problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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2. Let $S$ be a subset of $\{1,2, \cdots, 9\}$ such that the sum of any two distinct elements of $S$ is unique. How many elements can $S$ have at most?
(2002, Canadian Mathematical Olympiad) | (When $S=\{1,2,3,5,8\}$, $S$ meets the requirements of the problem. If $T \subseteq\{1,2, \cdots, 9\},|T| \geqslant 6$, then since the sum of any two different numbers in $T$ is between 3 and 17, at most 15 different sum numbers can be formed. And choosing any two numbers from $T$, there are at least $\mathrm{C}_{6}^{2... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $x=\frac{1}{\sqrt{2}-1}, a$ be the fractional part of $x$, and $b$ be the fractional part of $-x$. Then $a^{3}+b^{3}+3 a b=$ $\qquad$ . | ニ、1. 1 .
Since $x=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$, and $2<\sqrt{2}+1<3$, therefore,
$$
\begin{array}{l}
a=x-2=\sqrt{2}-1 . \\
\text { Also }-x=-\sqrt{2}-1, \text { and }-3<-\sqrt{2}-1<-2 \text {, so, } \\
b=-x-(-3)=2-\sqrt{2} .
\end{array}
$$
Then $a+b=1$.
Thus $a^{3}+b^{3}+3 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+3 a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given a right trapezoid $A B C D$ with side lengths $A B=2, B C=C D=10, A D=6$, a circle is drawn through points $B$ and $D$, intersecting the extension of $B A$ at point $E$ and the extension of $C B$ at point $F$. Then the value of $B E-B F$ is $\qquad$ | 3.4 .
As shown in Figure 3, extend $C D$ to intersect $\odot O$ at point $G$. Let the midpoints of $B E$ and $D G$ be $M$ and $N$, respectively. It is easy to see that $A M = D N$. Since $B C = C D = 10$, by the secant theorem, it is easy to prove that
$$
\begin{aligned}
B F = & D G. \text{ Therefore, } \\
& B E - B F... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let positive integers $a, b, k$ satisfy $\frac{a^{2}+b^{2}}{a b-1}=k$. Prove: $k=5$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | For every $k$ that satisfies the above equation, let $a_{0}, b_{0}$ satisfy $\frac{a_{0}^{2}+b_{0}^{2}}{a_{0} b_{0}-1}=k$, and $a_{0}+b_{0}$ is the smallest pair. Without loss of generality, assume $a_{0} \geqslant b_{0}$.
(1) If $a_{0}=b_{0}$, then
$$
k=\frac{a_{0}^{2}+b_{0}^{2}}{a_{0} b_{0}-1}=2+\frac{2}{a_{0}^{2}-1}... | 5 | Number Theory | proof | Yes | Yes | cn_contest | false |
4. If $100a+64$ and $201a+64$ are both four-digit numbers, and both are perfect squares, then the value of the integer $a$ is | 4.17.
Let $100 a+64=m^{2}, 201 a+64=n^{2}$, then
$$
32 \leqslant m, n<100 \text {. }
$$
Subtracting the two equations gives
$$
101 a=n^{2}-m^{2}=(n+m)(n-m) \text {. }
$$
Since 101 is a prime number, and $-101<n-m<101, 0<n$ $+m<200$, so, $n+m=101$.
Thus, $a=n-m=2 n-101$.
Substituting $a=2 n-101$ into $201 a+64=n^{2}$... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Let $a$ be a positive integer, the quadratic function $y=x^{2}+(a+17) x+38-a$, and the reciprocal function $y=\frac{56}{x}$. If the intersection points of the two functions are all integer points (points with both coordinates as integers), find the value of $a$.
---
The above text has been translat... | Three, by eliminating $y$ from the two equations, we get
$$
x^{2}+(a+17) x+38-a=\frac{56}{x} \text {, }
$$
which simplifies to $x^{3}+(a+17) x^{2}+(38-a) x-56=0$.
Factoring, we get
$$
(x-1)\left[x^{2}+(a+18) x+56\right]=0 \text {. }
$$
Clearly, $x_{1}=1$ is a root of equation (1), and $(1,56)$ is one of the intersect... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given that for all real numbers $x$, we have
$$
|x+1|+\sqrt{x-1} \geqslant m-|x-2|
$$
always holds. Then the maximum value that $m$ can take is $\qquad$ . | When $-1 \leqslant x \leqslant 2$, the minimum value of $|x+1|+|x-2|$ is
3. Since $\sqrt{x-1} \geqslant 0$, when $x=1$,
$$
|x+1|+\sqrt{x-1}+|x-2|
$$
the minimum value is 3, so, $3 \geqslant m$, hence the maximum value of $m$ is 3. | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
13. The military training base purchases apples to comfort the trainees. It is known that the total number of apples, when represented in octal (base 8), is $\overline{a b c}$, and when represented in septenary (base 7), it is $\overline{c b a}$. Therefore, the total number of apples, when represented in decimal (base ... | 13.220.
$$
\begin{array}{l}
\text { Given } 1 \leqslant a, b, c \leqslant 6, \\
a \times 8^{2}+b \times 8+c \\
=c \times 7^{2}+b \times 7+a, \\
63 a+b-48 c=0, \\
b=3(16 c-21 a),
\end{array}
$$
Therefore, $b=0,3,6$.
Upon verification, $b=3$ meets the condition. Hence, $b=3, c=4, a=3$. Thus, $3 \times 8^{2}+3 \times 8+4... | 220 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $x$ and $y$ are real numbers, and $x^{2}+x y+y^{2}=3$. Let the maximum and minimum values of $x^{2}-x y+y^{2}$ be $m$ and $n$, respectively. Then the value of $m+n$ is $\qquad$ | $=, 1.10$.
Given the equation is symmetric about $x, y$, let $x=a+b, y=a-b$, then $3=x^{2}+xy+y^{2}=3a^{2}+b^{2}$.
Therefore, $3a^{2}=3-b^{2}$.
So, $0 \leqslant b^{2} \leqslant 3$.
$$
\begin{array}{l}
\text { Also, } x^{2}-xy+y^{2}=a^{2}+3b^{2}=\frac{1}{3}\left(3a^{2}+b^{2}\right)+\frac{8}{3}b^{2} \\
=\frac{1}{3} \time... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 3, on a $2 \times 3$ rectangular grid paper, the vertices of each small square are called grid points. Then the number of isosceles right triangles with grid points as vertices is ( ).
(A) 24
(B) 38
(C) 46
(D) 50
This problem can be solved using the labeling method, making the solution clea... | Solution: As shown in Figure 4, at each grid point, mark the number of isosceles right triangles with it as the right-angle vertex. For example, $A_{2+0}$ indicates that there are 2 upright and 0 slanted isosceles right triangles with $A$ as the right-angle vertex; $B_{3+2}$ indicates that there are 3 upright and 2 sla... | 50 | Geometry | MCQ | Yes | Yes | cn_contest | false |
4. Let $a$ and $b$ be integers, and one root of the equation $x^{2} + a x + b = 0$ is $\sqrt{4 - 2 \sqrt{3}}$. Then the value of $a + b$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2 | 4.B.
Notice that $\sqrt{4-2 \sqrt{3}}=\sqrt{3}-1$.
According to the problem, we have $(\sqrt{3}-1)^{2}+a(\sqrt{3}-1)+b=0$, which simplifies to $(a-2) \sqrt{3}+4-a+b=0$.
Thus, $a-2=0$ and $4-a+b=0$.
Solving these, we get $a=2, b=-2$. Therefore, $a+b=0$. | 0 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. As shown in Figure 1, in the right trapezoid $A B C D$, $A B=B C=4$, $M$ is a point on the leg $B C$, and $\triangle A D M$ is an equilateral triangle. Then $S_{\triangle C D M}$ : $S_{\triangle A B M}=$ $\qquad$ . | 2.2.
As shown in Figure 6, draw $A E \perp$ $C D$ intersecting the extension of $C D$ at point $E$, then quadrilateral $A B C E$ is a square.
It is easy to prove
$\mathrm{Rt} \triangle A B M \cong \mathrm{Rt} \triangle A E D$.
Therefore, $B M=D E$.
Thus, $C M=C D$.
Let this value be $x$, then
$$
\begin{array}{l}
x^{2}... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. The quality requirements of a product are divided into four different levels from low to high, labeled as $1,2,3,4$. If the working hours remain unchanged, the workshop can produce 40 units of the lowest level (i.e., level 1) product per day, with a profit of 16 yuan per unit; if the level is increased by one, the p... | 3.3.
Let the profit obtained from producing products of the $x$-th grade in the workshop be $y$. According to the problem, we have
$$
\begin{array}{l}
y=[40-2(x-1)][16+(x-1)] \\
=-2 x^{2}+12 x+630=-2(x-3)^{2}+648 .
\end{array}
$$
Therefore, when $x=3$, the profit $y$ is maximized, at 648 yuan. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. The number of all different integer solutions to the equation $2 x^{2}+5 x y+2 y^{2}=2007$ is $\qquad$ groups.
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 note above is not part of the translation but is provided to clarify the instruction. The actual translation is above this note. | 4.4 .
Let's first assume $x \geqslant y$, the original equation can be transformed into $(2 x+y)(x+2 y)=2007$.
Since $2007=2007 \times 1=669 \times 3=223 \times 9$
$$
\begin{array}{l}
=(-1) \times 2007=(-3) \times(-669) \\
=(-9) \times(-223),
\end{array}
$$
Therefore, $\left\{\begin{array}{l}2 x+y=2007, \\ x+2 y=1 .\... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
(1) Find $f(2006)$;
(2) If $f(n) \geqslant 30$, find the minimum value of $n$.
For a positive integer $n$, let $f(n)$ be the sum of the digits of $n^2$. | (1) Since $2006^{2}=4024036$, we have
$$
f(2006)=19.
$$
(2) Since $f(n) \geqslant 30$, then $n^{2} \geqslant 3999$.
Thus, $n>63$.
$$
\begin{array}{l}
\text{And } f(64)=19, f(65)=13, f(66)=18, \\
f(67)=25, f(68)=16, f(69)=18, \\
f(70)=13, f(71)=10, f(72)=18, \\
f(73)=19, f(74)=22, f(75)=18, \\
f(76)=25, f(77)=25, f(78)=... | 83 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) How many real roots does the equation $\sqrt{x} \sin x^{2}-2=0$ have in the interval $[0,20]$? | Three, Solution 1: Let $y=\sqrt{x} \sin x^{2}$.
The solutions to the equation $y=0$ are $x=\sqrt{k \pi} \in[0,20], k \in \mathbf{Z}$. Therefore, $x=\sqrt{k \pi}(k=0,1, \cdots, 127)$.
When $\sqrt{x} \geqslant 2$, the graph of $y=\sqrt{x} \sin x^{2}$ may intersect with the graph of $y=2$, so $x \geqslant 4$.
Also, $\sqr... | 122 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the set of complex numbers $D$, a complex number $z \in D$ if and only if there exists a complex number $z_{1}$ with modulus 1, such that
$$
|z-2005-2006 \mathrm{i}|=\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \text {. }
$$
Then the number of complex numbers in $D$ whose real and imaginary parts are both integers is... | 8.49 .
Given $\left|z_{1}\right|=1$, let $z_{1}=\cos \theta+\mathrm{i} \sin \theta$.
$$
\begin{array}{l}
\text { Then }\left|z_{1}^{4}+1-2 z_{1}^{2}\right|=\left|\left(z_{1}^{2}-1\right)^{2}\right|=\left|z_{1}^{2}-1\right|^{2} \\
=|(\cos 2 \theta-1)+\mathrm{i} \sin 2 \theta|^{2} \\
=(\cos 2 \theta-1)^{2}+\sin ^{2} 2 \... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. For any 4 vertices $A, B, C, D$ of a cube that do not lie on the same plane, the number of cosine values of the dihedral angle $A-BC-D$ that are less than $\frac{1}{2}$ is
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 9.4 .
Among the cosines of the dihedral angles with the body diagonal as the edge,
$$
\begin{array}{l}
\cos \angle A_{1} O_{1} C_{1} \\
=\cos 120^{\circ}=-\frac{1}{2} .
\end{array}
$$
Among the cosines of the dihedral angles with the edge as the edge, only 0 is less than $\frac{1}{2}$.
Among the cosines of the dihedr... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $x_{1}, x_{2}, \cdots, x_{n}$ be numbers that can take one of the values $-3, 0, 1$, and
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=279, \\
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=-585 .
\end{array}
$$
Then the value of $x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}$ is ( ).
(A) 2005
(B) 2006
(C) 2007
(D) 200... | - 1.C.
Let $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ numbers, where $-3$ appears $b$ times, $1$ appears $b$ times, and $0$ appears $c$ times. Then
$$
\left\{\begin{array} { l }
{ 9 a + b = 2 7 9 , } \\
{ - 2 7 a + b = - 5 8 5 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=24, \\
b=63 .
\end{array}\right.\right.
$$
T... | 2007 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. Divide the sides of the equilateral $\triangle A B C$ into four equal parts, as shown in Figure 1. Then the number of pairs of congruent equilateral triangles in the figure is ( ).
(A) 100
(B) 121
(C) 144
(D) 169 | 3. C.
Let the side length of the equilateral triangle be 4. From Figure 1, we can see:
(1) There are 16 equilateral triangles with a side length of 1, all of which are congruent, and the number of pairs is $15+14+\cdots+1=120$;
(2) There are 7 equilateral triangles with a side length of 2, all of which are congruent, ... | 144 | Geometry | MCQ | Yes | Yes | cn_contest | false |
Three, (25 points) Given that $m$ and $n$ ($m>n$) are positive integers, and the last three digits of $3^{m}$ and $3^{n}$ are the same. Find the minimum value of $m-n$.
保留源文本的换行和格式，翻译结果如下：
```
Three, (25 points) Given that $m$ and $n$ ($m>n$) are positive integers, and the last three digits of $3^{m}$ and $3^{n}$ are... | Three, from the given information, $3^{m}-3^{n}$ is a multiple of 1000, i.e.,
$$
3^{m}-3^{n}=3^{n}\left(3^{m-n}-1\right)
$$
is a multiple of 1000.
Also, $\left(3^{n}, 1000\right)=1$, hence $3^{m-n}-1$ is a multiple of 1000.
Let $s=m-n$, then $3^{s}-1$ is a multiple of 1000.
We only need to find the smallest $s$ such t... | 100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a, b, c$ be three distinct positive integers. If
$$
\begin{array}{l}
\{a+b, b+c, c+a\} \\
=\left\{n^{2},(n+1)^{2},(n+2)^{2}\right\}\left(n \in \mathbf{N}_{+}\right),
\end{array}
$$
then the minimum value of $a^{2}+b^{2}+c^{2}$ is ( ).
(A) 2007
(B) 1949
(C) 1297
(D) 1000 | - 1.C.
Assume $a>b>c$, then $a+b>c+a>b+c$.
Since $(a+b)+(b+c)+(c+a)=2(a+b+c)$ is an even number, $n^{2} 、(n+1)^{2} 、(n+2)^{2}$ must be two odd and one even. Therefore, $n$ must be an odd number.
Also, since $b+c>1$, $n$ must be an odd number no less than 3.
If $n=3$, then $\{a+b, b+c, c+a\}=\left\{3^{2}, 4^{2}, 5^{2}\... | 1297 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
6. Let $m, n$ be positive integers. If there exists a positive integer $k$ such that $\log _{2} m, \log _{2} n$ and $\log _{2} k$ can be the lengths of the three sides of a triangle, then $k$ is called a “good number”. It is known that there are exactly 100 good numbers $k$. Then the maximum possible value of $m n$ is ... | 6.C.
Assume $m \geqslant n$. From the problem, we have
$$
\begin{array}{l}
\log _{2} m-\log _{2} n\frac{m}{n}, m n-101 \leqslant \frac{m}{n},
\end{array}
$$
which implies $\frac{100}{1-\frac{1}{n^{2}}}<m n \leqslant \frac{101}{1-\frac{1}{n^{2}}}$.
It is easy to see that when $n=2$, $\frac{101}{1-\frac{1}{n^{2}}}$ rea... | 134 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
Example 1 Figure 1 is a shape composed of 4 unit squares, with each vertex of the unit squares called a grid point. By taking any three grid points as vertices, how many different isosceles right triangles can be formed?
Regarding this problem, the answer given in [1] is 24, [2] supplements it to 30, and [3] provides ... | Solution: Through the above classification, mark the number of isosceles right triangles with each grid point as the right-angle vertex (Figure 2). For example: $B_{3+1}$ indicates that there are 3 upright and 1 slanted isosceles right triangles with $B$ as the right-angle vertex; $E_{1+2}$ indicates that there are 1 u... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. There are 5 rooms $A, B, C, D, E$ arranged in a circular pattern, with the number of people living in them being $17, 9, 14, 16, 4$ respectively. Now, adjustments are to be made so that the number of people in each room is the same, and it is stipulated that people can only move to the adjacent left or right room. H... | (Tip: Let the number of people moved from room $A$ to room $B$ be $x_{B}$, and so on, then we have
$$
\begin{array}{l}
9+x_{B}-x_{C}=14+x_{C}-x_{D}=16+x_{D}-x_{E} \\
=4+x_{E}-x_{A}=17+x_{A}-x_{B} \\
=\frac{1}{5}(17+9+14+16+4)=12 .
\end{array}
$$
This is transformed into finding
$$
\begin{array}{l}
y=\left|x_{B}-5\righ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
14. Given that the weights of $A$, $B$, $C$, and $D$ are all integers in kilograms, where $A$ is the lightest, followed by $B$, $C$, and $D$, the weights of each pair of them are as follows (unit: kg):
$45,49,54,55,60,64$.
Then the weight of $D$ is $\qquad$ kg. | 14.35.
Since $A+B=45, A+C=49, B+D=60, C+D$ $=64$, therefore, $C-B=4$.
Then $B+C=B+(B+4)=2B+4$ is an even number.
Among $54 \text{~kg}$ and $55 \text{~kg}$, only 54 is an even number, so $B=25$. Then $D=35$. | 35 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
16. (12 points) There are $a$ matchsticks of the same length. When placed as shown in Figure 7, they can form $m$ squares; when placed as shown in Figure 8, they can form $2 n$ squares.
(1) Express $m$ in terms of $n$;
(2) When these $a$ matchsticks can also be arranged as shown in Figure 9, find the minimum value of $... | 16. (1) The total number of matchsticks in Figure 7 is $3 m+1$, and the total number of matchsticks in Figure 8 is $5 n+2$.
Since the total number of matchsticks is the same, we have $3 m+1=5 n+2$.
Solving for $m$ gives $m=\frac{5 n+1}{3}$.
(2) Suppose there are $3 p$ squares in Figure 9, then the total number of match... | 52 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. The number of triangles with integer side lengths and a perimeter of 20 is $\qquad$ . | 2.8.
Let the three sides of a triangle be $a, b, c$, and $a \geqslant b \geqslant c$, $a+b+c=20$, then $a \geqslant 7$.
Also, from $b+c>a$, we get $2 a<a+b+c=20 \Rightarrow a<10$. Therefore, $7 \leqslant a \leqslant 9$. We can list
$$
\begin{array}{l}
(a, b, c)=(9,9,2),(9,8,3),(9,7,4),(9,6,5), \\
(8,8,4),(8,7,5),(8,6,... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange the four numbers $1, 2, 3, 4$ to form a four-digit number, such that this number is a multiple of 11. Then the number of such four-digit numbers is $\qquad$.
| 4.8.
Since $1+4=2+3$, we can place 1 and 4 in the even positions, and 2 and 3 in the odd positions, which gives us four arrangements; placing 2 and 3 in the even positions, and 1 and 4 in the odd positions also gives us four arrangements. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) If a number $a$ can be expressed as the sum of the squares of two natural numbers (allowing the same), then $a$ is called a "good number". Determine how many good numbers there are among the first 200 positive integers $1,2, \cdots, 200$.
Translate the above text into English, please retain the orig... | Three, the squares not exceeding 200 are $0^{2}, 1^{2}, \cdots, 14^{2}$.
Obviously, each number $k^{2}$ in $1^{2}, 2^{2}, \cdots, 14^{2}$ can be expressed in the form of $k^{2}+$ $0^{2}$, and there are 14 such numbers.
Each pair of numbers in $1^{2}, 2^{2}, \cdots, 10^{2}$ (which can be the same) has a sum not exceedi... | 79 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the sequence
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{2}=\frac{1}{3}+\frac{2}{3}, \cdots, \\
a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1}, \cdots .
\end{array}
$$
Let $S_{n}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}}$. Then the integer closest to $S_{2006}$ is $(\qua... | 3.C.
Since $a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1}=\frac{n}{2}$, we have
$$
\begin{array}{l}
S_{n}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}} \\
=\frac{4}{1 \times 2}+\frac{4}{2 \times 3}+\cdots+\frac{4}{n(n+1)} \\
=4\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\fr... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and the graph of $y=f(x)$ is symmetric about the line $x=\frac{1}{2}$. Then
$$
f(1)+f(2)+\cdots+f(2006)=
$$
$\qquad$ | 2.0.
Given that the graph of $y=f(x)$ is symmetric about the line $x=\frac{1}{2}$, we know that $f(x)=f(1-x)$.
Also, since $f(x)$ is an odd function defined on $\mathbf{R}$, it follows that $f(1-x)=-f(x-1)$.
Therefore, $f(x)+f(x-1)=0$.
Thus, $f(1)+f(2)+\cdots+f(2006)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let the increasing sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=6, n \in \mathbf{N}_{+}$, and when $n \geqslant 2$, $a_{n}+a_{n-1}=\frac{9}{a_{n}-a_{n-1}}+8$. Then $a_{x}=$ $\qquad$ | 4.29.
$$
\begin{array}{l}
\text { Given } a_{k}+a_{k-1}=\frac{9}{a_{k}-a_{k-1}}+8 \\
\Rightarrow a_{k}^{2}-a_{k-1}^{2}=8\left(a_{k}-a_{k-1}\right)+9 .
\end{array}
$$
Taking $k=2,3, \cdots, n$, summing up we get
$$
a_{n}^{2}-a_{1}^{2}=8\left(a_{n}-a_{1}\right)+9(n-1) \text {. }
$$
Thus, $a_{10}^{2}-6^{2}=8\left(a_{10}... | 29 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $\triangle A B C$ with three sides $A B=\sqrt{34}, B C$ $=5 \sqrt{10}, C A=2 \sqrt{26}$. Then the area of $\triangle A B C$ is $\qquad$ | 5.10.
As shown in Figure 4, with $BC$ as the hypotenuse, construct a right triangle $\triangle BCD$ on one side of $\triangle ABC$ such that $\angle BDC = 90^\circ$, $BD = 5$, and $CD = 15$.
Then construct a rectangle $DEA'F$ such that $DE = 2$ and $DF = 5$. Thus, $BE = 3$ and $CF = 10$. At this point,
$$
\begin{arra... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=3, a_{n+1}=$ $9 \sqrt[3]{a_{n}}(n \geqslant 1)$. Then $\lim _{n \rightarrow \infty} a_{n}=$ $\qquad$ . | 8.27.
Given $\log _{3} a_{n+1}=2+\frac{1}{3} \log _{3} a_{n}$, let $b_{n}=\log _{3} a_{n}$, then
$$
\begin{array}{l}
b_{n+1}=\frac{1}{3} b_{n}+2, b_{1}=1 \\
\Rightarrow b_{n+1}-3=\frac{1}{3}\left(b_{n}-3\right) \\
\Rightarrow b_{n}-3=\left(b_{1}-3\right)\left(\frac{1}{3}\right)^{n-1}=-2 \times\left(\frac{1}{3}\right)^... | 27 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. If $(2 x-1)^{8}=a_{8} x^{8}+a_{7} x^{7}+\cdots+a_{1} x$ $+a_{0}$, then $a_{8}+a_{6}+a_{4}+a_{2}=$ $\qquad$ | 9.3280 .
Let $f(x)=(2 x-1)^{8}$, then
$$
\begin{array}{l}
a_{0}=f(0)=1, \\
a_{8}+a_{7}+\cdots+a_{0}=f(1)=1, \\
a_{8}-a_{7}+a_{6}-a_{5}+a_{4}-a_{3}+a_{2}-a_{1}+a_{0} \\
=f(-1)=(-3)^{8}=6561 .
\end{array}
$$
Therefore, $a_{8}+a_{6}+a_{4}+a_{2}$
$$
=\frac{1}{2}(f(1)+f(-1))-f(0)=3280 \text {. }
$$ | 3280 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Let $n$ be a positive integer, and let $f(n)$ denote the number of positive integers not exceeding $\sqrt{n}$ (for example, $f(3)=1$, $f(9)=3$).
(1) Find $f(2007)$;
(2) Find the positive integer $n$ such that it satisfies
$$
f(1)+f(2)+\cdots+f(n)=2009 .
$$ | (1) From $44<\sqrt{2007}<45$, we have $f(2007)=44$.
(2) When $k=1,2,3$, $f(k)=1$, then
$$
f(1)+f(2)+f(3)=1 \times 3 \text {. }
$$
When $k=4,5, \cdots, 8$, $f(k)=2$, then
$$
f(4)+f(5)+\cdots+f(8)=2 \times 5 \text {. }
$$
When $k=9,10, \cdots, 15$, $f(k)=3$, then
$$
f(9)+f(10)+\cdots+f(15)=3 \times 7 \text {. }
$$
Whe... | 215 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Given three natural numbers $a, b, c$ where at least $a$ is a prime number, and they satisfy
$$
\left\{\begin{array}{l}
(4 a+2 b-4 c)^{2}=443(2 a-442 b+884 c), \\
\sqrt{4 a+2 b-4 c+886}-\sqrt{42 b-2 a+2 c-443}=\sqrt{443} .
\end{array}\right.
$$
Find the value of $a b c$. | Let $x=\frac{4 a+2 b-4 c}{443}, y=\frac{2 a-442 b+884 c}{443}$. Then
$4 a+2 b-4 c=443 x$,
$2 a-442 b+884 c=443 y$.
Hence $x^{2}=y$.
From equation (1), we know that $(4 a+2 b-4 c)^{2}$ is divisible by 443. Since 443 is a prime number, 443 divides $(4 a+2 b-4 c)$. Therefore, $x$ is an integer.
From equation (5), we know ... | 2007 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The area of the figure enclosed by the two curves $y=x^{3}$ and $x=y^{3}$ on the Cartesian plane is $\qquad$ . | 5.1.
Since the two curves are symmetric with respect to the origin, it is only necessary to calculate the area $A$ of the figure enclosed by the two curves in the first quadrant.
When $x>1$, $x^{3}>\sqrt[3]{x}$;
When $0<x<1$, $x^{3}<\sqrt[3]{x}$.
Therefore, the two curves have a unique intersection point $(1,1)$ in th... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
8.2. Write the 100 natural numbers from 1 to 100 in a $10 \times 10$ grid, with one number in each cell. In each operation, you can swap the positions of any two numbers. Prove that it is possible to ensure that the sum of any two numbers in cells that share a common edge is a composite number after just 35 operations. | 8.2. A vertical straight line $m$ divides the grid into two halves. In one of the halves, there are no more than 25 even numbers, let's assume it is the right half. Thus, in the left half, there is an equal number of odd numbers. By swapping the even numbers in the right half with the odd numbers in the left half one b... | 35 | Number Theory | proof | Yes | Yes | cn_contest | false |
8.3. On the side $BC$ of the rhombus $ABCD$, take a point $M$. Draw perpendiculars from $M$ to the diagonals $BD$ and $AC$, intersecting the line $AD$ at points $P$ and $Q$. If the lines $PB$ and $QC$ intersect $AM$ at the same point, find the ratio $\frac{BM}{MC}$. | 8.3. As shown in Figure 1, let the intersection of lines $P B$, $Q C$, and $A M$ be $R$.
From the given conditions, $P M \parallel A C$ and $M Q \parallel B D$, thus quadrilaterals $P M C A$ and $Q M B D$ are both parallelograms. Therefore,
$$
\begin{aligned}
M C &= P A, B M = D Q, \text{ and } \\
P Q &= P A + A D + D... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8.7. For a natural number $n(n>3)$, we use “$n$ ?" to denote the product of all prime numbers less than $n$. Solve the equation
$$
n ?=2 n+16
$$ | 8.7. The given equation is
$$
n ?-32=2(n-8) \text {. }
$$
Since $n$ ? cannot be divisible by 4, it follows from equation (1) that $n-8$ is odd.
Assume $n>9$, then $n-8$ has an odd prime factor $p$.
Also, $p2 \times 9+16$.
When $n=7$, it is clearly a root of the equation:
However, when $n=5$, we have $n ?=6<16$.
Theref... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9.8. Ji Ma calculated the reciprocal of the factorial of each integer from $80 \sim 99$, and printed the resulting decimal fractions on 20 infinitely long strips of paper (for example, Sasha cut a segment from one of the strips, which had exactly $n$ digits without a decimal point. If Sasha does not want Ji Ma to guess... | 9.8. The maximum value of $n$ is 155.
Assuming that on the slips of paper, $\frac{1}{k!}$ and $\frac{1}{l!}$ (where $k < l$) are written, the first 156 digits of the decimal expansion of $\frac{1}{k!} - \frac{1}{l!}$ are greater than $\frac{1}{10^{16}}$. Therefore, $n < 156$.
Thus, for any segment of 156 digits, we c... | 155 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10.4. A magician and his assistant perform the following act: First, the assistant asks the audience to write down $N$ numbers in a row on a blackboard, then the assistant covers up two adjacent numbers. After that, the magician comes on stage and guesses the two adjacent covered numbers (including their order). To ens... | 10.4. $N=101$.
For convenience, a sequence of $m$ digits is called an “$m$-digit number”. Suppose for some value of $N$, the magician can guess the result, so the magician can restore any two-digit number to the original $N$-digit number (the number of $N$-digit numbers that can be restored is denoted as $k_{1}$). Thi... | 101 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $N=99999$. Then $N^{3}=$ | $$
\text { Two, 1.999970000 } 299999 \text {. }
$$
Since $N=10^{5}-1$, then
$$
\begin{array}{l}
N^{3}=\left(10^{5}-1\right)^{3}=10^{15}-3 \times 10^{10}+3 \times 10^{5}-1 \\
=10^{10}\left(10^{5}-3\right)+3 \times 10^{5}-1 \\
=999970000299999 .
\end{array}
$$ | 999970000299999 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In the arithmetic sequence $\left\{a_{n}\right\}$, if $a_{2}+a_{4}+a_{6}+$ $a_{8}+a_{10}=80$, then $a_{7}-\frac{1}{2} a_{8}=(\quad)$.
(A) 4
(B) 6
(C) 8
(D) 10 | 4.C.
Since $a_{2}+a_{4}+a_{6}+a_{8}+a_{10}=5 a_{6}=80$, therefore, $a_{6}=16$.
Thus, $a_{7}-\frac{1}{2} a_{8}=a_{6}+d-\frac{1}{2}\left(a_{6}+2 d\right)=\frac{1}{2} a_{6}=8$. | 8 | Algebra | MCQ | Yes | Yes | cn_contest | false |
11. If the three medians $A D$, $B E$, $C F$ of $\triangle A B C$ intersect at point $M$, then $M A+M B+M C=$ | Ni, 11.0.
Diagram (omitted). Let the midpoint of $A B$ be $D$. By the parallelogram rule, we have
$$
\begin{array}{l}
M A+M B=2 M D=-M C \text {. } \\
\text { Therefore, } M A+M B+M C=0 \text {. } \\
\end{array}
$$ | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. In the sequence $\left\{a_{n}\right\}$, $a_{1}=2, a_{n}+a_{n+1}=1$ $\left(n \in \mathbf{N}_{+}\right)$, let $S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then
$$
\mathrm{S}_{2007}-2 \mathrm{~S}_{2006}+\mathrm{S}_{2000}=
$$
$\qquad$ | 13.3.
When $n$ is even, we have
$$
a_{1}+a_{2}=a_{3}+a_{4}=\cdots=a_{n-1}+a_{n}=1 .
$$
Thus, $S_{n}=\frac{n}{2}$.
When $n$ is odd, we have
$$
a_{1}=2, a_{2}+a_{3}=a_{4}+a_{5}=\cdots=a_{n-1}+a_{n}=1 \text {. }
$$
Thus, $S_{n}=2+\frac{n-1}{2}=\frac{n+3}{2}$.
Therefore, $S_{2007}-2 S_{2006}+S_{2005}$
$$
=1005-2 \times ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. In $\triangle A B C$, it is known that $A B=\sqrt{30}, A C=$ $\sqrt{6}, B C=\sqrt{15}$, point $D$ makes $A D$ bisect $B C$, and $\angle A D B=90^{\circ}$, the ratio $\frac{S_{\triangle A D B}}{S_{\triangle A B C}}$ can be written in the form $\frac{m}{n}$, where $m, n$ are coprime positive integers. Then $m+n=$ | 14.65.
Let the midpoint of $BC$ be $E$, and $AD=\frac{x}{2}$. By the median formula, we get $AE=\frac{\sqrt{57}}{2}$.
Thus, $(\sqrt{30})^{2}-\left(\frac{x}{2}\right)^{2}$
$$
=\left(\frac{\sqrt{15}}{2}\right)^{2}-\left(\frac{x}{2}-\frac{\sqrt{57}}{2}\right)^{2} \text {. }
$$
Solving for $x$ gives $x=\frac{81}{\sqrt{57... | 65 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. (12 points) In $\triangle A B C$, it is known that $\sin A \cdot \cos ^{2} \frac{C}{2}+\sin C \cdot \cos ^{2} \frac{A}{2}=\frac{3}{2} \sin B$. Find the value of $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}$. | Three, 15. From the given,
$\sin A \cdot \frac{1+\cos C}{2}+\sin C \cdot \frac{1+\cos A}{2}=\frac{3}{2} \sin B$.
Then, $\sin A+\sin C+\sin A \cdot \cos C+\cos A \cdot \sin C$ $=3 \sin B$.
Thus, $\sin A+\sin C+\sin (A+C)=3 \sin B$, which means $\sin A+\sin C=2 \sin B$.
Therefore, $2 \sin \frac{A+C}{2} \cdot \cos \frac{A... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Three people, A, B, and C, are playing a game. Each of them writes 100 different Chinese characters, and then they compare the characters written by each person. If a character is written by at least two people, it is deleted until there are no more identical characters. A, B, and C deleted 31, 27, and 39 characters... | 5.C.
Suppose there are $a$ Chinese characters that are the same for all three people, $b$ Chinese characters that are the same only for person A and B, $c$ Chinese characters that are the same only for person A and C, and $d$ Chinese characters that are the same only for person B and C. From the problem, we have:
$$
\... | 242 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
1. There are weights of $11 \mathrm{~g}$ and $17 \mathrm{~g}$ available in sufficient quantity. To weigh an object of mass $3 \mathrm{~g}$ on a balance, at least $\qquad$ such weights are needed. | Let the weights of 11 g and 17 g be used $x$ times and $y$ times, respectively. Then we have $11 x - 17 y = 3$ or $17 y - 11 x = 3$.
The solutions are $\left\{\begin{array}{l}x=8+17 t, \\ y=5+11 t\end{array}\right.$ or $\left\{\begin{array}{l}x=9+17 t, \\ y=6+11 t .\end{array}\right.$
where $t$ is an integer.
Since $x$... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For example, $\triangle ABC$ is an equilateral triangle with side length 8, $M$ is a point on side $AB$, $MP \perp AC$ at point $P$, $MQ \perp BC$ at point $Q$, and connect $PQ$.
(1) Find the minimum length of $PQ$;
(2) Find the maximum area of $\triangle CPQ$. | Solution: (1) Let the height of $\triangle ABC$ be $h$, then $h=4 \sqrt{3}$.
From $S_{\triangle C M}+S_{\triangle B C M}=S_{\triangle B B C}$, we get
$$
M P+M O=h=4 \sqrt{3} \text {. }
$$
As shown in Figure 11, draw perpendiculars from points $P$ and $Q$ to side $AB$, with the feet of the perpendiculars being $P_{1}$ ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2-1 Given that $x, y, z$ are positive numbers, and $xyz(x+y+z)=1$.
Find the minimum value of $(x+y)(y+z)$.
(1989, All-Soviet Union Mathematical Competition) | Solution 1: $(x+y)(y+z)$
$$
\begin{array}{l}
=x z+y(x+y+z) \\
\geqslant 2 \sqrt{x y z(x+y+z)}=2 .
\end{array}
$$
When $x=z=1, y=\sqrt{2}-1$, $y(x+y+z) = xz$, $(x+y)(y+z)$ takes the minimum value 2.
Solution 2: As shown in Figure 1, construct
$\triangle ABC$, with side lengths
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2-3 Let quadrilateral $A B C D$ be a rectangle with an area of 2, $P$ a point on side $C D$, and $Q$ the point where the incircle of $\triangle P A B$ touches side $A B$. The product $P A \cdot P B$ varies with the changes in rectangle $A B C D$ and point $P$. When $P A \cdot P B$ is minimized,
(1) Prove: $A B ... | Solution 1: (1) From the given information,
$S_{\triangle P A B}=\frac{1}{2} S_{\text {rectangle } A B C D}=1$.
Thus, $P A \cdot P B=\frac{2 S_{\triangle P A B}}{\sin \angle A P B} \geqslant 2$.
Equality holds if and only if $\angle A P B=90^{\circ}$. At this time, point $P$ lies on the circle with $A B$ as its diamete... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $p$ be a positive odd number. Then the remainder of $p^{2}$ divided by 8 is $\qquad$ . | Because $p$ is an odd positive integer, let $p=2k-1\left(k \in \mathbf{N}_{+}\right)$, so
$$
p^{2}=(2k-1)^{2}=4k^{2}-4k+1=4(k-1)k+1 \text{. }
$$
Since $(k-1)k$ is even, therefore, $p^{2}$ leaves a remainder of 1 when divided by 8. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given an isosceles triangle $\triangle A B C$ with side lengths $a$, $b$, and $c$ all being integers, and satisfying $a+b c+b+c a=24$. Then the number of such triangles is $\qquad$. | 4.3.
$$
\begin{array}{l}
\text { Since } a+b c+b+a a \\
=(a+b)(c+1)=24=12 \times 2=8 \times 3=6 \times 4, \text { and }
\end{array}
$$
$\triangle A B C$ is an isosceles triangle, so the length of the base can only be $c$.
Thus, there are 3 triangles that satisfy the conditions:
$$
c=1, a=b=6 ; c=2, a=b=4 ; c=3, a=b=3 \... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) Given that the graph of the linear function $y=a x+b$ passes through the points $A(\sqrt{3}, \sqrt{3}+2), B(-1, \sqrt{3})$, and $C(c, 2-c)$. Find the value of $a-b+c$.
---
The above text has been translated into English, preserving the original text's line breaks and format. | Three, from $\left\{\begin{array}{l}\sqrt{3}+2=\sqrt{3} a+b, \\ \sqrt{3}=-a+b\end{array} \Rightarrow\left\{\begin{array}{l}a=\sqrt{3}-1, \\ b=2 \sqrt{3}-1 .\end{array}\right.\right.$
Therefore, $2-c=a c+b=(\sqrt{3}-1) c+(2 \sqrt{3}-1)$.
Solving for $c$ gives $c=\sqrt{3}-2$.
Thus, $a-b+c=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (50 points)
As shown in Figure 3, in a $7 \times 8$
rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected." Now, some of the 56 chess pieces are to be removed ... | Second, at least 11 chess pieces must be taken out to possibly meet the requirement.
The reason is as follows:
If a square is in the $i$th row and the $j$th column, then this square is denoted as $(i, j)$.
Step 1 Proof: If any 10 chess pieces are taken, then the remaining chess pieces must have a five-in-a-row, i.e., ... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $\alpha, \beta$ are the roots of the equation $x^{2}+x-1=0$. Then the value of $\alpha^{16}-987 \beta+411$ is ( ).
(A) 2006
(B) 2007
(C) 2008
(D) 2009 | 5.C.
From the given, we have $\alpha^{2}=1-\alpha$ and $\alpha+\beta=-1$. Therefore,
$$
\begin{array}{l}
\alpha^{4}=(1-\alpha)^{2}=1-2 \alpha+\alpha^{2}=2-3 \alpha, \\
\alpha^{8}=(2-3 \alpha)^{2}=4-12 \alpha+9 \alpha^{2}=13-21 \alpha, \\
\alpha^{16}=(13-21 \alpha)^{2}=169-546 \alpha+441 \alpha^{2}=610-987 \alpha .
\en... | 2008 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Three, (25 points) Let $x, y, a, m, n$ be positive integers, and $x+y=a^{m}, x^{2}+y^{2}=a^{n}$. Find how many digits $a^{30}$ has.
保留源文本的换行和格式，直接输出翻译结果。 | Three, from the known we get
$$
a^{2 m}=x^{2}+y^{2}+2 x y=a^{n}+2 x y \text {. }
$$
From the problem and equation (1), we know that $a^{2 m}>a^{n}$. Therefore, $2 m>n$.
Dividing both sides of equation (1) by $a^{n}$, we get
$$
a^{2 m-n}=1+\frac{2 x y}{a^{n}} \text {. }
$$
Since the left side of equation (2) is a posi... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. In a convex quadrilateral $A B C D$, $A B+A C+C D=16$. When the diagonals $A C$ and $B D$ are what values, will the area of quadrilateral $A B C D$ be maximized? What is the maximum area? | (Hint: Let $A B=x, A C=y$, then $C D=16-x-y$. And
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=S_{\triangle A B C}+S_{\triangle C D} \\
\leqslant \frac{1}{2} x y+\frac{1}{2} y(16-x-y) \\
=-\frac{1}{2}(y-8)^{2}+32 .
\end{array}
$$
Therefore, when $\angle B A C=\angle A C D=90^{\circ}, A C=8, B D=8 \sqrt{2}$,... | 32 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $A=\{0,1,2,3,4,5,6,7\}, f: A \rightarrow$ $A$. If $i+j=7$, then $f(i) f(j)=i j$. Then, the number of mappings $f$ is $\qquad$ | For $0$ and $7$ elements, since $f(i) f(j)=0$, at least one of the images of $0$ and $7$ must be $0$. There are a total of $2 \times 8-1=15$ cases.
For the images of $1$ and $6$ elements, in this case, $f(i) f(j)=6=1 \times 6=2 \times 3$, so the images of $1$ and $6$ can be: $1, 6, 6, 1, 2, 3, 3, 2$, totaling 4 cases.... | 480 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. A five-digit number consists of only the three different digits $1, 3, 5$. Then the number of such five-digit numbers is $\qquad$.
| 3.150.
Considering overall, there are
$$
3^{5}-C_{3}^{2}\left(2^{5}-2\right)-3=150 \text { (cases). }
$$ | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. In a $4 \times 4$ grid, eighteen lines can be drawn, including eight lines (four horizontal and four vertical), and five diagonal lines from the top left to the bottom right and from the top right to the bottom left. These diagonal lines may pass through 2, 3, or 4 small squares. Ten chips are to be placed in the gr... | 2. As shown in Figure 11, the maximum score is 17.
The following proves that 18 points cannot be achieved.
Notice that, in Figure 11, there are always two opposite corner cells not covered by the same direction of five diagonal lines, and these two cells must be either both empty or both full. At the same time, there m... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Mary found that when a certain three-digit number is squared, the last three digits of the product are the same as the original three-digit number. What is the sum of all different three-digit numbers that satisfy this property? | 4. Let the three-digit number be $\overline{a b c}$, then
$$
\overline{a b c}^{2}=1000 k+\overline{a b c} \text {, }
$$
i.e., $\overline{a b c}(\overline{a b c}-1)=2^{3} \times 5^{3} k$.
Since $(\overline{a b c}, \overline{a b c}-1)=1$, therefore,
$2^{3} \mid \overline{a b c}$, and $5^{3} \mid(\overline{a b c}-1)$,
o... | 1001 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Arrange the $n$ positive integers from $1 \sim n (n>1)$ in a row so that the sum of any two adjacent numbers is a perfect square. What is the minimum value of $n$? | 9. The smallest positive integer $n$ is 15.
Since $n>1$, it includes 2. The smallest positive integer that, when added to 2, results in a perfect square is 7, represented as $2+7=9$ (the same notation applies below), so $n \geqslant 7$.
If $n=7$, we can get three non-adjacent segments: $(1,3,6),(2,7),(4,5)$.
Adding ... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Among the 2007 positive integers $1,2, \cdots, 2007$, what is the maximum number of integers that can be selected such that each selected number is coprime with 2007, and the sum of any three selected numbers is not a multiple of 7? | 3. When $1,2, \cdots, 2007$ are divided by 7, the remainders 1, 2, 3, 4, 5 each have $286+1=287$ numbers; the remainders 6, 0 each have 286 numbers.
Among $1,2, \cdots, 2007$, the numbers that are not coprime with 2007 are $3,2 \times 3,3 \times 3, \cdots, 669 \times 3$ and $223,2 \times 223,4 \times 223,5 \times 223,... | 386 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) As shown in Figure 5, there are 12 equally spaced points on the circumference of a clock face, marked with the numbers $1, 2, \cdots, 12$ in sequence. Please use these equally spaced points as vertices to form 4 triangles (dividing these 12 equally spaced points into 4 groups) such that the following... | Three, let 4 triangles be $\left(a_{i}, b_{i}, c_{i}\right), i=1,2, 3,4$, where $a_{i}=b_{i}+c_{i}, b_{i}>c_{i}$.
Let $a_{1}=27$.
Thus, $10 \leqslant a_{3} \leqslant 11$.
If $a_{3}=10$, then from $a_{1}+a_{2}=17, a_{2}a_{1}+a_{2}=17$, we get $a_{2}=9, a_{1}=8$, that is
$$
\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(8,9,10... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. If $a, b, c$ are 3 different positive integers, and $a b c=16$, then the maximum possible value of $a^{b}-b^{c}+c^{a}$ is ( ).
(A) 249
(B) 253
(C) 263
(D) 264 | $-1 . C$.
From the given, we easily know that $\{a, b, c\}=\{1,2,8\}$.
And when $b=1, c=2, a=8$, $a^{b}-b^{c}+c^{a}$ reaches the maximum value $8-1+2^{8}=263$. | 263 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Arrange positive integers starting from 1 in a number array according to the pattern shown in Figure 2, where 2 is at the 1st corner, 3 is at the 2nd corner, 5 is at the 3rd corner, 7 is at the 4th corner, $\cdots \cdots$. Then, the number at the 2007th corner is $\qquad$
Figure 2 | $$
\begin{array}{l}
a_{1}=2, a_{2 i}=a_{2 i-1}+i, \\
a_{2 i+1}=a_{2 i}+(i+1) .
\end{array}
$$
Since $2007=2 \times 1003+1$, we have,
$$
\begin{array}{l}
a_{2007}=1+2(1+2+\cdots+1003)+1004 \\
=1004^{2}+1=1008017 .
\end{array}
$$ | 1008017 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. In a $3 \times 3$ grid, the numbers $1,2,3,4,5,6,7,8,9$ are filled in, with one number per cell. Now, the cells containing the maximum number in each row are colored red, and the cells containing the minimum number in each row are colored green. Let $M$ be the smallest number in the red cells, and $m$ be the largest... | 2.8
Obviously, $3 \leqslant m, M \leqslant 7$, and $m \neq M$. Therefore,
$$
M-m \in\{-4,-3,-2,-1,1,2,3,4\} \text {. }
$$
As shown in Figure 7, all 8 values can be obtained
\begin{tabular}{|l|l|l|}
\hline 7 & 9 & 8 \\
\hline 6 & 5 & 4 \\
\hline 3 & 2 & 1 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 6 & 9 &... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Five. (15 points) If for any $n$ consecutive positive integers, there always exists a number whose sum of digits is a multiple of 8. Determine the minimum value of $n$. And explain the reason.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translati... | Five, first prove that when $n \leqslant 14$, the property in the problem does not hold.
When $n=14$, for
$9999993, 9999994, \cdots, 10000006$
these 14 consecutive integers, the sum of the digits of any number cannot be divisible by 8.
Therefore, when $n \leqslant 14$, the property in the problem does not hold.
Thus, t... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of positive integers $m$ that make $m^{2}+m+7$ a perfect square is $\qquad$ . | 3.2.
It has been verified that when $m=1$, $m^{2}+m+7=9$ is a perfect square;
when $m=2,3,4,5$, $m^{2}+m+7$ are not perfect squares;
when $m=6$, $m^{2}+m+7=49$ is a perfect square.
When $m>6$, $m^{2}+m+7$ are not perfect squares.
Therefore, there are only 2 positive integers $m$ that meet the condition. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (20 points) For any real numbers $x, y$, we have
$$
|x-2|+|x-4| \geqslant m\left(-y^{2}+2 y\right)
$$
Determine the maximum value of the real number $m$. | Three, by the geometric meaning of absolute value, $|x-2|+|x-4|$ has a minimum value of 2 when $x \in [2,4]$.
And $-y^{2}+2y=-(y-1)^{2}+1$ has a maximum value of 1 when $y=1$.
From the condition, $2 \geqslant m \times 1$, then $m \leqslant 2$.
Therefore, the maximum value of $m$ is 2. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Five. (25 points) Given the system of equations in $x$ and $y$
$$
\left\{\begin{array}{l}
x^{2}-y^{2}=p, \\
3 x y+p(x-y)=p^{2}
\end{array}\right.
$$
has integer solutions $(x, y)$. Find the prime number $p$ that satisfies the condition. | Five, from $p=x^{2}-y^{2}=(x-y)(x+y)$ and $p$ being a prime number, we have
$\left\{\begin{array}{l}x+y=p, \\ x-y=1\end{array}\right.$ or $\quad\left\{\begin{array}{l}x+y=-p, \\ x-y=-1\end{array}\right.$
or $\left\{\begin{array}{l}x+y=1, \\ x-y=p\end{array}\right.$ or $\left\{\begin{array}{l}x+y=-1, \\ x-y=-p .\end{ar... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
In the pentagram $A B C D E$, the intersection points of each line segment are shown in Figure 3, $A G=G F=F D, B G=$ $3 E G, F C=2 F E$. If the area of pentagon $F G H I J$ is 91, try to find the area of the pentagram $A B C D E$. | Solution: As shown in Figure 3, connect $F H, F I, G J, H J, A E,$ and $B C$.
From $\frac{S_{\triangle A E C}}{S_{\triangle A C F}} \cdot \frac{S_{\triangle C F A}}{S_{\triangle C A G}}$.
$\frac{S_{\triangle C A G}}{S_{\triangle C A E}}=1$, we get
$$
\frac{E C}{C F} \cdot \frac{F A}{A G} \cdot \frac{G H}{H E}=1,
$$
wh... | 349 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. From $1,2, \cdots, 2006$, at least
$\qquad$ odd numbers must be taken to ensure that there are definitely two numbers whose sum is 2008. | 4.503.
From $1,2, \cdots, 2006$, selecting two odd numbers whose sum is 2008, there are a total of 501 pairs as follows:
$$
3+2005,5+2003, \cdots, 1003+1005 \text {. }
$$
Since 1 added to any of these odd numbers will not equal 2008, therefore, at least 503 odd numbers must be selected to ensure that there are defini... | 503 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Real numbers $x, y, z, w$ satisfy $x \geqslant y \geqslant z \geqslant w \geqslant 0$, and $5 x+4 y+3 z+6 w=100$. Find the maximum and minimum values of $x+y+z+w$. | Let $z=w+a, y=w+a+b, x=w+$ $a+b+c$. Then $a, b, c \geqslant 0$, and
$$
x+y+z+w=4 w+3 a+2 b+c .
$$
Therefore, $100=5(w+a+b+c)+4(w+a+$
$$
\begin{array}{l}
\quad b)+3(w+a)+6 w \\
=18 w+12 a+9 b+5 c \\
=4(4 w+3 a+2 b+c)+(2 w+b+c) \\
\geqslant 4(x+y+z+w) .
\end{array}
$$
Thus, $x+y+z+w \leqslant 25$.
When $x=y=z=\frac{25}... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{r}
\text { Three. (25 points) In }\left[\frac{1^{2}}{2008}\right],\left[\frac{2^{2}}{2008}\right], \cdots, \\
{\left[\frac{2008^{2}}{2008}\right] \text {, how many different integers are there (where }[x]}
\end{array}
$$
indicates the greatest integer not greater than $x$)? | Three, let $f(n)=\frac{n^{2}}{2008}$.
When $n=2,3, \cdots, 1004$, we have
$$
\begin{array}{l}
f(n)-f(n-1) \\
=\frac{n^{2}}{2008}-\frac{(n-1)^{2}}{2008}=\frac{2 n-1}{2008}1 .
\end{array}
$$
And $f(1005)=\frac{1005^{2}}{2008}=\frac{(1004+1)^{2}}{2008}$
$$
=502+1+\frac{1}{2008}>503,
$$
Therefore, $\left[\frac{1005^{2}}{... | 1507 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Add the same integer $a(a>0)$ to the numerator and denominator of $\frac{2008}{3}$, making the fraction an integer. Then the integer $a$ added has $\qquad$ solutions. | $=1.3$.
From the problem, we know that $\frac{2008+a}{3+a}$ should be an integer, which means $\frac{2008+a}{3+a}=\frac{2005}{3+a}+1$ should be an integer. Therefore, $(3+a) \mid 2005=5 \times 401$.
Since 2005 has 4 divisors, at this point, $a$ can take the values $2002$, $398$, $2$, and one divisor corresponds to a va... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. In rectangle $A B C D$, $A B=56, A D=35$. Using lines parallel to $A B$ and $A D$, it is divided into $35 \times$ 56 unit squares. The number of unit squares that are internally crossed by diagonal $A C$ is ( ) .
(A) 84
(B) 85
(C) 91
(D) 92 | 5.A.
Take $A$ as the origin, the lines $AB$ and $AD$ as the $x$-axis and $y$-axis, respectively. Apart from the coordinate axes, there are 35 horizontal lines and 56 vertical lines. The diagonal $AC$ intersects with each of these lines, resulting in $35+56=91$ intersection points (including coincident points).
Next, ... | 84 | Geometry | MCQ | Yes | Yes | cn_contest | false |
1. A bicycle tire, if installed on the front wheel, will wear out after traveling $5000 \mathrm{~km}$; if installed on the back wheel, it will wear out after traveling $3000 \mathrm{~km}$. If the front and back tires are swapped after traveling a certain distance, so that a pair of new tires wear out simultaneously, th... | II. 1.3750.
Assume the total wear of each new tire when it is scrapped is $k$, then the wear per kilometer for a tire installed on the front wheel is $\frac{k}{5000}$, and the wear per kilometer for a tire installed on the rear wheel is $\frac{k}{3000}$. Let a pair of new tires travel $x \mathrm{~km}$ before swapping p... | 3750 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $x$ is a real number,
$$
\sqrt{x^{3}+2020}-\sqrt{2030-x^{3}}=54 \text {. }
$$
Then $28 \sqrt{x^{3}+2020}+27 \sqrt{2030-x^{3}}=$ | II. 1.2007.
Let $x^{3}+2020=a, 2030-x^{3}=b$. Then $a+b=4050$.
From the problem, we have
$$
\begin{array}{l}
\sqrt{a}-\sqrt{b}=54 \\
\Rightarrow a+b-2 \sqrt{a b}=54^{2}=2916 \\
\Rightarrow 2 \sqrt{a b}=(a+b)-2916 \\
\quad=4050-2916=1134 \\
\Rightarrow(\sqrt{a}+\sqrt{b})^{2}=(a+b)+2 \sqrt{a b} \\
\quad=4050+1134=5184 \\... | 2007 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $|a|>1$, simplify
$$
\left(a+\sqrt{a^{2}-1}\right)^{4}+2\left(1-2 a^{2}\right)\left(a+\sqrt{a^{2}-1}\right)^{2}+3
$$
the result is $\qquad$ . | Let $x_{0}=a+\sqrt{a^{2}-1}$. Clearly, $x_{0}$ is a root of the equation $x^{2}-2 a x+1=0$, and thus it is also a root of the equation
$$
\left(x^{2}-2 a x+1\right)\left(x^{2}+2 a x+1\right)=0
$$
Then $\left(x_{0}^{2}+1\right)^{2}-4 a^{2} x_{0}^{2}=0$, which means
$$
x_{0}^{4}+2\left(1-2 a^{2}\right) x_{0}^{2}+1=0 \te... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $n$ is a natural number, $9 n^{2}-10 n+2009$ can be expressed as the product of two consecutive natural numbers. Then the maximum value of $n$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.2007 .
Let $9 n^{2}-10 n+2009=m(m+1)$, where $m$ is a natural number. Then
$$
9 n^{2}-10 n+\left(2009-m^{2}-m\right)=0 \text {. }
$$
Considering equation (1) as a quadratic equation in the natural number $n$, its discriminant should be the square of a natural number. Let's assume $\Delta=t^{2}(t \in \mathbf{N})$, t... | 2007 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) During the New Year, the grandmother gave three grandsons a total of 400 yuan as New Year's money. There are 50 yuan, 20 yuan, and 10 yuan banknotes available in various quantities for the three grandsons to choose from, but each can only take banknotes of the same denomination. One of them takes a n... | Three people take money in the amounts of $x y$, $x$, and $y$, with denominations of $a$ yuan, $b$ yuan, and $c$ yuan respectively, where $x \geqslant 1, y \geqslant 1$. Then $a x y + b x + c y = 400$.
(1) When $a = b = c$, i.e., the denominations chosen by the three people are the same, then
$$
x y + x + y = \frac{400... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Within the range of the independent variable $x$, $59 \leqslant x \leqslant 60$, the number of integer values of the quadratic function $y=x^{2}+x+\frac{1}{2}$ is ( ).
(A) 59
(B) 120
(C) 118
(D) 60 | 5.B.
Notice that $y=x^{2}+x+\frac{1}{2}=\left(x+\frac{1}{2}\right)^{2}+\frac{1}{4}$. When $x>-\frac{1}{2}$, $y$ increases as $x$ increases.
Since $59 \leqslant x \leqslant 60$, then
$$
59^{2}+59+\frac{1}{2} \leqslant y \leqslant 60^{2}+60+\frac{1}{2},
$$
which means $3540 \frac{1}{2} \leqslant y \leqslant 3660 \frac{... | 120 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. $a_{1}, a_{2}, \cdots, a_{10}$ represent the ten digits $1,2,3,4,5,6$, $7,8,9,0$, respectively, to form two five-digit numbers
$$
m=\overline{a_{1} a_{2} a_{3} a_{4} a_{5}}, n=\overline{a_{6} a_{7} a_{8} a_{9} a_{10}}(m>n) .
$$
Then the minimum value of $m-n$ is | 2.247.
Since $m>n$, we have $a_{1}>a_{6}$.
To make $m-n$ as small as possible, we should take $a_{1}-a_{6}=1$, the last four digits of $m$ should be the smallest, and the last four digits of $n$ should be the largest. The largest four-digit number that can be formed from the different digits 1,2,3,4,5,6,7,8,9,0 is 987... | 247 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given
$$
\frac{y+z-x}{x+y+z}=\frac{z+x-y}{y+z-x}=\frac{x+y-z}{z+x-y}=p \text {. }
$$
Then $p^{3}+p^{2}+p=$ $\qquad$ . | Notice
$$
\begin{array}{l}
p^{2}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x}=\frac{z+x-y}{x+y+z}, \\
p^{3}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x} \cdot \frac{x+y-z}{z+x-y} \\
=\frac{x+y-z}{x+y+z}, \\
p^{3}+p^{2}+p \\
=\frac{x+y-z}{x+y+z}+\frac{z+x-y}{x+y+z}+\frac{y+z-x}{x+y+z} \\
=\frac{x+y+z}{x+y+z}=1 .
\end{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) A company spent 4.8 million yuan to purchase the production technology of a certain product, and then invested another 15.2 million yuan to buy production equipment to process and produce the product. It is known that the cost of producing each piece of this product is 40 yuan. Market research has foun... | $-、(1) y=\left\{\begin{array}{ll}-\frac{2}{25} x+28, & 100 \leqslant x \leqslant 200 \\ -\frac{1}{10} x+32, & 200<x \leqslant 300\end{array}\right.$
(2) When $100 \leqslant x \leqslant 200$, $w=x y-40 y-(1520+480)$.
Substituting $y=-\frac{2}{25} x+28$ into equation (1) gives $w=x\left(-\frac{2}{25} x+28\right)-40\left(... | 190 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that the two roots of the equation $x^{2}+x-1=0$ are $\alpha, \beta$. Then the value of $\frac{\alpha^{3}}{\beta}+\frac{\beta^{3}}{\alpha}$ is $\qquad$ | Let $A=\frac{\alpha^{3}}{\beta}+\frac{\beta^{3}}{\alpha}, B=\frac{\alpha^{3}}{\alpha}+\frac{\beta^{3}}{\beta}=\alpha^{2}+\beta^{2}$.
From the given information,
$$
\begin{array}{l}
\alpha+\beta=-1, \alpha \beta=-1 . \\
\text { Hence } B=(\alpha+\beta)^{2}-2 \alpha \beta=1+2=3 . \\
\text { Also, } \alpha^{3}+\beta^{3}=(... | -7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Divide the positive integers $1,2, \cdots, 2008$ into 1004 groups: $a_{1}, b_{1} ; a_{2}, b_{2} ; \cdots ; a_{1004}, b_{1004}$, and satisfy
$$
a_{1}+b_{1}=a_{2}+b_{2}=\cdots=a_{1004}+b_{1004} \text {. }
$$
For all $i(i=1,2, \cdots, 1004), a_{i} b_{i}$ the maximum value is | 2.1009020 .
Notice that
$$
\begin{array}{l}
a_{i} b_{i}=\frac{1}{4}\left[\left(a_{i}+b_{i}\right)^{2}-\left(a_{i}-b_{i}\right)^{2}\right], \\
a_{i}+b_{i}=\frac{(1+2008) \times 1004}{1004}=2009 .
\end{array}
$$
To maximize the value of $a_{i} b_{i}$, the value of $a_{i}-b_{i}$ must be minimized, and the minimum value ... | 1009020 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Among the following four propositions:
(1) A quadrilateral with one pair of opposite sides equal and one pair of opposite angles equal is a parallelogram;
(2) A quadrilateral with one pair of opposite sides equal and one diagonal bisecting the other diagonal is a parallelogram;
(3) A quadrilateral with one pair of o... | 4. (4).
Propositions (1), (2), and (3) can be refuted with the following counterexamples:
Proposition (1): The quadrilateral $ABCD$ in Figure 5(a), where $\triangle ABD \cong \triangle CDE$.
Proposition (2): As shown in Figure 5(b), construct an isosceles $\triangle ADE$, extend the base $ED$ to any point $O$, and us... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) In $\triangle A B C$, it is known that $\angle A>\angle B>$ $\angle C$, and $\angle A=2 \angle B$. If the three sides of the triangle are integers, and the area is also an integer, find the minimum value of the area of $\triangle A B C$. | Let $B C=a, C A=b, A B=c$.
As shown in Figure 6, construct the angle bisector $A D$ of $\angle B A C$, then
$$
\begin{array}{l}
\angle B A D=\angle D A C \\
= \angle B, \\
\angle A D C=\angle B+\angle B A D \\
= 2 \angle B .
\end{array}
$$
Figure 6
Therefore, $\triangle A C D \backsim \triangle B C A$. Thus,
$$
\frac... | 132 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. If 6 pieces of $1 \times 2$ paper are used to cover a $3 \times 4$ grid, the number of different ways to cover it is. | 4.11.
As shown in Figure 8, the cells of the grid are numbered. Let $M(a, b)$ denote the number of ways to cover the entire grid when the cells numbered $a$ and $b$ (which are adjacent) are covered by the same piece of paper. We focus on the covering of cell 8. It is given that $M(8,5) = 2, M(8,11) = 3, M(8,7) = M(8,9... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) There are 2008 students participating in a large public welfare activity. If two students know each other, then these two students are considered as a cooperative group.
(1) Find the minimum number of cooperative groups $m$, such that no matter how the students know each other, there exist three stud... | (1) Let $n=1004$.
The following proof: $m=n^{2}+1$.
Divide the students into two large groups, each with $n$ students, and no students in the same large group know each other, while each student knows every student in the other large group, thus forming $n^{2}$ cooperative groups, but there do not exist three students ... | 1008017 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Second question As shown in Figure 8, in a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center point of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected". Now, some of the 56 chess pieces are removed ... | Solution: The minimum number of chess pieces to be removed is 11.
Define the square at the $x$-th column and $y$-th row as $(x, y)$ $(1 \leqslant x \leqslant 8,1 \leqslant y \leqslant 7, x, y \in \mathbf{Z})$.
A group of 5 consecutive chess pieces is called a "good group". The problem requires that no chess pieces for... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. If the system of equations concerning $x$ and $y$ $\left\{\begin{array}{l}a x+b y=1, \\ x^{2}+y^{2}=10\end{array}\right.$ has solutions, and all solutions are integers, then the number of ordered pairs $(a, b)$ is $\qquad$ | $=7.32$.
Since the integer solutions of $x^{2}+y^{2}=10$ are
$$
\begin{array}{l}
(1,3),(3,1),(1,-3),(-3,1), \\
(-1,3),(3,-1),(-1,-3),(-3,-1),
\end{array}
$$
Therefore, the number of lines connecting these eight points that do not pass through the origin is 24, and the number of tangents passing through these eight poi... | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Equation
$$
\begin{array}{l}
\frac{(x-1)(x-4)(x-9)}{(x+1)(x+4)(x+9)}+ \\
\frac{2}{3}\left[\frac{x^{3}+1}{(x+1)^{3}}+\frac{x^{3}+4^{3}}{(x+4)^{3}}+\frac{x^{3}+9^{3}}{(x+9)^{3}}\right]=1
\end{array}
$$
The number of distinct non-zero integer solutions is | 10.4 .
Using $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$, the original equation is
$$
\begin{array}{l}
\frac{(x-1)(x-4)(x-9)}{(x+1)(x+4)(x+9)}+1+\frac{2}{3}\left[\frac{x^{3}+1}{(x+1)^{3}}-\right. \\
\left.1+\frac{x^{3}+4^{3}}{(x+4)^{3}}-1+\frac{x^{3}+9^{3}}{(x+9)^{3}}-1\right]=0 \\
\Leftrightarrow \frac{x^{3}+49 x... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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