector $B F$ and $E D$
is
$\qquad$ | 10.120.
As shown in Figure 5, establish a Cartesian coordinate system. Then
$$
\begin{array}{l}
A\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), \\
B(0,0), \\
C(1,0), \\
D\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) .
\end{array}
$$
Let $E(a, 0)(a>1)$. Then
$l_{c p}: y=-\sqrt{3}(x-1)$,
$$
l_{\text {AE }}: y=-\frac{\sq... | 120 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. As shown in Figure 3(a), the eight vertices of a cube are assigned values $a, b, c, d, e, f, g, h$, respectively. Then, the arithmetic mean of the values of the three adjacent vertices of each vertex, denoted as $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{e}, \bar{f}, \bar{g}, \bar{h}$, is placed at the corresponding... | 11.20.
According to the problem, we have
$$
\begin{array}{l}
\bar{a}=\frac{b+d+e}{3}, \bar{b}=\frac{a+c+f}{3}, \bar{c}=\frac{b+d+g}{3}, \\
\bar{d}=\frac{a+c+h}{3}, e=\frac{a+f+h}{3}, \bar{f}=\frac{b+e+g}{3}, \\
\bar{g}=\frac{c+f+h}{3}, \bar{h}=\frac{d+e+g}{3} .
\end{array}
$$
Then, $a=(\bar{b}+\bar{d}+\bar{e})-2 \bar... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 A coach can travel from Nanjing to Shanghai on the Shanghai-Nanjing Expressway in $4 \mathrm{~h}$. It is known that a coach departs from Nanjing to Shanghai every half hour, and at the same time, a coach also departs from Shanghai to Nanjing. If a passenger takes a bus from Nanjing to Shanghai, how many coach... | Explanation: This problem can be easily solved by just drawing a schematic diagram (Figure 6).
The traveler encounters 8 trains (including the one arriving at Nanjing station) in the first 2 hours (i.e., traveling half the distance). 2 hours after departure, they encounter exactly 1 train that departed from Shanghai t... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
13. Let two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ in a plane be perpendicular to each other, and $|\boldsymbol{a}|=2,|\boldsymbol{b}|=1$. Also, $k$ and $t(t \geqslant 0)$ are two real numbers that are not both zero. If the vectors
$$
\boldsymbol{x}=\boldsymbol{a}+(3-t) \boldsymbol{b} \text { and } \boldsymbol{y... | 13.1.
Given $\boldsymbol{a} \cdot \boldsymbol{b}=0$.
$$
\begin{array}{l}
\boldsymbol{x} \cdot \boldsymbol{y}=[\boldsymbol{a}+(3-t) \boldsymbol{b}] \cdot\left(-k \boldsymbol{a}+t^{2} \boldsymbol{b}\right) \\
=-k \boldsymbol{a}^{2}+\left[-k(3-t)+t^{2}\right] \boldsymbol{a} \cdot \boldsymbol{b}+t^{2}(3-t) \boldsymbol{b}^... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
19. (15 points) There are 800 points on a circle, labeled $1, 2, \cdots, 800$ in a clockwise direction. They divide the circle into 800 gaps. Arbitrarily choose one point and color it red, then color the remaining points red according to the following rule: if the $k$-th point has been colored red, then move clockwise ... | 19. Generally, for a circle with $n$ points, we denote the maximum number of points that can be colored red according to the given rules as $f(n)$.
If the circle has $2 n$ points, and the first point to be colored red is labeled $i$.
(1) If $i=2 k (k \geqslant 1)$ is an even number, then all the points that are colore... | 25 | Number Theory | proof | Yes | Yes | cn_contest | false |
6. Given the set $S=\{1,2,3,4,5,6\}$, a one-to-one mapping $f: S \rightarrow S$ satisfies the condition: for any $x \in S$, $f(f(f(x)))=x$. Then the number of mappings $f$ that satisfy the condition is ( ).
(A) 81
(B) 80
(C) 40
(D) 27 | 6.A.
It is clear that there does not exist $x \in S$ such that $f(x) \neq x$ and $f(f(x)) = x$. Otherwise, $f(f(f(x))) = f(x) \neq x$, which contradicts the given condition. Therefore, for any $x \in S$, either $f(x) = x$, or $f(x) = x_{1}$, $f(x_{1}) = x_{2}$, $f(x_{2}) = x$, and $x, x_{1}, x_{2}$ are distinct. Hence... | 81 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
5. If there exists a permutation $a_{1}$, $a_{2}, \cdots, a_{n}$ of $1,2, \cdots, n$, such that $k+a_{k}(k=1,2, \cdots, n)$ are all perfect squares, then $n$ is called a "middle number". Then, in the set $\{15,17,2000\}$, the number of elements that are middle numbers is $\qquad$ . | 5.3.
(1) 15 is a median number. Because in the arrangement of Table 1, $k+a_{k}$ $(k=1,2, \cdots, 15)$ are all perfect squares:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$k$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\
\hline$a_{k}$ & 15 & 14 & 13 & 12 & 11 & 10 & 9 & 8 & ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) A $98 \times$ 98 chessboard is displayed on a computer screen, colored in the usual way (i.e., with two colors alternating). A person can drag the mouse to select a rectangle with borders along the chessboard lines, then click the mouse, and all the colors within this frame will change (i.e., white t... | Three, we prove that for an $n \times n$ chessboard, if $n$ is odd, then at least $n-1$ mouse clicks are required to turn the entire chessboard into one color; if $n$ is even, then at least $n$ mouse clicks are required to turn the entire chessboard into one color.
Consider the $4(n-1)$ small squares along the border ... | 98 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Let $F$ be the set of real polynomials $f(x)$ satisfying
(1) the degree of $f(x)$ is less than or equal to 3;
(2) for any $x \in [0,1]$, $|f(x)| \leqslant 1$.
Find $\max _{f \in F} f(2)$. | Solution: Let $f(x)=a x^{3}+b x^{2}+c x+d$. Denote
$$
\left.\begin{array}{l}
A=f(1)=a+b+c+d, \\
B=f\left(\frac{3}{4}\right)=\frac{27}{64} a+\frac{9}{16} b+\frac{3}{4} c+d, \\
C=f\left(\frac{1}{4}\right)=\frac{1}{64} a+\frac{1}{16} b+\frac{1}{4} c+d, \\
D=f(0)=d .
\end{array}\right\}
$$
By condition (2), we know $A, B,... | 99 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Two positive integers $a_{1}, a_{2}, \cdots, a_{2006}$ (which can be the same) are such that $\frac{a_{1}}{a_{2}}, \frac{a_{2}}{a_{3}}, \cdots, \frac{a_{2005}}{a_{2006}}$ are all distinct. How many different numbers are there at least among $a_{1}$, $a_{2}, \cdots, a_{2006}$?
（Chen Yonggao, problem contributor） | Due to the fact that the pairwise ratios of 45 distinct positive integers are at most $45 \times 44 + 1 = 1981$, the number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{2000}$ is greater than 45.
Below is an example to show that 46 can be achieved.
Let $p_{1}, p_{2}, \cdots, p_{46}$ be 46 distinct prime numbers, an... | 46 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Six, let $X$ be a 56-element set. Find the smallest positive integer $n$, such that for any 15 subsets of $X$, if the union of any 7 of them has at least $n$ elements, then there must exist 3 of these 15 subsets whose intersection is non-empty.
(Cold Gangsong provided the problem) | Six, the minimum value of $n$ is 41.
First, prove that $n=41$ meets the requirement. Use proof by contradiction.
Assume there exist 15 subsets of $X$ such that the union of any 7 of them contains at least 41 elements, and the intersection of any 3 of them is empty. Since each element belongs to at most 2 subsets, we ca... | 41 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let set $A=\{-2,0,1\}, B=\{1,2,3,4, 5\}$, and the mapping $f: A \rightarrow B$ such that for any $x \in A$, $x + f(x) + x f(x)$ is an odd number. Then the number of such mappings $f$ is ( ).
(A) 45
(B) 27
(C) 15
(D) 11 | 6.A.
When $x=-2$, $x+f(x)+x f(x)=-2-f(-2)$ is odd, then $f(-2)$ can take $1,3,5$, which gives 3 possible values;
When $x=0$, $x+f(x)+x f(x)=f(0)$ is odd, then $f(0)$ can take $1,3,5$, which gives 3 possible values;
When $x=1$, $x+f(x)+x f(x)=1+2 f(1)$ is odd, then $f(1)$ can take $1,2,3,4,5$, which gives 5 possible ... | 45 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
10. Given the set $M=\left\{x \left\lvert\, x=\lim _{n \rightarrow \infty} \frac{2^{n+1}-2}{\lambda^{n}+2^{n}}\right.\right.$, $\lambda$ is a constant, and $\lambda+2 \neq 0\}$. Then the sum of all elements of $M$ is $\qquad$ . | 10.3.
When $|\lambda|>2$, $x=\lim _{n \rightarrow \infty} \frac{2\left(\frac{2}{\lambda}\right)^{n}-2\left(\frac{1}{\lambda}\right)^{n}}{1+\left(\frac{2}{\lambda}\right)^{n}}=0$;
When $\lambda=2$, $x=\lim _{n \rightarrow \infty}\left(1-\frac{1}{2^{n}}\right)=1$;
When $|\lambda|<2$, $x=\lim _{n \rightarrow \infty} \fra... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=2, a_{n}=\frac{1+a_{n-1}}{1-a_{n-1}}(n \geqslant 2) \text {. }
$$
Then the value of $a_{2006}$ is $\qquad$ . | 3.2.
Since $a_{1}=2, a_{2}=-3, a_{3}=-\frac{1}{2}, a_{4}=\frac{1}{3}, a_{5}=$ 2, we conjecture that $a_{n}$ is a periodic sequence with a period of 4. And
$$
a_{n+4}=\frac{1+a_{n+3}}{1-a_{n+3}}=\frac{1+\frac{1+a_{n+2}}{1-a_{n+2}}}{1-\frac{1+a_{n+2}}{1-a_{n+2}}}=-\frac{1}{a_{n+2}}=a_{n} .
$$
Therefore, $a_{2008}=a_{4 ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In the non-decreasing sequence of positive odd numbers $\{1,3,3,3,5,5,5, 5,5, \cdots\}$, each positive odd number $k$ appears $k$ times. It is known that there exist integers $b$, $c$, and $d$, such that for all integers $n$, $a_{n}=$ $b[\sqrt{n+c}]+d$, where $[x]$ denotes the greatest integer not exceeding $x$. The... | 5.2.
Divide the known sequence into groups as follows:
$$
\begin{array}{l}
(1),(3,3,3),(5,5,5,5,5), \cdots, \\
(\underbrace{2 k-1,2 k-1, \cdots, 2 k-1}_{2 k-1 \uparrow}),
\end{array}
$$
Let $a_{n}$ be in the $k$-th group, where $a_{n}=2 k-1$. Then we have
$$
\begin{array}{l}
1+3+5+\cdots+2 k-3+1 \\
\leqslant n0$, sol... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6.4. A five-digit number is a multiple of 54, and none of its digits are 0. After deleting one of its digits, the resulting four-digit number is still a multiple of 54; after deleting one of the digits of this four-digit number, the resulting three-digit number is still a multiple of 54; after deleting one of the digit... | 6.4. The original five-digit number is 59994.
Since the number faced after each deletion of the front or back is a multiple of 9, the number deleted each time is a multiple of 9. This means that only a 9 can be deleted each time, and the last remaining two-digit number can only be 54. Among the three-digit numbers 549... | 59994 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7.2. The sum of three positive integers (not necessarily distinct) is 100. By subtracting them pairwise (the larger minus the smaller), three difference numbers can be obtained. What is the maximum possible value of the sum of these three difference numbers? | 7.2. The maximum possible value of the sum of these three differences is 194.
Let the three positive integers be $a$, $b$, and $c$, and assume without loss of generality that $a \geqslant b \geqslant c$. The sum of their pairwise differences is
$$
(a-b)+(a-c)+(b-c)=2(a-c).
$$
Since $b \geqslant 1$ and $c \geqslant 1$... | 194 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7.4. Toma thought of a positive integer and found the remainders when it is divided by 3, 6, and 9. It is known that the sum of these three remainders is 15. Try to find the remainder when the number is divided by 18. | 7.4. The remainder when this number is divided by 18 is 17.
Since the remainders when divided by $3$, $6$, and $9$ do not exceed $2$, $5$, and $8$ respectively, the sum of these three remainders is always no more than $2+5+8=15$. Since their sum is equal to 15, these three remainders must be $2$, $5$, and $8$. Further... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6.4. Find all such four-digit numbers: they are all 83 times the sum of their digits. | 6.4. There is only one such four-digit number: $1494=18 \times 83$.
Solution 1: Let the number we are looking for be $x$, and the sum of its digits be $a$. From the problem, we know that $x=83a$. Since the remainder when $x$ is divided by 9 is the same as the remainder when $a$ is divided by 9, $x-a=82a$ must be divis... | 1494 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The inequality about $x$ $\left\{\begin{array}{l}x-1>0, \\ x^{3}-x^{2}+x \leqslant k\end{array}\right.$ has only the positive integer solutions 2 and 3. Then the value of $| k-21|+| k-52 \mid+$ 1975 is ( ).
(A) 2004
(B) 2005
(C) 2006
(D) 2007 | 2.C.
From the problem, we know that $x>1$.
Since $x=2, 3$ are solutions to $x^{3}-x^{2}+x \leqslant k$, we have,
$$
\left\{\begin{array}{l}
2^{3}-2^{2}+2 \leqslant k, \\
3^{3}-3^{2}+3 \leqslant k
\end{array}\right.
$$
Solving this, we get $k \geqslant 21$.
From the given information, $x=4,5,6, \cdots$ are solutions t... | 2006 | Inequalities | MCQ | Yes | Yes | cn_contest | false |
1. Given that $x$, $y$, $z$ are positive real numbers, and $x y z(x+y+z)=1$. Then the minimum value of $(x+y)(y+z)$ is $\qquad$ | \begin{array}{l}=1.2. \\ (x+y)(y+z)=y^{2}+(x+z) y+x z \\ =y(x+y+z)+x z=y \cdot \frac{1}{x y z}+x z \\ =\frac{1}{x z}+x z \geqslant 2 .\end{array} | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) In a certain competition, each player plays exactly one game against every other player. The winner of each game gets 1 point, the loser gets 0 points, and in the case of a draw, both get 0.5 points. After the competition, it is found that each player's score is exactly half from games played against t... | Given $n$ players, the $n$ players scored a total of $\frac{n(n-1)}{2}$ points. The 10 "math players" scored a total of $\frac{10 \times 9}{2}=45$ points through their matches with each other, which is half of their total score, so they scored a total of 90 points. The remaining $n-10$ players scored a total of $\frac{... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given the quadratic trinomial $a x^{2}+b x+c$ $(a>0)$.
(1) When $c<0$, find the maximum value of the function
$$
y=-2\left|a x^{2}+b x+c\right|-1
$$
(2) For any real number $k$, the line $y=k(x-$
1) $-\frac{k^{2}}{4}$ intersects the parabola $y=a x^{2}+b x+c$ at exactly one point, find the value of ... | (1) Given $a>0, c<0$, we know that $y^{\prime}=a x^{2}+b x+c$ intersects the $x$-axis, and $y_{\text {nuin }}^{\prime}<0$. Therefore,
$$
\left|y^{\prime}\right|=\left|a x^{2}+b x+c\right| \geqslant 0 \text {. }
$$
Thus, the minimum value of $\left|y^{\prime}\right|$ is 0.
At this point, $y_{\text {man }}=-2 \times 0-1... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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