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3. A math competition has a total of 15 questions. The table below shows the number of people who got $n$ $(n=0,1,2, \cdots, 15)$ questions correct:
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
$n$ & 0 & 1 & 2 & 3 & $\cdots$ & 12 & 13 & 14 & 15 \\
\hline Number of people who got $n$ questions correct & 7 & 8 & 10 & 21 & $\cdot... | (Tip: The number of people who got at most 3 questions correct is $7+8+10+21$ $=46$ people.
The number of people who got at least 12 questions correct is $15+6+3+1=25$ people.
The total number of questions correctly answered by those who got at most 3 questions correct is 91 questions,
The total number of questions cor... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Find all positive integers $n$ such that the sum $1+2+3+\cdots+n$ is a three-digit number composed of the same digit. | 5. $n=36$.
$1+2+\cdots+n=\frac{n(n+1)}{2}$, which means solving the equation $\frac{n(n+1)}{2} = 111,222, \cdots, 999$, or equivalently $n(n+1)=2 \times 3 \times 37 \times k, k = 1,2, \cdots, 9$. It can be seen that $n=36, k=6$ is the only integer solution. | 36 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Question: Among $1,2,3, \cdots, 1999,2000,2001$, what is the maximum number of numbers that can be chosen such that the sum of any three chosen numbers is divisible by 21? | 10. Take at most 95.
Let $a, b, c, d$ be 4 of the numbers taken. According to the problem, $a+b+c$ and $b+c+d$ are both multiples of 21. Therefore, the difference $a - d$ is also a multiple of 21. By the arbitrariness of $a$ and $d$, it can be concluded that the difference between any two of the numbers taken is a mul... | 95 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $\triangle A B C$ be an acute triangle, and construct isosceles triangles $\triangle D A C$, $\triangle E A B$, and $\triangle F B C$ outside $\triangle A B C$ such that $D A = D C$, $E A = E B$, and $F B = F C$, with $\angle A D C = 2 \angle B A C$, $\angle B E A = 2 \angle A B C$, and $\angle C F B = 2 \angle ... | Solution: Since $\angle ABC$ is an acute triangle, then $\angle ADC$, $\angle BEA$, $\angle CFB < \pi$. Therefore,
$$
\begin{array}{l}
\angle DAC = \frac{\pi}{2} - \frac{1}{2} \angle ADC = \frac{\pi}{2} - \angle BAC. \\
\angle BAE = \frac{\pi}{2} - \frac{1}{2} \angle BEA = \frac{\pi}{2} - \angle ABC. \\
\text{Thus, } \... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a_{1}=11^{11}, a_{2}=12^{12}, a_{3}=13^{13}$, and
$$
a_{n}=\left|a_{n-1}-a_{n-2}\right|+\left|a_{n-2}-a_{n-3}\right|, n \geqslant 4 \text {. }
$$
Find $a_{4^{4}}$. | For $n \geqslant 2$, define $s_{n}=\left|a_{n}-a_{n-1}\right|$. Then for $n \geqslant 5, a_{n}=s_{n-1}+s_{n-2}, a_{n-1}=s_{n-2}+s_{n-3}$. Thus, $s_{n}=$ $\left|s_{n-1}-s_{n-3}\right|$. Since $s_{n} \geqslant 0$, if $\max \left\{s_{n}, s_{n+1}, s_{n+2}\right\} \leqslant T$, then for all $m \geqslant n$, we have $s_{m} \... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 The three edge lengths of an isosceles tetrahedron are 3, $\sqrt{10}$, and $\sqrt{13}$. Then the radius of the circumscribed sphere of this tetrahedron is | Analysis: According to Basic Conclusion 10, the triangular pyramid can be expanded into a rectangular parallelepiped, making the known three edges the diagonals of the faces of the rectangular parallelepiped. At this point, the original triangular pyramid and the rectangular parallelepiped have the same circumscribed s... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Third Question: Before the World Cup, the coach of country $F$ plans to evaluate seven players, $A_{1}, A_{2}, \cdots, A_{7}$, by having them play in three training matches (each 90 minutes long). Assume that at any moment during the matches, exactly one of these players is on the field, and the total playing time (in ... | Solution: Let the times for $A_{1}, A_{2}, A_{3}, A_{4}$ be $7 k_{1}, 7 k_{2}$, $7 k_{3}, 7 k_{4}$, and their sum be $x: A_{5}, A_{6}, A_{7}$ have times $13 k_{5}$, $13 k_{6}, 13 k_{7}$, and their sum be $y$.
Then the original problem is to find the number of positive integer solutions $\left(k_{1}, k_{2}, \cdots, k_{... | 42244 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, try to find all positive integers $k$, such that for any positive numbers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a, b, c$. | Three, due to $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0$,
thus $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
It can be known that $k>5$. Noting that $k$ is a positive integer, therefore, $k \geqslant 6$.
Since there does not exist a triangle with side lengths $1,1,2$, according to the problem, we have
$$
k(1 \times 1+1 \... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Lift Your Veil
A 101-digit natural number $A=\underbrace{88 \cdots 8}_{\text {S0 digits }} \square \underbrace{99 \cdots 9}_{\text {S0 digits }}$ is divisible by 7. What is the digit covered by $\square$? | Explanation: β $\square$ β is a veil, covering the number to be found. To lift the veil, whether approaching from the front or the back, there are 50 digits in between, which is too many. The key is to find a way to reduce the number of digits.
The difference between two multiples of 7 is still a multiple of 7.
We alr... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Eight, let $A_{1}, A_{2}, \cdots, A_{8}$ be any 8 points taken on a plane. For any directed line $l$ taken on the plane, let the projections of $A_{1}, A_{2}, \cdots, A_{8}$ on this line be $P_{1}, P_{2}, \cdots, P_{8}$, respectively. If these 8 projections are pairwise distinct and are arranged in the direction of lin... | Proof 3: First, prove that $\triangle D E F$ is the triangle with the shortest perimeter among all inscribed triangles in the acute $\triangle A B C$.
Let point $D^{\prime}$ be any fixed point on side $B C$, as shown in Figure 4. Construct the reflection of point $D^{\prime}$ about $A B$ and $A C$.
Assume the number o... | 56 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, 17 football fans plan to go to Cao Guo to watch the World Cup football matches, and they have selected a total of 17 matches. The booking of tickets meets the following conditions:
(i) Each person can book at most one ticket per match;
(ii) The tickets booked by any two people have at most one match in common;
(... | Solution: Draw a $17 \times 17$ grid, with 17 columns representing 17 matches and 17 rows representing 17 people. If the $i$-th person has booked a ticket for the $j$-th match, then the center of the cell at the intersection of the $i$-th row and the $j$-th column is marked with a red dot. Thus, the problem is transfor... | 71 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. A bus, a truck, and a car are traveling in the same direction on the same straight line. At a certain moment, the truck is in the middle, the bus is in front, and the car is behind, and the distances between them are equal. After 10 minutes, the car catches up with the truck; after another 5 minutes, the car catches... | Ni.1.15.
As shown in Figure 4, let the distances between the truck and the bus, and the truck and the car be $S$. The speeds of the car, truck, and bus are approximately $a, b, c$. It is given that the truck catches up with the bus after $x$ minutes. Therefore, we have
$$
\begin{array}{l}
10(a-b)=S, \\
15(a-c)=2S, \\
x... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If the polynomial $P=2 a^{2}-8 a b+17 b^{2}-16 a-4 b$ +2070, then the minimum value of $P$ is | $$
\begin{array}{l}
2.2002 \\
P=2 a^{2}-8(b+2) a+17 b^{2}-4 b+2070 \\
=2\left[a^{2}-4(b+2) a+4(b+2)^{2}\right]-8(b+2)^{2} \\
\quad+17 b^{2}-4 b+2070 \\
=2(a-2 b-4)^{2}+9 b^{2}-36 b+36+2002 \\
=2(a-2 b-4)^{2}+9(b-2)^{2}+2002
\end{array}
$$
When $a-2 b-4=0$ and $b-2=0$, $P$ reaches its minimum value of 2002. | 2002 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure $1, \angle A O B=30^{\circ}$, within $\angle A O B$ there is a fixed point $P$, and $O P$ $=10$, on $O A$ there is a point $Q$, and on $O B$ there is a fixed point $R$. To make the perimeter of $\_P Q R$ the smallest, the minimum perimeter is | 3.10 .
As shown in Figure 5, construct the symmetric points $P_{1}, P_{2}$ of $P$ with respect to the sides $O A, O B$ of $\angle A O B$. Connect $P_{1} P_{2}$, intersecting $O A, O B$ at $Q, R$. Connect $P Q, P R$. It is easy to see that
$$
P Q=P_{1} Q, P R=P_{2} R,
$$
which implies
$$
\begin{aligned}
l_{\triangle P... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Given theorem: βFor any prime number $n$ greater than 3, $b$ and $c$ satisfy the equation $2a + 5b = c$. Then $a + b + c$ is how many times the integer $n$? Prove your conclusion.β
Translate the above text into English, please retain the original text's line breaks and format, and output the trans... | Three, the maximum possible value of $n$ is 9.
First, we prove that $a+b+c$ is divisible by 3.
In fact, $a+b+c=a+b+2a+5b=3(a+2b)$.
Thus, $a+b+c$ is a multiple of 3.
Let the remainders when $a$ and $b$ are divided by 3 be $r_{a}$ and $r_{b}$, respectively, with $r_{a} \neq 0$ and $r_{b} \neq 0$.
If $r_{a} \neq r_{b}$, ... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 7 Multiplicative Magic Square
Figure 1 shows a partially filled magic square. Fill in the following nine numbers: $\frac{1}{4}, \frac{1}{2}, 1,2,4,8,16,32,64$ in the grid so that the product of the numbers in each row, column, and diagonal is the same. The number that should be filled in the β $x$ β cell is $\q... | Explanation: Label the unfilled cells with letters, as shown in Figure 2.
The product of the nine known numbers is
$$
\frac{1}{4} \times \frac{1}{2} \times 8 \times 1 \times 2 \times 32 \times 4 \times 16 \times 64=64^{3} \text {. }
$$
Therefore, the fixed product of the three numbers in each row, each column, and eac... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
5. The value of the 1000th term in the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6, \cdots$ is ( ).
(A) 42
(B) 45
(C) 48
(D) 51 | 5. (B).
The key is to determine which number segment $a_{1000}$ falls into. Suppose $a_{1000}$ is in the $k$-th number segment, then $a_{100}=k$. Since the $k$-th number segment contains $k$ numbers, $k$ should satisfy the inequality $\frac{(k-1) k}{2}=1+2+3+\cdots+(k-1)<1000 \leqslant 1+2+3+\cdots+k=\frac{k(k+1)}{2}$... | 45 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
3. The remainder when the number $7^{99}$ is divided by 2550 is | 3.343 .
Notice that $2550=50 \times 51$. First, we find the remainders when $7^{*})$ is divided by 5 and 51, respectively.
First, $7^{199}=7 \times 7^{48}=7 \times 49^{19} \equiv 7 \times(-2)^{459}$
$\equiv-14 \times 16^{12} \equiv-14 \times 256^{6} \equiv-14 \times 1^{6}$
$\equiv-14(\bmod 51)$;
Additionally, $7^{(9)}... | 343 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
120. In a convex $n$-sided polygon, the difference between any two adjacent interior angles is $1^{\circ}$. Find the maximum value of the difference between the largest and smallest interior angles. | Solution: Let the largest interior angle be $\alpha$, and the smallest interior angle be $\alpha-k$ (omitting the degree symbol for simplicity, the same applies below). Therefore, what we are looking for is the maximum value of $k$. Hence, the interior angles should also include $\alpha-1, \alpha-2, \alpha-3, \cdots$, ... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. $x, y$ are real numbers, and $\left(x+\sqrt{x^{2}+1}\right)(y+$ $\left.\sqrt{y^{2}+1}\right)=1$. Then $x+y=$ $\qquad$ . | δΊγ1.0.
Multiply both sides of the original equation by $\sqrt{x^{2}+1}-x$, we get $y+\sqrt{y^{2}+1}=\sqrt{x^{2}+1}-x$; multiply both sides of the original equation by $\sqrt{y^{2}+1}-y$, we get $x+\sqrt{x^{2}+1}=\sqrt{y^{2}+1}-y$. Adding the two equations immediately yields $x+y=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. A triangle with all sides as integers, and the longest side being 11, has
$\qquad$ possibilities. | Ni, 1.36.
Let the three sides be $x, y, z$, and $x \leqslant y \leqslant z, z=11, x+$ $y>z$, then
$$
6 \leqslant y \leqslant 11 .
$$
When $y=6, x=6$, it forms 1 triangle;
When $y=7, x$ is $5, 6, 7$, it forms 3 triangles; .....
When $y=11, x$ is $1, 2, \cdots, 11$, it forms 11 triangles. Therefore, there are a total of... | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. If $\left|\log \frac{a}{\pi}\right|<2$, then the number of $a$ for which the function $y=\sin (x+a)+\cos (x-a)(x \in \mathbf{R})$ is an even function is $\qquad$. | 2.10.
$$
\begin{aligned}
\because-2 & <\log \frac{\alpha}{\pi}<2, \therefore \frac{1}{\pi}<\alpha<\pi^{3} . \\
f(x)= & \sin (x+\alpha)+\cos (x+\alpha) \\
& =\sqrt{2} \sin \left(x+\alpha+\frac{\pi}{4}\right) .
\end{aligned}
$$
Since $f(x)$ is an even function,
$$
\therefore \sin \left(\frac{\pi}{4}+\alpha+x\right)=\sin ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. From $\{1,2,3, \cdots, 20\}$, select 3 numbers such that no two numbers are adjacent, there are $\qquad$ different ways. | 4.816.
17 numbers produce 16 spaces plus 2 ends making a total of 18 spaces. According to the problem, there are $\mathrm{C}_{18}^{3}$ different ways to choose. | 816 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. The largest positive integer $n$ for which the inequality $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$ holds for a unique integer $k$ is $\qquad$ . | 6.112 .
Obviously, $\frac{13}{7}<\frac{n+k}{n}<\frac{15}{8} \Leftrightarrow \frac{6}{7}<\frac{k}{n}<\frac{7}{8}$
$$
\Leftrightarrow 48 n<56 k<49 n \text {. }
$$
There are at least $n-1$ integers between $48 n+1$ and $49 n-1$.
If $n-1 \geqslant 2 \times 56$, then there must be at least 2 multiples of 56 in this range, ... | 112 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Total 20 points) The sequence $\left\{x_{n}\right\}$ satisfies
$$
\begin{array}{l}
x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, n \in \mathbf{N}, y_{n}=\frac{1}{1+x_{n}}, \\
S_{n}=y_{1}+y_{2}+\cdots+y_{n}, P_{n}=y_{1} y_{2} \cdots y_{n} .
\end{array}
$$
Find $P_{n}+\frac{1}{2} S_{n}$. | $$
\begin{array}{l}
\text { Three, } \because x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, \\
\therefore x_{n+1}>x_{n}>0, x_{n+1}=x_{n}\left(1+x_{n}\right) . \\
\therefore y_{n}=\frac{1}{1+x_{n}}=\frac{x_{n}^{2}}{x_{n} x_{n+1}}=\frac{x_{n+1}-x_{n}}{x_{n} x_{n+1}}=\frac{1}{x_{n}}-\frac{1}{x_{n+1}} . \\
\therefore P_{n}=y... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Total 50 points) In a $13 \times 13$ square grid, select the centers of $k$ small squares such that no four of these points form the vertices of a rectangle (with sides parallel to those of the original square). Find the maximum value of $k$ that satisfies the above requirement. | Three, let the $i$-th column have $x_{i}$ points $(i=1,2, \cdots, 13)$, then $\sum_{i=1}^{13} x_{i}=k$, the $x_{i}$ points in the $i$-th column form $\mathrm{C}_{x_{i}}^{2}$ different point pairs (if $x_{i}<2$, then $\mathrm{C}_{x_{i}}^{2}=0$).
If an additional column is added to the side of a $13 \times 13$ square, a... | 52 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
II. Find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in $S$ is a positive integer not exceeding 100;
(2) For any two different elements $a, b$ in $S$, there exists an element $c$ in $S$ such that the greatest common divisor (gcd) of $a$ and $c$ is 1, and the gcd... | II. 72.
Express each positive integer $n$ not exceeding 100 as
$$
n=2^{\alpha_{1}} \cdot 3^{a_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4}} \cdot 11^{a_{5}} \cdot q \text {. }
$$
where $q$ is a positive integer not divisible by $2, 3, 5, 7, 11$, and $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}$ are non-negative ... | 72 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let real numbers $a, b, c, d$ satisfy $a^{2}+b^{2}+c^{2}+d^{2}=5$. Then the maximum value of $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}$ $+(c-d)^{2}$ is $\qquad$ . | 6. 20 | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four. (This question is worth 18 points)
On a piece of paper, there are $1,2, \cdots, n$ these $n$ positive integers. In the first step, the first 4 numbers $1,2,3,4$ are crossed out, and the sum of the 4 crossed-out numbers, 10, is written at the end of $n$; in the second step, the first 4 numbers $5,6,7,8$ are crosse... | (1) With each step, the number of numbers on the paper decreases by 3. If $n$ numbers are reduced to 1 number after $p$ steps, then $n-3p=1$, i.e., $n=3p+1$.
$\therefore$ The necessary and sufficient condition for $n$ is that $n$ leaves a remainder of 1 when divided by 3.
(2) Suppose there are $4^{k}$ numbers initially... | 12880878 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 A home appliance manufacturing company, based on market research analysis, has decided to adjust its production plan and is preparing to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours). Moreover, the production of refrigerators should be ... | Solution: Let the number of air conditioners produced each week be $x$, the number of color TVs be $y$, and the number of refrigerators be $z$, with the total output value being $w$ thousand yuan. According to the problem, we have
$$
\left\{\begin{array}{l}
x+y+z=360, \\
\frac{1}{2} x+\frac{1}{3} y+\frac{1}{4} z=120, \... | 1050 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. The number of sets $X$ that satisfy the condition $\{1,2,3\} \subseteq X \subseteq\{1,2,3,4,5,6\}$ is $\qquad$ . | ii. 7.8 items.
$X$ must include the 3 elements $1,2,3$, while the numbers 4,5,6 may or may not belong to $X$. Each number has 2 possibilities, so the total number of different $X$ is $2^{3}=8$. | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. The function $y=f(x)$ defined on $\mathbf{R}$ has the following properties:
οΌ1οΌFor any $x \in \mathbf{R}$, $f\left(x^{3}\right)=f^{3}(x)$;
(2) For any $x_{1} γ x_{2} \in \mathbf{R}, x_{1} \neq x_{2}$, $f\left(x_{1}\right)$ $\neq f\left(x_{2}\right)$.
Then the value of $f(0)+f(1)+f(-1)$ is $\qquad$ | 10.0 .
From $f(0)=f^{3}(0)$, we know $f(0)[1-f(0)][1+f(0)]=0$,
thus, $f(0)=0$ or $f(0)=1$, or $f(0)=-1$;
from $f(1)=f^{3}(1)$, similarly $f(1)=0$ or 1 or 1;
from $f(-1)=f^{3}(-1)$, similarly $f(-1)=0$ or 1 or -1.
However, $f(0)$, $f(1)$, and $f(-1)$ are pairwise distinct,
so $\{f(0), f(1), f(-1)\}=\{0,1,-1\}$.
Therefo... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. (13 points) Let the quadratic equation in $x$, $2 x^{2}-t x-2=0$, have two roots $\alpha, \beta(\alpha<\beta)$.
(1) If $x_{1} γ x_{2}$ are two different points in the interval $[\alpha, \beta]$, prove that $4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-4<0$;
(2) Let $f(x)=\frac{4 x-t}{x^{2}+1}$, and the maximum and minim... | $$
\begin{aligned}
14. (1) & \text{ From the conditions, we have } \alpha+\beta=\frac{t}{2}, \alpha \beta=-1. \\
& \text{ Without loss of generality, assume } \alpha \leqslant x_{1}4 x_{1} x_{2}-2(\alpha+\beta)\left(x_{1}+x_{2}\right)+4 \alpha \beta=4 x_{1} x_{2} \\
& -t\left(x_{1}+x_{2}\right)-4 .
\end{aligned}
$$
Th... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. (13 points) Given $a_{1}=1, a_{2}=3, a_{n+2}=(n+3) a_{n+1}$ $-(n+2) a_{n}$. If for $m \geqslant n$, the value of $a_{m}$ can always be divided by 9, find the minimum value of $n$.
| 15. From $a_{n+2}-a_{n+1}=(n+3) a_{n+1}-(n+2) a_{n}$
$$
\begin{array}{l}
-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)=(n+2)(n+1) \cdot \\
\left(a_{n}-a_{n-1}\right)=\cdots=(n+2) \cdot(n+1) \cdot n \cdots+4 \cdot 3 \cdot\left(a_{2}-\right. \\
\left.a_{1}\right)=(n+2)!, \\
\text { Therefore, } a_{n}=a_{1}+\left(a_{2}-a_{1}\r... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the sequence $\left\{a_{n}\right\}$, where $a_{n}$ is a real number, and for $n \geqslant 3, n \in$ $\mathbf{N}$, we have $a_{n}=a_{n-1}-a_{n-2}$. If the sum of the first 1985 terms is 1000, and the sum of the first 1995 terms is 4000, then the sum of the first 2002 terms is $\qquad$. | 8.3000
Let $a_{1}=a, a_{2}=b$, then the first 6 terms of the sequence are $a, b, b-a$, $-a, -b, a-b$. Also, $a_{7}=a, a_{8}=b$, it is easy to verify that the sum of any consecutive 6 terms of this sequence is 0.
$$
\begin{aligned}
\therefore S_{1995} & =S_{332 \times 6+3}=0+a+b+(b-a) \\
& =2 b=4000 . \\
S_{1 \text { g... | 3000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. The expansion of $(a+b+c)^{10}$, after combining like terms, has
$\qquad$ terms. | 12.66.
The form of each term after merging is $a^{k_{1}} b^{k_{2}} c^{k_{3}}$, where $k_{1}+k_{2}+k_{3}=10\left(k_{1} γ k_{2} γ k_{3} \in \mathrm{N}\right)$, and when $k_{1} γ k_{2}$ are determined, $k_{3}$ is uniquely determined.
If $k_{1}=10, k_{2}=0$, there is 1 way;
If $k_{1}=9, k_{2}=0$ or 1, there are 2 ways;
If... | 66 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Given $p$ is a prime number, $r$ is the remainder when $p$ is divided by 210. If $r$ is a composite number and can be expressed as the sum of two perfect squares, find
(52nd Belarusian Mathematical Olympiad (Final C Category)) | Let $p=210n+r$, then $07$.
Let $q$ be the smallest prime dividing $r$, i.e., $r=qm, q \leqslant m$, we get $210>r=qm \geqslant q^{2}$, so, $q \leqslant 13$.
On the other hand, $r$ cannot be divisible by 2, 3, 5, 7, otherwise $p$ would also be divisible by the same number, but $p$ is a prime, so $q>7$, hence $q=11$ or $... | 169 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 A certain project, if contracted by Team A and Team B, can be completed in $2 \frac{2}{5}$ days, costing 180000 yuan; if contracted by Team B and Team C, it can be completed in $3 \frac{3}{4}$ days, costing 150000 yuan; if contracted by Team A and Team C, it can be completed in $2 \frac{6}{7}$ days, costing ... | Solution: Let the time required for A, B, and C to complete the work alone be $x$, $y$, and $z$ days, respectively, then
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=\frac{5}{12} \\
\frac{1}{y}+\frac{1}{z}=\frac{4}{15} \\
\frac{1}{z}+\frac{1}{x}=\frac{7}{20}
\end{array} .\right.
$$
Solving, we get $x=4, y=6, z=1... | 177000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. In a mathematics competition,
(i) the number of problems is $n(n \geqslant 4)$;
(ii) each problem is solved by exactly 4 people;
(iii) for any two problems, exactly 1 person solves both problems.
If the number of participants is greater than or equal to $4 n$, find the minimum value of $n$ such that there always e... | Solution: Let $n \geqslant 14$, and represent each problem as a rectangle, with its 4 vertices representing the 4 contestants who solved the problem. Let $P$ be any problem.
Each problem different from $P$ shares a common vertex with $P$, since $n \geqslant 14$. By the pigeonhole principle, there is a vertex of $P$, d... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
17. 14 people participate in a Japanese chess round-robin tournament, where each person plays against the other 13 people, and there are no draws in the matches. Find the maximum number of "triangles" (here, a "triangle" refers to a set of three people where each person has one win and one loss).
(2002, Japan Mathemat... | Solution: Let the 14 people be represented as $A_{1}, A_{2}, \cdots, A_{14}$, and let $A_{4}$ win $a_{4}$ games. In any group of three people (i.e., selecting 3 out of 14 people), if there is no "double corner," then one person must have won the other two. Therefore, the number of groups of three that do not form a "tr... | 112 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
18. Given a set of 2002 points on the $x O y$ plane, denoted as $S$, and no two points in $S$ are connected by a line parallel to the coordinate axes. For any two distinct points $P, Q$ in $S$, consider the rectangle with $P Q$ as its diagonal, and whose sides are parallel to the coordinate axes. Let $W_{P Q}$ denote t... | Solution: Here we only discuss rectangles with sides parallel to the coordinate axes. For any two points $P, Q$ in $S$, $R_{P Q}$ denotes the rectangle with $P Q$ as its diagonal, and $W_{P Q}$ denotes the number of points in $S$ within the rectangle $R_{P Q}$. Here, $P$ and $Q$ do not have to be distinct; if $P$ and $... | 400 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. As shown in Figure 1, the area of square $A B C D$ is 256, point $F$ is on $A D$, and point $E$ is on the extension of $A B$. The area of right triangle $\triangle C E F$ is 200. Then the length of $B E$ is $(\quad)$.
(A) 10
(B) 11
(C) 12
(D) 15 | 2. (C).
It is easy to prove that $\mathrm{Rt} \triangle C D F \cong \mathrm{Rt} \triangle C B E$. Therefore, $C F=C E$.
Since the area of Rt $\triangle C E F$ is 200, i.e., $\frac{1}{2} \cdot C F \cdot C E=200$, hence $C E=20$.
And $S_{\text {square } B C D}=B C^{2}=256$, so $B C=16$.
By the Pythagorean theorem, $B E=... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false |
One. (20 points) In a square with an area of 1, construct a smaller square as follows: divide each side of the unit square into $n$ equal parts, then connect each vertex to the nearest division point of its opposite vertex, as shown in Figure 5. If the area of the smaller square is exactly $\frac{1}{3281}$, find the va... | I. Draw $C_{1} P \perp A_{1} C$, with the foot of the perpendicular being $P$. Since the area of the square $A B C D$ is 1, we have
$$
A B=B C=C D=1, A_{1} B=\frac{n-1}{n}, C C_{1}=\frac{1}{n} .
$$
By the Pythagorean theorem,
$$
A_{1} C=\sqrt{B C^{2}+A_{1} B^{2}}=\sqrt{1^{2}+\left(\frac{n-1}{n}\right)^{2}} .
$$
It is... | 41 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. In the sequence of positive integers, starting from 1, certain numbers are painted red according to the following rules. First, paint 1, then paint 2 even numbers $2, 4$; then paint the 3 nearest consecutive odd numbers after 4, which are $5, 7, 9$; then paint the 4 nearest consecutive even numbers after 9, which ar... | 6. (B).
The first time, 1 number is painted red: $1,1=1^{2}$;
The second time, 2 numbers are painted red: $2,4,4=2^{2}$;
The third time, 3 numbers are painted red: $5,7,9,9=3^{2}$.
It is easy to see that the last number painted red in the $k$-th time is $k^{2}$, then the $k+1$ numbers painted red in the $k+1$-th time ... | 3943 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
8. As shown in Figure $11, \angle A O B=$ $30^{\circ}, \angle A O B$ contains a fixed point $P$, and $O P=10, O A$ has a point $Q, O B$ has a fixed point $R$. If the perimeter of $\triangle P Q R$ is minimized, find its minimum value. | (Tip: Draw auxiliary lines with $O A$ and $O B$ as axes of symmetry. Answer:
10.) | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let the function $f_{0}(x)=|x|, f_{1}(x)=$ $\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed figure formed by the graph of $y$ $=f_{2}(x)$ and the $x$-axis is $\qquad$
(1989, National High School Mathematics Competition) | Analysis: If it is not easy to directly draw the graph of $y=f_{2}(x)$, but we know the relationship between the graphs of $y=f(x)$ and $y=|f(x)|$. If we follow the sequence
$$
\begin{array}{l}
f_{0}(x)=|x| \rightarrow y=f_{0}(x)-1 \\
\rightarrow f_{1}(x)=\left|f_{0}(x)-1\right| \rightarrow y=f_{1}(x)-2 \\
\rightarrow ... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 The sequence $a_{1}, a_{2}, a_{3}, \cdots, a_{2 n}, a_{2 n+1}$ forms an arithmetic sequence, and the sum of the terms with odd indices is 60, while the sum of the terms with even indices is 45. Then the number of terms $n=$ $\qquad$ | Analysis: From $\left\{\begin{array}{l}a_{1}+a_{3}+\cdots+a_{2 n+1}=60, \\ a_{2}+a_{4}+\cdots+a_{2 n}=45\end{array}\right.$ $\Rightarrow\left\{\begin{array}{l}(n+1) a_{n+1}=60, \\ n a_{n+1}=45\end{array} \Rightarrow \frac{n+1}{n}=\frac{4}{3}\right.$, solving gives $n=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 12 Given the three side lengths $a$, $b$, and $c$ of $\triangle ABC$ satisfy: (1) $a>b>c$, (2) $2b=a+c$, (3) $b$ is an integer, (4) $a^{2}+b^{2}+c^{2}=84$. Then the value of $b$ is $\qquad$ | Analysis: Starting from $2 b=a+c$, let $a=b+d$, $c=b-d(d>0)$, then
$$
(b+d)^{2}+b^{2}+(b-d)^{2}=84 .
$$
That is, $3 b^{2}+2 d^{2}=84$.
Obviously, $2 d^{2}$ is a multiple of 3. Also, by $b+c>a \Rightarrow$ $b>2 d$, substituting into (4) gives $2 d^{2}<12$. Therefore, $2 d^{2}=3,6,9$. Verification shows $b=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 13 On the coordinate plane, points with both integer horizontal and vertical coordinates are called integer points. For any natural number $n$, connect the origin $O$ with $A_{n}(n, n+3)$, and let $f(n)$ denote the number of integer points on the line segment $O A_{n}$, excluding the endpoints. Then
$$
f(1)+f(2... | Analysis: It is easy to obtain the equation of $O A_{n}$ as
$$
y=\frac{n+3}{n} x=x+\frac{3}{n} x \quad(0 \leqslant x \leqslant n) .
$$
Obviously, for integer points, $n$ must be a multiple of 3. Let $n=3 k$, then
$$
y=\frac{k+1}{k} x \text {. }
$$
When $x=k$ and $2 k$, we get the integer points $(k, k+1)$, $(2 k, 2(k... | 1334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 14 Let $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants. If $f(1)=10$, $f(2)=20$, $f(3)=30$, then $f(10)+f(-6)=$ $\qquad$
(1998, Zhongshan City, Guangdong Province Mathematics Competition) | Analysis: Based on $f(1)=10, f(2)=20, f(3)=$ 30, construct
$$
\begin{array}{c}
f(x)=(x-1)(x-2)(x-3)(x-t)+10 x . \\
\text { Hence } f(10)+f(-6)=9 \times 8 \times 7(10-t)+ \\
10 \times 10+[(-7) \times(-8) \times(-9)(-6-t)- \\
60]=8104 .
\end{array}
$$ | 8104 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 19 Let $f(x)=|1-2 x|, x \in[0,1]$. Then, the number of solutions to the equation $f\{f[f(x)]\}=\frac{1}{2} x$ is $\qquad$ . | Analysis: Let $y=f(x), z=f(y), w=f(z)$, use β $\rightarrow$ β to indicate the change, we have
$$
\begin{array}{l}
x: 0 \rightarrow 1, y: 1 \rightarrow 0 \rightarrow 1, \\
z: 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1, \\
w: 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarr... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 As shown in Figure 4, in rectangle $A B C D$, $A B=$ $20 \text{ cm}, B C=10 \text{ cm}$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value.
(1998, Beijing Junior High School Mathematics Competition) | Solution: As shown in Figure 4, construct $\angle B_{1} A C = \angle B A C$, draw $B Q \perp A C$ at $Q$, and extend $B Q$ to intersect $A B_{1}$ at $B_{1}$; draw $N P \perp A C$ at $P$, and extend $N P$ to intersect $A B_{1}$ at $N_{1}$, then points $B_{1}$ and $B$, and points $N_{1}$ and $N$ are symmetric with respec... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 If $x \in \mathbf{R}$, find the maximum value of $F(x)=\min \{2 x+1$, $x+2,-x+6\}$.
(38th AHSME) | Solution: As shown in Figure 1, draw the graph of $F(x)$ (the solid part). From the graph, we can see that the maximum value of $F(x)$ is equal to the y-coordinate of the intersection point of $y=x+2$ and $y=-x+6$.
Therefore, the maximum value of $F(x)$ is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}$, $\cdots, b_{n}$ satisfying
$$
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1 .
$$
Find the maximum value of $F=\min \left\{\frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \cdots, \frac{a_{n}}{b_{n}}\right\}$.
(1979, G... | Solution: It is easy to see that when all the letters are equal, the value of $F$ is 1.
Below we prove that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$; $b_{1}, b_{2}, \cdots, b_{n}$, we have $F \leqslant 1$.
If not, then $F>1$.
$$
\text { Hence } \frac{a_{1}}{b_{1}}>1, \frac{a_{2}}{b_{2}}>1, \cdots, \frac{... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
7. If real numbers $x, y, z$ satisfy $x+\frac{1}{y}=4, y+\frac{1}{z}=1, z+$ $\frac{1}{x}=\frac{7}{3}$, then the value of $x y z$ is $\qquad$. | 7.1 .
Since $4=x+\frac{1}{y}=x+\frac{1}{1-\frac{1}{z}}=x+\frac{z}{z-1}$
$=x+\frac{\frac{7}{3}-\frac{1}{x}}{\frac{7}{3}-\frac{1}{x}-1}=x+\frac{7 x-3}{4 x-3}$, then
$4(4 x-3)=x(4 x-3)+7 x-3$,
which simplifies to $(2 x-3)^{2}=0$.
Thus, $x=\frac{3}{2}$.
Therefore, $z=\frac{7}{3}-\frac{1}{x}=\frac{5}{3}, y=1-\frac{1}{z}=\f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Given the quadratic function $y=a x^{2}+b x+c$ (where $a$ is a positive integer) whose graph passes through the points $A(-1,4)$ and $B(2,1)$, and intersects the $x$-axis at two distinct points. Then the maximum value of $b+c$ is $\qquad$ . | 10. -4 .
From the given, we have $\left\{\begin{array}{l}a-b+c=4, \\ 4 a+2 b+c=1,\end{array}\right.$ solving this yields $\left\{\begin{array}{l}b=-a-1, \\ c=3-2 a .\end{array}\right.$
Since the graph of the quadratic function intersects the $x$-axis at two distinct points, we have,
$$
\begin{array}{l}
\Delta=b^{2}-4 ... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$. | 14. (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$. From the problem, we know $a>0$, and $b+c=2-a, bc=\frac{4}{a}$. Therefore, $b, c$ are the two real roots of the quadratic equation $x^{2}-(2-a)x+\frac{4}{a}=0$, then
$$
\begin{array}{l}
\Delta=(2-a)^{2}-4 \tim... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a_{1}, a_{2}, \cdots, a_{2002}$ have values of 1 or -1, let $S$ be the sum of the pairwise products of these 2002 numbers.
(1) Find the maximum and minimum values of $S$, and specify the conditions under which these maximum and minimum values can be achieved;
(2) Find the smallest positive value of $S$, ... | $$
\begin{array}{l}
\text { 3. (1) }\left(a_{1}+a_{2}+\cdots+a_{200}\right)^{2} \\
=a_{1}^{2}+a_{2}^{2}+\cdots+a_{2002}^{2}+2 m=2002+2 m, \\
m=\frac{\left(a_{1}+a_{2}+\cdots+a_{200}\right)^{2}-2002}{2} .
\end{array}
$$
When $a_{1}=a_{2}=\cdots=a_{200}=1$ or -1, $m$ reaches its maximum value of 2003001.
When $a_{1}, a... | 57 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $0 \leqslant a-b \leqslant 1,1 \leqslant a+b \leqslant 4$. Then, when $a-2 b$ reaches its maximum value, the value of $8 a+2002 b$ is $\qquad$ . | 3.8.
$$
\begin{array}{l}
\text { Let } 0 \leqslant a-b \leqslant 1, \\
1 \leqslant a+b \leqslant 4, \\
\text { and } m(a-b)+n(a+b)=a-2 b .
\end{array}
$$
By comparing the coefficients of $a$ and $b$ on both sides, we get the system of equations and solve to find
$$
m=\frac{3}{2}, n=-\frac{1}{2} \text {. }
$$
Thus, $a-... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. If a positive integer is equal to 4 times the sum of its digits, then we call this positive integer a quadnumber. The sum of all quadnumbers is $\qquad$ .
| 4.120.
Case analysis.
There are no four-composite numbers that are single-digit numbers.
For two-digit four-composite numbers, they satisfy
$$
\overline{a b}=4(a+b) \text {. }
$$
Thus, the four-composite numbers can be found to be $12, 24, 36, 48$, and their total sum is 120. For three-digit four-composite numbers, t... | 120 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $A$ be a $k$-element subset of the set $\{1,2,3, \cdots, 16\}$, and any two subsets of $A$ have different sums of elements. For any $(k+1)$-element subset $B$ of the set $\{1,2,3, \cdots, 16\}$ that contains $A$, there exist two subsets of $B$ whose sums of elements are equal.
(1) Prove: $k \leqslant 5$;
(2) Fin... | Proof: (1) Since $A$ has $2^{k}$ subsets (including the empty set), and the sum of elements of any two subsets is not equal, the sum of elements of $A$ must have at least $2^{k}-1$ different values.
If $k \geqslant 7$, then $2^{k}-1>16 k$, which is impossible.
If $k=6$, consider the one-, two-, three-, and four-element... | 66 | Combinatorics | proof | Yes | Yes | cn_contest | false |
8. A $9 \times 9$ grid of squares is colored in two colors, black and white, such that the number of black squares adjacent to each white square is greater than the number of white squares, and the number of white squares adjacent to each black square is greater than the number of black squares (squares sharing a commo... | Solution: A coloring scheme of a square matrix that satisfies the problem's conditions is called a good scheme. For a good coloring scheme, it is easy to prove the following conclusions:
(1) If two adjacent cells have the same color and are in different rows (or columns), then any two adjacent cells in these two rows (... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12.12 People are sitting around a table in a circle. How many ways are there for 6 pairs of people to shake hands so that no arms cross?
(2001, British Mathematical Olympiad (Round 1)) | Solution: According to the problem, handshakes only occur on the table, not behind other people.
Let $N_{p}$ denote the total number of ways for $2p$ people sitting around a circular table to shake hands without crossing arms, and each person shakes hands only once. Consider each person as a point. Without loss of gen... | 132 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
17. Given a wire of length $150 \mathrm{~cm}$, it is to be cut into $n(n>2)$ smaller segments, each of which has an integer length of no less than $1(\mathrm{~cm})$. If no three segments can form a triangle, find the maximum value of $n$, and how many ways there are to cut the wire into $n$ segments that satisfy the co... | 17. Since the sum of $n$ segments is a fixed value of $150(\mathrm{~cm})$, to make $n$ as large as possible, the length of each segment must be as small as possible. Given that the length of each segment is no less than $1(\mathrm{~cm})$, and no three segments can form a triangle, the lengths of these segments can only... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. For all positive integers $n$ greater than 2, the greatest common divisor of the number $n^{5}-5 n^{3}+4 n$ is
untranslated part:
untranslated part remains the same as it is a mathematical expression. | 4.120 .
$$
n^{5}-5 n^{3}+4 n=(n-2)(n-1) n(n+1)(n+2) \text {. }
$$
For any positive integer $n$ greater than 2, the number $n^{5}-5 n^{3}+4 n$ contains the common divisor $1 \times 2 \times 3 \times 4 \times 5=120$. Therefore, the greatest common divisor of these numbers is 120. | 120 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) A scientific expedition team went to the upper reaches of a certain river to investigate an ecological area. After setting out, they advanced at a speed of $17 \mathrm{~km}$ per day, traveling upstream along the riverbank for several days before reaching their destination. They then spent several days... | Let the expedition team took $x$ days to reach the ecological area, $y$ days to return, and spent $z$ days on the investigation. Then,
$$
x+y+z=60,
$$
and $17 x-25 y=-1$ or $25 y-17 x=1$.
(Here $x, y$ are positive integers)
First, find a special solution $\left(x_{0}, y_{0}\right)$ of equation (1) (here $x_{0}, y_{0}$... | 23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Mr. Huang's home phone number is an eight-digit number. The sum of the number formed by the first four digits and the number formed by the last four digits is 14405. The sum of the number formed by the first three digits and the number formed by the last five digits is 16970. Mr. Huang's home phone number is | 14. 82616144 .
Let the phone number be $100000 x+10000 y+z$, where $x$, $y$, and $z$ are all natural numbers, and $100 \leqslant x \leqslant 999, 0 \leqslant y \leqslant 9, 1000 \leqslant z \leqslant 9999$. Then
$$
\left\{\begin{array}{l}
10 x+y+z=14405, \\
x+10000 y+z=16970
\end{array} \Rightarrow 1111 y-x=285 .\righ... | 82616144 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. A bouncy ball falls from point $A$ to the ground, bounces up to point $B$, then falls to a platform $20 \mathrm{~cm}$ high, bounces up to point $C$, and finally falls to the ground. Each time it bounces, the height it reaches is $80 \%$ of the height from which it fell. It is known that point $A$ is $68 \mathrm{~cm... | Three, 15. Let the height of point $C$ above the ground be $x \mathrm{~cm}$, then $\frac{\frac{x-20}{80 \%}+20}{80 \%}-68=x$. Solving this, we get $x=132$. Answer:η₯.
Note: The word "η₯" at the end is not translated as it seems to be a placeholder or abbreviation in the original text, possibly indicating that the detail... | 132 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. The teachers and students of a township primary school went to the county town for a visit. It was stipulated that the bus would depart from the county town and arrive at the school at 7:00 AM to pick up the visiting teachers and students and immediately head to the county town. However, the bus broke down on its w... | 15. Suppose it took \( x \) minutes to fix the car. As shown in Figure 8, let point \( A \) be the location of the county town, point \( C \) be the location of the school, and point \( B \) be the place where the teachers and students met the car on the way.
In the 30 minutes that the teachers and students were late ... | 38 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $a$ and $b$ are both prime numbers, and $a^{2}+b=2003$, then the value of $a+b$ is ( ).
(A) 1999
(B) 2000
(C) 2001
(D) 2002 | $-γ 1 .(\mathrm{C})$.
Since 2003 is an odd number, one of $a$ or $b$ must be the even prime number 2. If $b=2$, then $a^{2}=2001$, but 2001 is not a perfect square. Therefore, $a=2, b=1999$, and we have $a+b=2001$. | 2001 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
3. The three sides of $\triangle A B C$, $a$, $b$, and $c$, satisfy $b+c=$ $8, bc=a^{2}-12a+52$. Then the perimeter of $\triangle A B C$ is ( ).
(A) 10
(B) 14
(C) 16
(D) cannot be determined | 3. (B).
Since $(b-c)^{2}=64-4\left(a^{2}-12 a+52\right)$
$$
=-4(a-6)^{2} \geqslant 0 \text {, }
$$
Therefore, $a=6$, and $b=c$.
Also, $b+c=8$, then $b=c=4$. Thus, $a+b+c=14$. | 14 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given that $a$, $b$, and $c$ are real numbers, and the polynomial $x^{3}+$ $a x^{2}+b x+c$ can be divided by $x^{2}+3 x-4$. Then $2 a-2 b$ $-c=$ $\qquad$ . | $=γ 1.14$.
From the problem, -4 and 1 are roots of the equation $x^{3}+a x^{2}+b x+c=0$, which means
$$
\begin{array}{l}
16 a-4 b+c=64, \\
a+b+c=-1 .
\end{array}
$$
(1) $\times 6$ - (2) gives $2 a-2 b-c=14$. | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If positive integers $a$, $b$, and $c$ satisfy $a b + b c = 518$, $a b - a c = 360$, then the maximum value of $a b c$ is $\qquad$ . | 2. 1008 .
From $ab+bc=518, ab-ac=360$, subtracting the two equations, we get $c(a+b)=2 \times 79$.
Upon verification, taking $c=2, a+b=79$, then $ab=518-bc=518-2b$.
Upon inspection, when $b=7$, $ab$ reaches its maximum value of 504. Therefore, the maximum value of $abc$ is 1008. | 1008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If $a b c=1, \frac{x}{1+a+a b}+\frac{x}{1+b+b c}+$ $\frac{x}{1+c+a c}=2003$, then $x=$ $\qquad$ | 3. 2003 .
Given $a b c=1$, we have
$$
\frac{1}{1+a+a b}+\frac{1}{1+b+b c}+\frac{1}{1+c+c a}=1,
$$
thus $x=2003$. | 2003 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.
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```
Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.
``` | Three, let $P=\frac{a+b+c}{a b c}$.
In equation (1), let $a$ and $b$ remain unchanged, and only let $c$ vary, where $c$ can take any integer from 1 to 9.
Then, from $P=\frac{a+b+c}{a b c}=\frac{1}{a b}+\frac{a+b}{a b c}$, we know that
when $c=1$, $P$ reaches its maximum value, so $c=1$.
Thus, $P=\frac{a+b+1}{a b}=\frac... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) In a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)(n$ $\geqslant 2)$ grid, each cell is filled with 1 or -1. If the number in any cell is equal to the product of the numbers in the cells that share an edge with it, then this filling method is called "successful". Find the total number of "successf... | Three, assuming there exists some successful filling method that contains -1. First, prove: if this successful filling method is symmetric about the middle column (row), then the middle column (row) is entirely 1.
Let $a_{0}=1, a_{2}^{n}=1$.
If $a_{1}=1$, by $a_{1}=a_{0} \times a_{2} \times 1$, we get $a_{2}=1$.
Simil... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c$ $=$ $\qquad$ . | -1 .
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$. Since $\triangle A B C$ is a right triangle, it follows that $x_{1} γ x_{2}$ must have opposite signs. Thus, $x_{1} x_{2}=\frac{c}{a}<0$.
By the projection theorem, we know $|O C|^{2}=|A O| \cdot|B O|$, i.e., $c^{2}=\left|x_{1}\right| \cdot\left|x_{2}\right|=\... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $m$ be an integer, and the two roots of the equation $3 x^{2}+m x-2=0$ are both greater than $-\frac{9}{5}$ and less than $\frac{3}{7}$. Then $m=$ $\qquad$ . | 2.4 .
From the problem, we have
$$
\left\{\begin{array}{l}
3 \times\left(-\frac{9}{5}\right)^{2}+m \times\left(-\frac{9}{5}\right)-2>0, \\
3 \times\left(\frac{3}{7}\right)^{2}+m \times\left(\frac{3}{7}\right)-2>0 .
\end{array}\right.
$$
Solving this, we get \(3 \frac{8}{21}<m<4 \frac{13}{45}\). Therefore, \(m=4\). | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given positive integers $a$ and $b$ differ by 120, their least common multiple is 105 times their greatest common divisor. The larger of $a$ and $b$ is $\qquad$ | 4.225.
Let $(a, b)=d$, and $a=m d, b=n d$, where $m>n, m$ and $n$ are coprime. Thus, the least common multiple of $a$ and $b$ is $m n d$. According to the problem, we have
$$
\left\{\begin{array}{l}
m d-n d=120, \\
\frac{m n d}{d}=105,
\end{array}\right.
$$
which simplifies to
$$
\left\{\begin{array}{l}(m-n) d=2^{3} ... | 225 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Given that the area of quadrilateral $ABCD$ is $32$, the lengths of $AB$, $CD$, and $AC$ are all integers, and their sum is $16$.
(1) How many such quadrilaterals are there?
(2) Find the minimum value of the sum of the squares of the side lengths of such quadrilaterals. | (1) As shown in Figure 7, let $AB = a$, $CD = b$, $AC = l$, and let the height from $A$ to $BC$ in $\triangle ABC$ be $h_1$, and the height from $D$ to $AC$ in $\triangle ADC$ be $h_2$. Then,
$$
\begin{array}{l}
S_{\text{quadrilateral } BCD} = S_{\triangle ABC} + S_{\triangle ADC} \\
= \frac{1}{2} (h_1 a + h_2 b) \leq ... | 192 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. An editor uses the digits $0 \sim 9$ to number the pages of a book. If a total of 636 digits were written, then the book has $\qquad$ pages. | $$
\text { Ni, 7. } 248 \text {. }
$$
Assuming the book has $x$ pages, then we have
$$
(x-99) \times 3+90 \times 2+9=636 \text {. }
$$
Solving for $x$ gives $x=248$. | 248 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, (18 points) On a plane, there are 7 points, and some line segments can be connected between them, so that any 3 points among the 7 points must have 2 points connected by a line segment. How many line segments are needed at least? Prove your conclusion. | (1) If one of the 7 points is isolated (i.e., it is not connected to any other points), then the remaining 6 points must be connected in pairs, which requires at least $\frac{6 \times 5}{2}=15$ lines.
(2) If one of the 7 points is connected to only one other point, then the remaining 5 points must be connected in pairs... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false |
2. (16 points) Two vegetable bases, A and B, supply the same type of vegetables to three farmers' markets, $A$, $B$, and $C$. According to the signed contract, 45 t should be supplied to $A$, 75 t to $B$, and 40 t to $C$. Base A can arrange for 60 t, and Base B can arrange for 100 t. The distance (in km) between A and ... | 2. Let base B supply $x \text{t}$ to $A$, $y \text{t}$ to $B$, and $[100-(x+y)] \text{t}$ to $C$. Then base A supplies $(45-x) \text{t}$ to $A$, $(75-y) \text{t}$ to $B$, and $[40-(100-x-y)] \text{t} = [(x+y)-60] \text{t}$ to $C$. According to the problem, the total transportation cost is
$$
\begin{aligned}
W= & 10(45-... | 960 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $\frac{3 x+4}{x^{2}-x-2}=\frac{A}{x-2}-\frac{B}{x+1}$, where $A$ and $B$ are constants. Then the value of $4 A-B$ is ( ).
(A) 7
(B) 9
(C) 13
(D) 5 | 2. (C).
From $\frac{A}{x-2}-\frac{B}{x+1}=\frac{(A-B) x+A+2 B}{x^{2}-x-2}$, we get $A-B=3, A+2 B=4$.
Thus, $4 A-B=3(A-B)+(A+2 B)=13$. | 13 | Algebra | MCQ | Yes | Yes | cn_contest | false |
8. If 4 lines in a plane intersect each other pairwise and no three lines are concurrent, then there are $\qquad$ pairs of consecutive interior angles. | 8.24.
Each line intersects with 3 other lines, forming 3 intersection points. Each 2 intersection points determine a line segment, making a total of 3 line segments. Since each line has a pair of consecutive interior angles on both sides, there are $3 \times 4=12$ line segments, and a total of 24 pairs of consecutive ... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. Given
$$
\left(x+\sqrt{x^{2}+2002}\right)\left(y+\sqrt{y^{2}+2002}\right)=2002 .
$$
Then $x^{2}-3 x y-4 y^{2}-6 x-6 y+58=$ $\qquad$ . | 11.58
$$
\begin{array}{l}
\text { Given }\left(x+\sqrt{x^{2}+2002}\right)\left(\sqrt{x^{2}+2002}-x\right) \\
=2002,
\end{array}
$$
we get $\sqrt{x^{2}+2002}-x=\sqrt{y^{2}+2002}+y$.
Similarly, $\sqrt{y^{2}+2002}-y=\sqrt{x^{2}+2002}+x$.
Adding these two equations gives $x+y=0$.
Therefore, $x^{2}-3 x y-4 y^{2}-6 x-6 y+58... | 58 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. As shown in Figure 4, line $AB$ intersects $\odot O$ at points $A$ and $B$, point $O$ is on $AB$, and point $C$ is on $\odot O$, with $\angle AOC=40^{\circ}$. Point $E$ is a moving point on line $AB$ (not coinciding with point $O$), and line $EC$ intersects $\odot O$ at another point $D$. The number of points $E$ t... | 12.3.
Consider the position of point $E$: Point $E$ can be on the extension of line segment $O A$; Point $E$ can be on line segment $O A$ (excluding point $O$); Point $E$ can be on the extension of line segment $O B$; but point $E$ cannot be on line segment $O B$ (excluding point $O$). Therefore, point $E$ has a total... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. There are two equations:
Good + Good $=$ Wonderful, Wonderful $\times$ GoodGood $\times$ ReallyGood $=$ WonderfulProblemProblemWonderful, where each Chinese character represents a digit from $0 \sim 9$, the same character represents the same digit, and different characters represent different digits. Therefore, th... | 13.16.
From the addition formula, we get βε₯½β $<5$, βε¦β $\neq 0$, so βε₯½β $=1$, βε¦β $=2$ or βε₯½β $=2$, βε¦β $=4$ or βε₯½β $=3$, βε¦β $=6$ or βε₯½β $=4$, βε¦β $=8$. Clearly, the middle two scenarios do not satisfy the multiplication formula, so it can only be:
(1) βε₯½β $=1$, βε¦β $=2$, thus the multiplication formula becomes
$2 \t... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
14. Given real numbers $a, b, c$, satisfying $a+b+c=0, a^{2}+b^{2}+c^{2}=6$. Then the maximum value of $a$ is | 14.2.
From the problem, we get $c=-(a+b)$, thus,
$$
a^{2}+b^{2}+[-(a+b)]^{2}=6,
$$
which simplifies to $b^{2}+a b+a^{2}-3=0$.
Since $b$ is a real number, the above equation, which is a quadratic equation in $b$, must have real roots, hence,
$$
\Delta=a^{2}-4\left(a^{2}-3\right) \geqslant 0, \quad -2 \leqslant a \leqs... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. A group of 17 middle school students went to several places for a summer social survey, with a budget for accommodation not exceeding $x$ yuan per person per day. One day, they arrived at a place with two hostels, $A$ and $B$. $A$ has 8 first-class beds and 11 second-class beds; $B$ has 10 first-class beds, 4 second... | 4. $x=10$.
If staying at location $A$, even choosing the most economical beds, the average accommodation cost per person will exceed 10 yuan (since $8 \times 11 + 14 \times 6 = 172$ (yuan), $172 \div 17 \approx 10.12$ (yuan)).
If staying at location $B$, with a reasonable choice of beds, the budget can be met, and th... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that the lengths of the four sides of a rectangle are all integers less than 10, these four length numbers can form a four-digit number, where the thousand's digit and the hundred's digit of this four-digit number are the same, and this four-digit number is a perfect square. Find the area of this rectangle. | (Tip: Let the two adjacent sides of the rectangle be $a$ and $b$, then $\overline{a a b b}=$ $11(100 a+b)$ is a perfect square, which implies 11 divides $(a+b)$. Thus, $a + b = 11$. Upon verification, only $a=7, b=4$. Answer: 28.) | 28 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. If the polynomial $x^{2}-x+1$ can divide another polynomial $x^{3}+x^{2}+a x+b(a, b$ are constants). Then $a+b$ equals ( ).
(A) 0
(B) -1
(C) 1
(D) 2 | 2. (C).
By synthetic division, it is easy to obtain the remainder $r(x)=(a+1) x+b-2$. Since it can be divided exactly, we have $a+1=0$ and $b-2=0$, i.e., $a=-1, b=2$. Therefore, $a+b=1$. | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false |
5. In decimal, if a positive integer with at least two digits has all its digits, except the leftmost one, smaller than the digit to their left, then it is called a descending positive integer. The total number of such descending positive integers is ( ).
(A) 1001
(B) 1010
(C) 1011
(D) 1013 | 5.(D).
Let the leftmost digit be $a$, then the digits $a-1$, $a-2, \cdots, 2,1,0$ that are smaller than $a$ can each appear at most once to the right of $a$. Therefore, the number of decreasing positive integers starting with $a$ is equal to the number of non-empty subsets of $\{a-1, a-2, \cdots, 2,1,0\}$, which is $2... | 1013 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
4. $f(x)=\frac{x^{2}}{8}+x \cos x+\cos (2 x)(x \in \mathbf{R})$'s minimum value is $\qquad$ . | 4. -1 .
$$
\begin{array}{l}
f(x)=\frac{x^{2}}{8}+x \cos x+2 \cos ^{2} x-1 \\
=\frac{1}{8}(x+4 \cos x)^{2}-1 \geqslant-1 .
\end{array}
$$
Since the equation $\cos x=-\frac{x}{4}$ (as can be seen from the graph) has a solution, we have
$$
f(x)_{\min }=-1
$$ | -1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Try to find the smallest possible value of the positive integer $k$, such that the following proposition holds: For any $k$ integers $a_{1}, a_{2}, \cdots, a_{k}$ (equality is allowed), there must exist corresponding $k$ integers $x_{1}, x_{2}, \cdots, x_{k}$ (equality is also allowed), and $\left|x_... | Three, first prove that the proposition holds when $k=7$. For this, consider the sum
$$
S\left(y_{1}, y_{2}, \cdots, y_{7}\right)=y_{1} a_{1}+y_{2} a_{2}+\cdots+y_{7} a_{7} \text {, }
$$
where $y_{i} \in\{-1,0,1\}$.
There are $3^{7}=2187$ such sums, and since $2187>2003$, by the pigeonhole principle, there must be two... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 A factory plans to arrange 214 workers to produce 6000 $A$ components and 2000 $B$ components. It is known that the time it takes for each worker to produce 5 $A$ components can produce 3 $B$ components. Now the workers are divided into two groups to produce these two types of components, and they start at th... | Solution: Let $x$ be the number of people producing component $A$, then the number of people producing component $B$ is $(214-x)$. Taking the time for one person to produce 5 $A$ components as the unit, the time required to produce 6000 $A$ components and 2000 $B$ components is respectively
$$
y_{1}=\frac{6000}{5 x} \t... | 137 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $n$ be a given positive integer. Try to find non-negative integers $k, l$, satisfying $k+l \neq 0$, and $k+l \neq n$, such that
$$
s=\frac{k}{k+l}+\frac{n-k}{n-(k+l)}
$$
takes the maximum value. | If $l=0$, then $s=2$.
If $l>0$, let $x=k+l$, then $00 .
\end{array}
$
Therefore, $f(1)1$ when, $s$ reaches the maximum value 2;
(2) If $n=2$, then when $l=1, k=0$ or $l=0, 0 < k \neq 2$, $s$ reaches the maximum value 2;
(3) If $n \geqslant 3$, then when $l=n-1, k=0$, $s$ reaches the maximum value $n$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 There are 1988 unit cubes, and they (all or part of them) are arranged into 3 "squares" (i.e., 3 one-layer rectangular prisms with dimensions $a \times a \times 1, b \times b \times 1, c \times c \times 1$ where $a \leqslant b \leqslant c$) $A, B$, and $C$. Now, place square $C$ in the first quadrant of the $... | Solution: According to the problem, the condition for $B$ to be placed on $C$ is $b \leqslant c - 2$, and there are $(c-b-1)^{2}$ different ways to place it; the condition for $A$ to be placed on $B$ is $a \leqslant b-2$, and there are $(b-a-1)^{2}$ different ways to place it. Thus, the problem is transformed into:
Fin... | 345 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. There are 1989 points in space, where no three points are collinear. Divide them into 30 groups with different numbers of points. From any 3 different groups, take one point each to form a triangle. To maximize the total number of such triangles, how many points should each group have? | (Tip: Let the number of points in the $i$-th group be $x_{i}$, then the problem is transformed into: when $x_{1}$ $+x_{2}+\cdots+x_{30}=1989$, and $x_{1}, x_{2}, \cdots, x_{30}$ are all different, find the maximum value of $M=\sum_{1<i<j<k \leqslant 30} x_{i} x_{j} x_{k}$. Without loss of generality, assume $x_{1}<x_{2... | 51 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 In $\triangle A B C$, $A B=37, A C=58$, a circle with center $A$ and radius $A B$ intersects $B C$ at point $D$, and $D$ is between $B$ and $C$. If the lengths of $B D$ and $D C$ are both integers, find the length of $B C$.
In $\triangle A B C$, $A B=37, A C=58$, a circle with center $A$ and radius $A B$ int... | Explanation: A triangle with two known sides is indeterminate, but note that the lengths of $B D$ and $D C$ are both integers, so the length of $B C$ is also an integer. Therefore, we can start by determining the range of possible values for $B C$, and then find the value of $B C$ that meets the requirements.
As shown... | 57 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
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