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Example 3 As shown in Figure 2, the three altitudes of acute $\triangle A B C$ intersect at point $H$. How many triangles are there in Figure 2?
The text above is translated into English, preserving the original text's line breaks and format. | Solution 1: Let the set of triangles with vertex $A$ be $P_{1}$. Then, the triangles with vertices on $A B$ and $A D$ are 3, the triangles with vertices on $A D$ and $A C$ are 3, and the triangles with vertices on $A B$ and $A C$ are 3, giving $\left|P_{1}\right|=9$.
Similarly, the sets of triangles with vertices at $... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. There are four numbers, among which the sum of every three numbers is $24, 36, 28, 32$. Then the average of these four numbers is $\qquad$ . | 10.10 .
The sum of these four numbers is $(24+36+28+32) \div 3=40$, so the average of these four numbers is 10 . | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. If $a^{4}+b^{4}=a^{2}-2 a^{2} b^{2}+b^{2}+6$, then $a^{2}+b^{2}=$ $\qquad$ . | 11.3.
Given $a^{4}+b^{4}=a^{2}-2 a^{2} b^{2}+b^{2}+6$, we have $\left(a^{2}+b^{2}\right)^{2}-\left(a^{2}+b^{2}\right)-6=0$.
Therefore, $a^{2}+b^{2}=3$ or -2.
Since $a^{2}+b^{2} \geqslant 0$, we have $a^{2}+b^{2}=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. If real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x y+x+y+7=0, \\
3 x+3 y=9+2 x y,
\end{array}\right.
$$
then $x^{2} y+x y^{2}=$ | 13.6.
From $\left\{\begin{array}{l}x y+x+y+7=0, \\ 3 x+3 y=9+2 x y,\end{array}\right.$ we get $x y=-6, x+y=-1$.
Therefore, $x^{2} y+x y^{2}=6$. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. A person's 5 trips to work (unit: $\mathrm{min}$) are $a, b, 8, 9, 10$. It is known that the average of this set of data is 9, and the variance is 2. Then the value of $|a-b|$ is $\qquad$. | 15.4.
Since the average of $a, b, 8, 9, 10$ is 9 and the variance is 2, it follows that $a, b, 8, 9, 10$ are 5 consecutive integers, so $a=7, b=11$ or $a=11, b=7$. Therefore, the value of $|a-b|$ is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
16. If the integer $m$ makes the equation
$$
x^{2}-m x+m+2006=0
$$
have non-zero integer roots, then the number of such integers $m$ is
$\qquad$. | 16.5.
Let the two integer roots of the equation be $\alpha, \beta$, then
$$
\begin{array}{l}
\alpha+\beta=m, a \beta=m+2006, \\
\text { i.e., } \alpha \beta-(\alpha+\beta)+1=2006+1 \\
=2007=(\alpha-1)(\beta-1) .
\end{array}
$$
Thus, $\alpha-1= \pm 1, \pm 3, \pm 9$;
$$
\beta-1= \pm 2007, \pm 669, \pm 223 \text {. }
$$... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
17. In a math test, there are 20 questions. Each correct answer earns 5 points, no answer earns 0 points, and each wrong answer deducts 2 points. If Xiaoli's score in this test is a prime number, then the maximum number of questions Xiaoli answered correctly is $\qquad$. | 17.17.
Let Xiaoli answer $a$ questions correctly and $b$ questions incorrectly, with the score being $5a-2b<100$.
Since the score is a prime number, we have $5a-2b=97, 91, 89, 83, \cdots$. Only when $a=17, b=1$, $5a-2b=83$ is the largest prime number.
Therefore, Xiaoli answered at most 17 questions correctly this time... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
18. Let the perimeter of a surface development of a $5 \mathrm{~cm} \times 4 \mathrm{~cm} \times 3 \mathrm{~cm}$ rectangular prism be $n \mathrm{~cm}$. Then the minimum value of $n$ is $\qquad$ . | 18.50.
As shown in Figure 9, the perimeter of the unfolded rectangular prism is
$$
8 c+4 b+2 a \text {. }
$$
Therefore, the minimum value of the perimeter is
$$
\begin{array}{l}
8 \times 3+4 \times 4+2 \times 5 \\
=50 .
\end{array}
$$ | 50 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
20. There is a pentagon $A B C D E$. If the vertices $A$, $B$, $C$, $D$, $E$ are colored with one of the three colors: red, yellow, green, such that adjacent vertices are colored differently, then there are a total of different coloring methods. | 20.30.
If point $A$ is colored red (as shown in Figure 11), there are 10 coloring methods.
Figure 11
Similarly, if point $A$ is colored yellow or green, there are also 10 coloring methods each. Therefore, there are a total of 30 different coloring methods. | 30 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $x, y$ be real numbers, the algebraic expression
$$
5 x^{2}+4 y^{2}-8 x y+2 x+4
$$
has a minimum value of | 9.3.
$$
\begin{array}{l}
\text { Since } 5 x^{2}+4 y^{2}-8 x y+2 x+4 \\
=5\left(x-\frac{4}{5} y+\frac{1}{5}\right)^{2}+\frac{4}{5}(y+1)^{2}+3,
\end{array}
$$
Therefore, when $y=-1, x=-1$, the original expression has a minimum value of 3. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. In a $3 \times 3$ grid filled with the numbers $1 \sim 9$, the largest number in each row is colored red, and the smallest number in each row is colored green. Let $M$ be the smallest number among the three red squares, and $m$ be the largest number among the three green squares. Then $M-m$ can have $\qquad$ differ... | 14.8.
From the conditions, it is known that the values of $m$ and $M$ are positive integers from 3 to 7 (inclusive of 3 and 7).
Obviously, $M \neq m$, so the value of $M-m$ is
$$
1,2,3,4,-1,-2,-3,-4 \text {. }
$$
Therefore, $M-m$ can have 8 different values. | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
17. In the $7 \times 7$ unit square grid shown in Figure 10, there are 64 grid points, and there are many squares with these grid points as vertices. How many different values of the areas of these squares are there? | 17. Since the vertices of the square are grid points in the grid, as shown in Figure 13, the area of the shaded square is $a^{2}+b^{2}$.
Where, $0 \leqslant a+b \leqslant 7$.
Without loss of generality, let $a \geqslant b$. We can enumerate all possible values of $(a, b)$:
$$
\begin{array}{l}
(0,0),(1,0),(2,0),(3,0),(4... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
18. Let $k$, $a$, $b$ be positive integers, and the quotients when $k$ is divided by $a^2$ and $b^2$ are $m$ and $m+116$ respectively.
(1) If $a$ and $b$ are coprime, prove that $a^2 - b^2$ is coprime with both $a^2$ and $b^2$;
(2) When $a$ and $b$ are coprime, find the value of $k$;
(3) If the greatest common divisor ... | 18. (1) Let $s$ be the greatest common divisor of $a^{2}-b^{2}$ and $a^{2}$.
Then $a^{2}-b^{2}=s u, a^{2}=s v\left(s, v \in \mathbf{N}_{+}\right)$.
Thus, $a^{2}-\left(a^{2}-b^{2}\right)=b^{2}=s(v-u)$.
It is clear that $s$ is a divisor of $b^{2}$.
Since $a, b$ are coprime, $a^{2}, b^{2}$ are also coprime.
Therefore, $s... | 4410000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that $a, b, c, d$ are even numbers, and $0<a<b<c<d, d-a=90, a, b, c$ form an arithmetic sequence, $b, c, d$ form a geometric sequence. Then $a+b+c+d=$ ( ).
(A) 384
(B) 324
(C) 284
(D) 194 | 9.D.
Let $a, b, c, d$ be $b-m, b, b+m, \frac{(b+m)^{2}}{b}$, respectively.
Also, $\frac{(b+m)^{2}}{b}-(b-m)=90$, which means
$$
b=\frac{m^{2}}{3(30-m)} \text {. }
$$
Since $a, b, c, d$ are even numbers, and $0<a<b<c<d$, we know that $m$ is a multiple of 6, and $m<30$.
Let $m=6k$, substituting into equation (1) yields... | 194 | Algebra | MCQ | Yes | Yes | cn_contest | false |
13. Given $\frac{\sin (\alpha+2 \beta)}{\sin \alpha}=3$, and $\beta \neq \frac{1}{2} k \pi$, $\alpha+\beta \neq n \pi+\frac{\pi}{2}(n, k \in \mathbf{Z})$. Then the value of $\frac{\tan (\alpha+\beta)}{\tan \beta}$ is $\qquad$ | $\begin{array}{l}13.2 \text {. } \\ \frac{\tan (\alpha+\beta)}{\tan \beta}=\frac{\sin (\alpha+\beta) \cdot \cos \beta}{\cos (\alpha+\beta) \cdot \sin \beta} \\ =\frac{\frac{1}{2}[\sin (\alpha+2 \beta)+\sin \alpha]}{\frac{1}{2}[\sin (\alpha+2 \beta)-\sin \alpha]}=\frac{\frac{\sin (\alpha+2 \beta)}{\sin \alpha}+1}{\frac{... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Let $\left\{a_{n}\right\}$ be a sequence of positive numbers, and let the sum of the first $n$ terms be $b_{n}$. The product of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $c_{n}$, and $b_{n}+c_{n}=1$. Then the number in the sequence $\left\{\frac{1}{a_{n}}\right\}$ that is closest to 2000 is $\qq... | 14.1980.
According to the problem, we have $b_{n}=\frac{c_{n}}{c_{n-1}}(n \geqslant 2)$.
Also, $b_{n}+c_{n}=1$, so $\frac{c_{n}}{c_{n-1}}+c_{n}=1$, which means $\frac{1}{c_{n}}-\frac{1}{c_{n-1}}=1$. Given $c_{1}=b_{1}, c_{1}+b_{1}=1$, we can get $c_{1}=b_{1}=\frac{1}{2}$. Therefore, $c_{n}=\frac{1}{n+1}, b_{n}=\frac{n... | 1980 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Given a finite set of planar vectors $M$, for any three elements chosen from $M$, there always exist two elements $\boldsymbol{a}, \boldsymbol{b}$ such that $\boldsymbol{a}+\boldsymbol{b} \in M$. Try to find the maximum number of elements in $M$.
| Three, the maximum number of elements in the set $M$ is 7.
Let points $A, B, C$ be any three points on a plane. Consider the 7-element set $M=\{\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C A}, \boldsymbol{B A}, \boldsymbol{C B}, \boldsymbol{A C}, \mathbf{0}\}$, which clearly satisfies the conditions. We will now p... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given real numbers $a, b, c, d, e, f$ satisfy the system of equations:
$$
\left\{\begin{array}{l}
2 a+b+c+d+e+f=20, \\
a+2 b+c+d+e+f=40, \\
a+b+2 c+d+e+f=80, \\
a+b+c+2 d+e+f=160, \\
a+b+c+d+2 e+f=320, \\
a+b+c+d+e+2 f=640 .
\end{array}\right.
$$
Then the value of $f-e+d-c+b-a$ is $\qquad$ . | 2. 420 .
$$
\begin{array}{l}
\text { (6) - (5) + (4) - (3) + (2)-(1) gives } \\
f-e+d-c+b-a \\
=640-320+160-80+40-20=420 .
\end{array}
$$ | 420 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The number of positive integers $n$ such that $n+1$ divides $n^{2006}+2006$ is $\qquad$.
Makes the positive integer $n$ such that $n+1$ can divide $n^{2006}+2006$ total $\qquad$.
Note: The second sentence seems to be a repetition or a different phrasing of the first. If it's meant to be a different statement, plea... | 5. 5 .
From the problem, we have
$$
\begin{array}{l}
n^{2006}+2006 \equiv(-1)^{2006}+2006=2007 \\
\equiv 0(\bmod (n+1)) .
\end{array}
$$
Since $2007=3 \times 3 \times 223$, then
$$
n+1=3,9,223,669,2007 \text {. }
$$
Therefore, $n=2,8,222,668,2006$. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given $n(n>1)$ integers (which can be the same) $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2007 .
$$
Then the minimum value of $n$ is $\qquad$ | 9. 5 .
Given $a_{1} a_{2} \cdots a_{n}=2007$, we know that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. Also, $a_{1}+a_{2}+\cdots+a_{n}=2007$ is an odd number, so $n$ is odd.
If $n=3$, i.e., $a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3}=2007$, without loss of generality, assume $a_{1} \geqslant a_{2} \geqslant a_{3}$, t... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8.3. There are 19 cards. Can a non-zero digit be written on each card so that these 19 cards can be arranged to form a 19-digit number divisible by 11? | 8.3. Answer: Yes.
Write a 2 on each of 10 cards, and a 1 on each of the remaining cards. It is well known that a positive decimal integer is divisible by 11 if and only if the difference $S$ between the sum of the digits in the odd positions and the sum of the digits in the even positions is a multiple of 11. Under al... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $A_{1}, A_{2}, \cdots, A_{n}$ are 11 stations sequentially on a straight highway, and
$$
\begin{array}{l}
A_{i} A_{i+2} \leqslant 12(i=1,2, \cdots, 9), \\
A_{i} A_{i+3} \geqslant 17(i=1,2, \cdots, 8) .
\end{array}
$$
If $A_{1} A_{11}=56 \mathrm{~km}$, then $A_{2} A_{7}=$ $\qquad$ $\mathrm{km}$. | Ni, 1.29.
According to the problem, we have $A_{1} A_{10} \geqslant 3 \times 17=51(\mathrm{~km})$, then
$$
A_{10} A_{11} \leqslant 5 \mathrm{~km} \text {; }
$$
Also, $A_{8} A_{11} \geqslant 17 \mathrm{~km}, A_{8} A_{10} \leqslant 12 \mathrm{~km}$, then $A_{10} A_{11} \geqslant 5 \mathrm{~km}$.
Therefore, $A_{10} A_{11... | 29 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of positive integer pairs $(x, y)$ that satisfy the equation
$$
\begin{array}{l}
x \sqrt{y}+y \sqrt{x}-\sqrt{2006 x}-\sqrt{2006 y}+\sqrt{2006 x y} \\
\quad=2006
\end{array}
$$ | 3.8 .
Given the equation can be transformed into
$$
(\sqrt{x}+\sqrt{y}+\sqrt{2006})(\sqrt{x y}-\sqrt{2006})=0 \text {. }
$$
Since $\sqrt{x}+\sqrt{y}+\sqrt{2006}>0$, we have
$$
\sqrt{x y}=\sqrt{2006} \text {, }
$$
which means $\square$
$$
\begin{array}{l}
x y=2006=1 \times 2006=2 \times 1003 \\
=17 \times 118=34 \tim... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The number of five-digit numbers where the sum of any two adjacent digits is divisible by 3 is $\qquad$ .
| 2.1254.
Classify the digits by their remainders, then the five-digit number $\overline{a b c d e}$ that meets the condition satisfies
$$
a \equiv c=e \equiv 2(\bmod 3), b \equiv d \equiv 1(\bmod 3),
$$
or $a=c=e=1(\bmod 3), b=d=2(\bmod 3)$,
or $a=b=c=d=e=0(\bmod 3)$.
If it is equation (1), then there are $3^{5}$ such... | 1254 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. A three-digit number $\overline{x y z}, 1 \leqslant x \leqslant 9,0 \leqslant y, z \leqslant 9$, and $x!+y!+z!=\overline{x y z}$. Then the value of $x+y+z$ is $\qquad$ | 5.10.
From $\overline{x y z}=x!+y!+z!$, we get
$$
100 x+10 y+z=x!+y!+z!.
$$
It is easy to see that $x, y, z \leqslant 6$. Otherwise,
$$
x!+y!+z!\geqslant 7!>1000.
$$
If $x=6$, then the left side of equation (1) is $700$, which is a contradiction.
Therefore, $x \leqslant 5$.
Similarly, $y \leqslant 5, z \leqslant 5$.... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Use several small cubes of the same size, with surfaces all white or all red, to form a large cube $A B C D-E F G H$ as shown in Figure 2. If the small cubes used on the diagonals $A G$, $B H$, $C E$, and $D F$ of the large cube are all red, and a total of 41 are used, and the rest of the large cube is made up of sm... | 5.C.
For a large cube, the number of small cubes used on each of the four space diagonals is the same, so the total number of small cubes used on the four space diagonals should be a multiple of 4. However, according to the given condition, the four space diagonals actually use 41 small cubes, which is an odd number. ... | 1290 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
7. If the graph of the inverse proportion function $y=\frac{k}{x}$ intersects the graph of the linear function $y=a x+b$ at points $A(-2, m)$ and $B(5, n)$, then the value of $3 a+b$ is
保留了源文本的换行和格式。 | Ni.7.0.
Since points $A(-2, m)$ and $B(5, n)$ lie on the graph of an inverse proportion function, we have
$$
\left\{\begin{array}{l}
m=-\frac{k}{2}, \\
n=\frac{k}{5} .
\end{array}\right.
$$
Since points $A(-2, m)$ and $B(5, n)$ also lie on the line $y=a x+b$, we get
$$
\left\{\begin{array}{l}
m=-2 a+b \\
n=5 a+b
\end... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $n$ be a positive integer, and $n^{2}+1085$ is a positive integer power of 3. Then the value of $n$ is $\qquad$ | 11.74.
Generally, the unit digit of $n^{2}$ (where $n$ is a positive integer) can only be 0, 1, 4, 5, 6, 9, so the unit digit of $n^{2}+1085$ can only be 5, 6, 9, 0, 1, 4; while the unit digit of $3^{m}$ (where $m$ is a positive integer) can only be 1, 3, 7, 9.
From the given, let $n^{2}+1085=3^{m}$ (where $n, m$ are... | 74 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. Given that $a$, $b$, and $c$ are positive integers, and the graph of the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from points $A$ and $B$ to the origin $O$ are both less than 1, find the minimum value of $a+b+c$. | 14. Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, where $x_{1}$ and $x_{2}$ are the roots of the equation $a x^{2}+b x+c=0$.
Given that $a$, $b$, and $c$ are positive integers, we have
$x_{1}+x_{2}=-\frac{b}{a} < 0$.
Thus, the equation $a x^{2}+b x+c=0$ has two negative real roots, i.e., $x_{1} < 0$.
Solvi... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. At a party, 9 celebrities performed $n$ "trio dance" programs. If in these programs, any two people have collaborated exactly once, then $n=$ $\qquad$ | 12.12.
Consider 9 people as 9 points. If two people have performed in the same group, then a line segment is drawn between the corresponding two points. Thus, every pair of points is connected, resulting in a total of 36 line segments. Each trio dance corresponds to a triangle, which has three sides. When all sides ap... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
13. (i) (Grade 11) In the arithmetic sequence $\left\{a_{n}\right\}: a_{n}=4 n -1\left(n \in \mathbf{N}_{+}\right)$, after deleting all numbers that can be divided by 3 or 5, the remaining numbers are arranged in ascending order to form a sequence $\left\{b_{n}\right\}$. Find the value of $b_{2006}$.
(ii) (Grade 12) Gi... | (i) Since $a_{n+15}-a_{n}=60$, $a_{n}$ is a multiple of 3 or 5 if and only if $a_{n+15}$ is a multiple of 3 or 5.
Now, divide the positive direction of the number line into a series of intervals of length 60:
$$
(0,+\infty)=(0,60] \cup(60,120] \cup(120,180] \cup \cdots \text {. }
$$
Note that the first interval contai... | 15043 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Find the smallest positive real number $k$ such that for any 4 distinct real numbers $a, b, c, d$ not less than $k$, there exists a permutation $p, q, r, s$ of $a, b, c, d$ such that the equation $\left(x^{2}+p x+q\right)\left(x^{2}+r x+s\right)=0$ has 4 distinct real roots. (Feng Zhigang) | 2. On one hand, if $k4(d-a)>0, c^{2}-4 b>4(c-b)$ $>0$, therefore, equations (1) and (2) both have two distinct real roots.
Secondly, if equations (1) and (2) have a common real root $\beta$, then
$$
\left\{\begin{array}{l}
\beta^{2}+d \beta+a=0, \\
\beta^{2}+\phi \beta+b=0 .
\end{array}\right.
$$
Subtracting the two e... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Arrange the numbers $1,2, \cdots, 13$ in a row $a_{1}, a_{2}$, $\cdots, a_{13}$, where $a_{1}=13, a_{2}=1$, and ensure that $a_{1}+a_{2}+$ $\cdots+a_{k}$ is divisible by $a_{k+1}(k=1,2, \cdots, 12)$. Then the value of $a_{4}$ $+a_{5}+\cdots+a_{12}$ is $\qquad$ . | II. 1.68.
Since $a_{1}+a_{2}+\cdots+a_{12}$ is divisible by $a_{13}$, then $a_{1}+a_{2}+\cdots+a_{12}+a_{13}$ is also divisible by $a_{13}$, meaning $a_{13}$ is a factor of $a_{1}+a_{2}+\cdots+a_{12}+a_{13}=13 \times 7$. Given that $a_{1}=13$, $a_{2}=1$, we have $a_{13}=7$.
Also, since $a_{1}+a_{2}$ is divisible by $a... | 68 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $a$, $b$, $c$ be positive integers, and satisfy
$$
a^{2}+b^{2}+c^{2}-a b-b c-c a=19 \text {. }
$$
Then the minimum value of $a+b+c$ is $\qquad$ | 4. 10 .
By the cyclic symmetry of $a, b, c$, without loss of generality, assume $a \geqslant b \geqslant c$.
Let $a-b=m, b-c=n$, then $a-c=m+n$, and
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[m^{2}+n^{2}+(m+n)^{2}\right],
\end{array}
$$
which is $m^{2}+m n+n^{2}-19=0$.
For the equation in ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) In a population of 1000 individuals numbered $0,1,2, \cdots, 999$, they are sequentially divided into 10 groups: Group 0 includes numbers $0,1, \cdots, 99$, Group 1 includes numbers $100,101, \cdots, 199$, and so on. Now, one number is to be selected from each group to form a sample of size 10, with ... | (1) When $x=24, a=33$, the values of $x+a k$ corresponding to $k$ taking the values $0,1, \cdots, 9$ are
$$
24,57,90,123,156,189,222,255,288,321 \text {. }
$$
Therefore, the 10 numbers of the sample drawn are
$$
24,157,290,323,456,589,622,755,888,921 \text {. }
$$
(2) Since a number with the last two digits 87 can app... | 585 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 1, label the six vertices of a regular hexagon with the numbers $0,1,2,3,4,5$ in sequence. An ant starts from vertex 0 and crawls counterclockwise, moving 1 edge on the first move, 2 edges on the second move, $\cdots \cdots$ and $2^{n-1}$ edges on the $n$-th move. After 2005 moves, the value at th... | Ni.1.1.
According to the problem, after 2005 movements, the ant has crawled over
$$
1+2+\cdots+2^{2004}=2^{2005}-1
$$
edges.
Let $x=2^{2005}-1$. Consider the remainder of $x$ modulo 6.
Since $x \equiv 1(\bmod 2), x \equiv 1(\bmod 3)$, we have
$$
3 x \equiv 3(\bmod 6), 2 x \equiv 2(\bmod 6), 5 x \equiv 5(\bmod 6) \text... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $x \in\left(0, \frac{\pi}{2}\right), \sin ^{2} x, \sin x \cdot \cos x$, and $\cos ^{2} x$ cannot form a triangle, and the length of the interval of $x$ that satisfies the condition is $\arctan k$. Then $k$ equals $\qquad$ (Note: If $x \in(a, b), b>a$, then the length of such an interval is $b-a$). | 2.2.
First, find the range of $x$ that can form a triangle.
(1) If $x=\frac{\pi}{4}$, then $\sin ^{2} \frac{\pi}{4} 、 \cos ^{2} \frac{\pi}{4} 、 \sin \frac{\pi}{4} \cdot \cos \frac{\pi}{4}$ can form a triangle.
(2) If $x \in\left(0, \frac{\pi}{4}\right)$, at this time,
$$
\sin ^{2} x\cos ^{2} x$, that is,
$$
\tan ^{2} ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Divide each side of a square into 8 equal parts, and take the division points as vertices (excluding the vertices of the square), the number of different triangles that can be formed is ( ).
(A) 1372
(B) 2024
(C) 3136
(D) 4495 | 6. C.
Solution 1: First, note that the three vertices of the triangle are not on the same side of the square. Choose any three sides of the square so that each of the three vertices lies on one of them, which can be done in 4 ways.
Next, select a point on each of the chosen three sides, which can be done in $7^{3}$ w... | 3136 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
7. The sum of the first $m$ terms of the arithmetic sequence $\left\{a_{n}\right\}$ is 90, and the sum of the first $2 m$ terms is 360. Then the sum of the first $4 m$ terms is $\qquad$ . | $=\mathbf{7 . 1 4 4 0}$.
Let $S_{k}=a_{1}+a_{2}+\cdots+a_{k}$, it is easy to know that $S_{m} 、 S_{2 m}-S_{m}$ 、 $S_{3 m}-S_{2 m}$ form an arithmetic sequence. Therefore, $S_{3 m}=810$.
It is also easy to know that $S_{2 m}-S_{m} 、 S_{3 m}-S_{2 m} 、 S_{4 m}-S_{3 m}$ form an arithmetic sequence, then $S_{4 m}=3 S_{3 m}... | 1440 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
4 y^{3}+\frac{1}{2} \sin 2 y+a=0 .
\end{array}\right.
$$
then the value of $\cos (x+2 y)$ is | 8.1.
Let $f(t)=t^{3}+\sin t$. Then $f(t)$ is monotonically increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
From the original system of equations, we get $f(x)=f(-2 y)=2 a$, and since $x$ and $-2 y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, it follows that $x=-2 y$, hence $x+2 y=0$. Therefore, $\cos (... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If the six edges of a tetrahedron are $2,3,4,5$, 6,7, then there are $\qquad$ different shapes (if two tetrahedra can be made to coincide by appropriate placement, they are considered the same shape). | 6.10.
Let the edge of length $k$ be denoted as $l_{k}(k \in\{2,3,4,5,6,7\})$. Consider $l_{2}$ and $l_{3}$.
(1) If $l_{2}$ and $l_{3}$ are coplanar, then the other side of the plane must be $l_{4}$.
(i) If $l_{2}$, $l_{3}$, and $l_{4}$ form a triangle in a clockwise direction (all referring to the direction of the thr... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9.100 Arrange chairs in a circle, with $n$ people sitting on the chairs, such that when one more person sits down, they will always be sitting next to one of the original $n$ people. The minimum value of $n$ is $\qquad$ . | 9.34.
From the problem, we know that after $n$ people sit down, there are at most two empty chairs between any two people. If we can arrange it so that there are exactly two empty chairs between every two people, then $n$ is minimized. In this case, if we number the chairs where people are sitting, we get an arithmeti... | 34 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. In $\triangle A B C$, $A B=\sqrt{30}, A C=\sqrt{6}, B C$ $=\sqrt{15}$, there is a point $D$ such that $A D$ bisects $B C$ and $\angle A D B$ is a right angle, the ratio $\frac{S_{\triangle A D B}}{S_{\triangle A B C}}$ can be written as $\frac{m}{n}$. ($m, n$ are coprime positive integers). Then $m+n=$ | 10.65.
Let the midpoint of $BC$ be $E$, and $AD=\frac{x}{2}$.
By the median formula, we get $AE=\frac{\sqrt{57}}{2}$.
By the Pythagorean theorem, we get $120-15+57=2 \sqrt{57} x$, solving for $x$ gives $x=\frac{81}{\sqrt{57}}$.
Thus, $\frac{m}{n}=\frac{AD}{2AE}=\frac{x}{2\sqrt{57}}=\frac{27}{38}$.
Therefore, $m+n=27+3... | 65 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. (16 points) Does there exist a smallest positive integer $t$, such that the inequality
$$
(n+t)^{n+t}>(1+n)^{3} n^{n} t^{t}
$$
holds for any positive integer $n$? Prove your conclusion. | $$
\text { Three, 13. Take }(t, n)=(1,1),(2,2),(3,3) \text {. }
$$
It is easy to verify that when $t=1,2,3$, none of them meet the requirement.
When $t=4$, if $n=1$, equation (1) obviously holds.
If $n \geqslant 2$, we have
$$
\begin{array}{l}
4^{4} n^{n}(n+1)^{3}=n^{n-2}(2 n)^{2}(2 n+2)^{3} \times 2^{3} \\
\leqslant\... | 4 | Inequalities | proof | Yes | Yes | cn_contest | false |
15. (22 points) Let $A$ and $B$ be the common left and right vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Let $P$ and $Q$ be moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
... | 15. (1) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}}.
$$
Similarly,
$$
k_{3}+k_{4}=-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}}.
$$
Let $O$ be the... | 8 | Geometry | proof | Yes | Yes | cn_contest | false |
3. Points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the parabola $y=2 x^{2}$ are symmetric with respect to the line $y=x+m$. If $2 x_{1} x_{2}=-1$, then the value of $2 m$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 3. A.
$$
\begin{array}{l}
\text { Given } \frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-1, \\
\frac{y_{1}+y_{2}}{2}=\frac{x_{1}+x_{2}}{2}+m, \\
2 x_{1} x_{2}=-1 \text { and } y_{1}=2 x_{1}^{2}, y_{2}=2 x_{2}^{2},
\end{array}
$$
we get
$$
\begin{array}{l}
x_{2}-x_{1}=y_{1}-y_{2}=2\left(x_{1}^{2}-x_{2}^{2}\right) \Rightarrow x_{1}+x... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
7. If $\frac{1-\cos \theta}{4+\sin ^{2} \theta}=\frac{1}{2}$, then
$$
\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=
$$ | ニ、7.9.
From the condition, we get $2-2 \cos \theta=4+\sin ^{2} \theta$, then
$$
\begin{array}{l}
(\cos \theta-1)^{2}=4 \Rightarrow \cos \theta=-1 . \\
\text { Therefore, }\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=9 .
\end{array}
$$ | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. The sequence $\left\{x_{n}\right\}: 1,3,3,3,5,5,5,5,5, \cdots$ is formed by arranging all positive odd numbers in ascending order, and each odd number $k(k=1,3,5, \cdots)$ appears consecutively $k$ times. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$ (where $[x]$ denotes the greatest inte... | 8.3.
Given $x_{k^{2}+1}=x_{k^{2}+2}=\cdots=x_{(k+1)^{2}}=2 k+1$, that is, when $k^{2}+1 \leqslant n \leqslant(k+1)^{2}$, we have $x_{n}=2 k+1(k=[\sqrt{n-1}])$.
Therefore, $x_{n}=2[\sqrt{n-1}]+1$.
Hence, $(a, b, c, d)=(2,1,-1,1), a+b+c+d=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. If each element in set $A$ can be expressed as the product of two different numbers from 1, $2, \cdots, 9$, then the maximum number of elements in set $A$ is $\qquad$. | 10.31.
From $1,2, \cdots, 9$ each time take a pair of numbers to form a product, a total of $C_{G}^{2}=$ 36 values are obtained. However, there are repetitions, the repeated cases are
$$
\begin{array}{l}
1 \times 6=2 \times 3,1 \times 8=2 \times 4,2 \times 9=3 \times 6, \\
2 \times 6=3 \times 4,3 \times 8=4 \times 6,
... | 31 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
12. Use five different colors to color the five vertices of the "pentagram" in Figure 1 (each vertex is colored with one color, and some colors may not be used), so that the two vertices on each line segment are of different colors. Then the number of different coloring methods is . | 12.1020.
Convert it into a coloring problem of a disk with 5 sectors, each to be colored with one of 5 colors, such that no two adjacent sectors have the same color (as shown in Figure 4).
Let the number of ways to color a disk with $k$ sectors using 5 colors be $x_{k}$. Then, we have $x_{k}+x_{k-1}=5 \times 4^{k-1}$... | 1020 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. The sequence $\left\{a_{n}\right\}$ satisfies: $a_{0}=1, a_{n}=\left[\sqrt{S_{n-1}}\right]$ $(n=1,2, \cdots,[x]$ represents the greatest integer not greater than $x$, $\left.S_{k}=\sum_{i=0}^{k} a_{i}(k=0,1, \cdots)\right)$. Find the value of $a_{2006}$. | 15. Observe the initial terms of the sequence (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$a_{n}$ & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & 4 & 4 \\
\hline$S_{n}$ & 1 & 2 & 3 & 4 & 6 & 8 & 10 & 13 & 16 & 20 & 24 \\
\hline \hline$n$ & 11 &... | 998 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. 5 people participate in 4 groups, each group has two people, and each person must participate in at least one group. Then, the number of different groupings is ( ).
(A) 135
(B) 130
(C) 125
(D) 120 | 10.A.
5 people can form $C_{5}^{2}=10$ pairs. Every 4 pairs are called a grouping method. A valid grouping method should include all 5 people. Since fewer than 4 people cannot form 4 pairs, all invalid grouping methods, i.e., those containing only 4 people, total $\mathrm{C}_{5}^{4} \mathrm{C}_{6}^{4}=75$ (each 4 peopl... | 135 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
11. Let $a_{n}$ be the coefficient of $x$ in the binomial expansion of $(3-\sqrt{x})^{n}(n=2,3, \cdots)$. Then the value of $\frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{18}}{a_{18}}$ is . $\qquad$ | $$
\begin{array}{l}
\text { Since } a_{n}=\mathrm{C}_{n}^{2} \cdot 3^{n-2}, \text { then, } \\
\frac{3^{n}}{a_{n}}=\frac{3^{n}}{\mathrm{C}_{n}^{2} \cdot 3^{n-2}}=3^{2} \times \frac{2}{n(n-1)}=\frac{18}{n(n-1)} . \\
\text { Therefore, } \frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{18}}{a_{18}} \\
=18\left(\fr... | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. (12 points) Find the area of the figure formed by the set of points on the right-angle coordinate plane $b O a$
$$
S=\left\{(b, a) \mid f(x)=a x^{3}+b x^{2}-3 x\right.
$$
is a monotonic function on $\mathbf{R}$, and $a \geqslant-1\}$. | Three, 15. When $a=0$, by $f(x)$ being monotonic on $\mathbf{R}$, we know $b=0$. $f(x)$ being monotonic on $\mathbf{R}$ $\Leftrightarrow f^{\prime}(x)$ does not change sign on $\mathbf{R}$.
Since $f^{\prime}(x)=3 a x^{2}+2 b x-3$, therefore, by $\Delta=4 b^{2}+36 a \leqslant 0$,
we get, $a \leqslant-\frac{1}{9} b^{2}$.... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any $n \in \mathbf{N}_{+}$, $a_{n}+a_{n+1}=2007$, and $31\left(a_{n}^{2}+a_{n} a_{n+1}+a_{n+1}^{2}\right)$. Then the sum of all different values of the sum of the first 2007 terms of the sequence $\left\{a_{n}\right\}$ is $\qquad$. | 3. 1345368366 .
From $31\left(a_{n}^{2}+a_{n} a_{n+1}+a_{n+1}^{2}\right)$, we get
$$
31\left[a_{n}^{2}+a_{n+1}\left(a_{n}+a_{n+1}\right)\right] \text {. }
$$
Also, $31\left(a_{n}+a_{n+1}\right)$, so, $31 a_{n}$.
Since $a_{1}=2007-a_{2}<2006$, we have
$$
\begin{array}{l}
a_{1} \leqslant 2004, a_{1}=3 k(k=1,2, \cdots, ... | 1345368366 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The number of prime pairs $(p, q)$ that satisfy $\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=q$ is $\qquad$ . | 5.2.
(1) When $p=2$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=1$, which is not a prime number, contradiction.
(2) When $p=3$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=1+1=2$ is a prime number.
(3) When $p=5$, $q=\left[\frac{p}{2}\right]+\left[\fr... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $f(x)$ is a function defined on $\mathbf{R}$, $f\left(\frac{\pi}{4}\right)=0$, and for any $x, y \in \mathbf{R}$, we have
$$
f(x)+f(y)=2 f\left(\frac{x+y}{2}\right) f\left(\frac{x-y}{2}\right) .
$$
Then $f\left(\frac{\pi}{4}\right)+f\left(\frac{3 \pi}{4}\right)+f\left(\frac{5 \pi}{4}\right)+\cdots+f\left... | 6.0 .
Let $\frac{x-y}{2}=\frac{\pi}{4}$, then
$$
\begin{array}{l}
f(x)+f\left(x-\frac{\pi}{2}\right)=2 f\left(x-\frac{\pi}{4}\right) f\left(\frac{\pi}{4}\right)=0 . \\
\text { Therefore } f\left(\frac{\pi}{4}\right)+f\left(\frac{3 \pi}{4}\right)+f\left(\frac{5 \pi}{4}\right)+\cdots+f\left(\frac{2007 \pi}{4}\right)=0 .... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (50 points) Find the maximum value of a prime number $p$ with the following property: there exist two permutations (which can be the same) of $1, 2, \cdots, p$, $a_{1}, a_{2}, \cdots, a_{p}$ and $b_{1}, b_{2}, \cdots, b_{p}$, such that the remainders of $a_{1} b_{1}$, $a_{2} b_{2}, \cdots, a_{p} b_{p}$ when divided... | Lemma (Wilson's Theorem): For any prime $p$, we have $(p-1)! \equiv -1 \pmod{p}$.
Proof of the Lemma: When $p=2,3$, equation (1) is obviously true.
When $p>3$, we prove: For any $2 \leqslant k \leqslant p-2$, there must exist $k^{\prime}\left(2 \leqslant k^{\prime} \leqslant p-2, k^{\prime} \neq k\right)$, such that $... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) $n$ people exchange greetings by phone during a holiday. It is known that each person makes calls to at most three friends, any two people make at most one call to each other, and among any three people, at least two of them, one person has called the other. Find the maximum value of $n$.
| Three, first prove the lemma.
Lemma In a simple graph $G$ of order $n$ without $K_{3}$, then
$$
f(G) \leqslant\left[\frac{n^{2}}{4}\right],
$$
where $f(G)$ represents the number of edges in $G$.
Proof of the lemma: Let $A$ be the vertex with the maximum degree, and let the set of vertices adjacent to $A$ be $M=\left\{... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Square $A B C D$ and square $A B E F$ are in planes that form a $120^{\circ}$ angle, $M$ and $N$ are points on the diagonals $A C$ and $B F$ respectively, and $A M=F N$. If $A B=1$, then the maximum value of $M N$ is $\qquad$ | 5.1.
As shown in Figure 2, draw $M P \perp A B$ at $P$, and connect $P N$. It can be proven that $P N \perp A B$. Thus, $\angle M P N=120^{\circ}$.
Let $A M=F N=x$.
Therefore, $\frac{M P}{1}=\frac{A M}{\sqrt{2}}$,
which means $M P=\frac{\sqrt{2}}{2} x$.
Hence, $P N=\frac{\sqrt{2}-x}{\sqrt{2}}\left(\frac{P N}{1}=\frac... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $\cos \beta+\sin \beta \cdot \cot \theta=\tan 54^{\circ}$, $\cos \beta-\sin \beta \cdot \cot \theta=\tan 18^{\circ}$.
Then the value of $\tan ^{2} \theta$ is $\qquad$ | 6.1
From the given conditions, we have
$\cos \beta=\frac{1}{2}\left(\tan 54^{\circ}+\tan 18^{\circ}\right)=\frac{1}{2 \cos 54^{\circ}}$,
$\sin \beta=\frac{\tan \theta}{2}\left(\tan 54^{\circ}-\tan 18^{\circ}\right)=\frac{\tan \theta}{2 \cos 18^{\circ}}$.
Since $\sin ^{2} \beta+\cos ^{2} \beta=1$, we have
$\frac{1}{\co... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { 2. Let } M=\frac{2 \cos 34^{\circ}-\cos 22^{\circ}}{\cos 14^{\circ}} \text {, } \\
N=\sin 56^{\circ} \cdot \sin 28^{\circ} \cdot \sin 14^{\circ} \text {. } \\
\text { Then } \frac{M}{N}=
\end{array}
$$ | 2.8.
$$
\begin{array}{l}
M=\frac{2 \cos 34^{\circ}-\sin 68^{\circ}}{\cos 14^{\circ}}=\frac{2 \cos 34^{\circ}\left(1-\sin 34^{\circ}\right)}{\cos 14^{\circ}} \\
=\frac{2 \cos 34^{\circ}\left(1-\cos 56^{\circ}\right)}{\cos 14^{\circ}}=\frac{4 \cos 34^{\circ} \cdot \sin ^{2} 28^{\circ}}{\cos 14^{\circ}} \\
=8 \cos 34^{\ci... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. From the 6 face diagonals of the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, two tetrahedra $A-B_{1} C D_{1}$ and $A_{1}-B C_{1} D$ can be formed. If the combined volume of the two tetrahedra is 1 (overlapping parts are counted only once), then the volume of the rectangular prism is $\qquad$ . | 3.2.
Let the length, width, and height of the rectangular prism be $a$, $b$, and $c$ respectively, then the volume of the rectangular prism $V_{0}=a b c$. The volumes of the two tetrahedra (as shown in Figure 4) are both
$$
\begin{array}{l}
V_{1}=V_{0}-V_{A_{1}-A B D}-V_{A_{1}-B B C_{1}}-V_{A_{1}-D_{1} D C_{1}}-V_{C_{... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Arrange all positive divisors of 8128 in ascending order as $a_{1}, a_{2}, \cdots, a_{n}$, then $\sum_{k=1}^{n} k a_{k}=$ $\qquad$ . | 5.211335 .
Given $8128=2^{6}\left(2^{7}-1\right)$, we have
$$
\begin{array}{l}
a_{n}=\left\{\begin{array}{ll}
2^{n-1}, & 1 \leqslant n \leqslant 7 ; \\
2^{n-8}\left(2^{7}-1\right), & 8 \leqslant n \leqslant 14 .
\end{array}\right. \\
\text { Then } \sum_{k=1}^{n} k a_{k}=\sum_{k=1}^{14} k \cdot 2^{k-1}-\sum_{k=1}^{7}(... | 211335 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Among the 8 vertices, 6 centers of the faces (bottom faces), and the center of the body, a total of 15 points, if a plane formed by any 3 different and non-collinear points is perpendicular to a line formed by another 2 different points, then these 5 points are called an "orthogonal 5-point group". Therefore, the to... | 6.D.
As shown in Figure 6, let $O$ be the center of the body, and $O_{i} (i = 1,2, \cdots, 6)$ be the centers of the respective faces.
Classify and discuss according to the plane formed by the 3 points (referred to as the "3-point plane").
Observation shows that there are only 4 possible 3-point planes: side faces (... | 1824 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
3. Among all the eight-digit numbers formed by the digits $1,2, \cdots, 8$ without repetition, the number of those divisible by 11 is. $\qquad$ | 3.4608 .
Since there are 4 odd numbers in $1,2, \cdots, 8$, the algebraic sum after arbitrarily adding “+” and “-” signs is always even. Because the difference between the sum of the largest four numbers and the sum of the smallest four numbers in $1,2, \cdots, 8$ is no more than 16, for each eight-digit number that m... | 4608 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
II. (50 points) Define a "Hope Set" (Hope Set) abbreviated as HS as follows: HS is a non-empty set that satisfies the condition "if $x \in \mathrm{HS}$, then $2 x \notin \mathrm{HS}$". How many "Hope Subsets" are there in the set $\{1,2, \cdots, 30\}$? Please explain your reasoning. | Sure, here is the translation:
```
II. Below, “ $a \longmapsto 2 a$ ” represents the doubling relationship between $a$ and $2 a$. Notice:
$\left\{\begin{array}{ll}\text { (1) } & 15 \longmapsto 30, \\ \text { (2) } & 13 \longmapsto 26, \\ \text { (3) } & 11 \longmapsto 22, \\ \text { (4) } & 9 \longmapsto 18,\end{arra... | 26956799 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the sum of the squares of the first 101 positive integers starting from a positive integer $k$ be equal to the sum of the squares of the next 100 positive integers. Then the value of $k$ is $\qquad$ . | 2.20100.
From the problem, we get
$$
k^{2}+\sum_{i=1}^{100}(k+i)^{2}=\sum_{i=1}^{100}(k+100+i)^{2} \text {. }
$$
Then $k^{2}=\sum_{i=1}^{100}\left[(k+100+i)^{2}-(k+i)^{2}\right]$
$$
=200 \sum_{i=1}^{100}(k+50+i)=10000(2 k+201) \text {. }
$$
Thus, $k^{2}-20000 k-2010000=0$.
Taking the positive root, we get $k=20100$. | 20100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. For any real numbers $a, b$, the inequality
$$
\max \{|a+b|,|a-b|,|2006-b|\} \geqslant c
$$
always holds, then the maximum value of the constant $c$ is $\qquad$ (where, $\max \{x, y, z\}$ denotes the maximum of $x, y, z$). | 5.1003 .
Let $\max \{|a+b|,|a-b|,|2006-b|\}=M$, then $M \geqslant|a+b|, M \geqslant|a-b|, M \geqslant|2006-b|$.
Therefore, $4 M \geqslant|a+b|+|b-a|+2|2006-b|$
$$
\geqslant|(a+b)+(b-a)+2(2006-b)|=4012,
$$
which means $M \geqslant 1003$.
When $a=0, b=1003$, $M=1003$.
Thus, the minimum value of $M$ is 1003.
Therefore, ... | 1003 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
10. (14 points) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=1$, and for $n \geqslant 2$,
$a_{n}=\left\{\begin{array}{ll}a_{\frac{n}{2}}+1, & \text { when } n \text { is even; } \\ \frac{1}{a_{n-1}}, & \text { when } n \text { is odd. }\end{array}\right.$
It is known that $a_{n}=\frac{30}{19}$. Fin... | 10. From the given conditions, it is easy to know that $a_{n}>0(n=1,2, \cdots)$.
From $a_{1}=1$, we know that:
when $n$ is even, $a_{n}>1$;
when $n(n>1)$ is odd, $a_{n}=\frac{1}{a_{n-1}}1$, so $n$ is even.
Thus, $a_{\frac{n}{2}}=\frac{30}{19}-1=\frac{11}{19}1, \frac{n}{2}-1$ is even;
$a_{\frac{n-2}{4}}=\frac{19}{11}-1... | 238 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 There is a sequence of numbers, the 1st number is 105, the 2nd number is 85, starting from the 3rd number, each number is the average of the two numbers before it. What is the integer part of the 19th number? | Notice that "the average of any two numbers $a$ and $b$, $\frac{a+b}{2}$, corresponds to the midpoint of the segment between the points representing $a$ and $b$ on the number line," we can consider placing these 19 numbers on the number line, as shown in Figure 2.
From the property mentioned above, starting from the 6... | 91 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $x=\frac{1}{2-\sqrt{5}}$. Then $x^{3}+3 x^{2}-5 x+1=$ | From $x=\frac{1}{2-\sqrt{5}}=-2-\sqrt{5}$, we get $x+2=-\sqrt{5}$. Therefore, $(x+2)^{2}=5$, which means $x^{2}+4 x=1$.
Then $x^{3}+3 x^{2}-5 x+1=\left(x^{2}+4 x-1\right)(x-1)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The quadratic equation in $x$
$$
6 x^{2}-(2 m-1) x-(m+1)=0
$$
has a root $\alpha$, given that $\alpha$ satisfies $|\alpha| \leqslant 2000$, and makes $\frac{3}{5} \alpha$ an integer. Then the number of possible values for $m$ is $\qquad$. | 3.2 401 .
From the quadratic formula, we get
$$
x_{1,2}=\frac{(2 m-1) \pm(2 m+5)}{12},
$$
which gives the roots as $-\frac{1}{2}$ and $\frac{m+1}{3}$.
If $\alpha=-\frac{1}{2}$, then $\frac{3}{5} \alpha=-\frac{3}{10}$ is not an integer. Therefore,
$$
\alpha=\frac{m+1}{3} \text {. }
$$
Given $|\alpha| \leqslant 2000$,... | 2401 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) Sustainable development has become a common concern for human development, and China has made it a basic national policy. To protect the environment and maintain ecological balance, a policy of zoned grazing has been implemented in the northern pastoral areas of China. Mr. Li, a herdsman, has a grasslan... | Let the amount of grass consumed by each cow and each sheep per day be $x$ and $y$, respectively, the amount of grass that grows on the pasture each day be $z$, and the original amount of grass on the pasture be $A$. According to the problem, we have
$$
\left\{\begin{array}{l}
200 \times 10 x + 200 \times 10 y = 10 z +... | 83100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $a, b, x, y$ satisfy
$$
\begin{array}{l}
a x+b y=3, a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, a x^{4}+b y^{4}=42 .
\end{array}
$$
Then $a x^{5}+b y^{5}=$ $\qquad$ | 4.20 .
$$
\begin{array}{l}
\text { Given } a x^{3}+b y^{3}=16 \\
\Rightarrow\left(a x^{3}+b y^{3}\right)(x+y)=16(x+y) \\
\Rightarrow\left(a x^{4}+b y^{4}\right)+x y\left(a x^{2}+b y^{2}\right)=16(x+y) \\
\Rightarrow 42+7 x y=16(x+y), \\
a x^{2}+b y^{2}=7 \\
\Rightarrow\left(a x^{2}+b y^{2}\right)(x+y)=7(x+y) \\
\Righta... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let for any natural numbers $m$, $n$ satisfying $\frac{m}{n}<\sqrt{7}$, the inequality $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\lambda}{n^{2}}$ always holds. Then the maximum value of $\lambda$ is $\qquad$. | 5.3.
The original inequality $\Leftrightarrow 7 n^{2}-m^{2} \geqslant \lambda$.
Since $7 n^{2} \equiv 0(\bmod 7), m^{2} \equiv 0,1,2,4(\bmod 7)$, then $\left(7 n^{2}-m^{2}\right)_{\text {min }}=3$, so, $\lambda_{\text {max }}=3$. | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) Given the semi-ellipse $\frac{x^{2}}{4}+y^{2}=1(y>0)$, two perpendicular lines are drawn through a fixed point $C(1,0)$ intersecting the ellipse at points $P$ and $Q$, respectively. Here, $O$ is the origin, and $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse.
(1) Find the minimum value ... | (1) Since $P O$ is the median of $\triangle P F_{1} F_{2}$, we have
$$
P F_{1}+P F_{2}=2 P O, \left|P F_{1}+P F_{2}\right|=2|P O| \text {. }
$$
Therefore, when $P$ is at the vertex of the minor axis, $\left|P F_{1}+P F_{2}\right|$ achieves its minimum value of 2.
(2) From the problem analysis, the slopes of the lines ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
204 Given that $a, b, c$ are positive numbers satisfying $a+b+c=1$. Prove:
$$
\frac{a^{3}+b}{a(b+c)}+\frac{b^{3}+c}{b(c+a)}+\frac{c^{3}+a}{c(a+b)} \geqslant 5 .
$$ | Prove: $\frac{a^{3}+b}{a(b+c)}=\frac{a^{2}(1-b-c)+b}{a(b+c)}$
$$
\begin{array}{l}
=\frac{a^{2}+b}{a(b+c)}-a=\frac{a(1-b-c)+b}{a(b+c)}-a \\
=\frac{a+b}{a(b+c)}-1-a .
\end{array}
$$
Similarly, $\frac{b^{3}+c}{b(c+a)}=\frac{b+c}{b(c+a)}-1-b$,
$$
\frac{c^{3}+a}{c(a+b)}=\frac{c+a}{c(a+b)}-1-c \text {. }
$$
Adding the abov... | 5 | Inequalities | proof | Yes | Yes | cn_contest | false |
Example 3 For all $a, b, c \in \mathbf{R}_{+}$, find
$$
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}}
$$
the minimum value. | Explanation: Make the substitution
$$
\begin{array}{l}
x=\frac{a}{\sqrt{a^{2}+8 b c}}, y=\frac{b}{\sqrt{b^{2}+8 a c}}, \\
z=\frac{c}{\sqrt{c^{2}+8 a b}} .
\end{array}
$$
Then $x, y, z \in (0, +\infty)$.
Thus, $x^{2}=\frac{a^{2}}{a^{2}+8 b c}$, which means $\frac{1}{x^{2}}-1=\frac{8 b c}{a^{2}}$.
Similarly, $\frac{1}{y... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Find
$$
\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}
$$
the minimum value. | Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}(x, y, z \in \mathbf{R}_{+})$, then
$$
\begin{array}{l}
\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1} \\
=\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{[y(y+2 x)+z(z+2 y)+x(x+2 z)] \cdot}... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Given $x>y>0$, and $x y=1$. Find the minimum value of $\frac{3 x^{3}+125 y^{3}}{x-y}$. | Let the minimum value (target) be a positive number $t$, then
$$
\frac{3 x^{3}+125 y^{3}}{x-y} \geqslant t \text {, }
$$
which means $3 x^{3}+125 y^{3}+y t \geqslant x t$.
By $x y=1$ and the AM-GM inequality, we have
$3 x^{3}+125 y^{3}+y t$
$=x^{3}+x^{3}+x^{3}+125 y^{3}+y t$
$\geqslant 5 \sqrt[5]{x^{9} \cdot 125 y^{4}... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a, b, c \in \mathbf{R}_{+}$, and $a+b+c=1$. Find
$$
\frac{3 a^{2}-a}{1+a^{2}}+\frac{3 b^{2}-b}{1+b^{2}}+\frac{3 c^{2}-c}{1+c^{2}}
$$
the minimum value. | ( Hint: First prove $\frac{3 a^{2}-a}{1+a^{2}} \geqslant \frac{9}{10}\left(a-\frac{1}{3}\right)$. Similarly, obtain the other two inequalities. Then add the three inequalities, to get the minimum value of 0. ) | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 The inequality $x^{2}+|2 x-4| \geqslant p$ holds for all real numbers $x$. Find the maximum value of the real number $p$.
保留了原文的换行和格式，如上所示。 | Explanation 1: Transform the original inequality into
$|x-2| \geqslant-\frac{x^{2}}{2}+\frac{p}{2}$.
Let $y_{1}=|x-2|, y_{2}=-\frac{x^{2}}{2}+\frac{p}{2}$.
Thus, solving inequality (1) is equivalent to determining the range of $x$ values for which the graph of the function $y_{1}=|x-2|$ is above the graph of the functi... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Color each vertex of a square pyramid with one color, and make the endpoints of the same edge different colors. If only 5 colors are available, then the total number of different coloring methods is $\qquad$ | Solution: According to the problem, for the quadrilateral pyramid $S-ABCD$, the vertices $S$, $A$, and $B$ are colored with different colors, which gives a total of $5 \times 4 \times 3=60$ coloring methods.
When $S$, $A$, and $B$ are already colored, let's assume their colors are $1$, $2$, and $3$ respectively. If $C... | 420 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Algebraic expression
$$
\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}+9}
$$
The minimum value is $\qquad$ | Notice:
$$
\begin{array}{l}
\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}-9} \\
=\sqrt{(0-x)^{2}+[(-2)-0]^{2}}+ \\
\sqrt{(12-x)^{2}+(3-0)^{2}} .
\end{array}
$$
Establish a rectangular coordinate system as shown in Figure 8.
Let $A(0,-2)$, $B(12,3)$, and $P(x, 0)$. Then,
$$
\begin{array}{l}
|P A|=\sqrt{x^{2}+4}, \\
|P B|=\sqrt{(12-x... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. As shown in Figure 1, in the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $CA=4$, $P$ is the midpoint of the semicircular arc $\overparen{AC}$, connect $BP$, the line segment $BP$ divides the figure $APCB$ into two parts. The absolute value of the difference in the areas of these two parts is $\qquad$. | Ni.6.4.
As shown in Figure 8, let $A C$ and $B P$ intersect at point $D$, and the point symmetric to $D$ with respect to the circle center $O$ is denoted as $E$. The line segment $B P$ divides the figure $A P C B$ into two parts, and the absolute value of the difference in the areas of these two parts is the area of $\... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12.B. The real numbers $a, b, c$ satisfy $a \leqslant b \leqslant c$, and $ab + bc + ca = 0, abc = 1$. Find the largest real number $k$ such that the inequality $|a+b| \geqslant k|c|$ always holds. | 12. B. When $a=b=-\sqrt[3]{2}, c=\frac{\sqrt[3]{2}}{2}$, the real numbers $a, b, c$ satisfy the given conditions, at this time, $k \leqslant 4$.
Below is the proof: The inequality $|a+b| \geqslant 4|c|$ holds for all real numbers $a, b, c$ that satisfy the given conditions.
From the given conditions, we know that $a, ... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11.B. Given the parabolas $C_{1}: y=-x^{2}-3 x+4$ and $C_{2}: y=x^{2}-3 x-4$ intersect at points $A$ and $B$, point $P$ is on parabola $C_{1}$ and lies between points $A$ and $B$; point $Q$ is on parabola $C_{2}$, also lying between points $A$ and $B$.
(1) Find the length of segment $A B$;
(2) When $P Q \parallel y$-ax... | 11. B. (1) Solve the system of equations $\left\{\begin{array}{l}y=-x^{2}-3 x+4, \\ y=x^{2}-3 x-4,\end{array}\right.$ to get
$$
\left\{\begin{array} { l }
{ x _ { 1 } = - 2 , } \\
{ y _ { 1 } = 6 , }
\end{array} \left\{\begin{array}{l}
x_{2}=2, \\
y_{2}=-6 .
\end{array}\right.\right.
$$
Therefore, $A(-2,6)$ and $B(2,... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $a b=1$, and $\frac{1}{1-2^{x} a}+\frac{1}{1-2^{y+1} b}=1$, then the value of $x+y$ is $\qquad$. | 8. -1 .
Given that both sides of the equation are multiplied by $\left(1-2^{x} a\right)\left(1-2^{y+1} b\right)$, we get
$$
2-2^{x} a-2^{y+1} b=1-2^{x} a-2^{y+1} b+2^{x+y+1} a b \text {, }
$$
which simplifies to $1=2 \times 2^{x+y}$.
Therefore, $x+y=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Let the sum of 10 consecutive positive integers not greater than 2006 form the set $S$, and the sum of 11 consecutive positive integers not greater than 2006 form the set $T$, then the number of elements in $S \cap T$ is $\qquad$ . | 11.181.
$S$ is the set of all integers ending in 5, starting from 55 to
$$
20015=\sum_{i=1}^{10}(1996+i)=10 \times 1996+55
$$
Similarly, $T$ is the set of integers starting from 66 and increasing by 11 each time, with the largest number being
$$
22011=\sum_{i=1}^{11}(1995+i)=11 \times 1995+66 \text {. }
$$
In $T$, one... | 181 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
15. Let $a>1, b>1$. Prove:
$$
\frac{a^{4}}{(b-1)^{2}}+\frac{b^{4}}{(a-1)^{2}} \geqslant 32 .
$$ | 15. Let $\left\{\begin{array}{l}x=a-1, \\ y=b-1,\end{array}\right.$ then $\left\{\begin{array}{l}a=x+1, \\ b=y+1 .\end{array}\right.$
The left side of the inequality
$$
\geqslant \frac{(2 \sqrt{x})^{4}}{y^{2}}+\frac{(2 \sqrt{y})^{4}}{x^{2}}=16\left(\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}\right) \geqslant 32 \text {. }... | 32 | Inequalities | proof | Yes | Yes | cn_contest | false |
10. As shown in Figure 2, in Pascal's Triangle, the numbers above the diagonal form the sequence: $1,3,6,10, \cdots$, let the sum of the first $n$ terms of this sequence be $S_{n}$. Then, as $n \rightarrow +\infty$, the limit of $\frac{n^{3}}{S(n)}$ is $\qquad$ | 10.6 .
It is known that $S_{n}=\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\cdots+\mathrm{C}_{n+1}^{2}=\mathrm{C}_{\mathrm{n}+2}^{3}$. Therefore, $\lim _{n \rightarrow+\infty} \frac{n^{3}}{S(n)}=\lim _{n \rightarrow+\infty} \frac{n^{3} \times 6}{(n+2)(n+1) n}=6$. | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. The coefficient of $x^{150}$ in the expansion of $\left(1+x+x^{2}+\cdots+x^{100}\right)^{3}$, after combining like terms, is $\qquad$ | 5.7651 .
By the polynomial multiplication rule, the problem can be transformed into finding the number of natural number solutions of the equation
$$
s+t+r=150
$$
that do not exceed 100.
Obviously, the number of natural number solutions of equation (1) is $\mathrm{C}_{152}^{2}$.
Next, we find the number of natural nu... | 7651 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Given sets $A_{1}, A_{2}, \cdots, A_{n}$ are different subsets of the set $\{1,2, \cdots, n\}$, satisfying the following conditions:
(i) $i \notin A_{i}$ and $\operatorname{Card}\left(A_{i}\right) \geqslant 3, i=1,2, \cdots, n$;
(ii) $i \in A_{j}$ if and only if $j \notin A_{i}(i \neq j$, $i, j=1,2, ... | (1) Let $A_{1}, A_{2}, \cdots, A_{n}$ be $n$ sets, and let $r_{i}$ be the number of sets that contain element $i$ for $i=1,2, \cdots, n$. Then,
$$
\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)=\sum_{i=1}^{n} r_{i} \text {. }
$$
Assume the $r_{1}$ sets that contain element 1 are
$$
A_{2}, A_{3}, \cdots, A_{r_{1}... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Take any 21 points on a circle. Prove: Among all the arcs with these points as endpoints, there are no fewer than 100 arcs that do not exceed $120^{\circ}$. | (Assuming among all points, the chord with $A_{1}$ as an endpoint is the least, and denoting the chords with $A_{1}$ as an endpoint as $A_{1} A_{2}, A_{1} A_{3}, \cdots$, $A_{1} A_{n}$, a total of $n-1$ chords, and the chords with $A_{2}, A_{3}, \cdots, A_{n}$ as endpoints are no less than $n-1$ each. Therefore, these ... | 100 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 1 Let $S$ be a subset of the set $\{1,2, \cdots, 50\}$ with the following property: the sum of any two distinct elements of $S$ cannot be divisible by 7. Then, what is the maximum number of elements that $S$ can have?
(43rd American High School Mathematics Examination) | For two different natural numbers $a$ and $b$, if $7 \times (a+b)$, then the sum of their remainders when divided by 7 is not 0. Therefore, the set $\{1,2, \cdots, 50\}$ can be divided into 7 subsets based on the remainders when divided by 7. Among them, each element in $K_{i}$ has a remainder of $i$ when divided by 7 ... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let $X=\{1,2, \cdots, 2001\}$. Find the smallest positive integer $m$ that satisfies the following condition: for any $m$-element subset $W$ of $X$, there exist $u, v \in W$ (where $u$ and $v$ can be the same), such that $u+v$ is a power of 2.
(2001, China Mathematical Olympiad) | Explanation: When $u$ and $v$ are $2^{r} + a$ and $2^{r} - a$ respectively, $u + v = 2 \times 2^{r} = 2^{r+1}$ is a power of 2. Based on this, $X$ is divided into the following 5 subsets:
$$
\begin{array}{l}
2001 = 2^{10} + 977 \geqslant x \geqslant 2^{10} - 977 = 47, \\
46 = 2^{5} + 14 \geqslant x \geqslant 2^{5} - 14... | 999 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Let $M=\{1,2, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: when $x \in A$, $15 x \notin A$. Then the maximum number of elements in $A$ is $\qquad$
(1995, National High School Mathematics Competition) | Construct the subset $A$ as follows:
Notice that $1995 \div 15=133$, and let
$$
A_{1}=\{134,135, \cdots, 1995\} \text {, }
$$
then $\left|A_{1}\right|=1862$.
Also, $133=15 \times 8+13$, and let
$A_{2}=\{9,10, \cdots, 133\}$,
then $\left|A_{2}\right|=125$.
Let $A_{3}=\{1,2, \cdots, 8\}$, then $\left|A_{3}\right|=8$.
T... | 1870 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let $S=\{1,2, \cdots, 2005\}$. If any set of $n$ pairwise coprime numbers in $S$ contains at least one prime number, find the minimum value of $n$.
(2005, China Western Mathematical Olympiad) | Explanation: First, take a subset of $S$
$$
A_{0}=\left\{1,2^{2}, 3^{2}, 5^{2}, \cdots, 41^{2}, 43^{2}\right\},
$$
then $\left|A_{0}\right|=15, A_{0}$ contains any two numbers that are coprime, but there are no primes in it. This indicates that $n \geqslant 16$.
Second, it can be proven: For any $A \subseteq S, n=|A|... | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Let sets $A$ and $B$ both consist of positive numbers, with $|A|=10, |B|=9$, and set $A$ satisfies the following condition: If $x, y, u, v \in A, x+y=u+v$, then \begin{aligned}\{x, y\} & =\{u, v\} . \text{ Let } \\ A & +B=\{a+b \mid a \in A, b \in B\} .\end{aligned}
Prove: $|A+B| \geqslant 50$ ($|X|$ denotes... | Explanation: Consider the general case.
Let $|A|=m,|B|=n$,
$A+B=\left\{s_{1}, s_{2}, \cdots, s_{k}\right\}$.
Suppose $s_{i}(1 \leqslant i \leqslant k)$ can be expressed in $f(i)$ ways as $a+b$, where $a \in A, b \in B$, i.e.,
$$
s_{i}=a_{i 1}+b_{i 1}=a_{i 2}+b_{i 2}=\cdots=a_{i(i)}+b_{i f(i)}.
$$
Then $f(1)+f(2)+\cdot... | 50 | Combinatorics | proof | Yes | Yes | cn_contest | false |
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