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Example 13 Unload several boxes from a cargo ship, with a total weight of $10 \mathrm{t}$, and the weight of each box does not exceed $1 \mathrm{t}$. To ensure that these boxes can be transported away in one go, how many trucks with a carrying capacity of $3 \mathrm{t}$ are needed at least?
(1990, Jiangsu Province Juni... | First, note that the weight of each box does not exceed $1 \mathrm{t}$, so the weight of the boxes that each vehicle can carry at one time will not be less than $2 \mathrm{t}$; otherwise, another box can be added.
Let $n$ be the number of trucks needed, and the weights of the boxes they carry be $a_{1}, a_{2}, \cdots,... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Three, (16 points) 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 $|a|+|b|+|c|$. | (1) Let's assume $a=\max \{a, b, c\}$. From the given conditions, we have
$$
a>0, b+c=2-a, bc=\frac{4}{a} \text{. }
$$
Therefore, $b$ and $c$ are the two real roots of the quadratic equation $x^{2}-(2-a)x+\frac{4}{a}=0$. Thus,
$$
\begin{array}{l}
\Delta=(2-a)^{2}-4 \times \frac{4}{a} \geqslant 0 \\
\Leftrightarrow a^{... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Observe the following expressions:
$$
97^{2}=9409,997^{2}=994009,9997^{2}=99940009 \text {. }
$$
Conjecture $999998^{2}=$ $\qquad$ | $1.999996000004$ | 999996000004 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $k$ is an integer. If the quadratic equation $k x^{2}+(2 k+3) x+1$ $=0$ has rational roots, then the value of $k$ is $\qquad$ | 2. -2 .
From the given, $\Delta_{1}=(2 k+3)^{2}-4 k$ is a perfect square. Let $(2 k+3)^{2}-4 k=m^{2}(m$ be a positive integer), that is,
$$
4 k^{2}+8 k+9-m^{2}=0 \text {. }
$$
Considering equation (1) as a quadratic equation in $k$, by the problem's condition, it has integer roots, so the discriminant of equation (1)... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In trapezoid $A B C D$; $A D / / B C, A C 、 B D$ intersect at point $O, \triangle A O D$ and $\triangle A O B$ have areas of 9 and 12, respectively. Then the area of trapezoid $A B C D$ is $\qquad$ | 3. 49.
As shown in Figure 10, since
$$
\begin{array}{l}
\frac{S_{\triangle A O D}}{S_{\triangle A U B}}=\frac{O D}{O B}=\frac{O A}{O C} \\
=\frac{S_{\triangle A O B}}{S_{\triangle B C C}},
\end{array}
$$
thus $S_{\triangle A B B}^{2}=S_{\triangle M D} \cdot S_{\triangle B O C}$,
which means $S_{\triangle B O C}=\frac... | 49 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $m$ be an integer, and the two roots of the equation $3 x^{2}+m x-2=0$ are both greater than $-\frac{9}{5}$ and less than $\frac{3}{7}$. Then $m=$ $\qquad$ .
(2003, National Junior High School Mathematics League) | (Solution: $m=4$ ) | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let $\left\{a_{n}\right\}$ be the sequence of all numbers in the set $\left\{2^{s}+2^{t} \mid 10 \leqslant s<t\right.$ and $s, t \in \mathbf{Z}\}$ arranged in ascending order, i.e., $\square$
$$
\begin{aligned}
& a_{1}=3, a_{2}=5, a_{3}=6, a_{4}=9, a_{5}=10, \\
a_{6}= & 12, \cdots .
\end{aligned}
$$
Arrange ... | (1) Arrange the above triangular number table into a right-angled triangular number table:
$$
\begin{array}{l}
a_{1}=2^{0}+2^{1} \\
a_{2}=2^{0}+2^{2} \quad a_{3}=2^{1}+2^{2} \\
a_{4}=2^{0}+2^{3} \quad a_{5}=2^{1}+2^{3} \quad a_{6}=2^{2}+2^{3} \\
a_{7}=2^{0}+2^{4} \quad a_{8}=2^{1}+2^{4} \quad a_{9}=2^{2}+2^{4} \quad a_... | 16640 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 In the non-decreasing sequence of positive odd numbers $\{1,3,3,3,5,5,5,5,5, \cdots\}$, each positive odd number $k$ appears $k$ times. It is known that there exist integers $b$, $c$, and $d$ such that for all integers $n$, $a_{n} = b[\sqrt{n+c}]+d$, where $[x]$ denotes the greatest integer not exceeding $x$.... | Explanation: Divide the sequence $\{1,3,3,3,5,5,5,5,5, \cdots\}$ into groups:
(1), $(3,3,3),(5,5,5,5,5), \cdots,(2 k-1, 2 k-1, \cdots, 2 k-1), \cdots$,
where the $k$-th group consists of $2 k-1$ occurrences of $2 k-1$. If $a_{n}$ is in the $k$-th group, then $a_{n}=2 k-1$. Therefore,
$$
\begin{array}{l}
1+3+\cdots+(2 k... | 2 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
2. The sequence $1,1,2,2,2,2,2, \cdots, k, k, \cdots, k, \cdots$, where each positive integer $k$ appears $3 k-1$ times. Then the 2004th term of this sequence is $\qquad$ | (Answer: 37) | 37 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find all positive integers $k$ such that for any positive numbers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a, b, c$.
(First China Girls Mathematical Olympiad)
Analysis: To find $k$, we can first determine the upper... | Solution: It is easy to know that $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
Therefore, the required $k>5$.
Take $a, b, c$ that do not form a triangle ($a=1, b=1, c=2$), then we have
$$
k(1 \times 1+1 \times 2+2 \times 1) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right).
$$
This gives $k \leqslant 6$.
Therefore, if $55\left(... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Let the lengths of the three sides of a triangle be integers $l$, $m$, $n$, and $l > m > n$. It is known that
$$
\left\{\frac{3^{l}}{10^{4}}\right\}=\left\{\frac{3^{m}}{10^{4}}\right\}=\left\{\frac{3^{n}}{10^{4}}\right\},
$$
where $\{x\}=x-[x]$, and $[x]$ represents the greatest integer not exceeding $x$. Find the min... | 1 Another Solution to the Test Question
Solution: From the given, we have
$$
\begin{array}{l}
3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod 10^{4}\right) . \\
\text { Equation (1) } \Leftrightarrow\left\{\begin{array}{l}
3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod 2^{4}\right), \\
3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod... | 3003 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$, and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ .
(1999, National Junior High School Mathematics League) This problem has a resemblance to a 1979 college entrance exam question. | Example $\mathbf{3}^{\prime}$ If $(z-x)^{2}-4(x-y)(y-z)=0$, prove that $x, y, z$ form an arithmetic sequence.
In Example 3, taking $a=b=c$ can guess the answer to be 2. But this poses a risk of "root reduction," because the condition is a quadratic expression, while the conclusion
$$
\frac{b+c}{a}=2 \Leftrightarrow b+... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the three-digit number $n=\overline{a b c}$. If the lengths of the sides of a triangle can be formed by $a, b, c$ to form an isosceles (including equilateral) triangle, then the number of such three-digit numbers $n$ is ( ) .
(A) 45
(B) 81
(C) 165
(D) 216 | 5.C.
$a, b, c$ must be able to form the sides of a triangle, clearly none of them can be 0, i.e., $a, b, c \in\{1,2, \cdots, 9\}$.
(1) If they form an equilateral triangle, let the number of such three-digit numbers be $n_{1}$. Since all three digits in the three-digit number are the same, then $n_{1}=\mathrm{C}_{9}^{1... | 165 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
12. In the Cartesian coordinate system $x 0 y$, given two points $M(-1,2)$ and $N(1,4)$, point $P$ moves on the $x$-axis. When $\angle M P N$ takes the maximum value, the x-coordinate of point $P$ is $\qquad$ $-$.
| 12.1.
The center of the circle passing through points $M$ and $N$ lies on the perpendicular bisector of line segment $MN$, which is $y=3-x$. Let the center of the circle be $S(a, 3-a)$, then the equation of circle $S$ is $(x-a)^{2}+(y-3+a)^{2}=2\left(1+a^{2}\right)$.
For a fixed-length chord, the inscribed angle subt... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given $a, b, c \in \mathbf{N}_{+}$, and the parabola $f(x) = ax^{2} + bx + c$ intersects the $x$-axis at two different points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a + b + c$.
(1996, National Junior High School Mathematics Competition) | Let $f(x)=a x^{2}+b x+c, A\left(x_{1}, 0\right)$, and $B\left(x_{2}, 0\right)$. According to the problem, we have
$$
\left\{\begin{array}{l}
\Delta=b^{2}-4 a c>0, \\
f(1)=a+b+c>0, \\
f(-1)=a-b+c>0 .
\end{array}\right.
$$
From (1), we get $b>2 \sqrt{a c}$; from (3), we get $a+c>b$.
$$
\begin{array}{l}
a+c>b \Rightarrow... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. A square of size $3 \times 3$ with a unit square removed from each corner is called a "cross shape". On a $10 \times 11$ chessboard, what is the maximum number of non-overlapping "cross shapes" (each "cross shape" exactly covers 5 small squares on the chessboard)?
(Supplied by Feng Zuming) | 8. 15.
First, prove that the maximum number of "crosses" that can be placed is 15.
Proof by contradiction. Assume that 16 "crosses" can be placed.
For each "cross", we call the central square the "center" (denoted as *). As shown in Figure 3, remove the outermost layer of squares from the $10 \times 11$ chessboard, re... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$ and $b$ are positive integers, and satisfy $\frac{a+b}{a^{2}+a b+b^{2}}=\frac{4}{49}$. Then the value of $a+b$ is $\qquad$ ـ. | II. 1. 16.
Let $a+b=4k$ ($k$ is a positive integer), then
$$
a^{2}+a b+b^{2}=49 k,
$$
which means $(a+b)^{2}-a b=49 k$.
Therefore, $a b=16 k^{2}-49 k$.
It is easy to see that $a, b$ are the two positive integer roots of the quadratic equation in $x$
$$
x^{2}-4 k x+\left(16 k^{2}-49 k\right)=0
$$
By $\Delta=16 k^{2}-4... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a, b$ are the real roots of the equation $x^{2}+2 p x+3=0$, $c, d$ are the real roots of the equation $x^{2}+2 q x+3=0$, and $e, v$ are the real roots of the equation $x^{2}+2\left(p^{2}-q^{2}\right) x+3=0$, where $p, q$ are real numbers. Then, the value of $\frac{(a-c)(b-c)(a+d)(b+d)}{e+v}$ is $\qquad$ ... | 3.6 .
From the relationship between roots and coefficients, we get
$$
\begin{array}{l}
a+b=-2 p, a b=3 ; c+d=-2 q, c d=3 ; \\
e+f=-2\left(p^{2}-q^{2}\right), e f=3 . \\
\text { Therefore } \frac{(a-c)(b-c)(a+d)(b+d)}{e+f} \\
=\frac{\left[a b-(a+b) c+c^{2}\right]\left[a b+(a+b) d+d^{2}\right]}{e+f} \\
=\frac{\left(c d+... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) In the quadratic function $f(x)=a x^{2}+b x$ $+c$, $a$ is a positive integer, $a+b+c \geqslant 1, c \geqslant 1$, and the equation $a x^{2}+b x+c=0$ has two distinct positive roots less than 1. Find the minimum value of $a$.
保留源文本的换行和格式，直接输出翻译结果如下：
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One, (20 points) In the quadratic function $f(x)... | Given the equation $a x^{2}+b x+c=0$ has two roots $x_{1} 、 x_{2}(0<x_{1}<x_{2})$, and $x_{1} x_{2}=\frac{c}{a}=\frac{3}{5}$, we have $a x_{1} x_{2}=3$.
Since $x_{1}+x_{2}=\frac{11}{10}$, we have $a(x_{1}+x_{2})=\frac{11}{10}a>4.4$.
Given $a>4$, and $a$ is a positive integer, so, $a \geqslant 5$.
Taking $f(x)=5\left... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The set $A=\left\{x \mid x \in \mathbf{Z}\right.$ and $\left.\frac{600}{5-x} \in \mathbf{Z}\right\}$ has $\qquad$ elements, and the sum of all elements is $\qquad$ . | 二、1. 48,240.
Since $600=2^{3} \times 3 \times 5^{2}$, the number of positive divisors of 600 is $(3+1)(1+1)(2+1)=24$.
Therefore, the set $A$ contains $2 \times 24=48$ elements.
Considering that the sum of the elements in $A$ corresponding to two opposite divisors of 600 is 10, the sum of the elements in $A$ is $24 \tim... | 240 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the set $A=\{1,2,3,4,5,6\}$, and the mapping $f: A$
$\rightarrow A$ satisfies $f(f(x))=x$. Then the number of mappings $f$ is
$\qquad$ ـ. | 2.76.
Let $f(x)=a$, then $f(a)=x$, i.e., $x \rightarrow a, a \rightarrow x$.
If $a=x$, i.e., $x \rightarrow x$ is called self-correspondence;
If $a \neq x$, i.e., $x \rightarrow a, a \rightarrow x$ is called a pair of cyclic correspondence.
It is also known that the number of self-correspondences should be even, so th... | 76 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 In quadrilateral $A B C D$, $A B=30, A D$ $=48, B C=14, C D=40, \angle A B D+\angle B D C=$ $90^{\circ}$. Find the area of quadrilateral $A B C D$. | Solution: As shown in Figure 2, we have
$$
\begin{array}{l}
S_{\triangle A B D}=S_{\triangle A_{1} D B}, \\
A_{1} D=A B=30, \\
A_{1} B=A D=48, \\
\angle A_{1} D B=\angle A B D .
\end{array}
$$
Thus, we have
$$
\begin{array}{l}
\angle A_{1} D C=\angle A_{1} D B+\angle B D C \\
=\angle A B D+\angle B D C=90^{\circ} .
\e... | 936 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 If numbers $a_{1}, a_{2}, a_{3}$ are taken in ascending order from the set $1,2, \cdots, 14$, such that both $a_{2}-a_{1} \geqslant 3$ and $a_{3}-a_{2} \geqslant 3$ are satisfied. Then, the number of all different ways to select the numbers is $\qquad$ kinds. | Solution: Given $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$, $14-a_{3} \geqslant 0$, let $a_{1}=x_{1}, a_{2}-a_{1}=x_{2}, a_{3}-a_{2}=x_{3}, 14-a_{3}=x_{4}$, then
$$
x_{1}+x_{2}+x_{3}+x_{4}=14 .
$$
Thus, the problem is transformed into finding the number of integer solutions to the equation u... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Try to determine, for any $n$ positive integers, the smallest positive integer $n$ such that at least 2 of these numbers have a sum or difference that is divisible by 21.
| Three, let $A_{i}$ represent all numbers that leave a remainder of $i$ when divided by 21; $i=0,1$, $\cdots, 20$; let $B_{j}=A_{j} \cup A_{21-j}, j=1,2, \cdots, 10, B_{11}=A_{0}$.
By taking 12 numbers, there will be at least 2 numbers belonging to the same category among the 11 categories $B_{1}, B_{2}$, $\cdots, B_{1... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given points $A(1,1)、B(3,2)、C(2,3)$ and line $l: y=k x(k \in \mathbf{R})$. Then the maximum value of the sum of the squares of the distances from points $A、B、C$ to line $l$ is $\qquad$ | Let $A$, $B$, $C$ be three points, and the sum of the squares of their distances to the line $k x - y = 0$ is $d$. Then,
$$
\begin{aligned}
d & =\frac{(k-1)^{2}}{k^{2}+1}+\frac{(3 k-2)^{2}}{k^{2}+1}+\frac{(2 k-3)^{2}}{k^{2}+1} \\
& =\frac{14 k^{2}-26 k+14}{k^{2}+1} .
\end{aligned}
$$
Therefore, $(d-14) k^{2}+26 k+(d-1... | 27 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Consider the binomial coefficients $\mathrm{C}_{n}^{0}, \mathrm{C}_{n}^{1}, \cdots, \mathrm{C}_{n}^{n}$ as a sequence. When $n \leqslant 2004\left(n \in \mathbf{N}_{+}\right)$, the number of sequences in which all terms are odd is $\qquad$ groups. | 2.10.
From $\mathrm{C}_{n}^{r}=\mathrm{C}_{n-1}^{r-1}+C_{n-1}^{r}$, we know that the 2004 numbers $n=1,2, \cdots, 2004$ can be arranged in the shape of Pascal's Triangle, and the even and odd numbers can be represented by 0 and 1, respectively, as shown in Figure 3.
The pattern of the changes in the parity of each ter... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Place 7 goldfish of different colors into 3 glass fish tanks numbered $1,2,3$. If the number of fish in each tank must be no less than its number, then the number of different ways to place the fish is $\qquad$ kinds. | 3.455.
(1) If the number of goldfish placed in tanks 1, 2, and 3 are $2, 2, 3$ respectively, then there are $C_{5}^{2} \cdot C_{5}^{2}=210$ different ways;
(2) If the number of goldfish placed in tanks 1, 2, and 3 are $1, 3, 3$ respectively, then there are $\mathrm{C}_{6}^{1} \cdot \mathrm{C}_{6}^{3}=140$ different way... | 455 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=a_{n}^{2}+2 a_{n}+3, n \in \mathbf{N}_{+}$, try to find
$$
M=a_{m}^{2}+b_{n}^{2}+m^{2}+n^{2}-2\left(a_{m} b_{n}+m n\right)
$$
$\left(m 、 n \in \mathbf{N}_{+}\right)$ the minimum value. | (1) It is easy to know that when $n \geqslant 2, n \in \mathbf{N}_{+}$, we have
$4\left(S_{n}-S_{n-1}\right)=\left(a_{n}+1\right)^{2}-\left(a_{n-1}+1\right)^{2}$.
Thus, $\left(a_{n}+a_{n-1}\right)\left(a_{n}-a_{n-1}-2\right)=0$.
Since $a_{n}+a_{n-1}>0$, it follows that,
$$
a_{n}-a_{n-1}=2 \text {. }
$$
It is easy to s... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If: (1) $a, b, c, d$ all belong to $\{1,2,3,4\}$;
(2) $a \neq b, b \neq c, c \neq d, d \neq a$;
(3) $a$ is the smallest number among $a, b, c, d$.
Then, the number of different four-digit numbers $\overline{a b c d}$ that can be formed is $\qquad$. | (提示: $\overline{a b c d}$ consists of 2 different digits, then it must be that $a=$ $c, b=d$. Therefore, there are $\mathrm{C}_{4}^{2}=6$ such numbers. $\overline{a b c d}$ consists of 3 different digits, then $a$ is uniquely determined, when $c=a$, $b, d$ have 2 ways to be chosen; when $c \neq$ $a$, $c$ has 2 ways to ... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) On a plane, there are 12 points, and no three points are collinear. Using any one of these points as the starting point and another as the endpoint, draw all possible vectors. A triangle whose three side vectors sum to the zero vector is called a "zero triangle." Find the maximum number of "zero tria... | Three, let these 12 points be $P_{1}, P_{2}, \cdots, P_{12}$. The number of triangles determined by these 12 points is $\mathrm{C}_{12}^{3}$. Let the number of vectors starting from $P_{1}(i=1,2, \cdots, 12)$ be $x_{i}, 0 \leqslant x_{i} \leqslant 11$. If a triangle with 3 points as vertices is a "non-zero triangle", t... | 70 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Color each vertex of a square pyramid with one color, and make the endpoints of the same edge have different colors. If only 5 colors are available, the number of different coloring methods is $\qquad$ . | (Given that there are 5 colors, 1, 2, 3, 4, 5, to color the pyramid $S-ABCD$. Since $S$, $A$, and $B$ are colored differently, there are $5 \times 4 \times 3=60$ coloring methods. Assuming $S$, $A$, and $B$ are already colored, we then categorize and discuss the coloring of $C$ and $D$, which have 7 coloring methods. T... | 420 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given $a>0$, and
$$
\sqrt{b^{2}-4 a c}=b-2 a c \text{. }
$$
Find the minimum value of $b^{2}-4 a c$. | Solution 1: The given condition is
$$
\frac{-b+\sqrt{b^{2}-4 a c}}{2 a c}=-1(a c \neq 0),
$$
This indicates that the quadratic equation
$$
a c x^{2}+b x+1=0
$$
has a real root $x=-1$. Substituting into equation (1) yields
$$
a c=b-1 \text {. }
$$
Transform equation (3) into the overall structure of the discriminant
... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: From the 10 numbers $0,1,2,3,4,5,6,7,8,9$, select 3 numbers such that their sum is an even number not less than 10. The number of different ways to do this is $\qquad$ | Solution: To select 3 numbers from 10 numbers such that their sum is even, the 3 numbers must all be even or 1 even and 2 odd.
When all 3 numbers are even, there are $\mathrm{C}_{5}^{3}$ ways to choose;
When there is 1 even and 2 odd numbers, there are $\mathrm{C}_{5}^{1} \mathrm{C}_{5}^{2}$ ways to choose. Since their... | 51 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
16. Rearrange the digits of any three-digit number to form the largest and smallest numbers possible, and their difference will form another three-digit number (allowing the hundreds digit to be $0$). Repeat this process. What is the number obtained after 2003 repetitions? Prove your conclusion. | 16. (1) If all three digits are the same, the resulting number is 0.
(2) If the three digits are not all the same, the resulting number is 495.
(1) This is obviously true.
Below is the proof of (2).
If the three digits are not all the same, let's assume this three-digit number is $\overline{a b c}$, where $a \geqslant ... | 495 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let ABCDEF be a regular hexagon. A frog starts at vertex $A$, and each time it can randomly jump to one of the two adjacent vertices. If it reaches point $D$ within 5 jumps, it stops jumping; if it cannot reach point $D$ within 5 jumps, it also stops after 5 jumps. Then, the number of different possible jumpi... | Solution: As shown in Figure 1, the frog cannot reach point $D$ by jumping 1 time, 2 times, or 4 times. Therefore, the frog's jumping methods are only in the following two cases:
(1) The frog jumps 3 times to reach point $D$, with two jumping methods: $A B C D$ and $A F E D$.
(2) The frog stops after a total of 5 jumps... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. The license plates of motor vehicles in a certain city are numbered consecutively from "10000" to "99999". Then, among these 90000 license plates, the number of plates where the digit 9 appears at least once, and the sum of the digits is a multiple of 9, is $\qquad$ . $\qquad$ | 9. 4168 | 4168 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Choose several colors from the given 6 different colors to color the 6 faces of a cube, with each face being colored with exactly 1 color, and any 2 faces sharing a common edge must be colored differently. Then the number of different coloring schemes is $\qquad$ (Note: If we color two identical cubes and can... | Solution: This problem is relatively complex and requires a classification discussion based on the number of colors used.
(1) If only 3 colors are used, there are $\mathrm{C}_{6}^{3}$ ways to choose 3 colors from 6 different colors. Since every 2 faces sharing a common edge are painted in different colors, the opposite... | 230 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9.3. There are 2004 small boxes on the table, each containing 1 ball. It is known that some of the balls are white, and there are an even number of white balls. You are allowed to point to any 2 boxes and ask: "Do they contain at least 1 white ball?" How many times do you need to ask, at a minimum, to determine 2 boxes... | 9.3. 4005.
We will number the boxes (and the balls inside them) from 1 to 2004, and refer to the problems by the numbers of the box pairs. We will call all non-white balls black balls.
First, we prove that 2 white balls can be found with 4005 questions. We ask the questions $(1,2),(1,3), \cdots,(1,2004)$, $(2,3),(2,4... | 4005 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9.7. Positive integers from 1 to 100 are arranged in a circle in such a way that each number is either greater than both of its neighboring numbers or less than both of its neighboring numbers. A pair of adjacent numbers is called "good" if, after removing them, the above property still holds. How many "good" pairs of ... | 9.7. The minimum possible number of "good" neighboring pairs is 51.
For example, first arrange the numbers from 1 to 100 in a clockwise direction in increasing order on a circle, then swap the positions of 2 and 3, 4 and 5, ..., 98 and 99. In the resulting arrangement \(1, 3, 2, 5, 4, \cdots, 99, 98, 100\), there are ... | 51 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.2. There are 2004 small boxes on the table, each containing 1 ball. It is known that some of the balls are white, and there are an even number of white balls. You are allowed to point to any 2 boxes and ask: "Do they contain at least 1 white ball?" How many times do you need to ask, at a minimum, to determine a box ... | 10.2. 2003.
Number the boxes (and the balls inside them) from 1 to 2004, and refer to the problems by the numbers of the box pairs, and call all non-white balls black balls.
First, we prove that a white ball can be found with 2003 questions. Ask the questions $(1,2),(1,3), \cdots,(1,2004)$ in sequence. If all the ans... | 2003 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11.4. There is a $9 \times 2004$ grid, in which the positive integers from 1 to 2004 are each filled in 9 times, and the difference between the numbers in each column does not exceed 3. Find the minimum possible value of the sum of the numbers in the first row. | 11.4. $C_{2003}^{2}+1=2005004$.
Consider a $9 \times n$ grid, where the integers from 1 to $n$ are each filled in 9 times, and the difference between the numbers in each column does not exceed 3. We will prove by mathematical induction that the sum of the numbers in the first row is not less than $C_{n-1}^{2}+1$.
Whe... | 2005004 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $m=\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\cdots+$ $\frac{1}{\sqrt{2003}+\sqrt{2005}}$. Then $50[2 m]-142=$ $\qquad$ .
(Where $[x]$ denotes the greatest integer not exceeding $x$) | 2. 2008 .
From the known equation, we get
$$
m=\frac{\sqrt{2005}-1}{2} \text {. }
$$
Thus, $[2 m]=43$.
Therefore, $50[2 m]-142=2008$. | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. As shown in Figure 1, PB is a secant of circle $\odot O$ with radius 5, and the lengths of $PA$ and $PB$ are the two roots of the equation $x^{2}-10 x+16=0$ $(PA<PB)$, $PC$ is a tangent to $\odot O$, and $C$ is the point of tangency. Then the area of quadrilateral $PAOC$ is $\qquad$ | 2.14.
As shown in Figure 5, connect $O P$, and draw $O D \perp A B$, with $D$ being the foot of the perpendicular. Solving the equation $x^{2}-10 x+16=0$, we get $P A=2, P B=8$. Thus, $A D=3$, $O D=4$, and $P C=\sqrt{P A \cdot P B}=$
4. Therefore,
$S_{\text {quadrilateral PAOC }}$
$$
\begin{array}{l}
=S_{\triangle P O... | 14 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $|y-2| \leqslant 1$, and $x+y=3$. Then the minimum value of $2 x^{2}-$ $12 x+y^{2}+1$ is $\qquad$ .
| 3. -14 .
It is easy to get $1 \leqslant y \leqslant 3, y=3-x$.
Also, $M=2(x-3)^{2}+y^{2}-17=3 y^{2}-17$, it is clear that when $y=1$, $M$ is the smallest, and the minimum value is -14. | -14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$8.6 .7 \times 7$ grid squares are colored with two different colors. Prove: at least 21 rectangles can be found, whose vertices are the centers of squares of the same color, and whose sides are parallel to the grid lines. | 8.6. Consider any one column and estimate the number of same-color "square pairs" in it. Suppose there are $k$ squares of one color and $7-k$ squares of the other color in this column. Then, the total number of same-color "square pairs" in this column is
$$
\begin{array}{l}
\mathrm{C}_{k}^{2}+\mathrm{C}_{7-k}^{2}=\frac... | 21 | Combinatorics | proof | Yes | Yes | cn_contest | false |
1. Let $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}, y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$. Then $x^{2}+x y$ $+y^{2}+1=$ $\qquad$ | 1.100 .
$$
\begin{array}{l}
\text { Given } x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=(\sqrt{3}+\sqrt{2})^{2}=5+2 \sqrt{6}, \\
y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=(\sqrt{3}-\sqrt{2})^{2}=5-2 \sqrt{6},
\end{array}
$$
Therefore, $x^{2}+x y+y^{2}+1=(x+y)^{2}-x y+1=100$. | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If the decimal parts of $7+\sqrt{7}$ and $7-\sqrt{7}$ are $a$ and $b$ respectively, then $a b-3 a+2 b+1=$ $\qquad$ . | 2.0.
Since $2<\sqrt{7}<3$, therefore, the decimal part of $7+\sqrt{7}$ is $a=\sqrt{7}-2$. Also, because $-3<-\sqrt{7}<-2$, then $0<3-\sqrt{7}<1$. Therefore, the decimal part of $7-\sqrt{7}$ is $b=3-\sqrt{7}$. Hence
$$
\begin{array}{l}
a b-3 a+2 b+1 \\
=(a+2)(b-3)+7=\sqrt{7} \times(-\sqrt{7})+7=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given that $a, b, c, d$ take certain real values, the equation $x^{4}+a x^{3}+b x^{2}+c x+d=0$ has 4 non-real roots, where the product of 2 of the roots is $13+i$, and the sum of the other 2 roots is $3+4i$, where $i$ is the imaginary unit. Find $b$.
(13th American Invitational Mathematics Examination) | Explanation: Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the 4 roots of the equation. By the problem, we can assume $x_{3}=\overline{x_{1}}, x_{4}=\overline{x_{2}}$, then we have $x_{1} x_{2}=$
$$
\begin{aligned}
13+\mathrm{i}, & x_{3}+x_{4}=3+4 \mathrm{i} \text {. Therefore, } \\
b & =x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2}... | 51 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. A bookstore is holding a summer book fair, where book buyers can enjoy the following discounts:
(1) For a single purchase not exceeding 50 yuan, no discount is given;
(2) For a single purchase exceeding 50 yuan but not exceeding 200 yuan, a 10% discount is given on the marked price;
(3) For a single purchase exceedi... | 4.3.
Obviously, 81 and 126 are both the prices after the discount. If not discounted, the combined price should be
$$
(81+126) \div 0.9=230 \text { (yuan). }
$$
Since 230 yuan exceeds 200 yuan, according to (3), it should be
$$
200 \times 0.9+30 \times 0.8=204 \text { (yuan) } \text {. }
$$
Thus, we have $81+126-204... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) As shown in Figure 3, in the right triangle $\triangle ABC$, $\angle B=90^{\circ}$, $AB=BM=12$, $DM \parallel AB$. Also, $N$ is the midpoint of $BM$, and $\angle ADN=\angle BAD$.
(1) Prove:
$$
CN \cdot AB=CB \cdot DN \text {; }
$$
(2) Find $S_{\triangle DNM}$. | II. As shown in Figure 5, construct $\angle T B A = \angle C$.
Since $\angle A D N = \angle B A D$, then $\angle C D N = \angle B A T$.
Therefore, $\triangle C D N \sim \triangle B A T$.
Thus, $\frac{C N}{T B} = \frac{C D}{A B}$, which means
$C N \cdot A B = T B \cdot C D$.
Also, $\angle C = \angle C$,
$\angle C N D = ... | 24 | Geometry | proof | Yes | Yes | cn_contest | false |
5. Let $\{x\}$ denote the fractional part of the real number $x$. If $a=$ $(5 \sqrt{13}+18)^{2005}$, then $a \cdot\{a\}=$ $\qquad$ | 5.1.
Let $b=(5 \sqrt{13}-18)^{2005}$.
Since $0<5 \sqrt{13}-18<1$, we have $0<b<1$.
By the binomial theorem, we know
$$
\begin{aligned}
a= & (5 \sqrt{13})^{2005}+\mathrm{C}_{2005}^{1}(5 \sqrt{13})^{2004} \times 18^{1}+\cdots+ \\
& \mathrm{C}_{2005}^{r}(5 \sqrt{13})^{2005-r} \times 18^{r}+\cdots+18^{2005}, \\
b= & (5 \s... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let $f(n)$ be a function defined on $\mathbf{N}_{+}$ taking non-negative integer values, and for all $m, n \in \mathbf{N}_{+}$, we have
$$
\begin{array}{l}
f(m+n)-f(m)-f(n)=0 \text{ or } 1, \\
f(2)=0, f(3)>0, f(6000)=2000 . \\
\text{Find } f(5961) .
\end{array}
$$ | Solution: From the analysis, we know that $f(1)=0$.
From $f(3)-f(2)-f(1)=0$ or 1, we get $0 \leqslant f(3) \leqslant 1$.
But $f(3)>0$, so $f(3)=1$.
By the problem statement, we have
$$
f(3 n+3)=f(3 n)+f(3)+0 \text { or } 1 \text {, }
$$
which means $f(3 n+3)=f(3 n)+1$ or 2.
Therefore, $f(3(n+1)) \geqslant f(3 n)+1$.
L... | 1987 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given $a \neq b$, and
$$
a^{2}-4 a+1=0, b^{2}-4 b+1=0 \text {. }
$$
Find the value of $\frac{1}{a+1}+\frac{1}{b+1}$. | Solution 1: Since
$$
\frac{1}{a+1}+\frac{1}{b+1}=\frac{(a+b)+2}{a b+(a+b)+1},
$$
we only need to find the values of $a b$ and $a+b$.
From the given conditions, $a$ and $b$ are the roots of the quadratic equation $x^{2}-4 x+1=0$, so
$$
a+b=4, a b=1 \text {. }
$$
Therefore, $\frac{1}{a+1}+\frac{1}{b+1}=\frac{(a+b)+2}{a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example $9 . \angle P O Q=30^{\circ}, A$ is a point on $O Q$, $B$ is a point on $O P$, and $O A=5, O B=12$. Take point $A_{1}$ on $O B$, and take point $A_{2}$ on $A Q$, let $l=A A_{1}+A_{1} A_{2}+A_{2} B$. Find the minimum value of $l$. | Solution: As shown in Figure 13, with the line $O P$ as the axis of symmetry, construct the axis-symmetric figure $\angle P O Q_{0}$ of $\angle P O Q$; with the line $O Q$ as the axis of symmetry, construct the axis-symmetric figure $\angle Q O P_{0}$ of $\angle Q O P$. At this point, the symmetric point of point $A$ a... | 13 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 As shown in Figure 14, villages $A$, $B$, and $C$ are located along an east-west oriented highway, with $AB=2 \text{ km}$, $BC=3 \text{ km}$. To the due north of village $B$ is village $D$, and it is measured that $\angle ADC=45^{\circ}$. Now the $\triangle ADC$ area is planned to be a development zone, exce... | Solution: As shown in Figure 15, construct the axisymmetric figure of Rt $\triangle A D B$ with respect to the line containing $D A$, which is Rt $\triangle A D B_{1}$. It is easy to know that
Rt $\triangle A D B \cong$
Rt $\triangle A D B_{1}$.
Construct the axisymmetric figure of Rt $\triangle C D E$ with respect to ... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 5, in isosceles $\triangle A B C$, the base angle $\angle B=$ $15^{\circ}$, and the length of the leg $A B=10$. Then the area of this triangle is $\qquad$ | 8. 25 .
Draw the altitude $C D$ from $C$ to $B A$ extended, meeting at point $D$. In the right triangle $\triangle A C D$,
$$
\angle C A D=\angle B+\angle C=2 \angle B=30^{\circ} \text {. }
$$
Then $C D=\frac{1}{2} A C=\frac{1}{2} \times 10=5$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} A B \cdot C D=\frac{1}{2} \ti... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. Given real numbers $a$, $b$, $x$, $y$ satisfy $a x+b y=3$, $a y-b x=5$. Then the value of $\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right)$ is $\qquad$. | 9. 34 .
$$
\begin{array}{l}
\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right) \\
=a^{2} x^{2}+a^{2} y^{2}+b^{2} x^{2}+b^{2} y^{2} \\
=\left(a^{2} x^{2}+b^{2} y^{2}+2 a b x y\right)+\left(a^{2} y^{2}+b^{2} x^{2}-2 a b x y\right) \\
=(a x+b y)^{2}+(a y-b x)^{2} \\
=3^{2}+5^{2}=34 .
\end{array}
$$ | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Given $f(x)=\frac{x}{1+x}$. Find the value of the following expression:
$$
\begin{array}{l}
f\left(\frac{1}{2004}\right)+f\left(\frac{1}{2003}\right)+\cdots+f\left(\frac{1}{2}\right)+f(1)+ \\
f(0)+f(1)+f(2)+\cdots+f(2003)+f(2004) .
\end{array}
$$ | Three, 13. Since $f(x)=\frac{x}{1+x}$, then
$$
f\left(\frac{1}{n}\right)=\frac{\frac{1}{n}}{1+\frac{1}{n}}=\frac{1}{n+1}, f(n)=\frac{n}{1+n} \text {. }
$$
Therefore, $f\left(\frac{1}{n}\right)+f(n)=\frac{1}{n+1}+\frac{n}{n+1}=1$.
Thus, the original expression is
$$
\begin{array}{l}
=f\left(\frac{1}{2004}\right)+f(2004... | 2004 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
19. (15 points) As shown in Figure 4, there is a pointer on a disk, initially pointing to the top of the disk. The pointer rotates clockwise around the center of the disk by an angle $\alpha$ each time, and $3.6^{\circ}<\alpha<180^{\circ}$. After 2,004 rotations, it returns to its initial position for the first time, p... | 19. Clearly, if
$3.6^{\circ}1$, then let
$n_{1}=\frac{n}{d}, n_{2}=\frac{2004}{d}$.
At this point, we have $\alpha=\frac{n_{1} \times 360^{\circ}}{n_{2}}$.
This means that after the pointer rotates $n_{2}$ times, each time rotating by $\alpha$, the pointer will rotate $n_{1}$ full circles and return to its initial posi... | 325 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7.2. There are 12 people in the room,
some of whom always lie, while the rest always tell the truth. One of them says: “There are no honest people here”; another says: “There is at most 1 honest person here”; the third person says: “There are at most 2 honest people here”; and so on, until the 12th person says: “There... | 7.2. There are 6 honest people in the room.
Obviously, the first few people are not honest. If the 6th person is telling the truth, then there are no more than 5 honest people in the room. On the other hand, from him onwards, the next 7 people are telling the truth, and thus, they should all be honest, which is imposs... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
6.2. The shape of a certain castle is a heptagon, and there is a bell tower at each of its vertices. Each side of the castle wall is guarded by the soldiers in the bell towers at the two endpoints of that side. How many soldiers are needed in total to ensure that each side of the wall is guarded by at least 7 soldiers? | 6.2. If there are no more than 3 guards on each tower, then no wall will be guarded by 7 guards. Observe any tower with at least 4 guards. At this point, the remaining 6 towers can be divided into 3 "adjacent pairs," with each "adjacent pair" corresponding to one wall, so each "adjacent pair" has at least 7 guards. The... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given the equation $x^{10}+(13 x-1)^{10}=0$ has 10 complex roots $r_{i}, \overline{r_{i}}(i=1,2,3,4,5)$, where $\overline{r_{i}}$ is the conjugate of $r_{i}$. Find the value of $\sum_{i=1}^{5} \frac{1}{r_{i} \bar{r}_{i}}$.
(12th American Invitational Mathematics Examination) | Explanation: The original equation can be transformed into
$$
\left(\frac{13 x-1}{x}\right)^{10}=-1 \text {. }
$$
Let the 10 complex roots of $y^{10}=-1$ be $\omega_{i}$ and $\overline{\omega_{i}}$ $(i=1,2,3,4,5)$, then we have $\omega_{i}=\frac{13 r_{i}-1}{r_{i}}$, which means
$$
\begin{array}{l}
\frac{1}{r_{i}}=13-\... | 850 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7.2. Does there exist a number of the form $1000 \cdots 001$ that can be divided by a number of the form $111 \cdots 11$? | 7.2. Only for 11
In fact, $1000 \cdots 001=99 \cdots 9+2$, and the remainder of $99 \cdots 9$ divided by $111 \cdots 11$ is a number of the form $99 \cdots 9$ with fewer digits than the divisor. Therefore, if $99 \cdots 9+2$ can be divided by $111 \cdots 11$, then $99 \cdots 9+2$ must equal $111 \cdots 11$. This is on... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $k$ be a real number, and the quadratic equation $x^{2}+k x+k+1=0$ has two real roots $x_{1}$ and $x_{2}$. If $x_{1}+2 x_{2}^{2}=k$, then $k$ equals $\qquad$ . | 2.5.
From $\Delta \geqslant 0$ we get
$$
k^{2}-4 k-4 \geqslant 0 \text {. }
$$
Since $x_{2}^{2}=-k x_{2}-k-1$, then,
$$
\begin{array}{l}
x_{1}+2 x_{2}^{2}=k \Rightarrow x_{1}-2 k x_{2}=3 k+2 \\
\Rightarrow\left(x_{1}+x_{2}\right)-(2 k+1) x_{2}=3 k+2 \\
\Rightarrow-k-(2 k+1) x_{2}=3 k+2 \\
\Rightarrow-(2 k+1) x_{2}=2(... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of integers $x$ that make $x^{2}+x-2$ a perfect square is $\qquad$.
Make the above text in English, please keep the original text's line breaks and format, and output the translation result directly. | 3.4 .
Let $x^{2}+x-2=y^{2}, y \in \mathbf{N}$, then $4 x^{2}+4 x-8=(2 y)^{2}$,
i.e., $(2 x-2 y+1)(2 x+2 y+1)=3^{2}$.
Therefore, we have $\left\{\begin{array}{l}2 x-2 y+1=1, \\ 2 x+2 y+1=9 ;\end{array}\left\{\begin{array}{l}2 x-2 y+1=3, \\ 2 x+2 y+1=3 ;\end{array}\right.\right.$ $\left\{\begin{array}{l}2 x-2 y+1=-3, \\... | 4 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
Five, (20 points) Find the smallest natural number $k$, such that for any $x \in [0,1]$ and $n \in \mathbf{N}_{+}$, the inequality
$$
x^{k}(1-x)^{n}<\frac{1}{(1+n)^{3}}
$$
always holds. | Given $x \in [0,1]$, by the AM-GM inequality, we have
$$
\begin{array}{l}
x^{k}(1-x)^{n} \\
=\underbrace{x \cdots \cdots x}_{k \uparrow} x \cdot \underbrace{(1-x) \cdot(1-x) \cdots \cdots(1-x)}_{n \uparrow} \\
=\left(\frac{n}{k}\right)^{n} \underbrace{x \cdots \cdots x}_{k \uparrow} \cdot \underbrace{\frac{k(1-x)}{n} \... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 4, all quadrilaterals are squares, and all triangles are right triangles, where the side length of the largest square is $13 \mathrm{~cm}$. Then the sum of the areas of the four shaded squares is $\qquad$ | 3.169.
As shown in Figure 12, by repeatedly applying the Pythagorean theorem, we get
$$
\begin{array}{l}
S_{A}+S_{B}+S_{C}+S_{D} \\
=S_{E}+S_{F}=S_{G} \\
=13^{2}=169 .
\end{array}
$$ | 169 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $\frac{y+z-x}{x+y+z}=\frac{z+x-y}{y+z-x}=\frac{x+y-z}{z+x-y}$ $=p$. Then $p^{3}+p^{2}+p=$ $\qquad$ | 4.1 .
From the given, we have
$$
\begin{array}{l}
p^{3}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x} \cdot \frac{x+y-z}{z+x-y}=\frac{x+y-z}{x+y+z}, \\
p^{2}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x}=\frac{z+x-y}{x+y+z} .
\end{array}
$$
Then \( p^{3}+p^{2}+p=1 \). | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If the sum of the areas of three square pieces of paper with integer side lengths is 2004, and the area of the largest square piece of paper is $S_{1}$, and the area of the smallest square piece of paper is $S_{2}$, then the maximum value of $\frac{S_{1}}{S_{2}}$ is $\qquad$ | 6.484.
Let the side lengths of the squares be $a$, $b$, and $c$, with $a<b<c$. Then, $a^{2}+b^{2}+c^{2}=2004$.
Obviously, $S_{2}=a^{2}, S_{1}=c^{2} \Rightarrow \frac{S_{1}}{S_{2}}=\frac{c^{2}}{a^{2}}$.
It is easy to see that $40^{2}+20^{2}+2^{2}=2004$, $44^{2}+8^{2}+2^{2}=2004$.
To maximize $\frac{S_{1}}{S_{2}}$, $S_... | 484 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. There are 8 consecutive positive
integers, the sum of which can be expressed as the sum of 7
consecutive positive integers, but cannot be expressed as the sum of 3 consecutive positive integers. Then, the minimum value of the largest number among these 8 consecutive positive integers is $\qquad$ | 4.21.
Let these 8 consecutive positive integers be
$$
a, a+1, a+2, a+3, a+4, a+5, a+6, a+7 \text {. }
$$
Their sum is $S=8a+28$, which means $S$ leaves a remainder of 4 when divided by 8.
Since $S$ is divisible by 7, we have $8a+28=7n$.
Therefore, $41n$.
Let $n=4k$, then $2a+7=7k$.
Thus, $a=\frac{7(k-1)}{2}$.
For $a$... | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. As shown in Figure 9, in $\triangle A B C$, $\angle C=$ $90^{\circ}, I$ is the intersection of the angle bisectors $A D$ and $B E$ of $\angle A$ and $\angle B$. Given that the area of $\triangle A B I$ is 12. Then the area of quadrilateral $A B D E$ is $\qquad$ | 5.24
As shown in Figure 16, construct the symmetric point $F$ of $E$ with respect to $AD$, and the symmetric point $G$ of $D$ with respect to $BE$. Then $F$ and $G$ lie on $AB$, and $AF = AE$, $BG = BD$. Therefore, $\angle AIB = \angle EID = 135^{\circ}$, $\angle DIB = \angle EIA = \angle AIF = \angle BIG = 45^{\circ}... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (15 points) Given that $a$ is a positive integer, and $a^{2}+$ $2004 a$ is a perfect square of a positive integer. Find the maximum value of $a$. | Let $a^{2}+2004 a=m^{2}$, where $m$ is a positive integer.
By completing the square and constant transformation, we get
$$
(a+1002)^{2}-m^{2}=1002^{2}=2^{2} \times 3^{2} \times 167^{2} \text{. }
$$
It is easy to see that $a+1002+m$ and $a+1002-m$ are both even numbers, and
$$
a+1002+m>a+1002-m>0 \text{. }
$$
To find ... | 250000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Given
$$
f(x)=(x-1)(x-2) \cdots \cdots(x-2004) \text {. }
$$
Ask: How many real numbers $x$ satisfy $|f(x)|=1$? | 一、As shown in Figure 7, draw a sketch of the function $f(x)$.
From the graph, we know that between 1 and 2004, for every 2 consecutive integers, there are 2 real numbers such that $|f(x)|=1$.
In the intervals $(0,1)$ and $(2004,+\infty)$, there is 1 real number $x$ in each interval such that $|f(x)|=1$.
Therefore, th... | 4008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) Find all positive integers $n$ such that $n=$ $p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}$, where $p_{1}, p_{2}, p_{3}, p_{4}$ are the four smallest distinct positive divisors of $n$. | If $n$ is odd, then all factors of $n$ are odd, i.e., $n \not\equiv 0(\bmod 4)$.
But $n=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2} \equiv 0(\bmod 4)$, which is a contradiction.
Therefore, $21 \nmid n$.
If $4 \mid n$, then $p_{1}=1, p_{2}=2$.
Thus, $n=1+0+p_{3}^{2}+p_{4}^{2} \not\equiv 0(\bmod 4)$, which is a contradiction... | 130 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find the minimum value of the function with real variables $x$ and $y$
$$
u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}
$$
(2nd "Hope Cup" National Mathematics Invitational Competition) | Explanation: The original formula is transformed to
$$
u(x, y)=\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}+(x-y)^{2}-2 \text {. }
$$
Consider the points $P_{1}\left(x, \frac{9}{x}\right), P_{2}\left(y,-\sqrt{2-y^{2}}\right)$. When $x \in \mathbf{R}(x \neq 0)$, point $P_{1}$ lies on a hyperbola with the coordinate axes... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7.5. If $a, b, c, d, e, f, g, h, k$ are all 1 or -1, try to find the maximum possible value of
$$
a e k - a f h + b f g - b d k + c d h - c e g
$$ | 7.5. Since each term in the expression $a e k - a f h + b f g - b d k + c d h - c e g$ is either 1 or -1, the value of the expression is even. However, the expression cannot equal 6. If it did, then $a e k$, $b f g$, and $c d h$ would all have to be 1, making their product 1, while $a f h$, $b d k$, and $c e g$ would a... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8.7. Squares in a grid that share a common edge are called adjacent. In a $7 \times 7$ grid, one Go stone is placed in one of the squares. It is allowed to place new stones in empty squares, but only in those that have at most one adjacent square occupied by a stone. How many stones can be placed in the grid at most? | 8.7. Place chess pieces according to the rules while calculating the number of sides of the squares that have pieces (each side is counted only once, even if both squares on either side of the side have pieces. After placing the 1st piece, 4 sides are obtained, and each subsequent piece can add at most 3 new sides. The... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $x, y$ satisfy $x^{3}+y^{3}=2$. Then the maximum value of $x+y$ is $\qquad$ . | Ni.1.2.
Let $x+y=k$, it is easy to know that $k>0$.
From $x^{3}+y^{3}=2$, we get
$$
(x+y)\left(x^{2}-x y+y^{2}\right)=2 \text {. }
$$
Thus, $x y=\frac{1}{3}\left(k^{2}-\frac{2}{k}\right)$.
From this, we know that $x, y$ are the two real roots of the equation about $t$
$$
t^{2}-k t+\frac{1}{3}\left(k^{2}-\frac{2}{k}\ri... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given a three-digit number $\overline{a b c}$, the last three digits of its square are also $\overline{a b c}$. Then, the sum of all such three-digit numbers is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4.1001 .
Let $x=\overline{a b c}$, then the last three digits of $x^{2}-x$ are all 0. Therefore, $x(x-1)$ is a multiple of 1000.
Since $1000=2^{3} \times 5^{3}$, and $x$ and $x-1$ are coprime, thus, $x-1$ is a multiple of $5^{3}$ or $x$ is a multiple of $5^{3}$.
When $x-1=125,125 \times 3,125 \times 5,125 \times 7$,... | 1001 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Three piles of stones have the numbers 21, 10, and 11, respectively. Now, the following operation is performed: each time, 1 stone is taken from any two piles, and then these 2 stones are added to the other pile. The question is:
(1) Can the number of stones in the three piles be 4, 14, and 24 after several such opera... | (1) It can be achieved.
The minimum number of operations required is 6, for example:
$$
\begin{array}{l}
(21,10,11) \rightarrow(23,9,10) \rightarrow(22,8,12) \\
\rightarrow(24,7,11) \rightarrow(23,6,13) \rightarrow(25,5,12) \\
\rightarrow(24,4,14) .
\end{array}
$$
Since the minimum number of stones in one pile decreas... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the sequence
$$
\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} .
\end{array}
$$
Then $\lim _{n \rightarrow \infty}\left(\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}}\right)=$ ( ).
(A) 2
(B) 3
(C)... | 6.C.
$$
\begin{array}{l}
\text { Because } a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1} \\
=\frac{n(n+1)}{2(n+1)}=\frac{n}{2}, \\
\text { so } \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-\fr... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
6. Given that the odd number $n$ is a three-digit number, and the sum of the last digits of all its factors (including 1 and $n$) is 33. Then $n$ $=$ . $\qquad$ | 6.729 (or $27^{2}$).
From $n$ being an odd number, we know that each of its factors is also odd, and of course, the unit digit of each factor is odd. Since the sum of the unit digits of all factors is 33, which is an odd number, $n$ must have an odd number of factors. Therefore, $n$ is a perfect square.
Since a prime... | 729 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) Let $a_{n}$ be the number of subsets of the set $\{1,2, \cdots, n\}$ $(n \geqslant 3)$ that have the following property: each subset contains at least 2 elements, and the difference (absolute value) between any 2 elements in each subset is greater than 1. Find $a_{10}$. | Solution 1: Consider the recurrence relation of $a_{n}$, dividing the subsets that satisfy the condition into two categories:
The first category contains $n$. This type of subset, apart from the $a_{n-2}$ subsets that satisfy the condition in $\{1,2, \cdots, n-2\}$ combined with the element $n$, also includes $n-2$ tw... | 133 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
II. (50 points) An old man divides his savings of $m$ gold coins among his $n$ children ($m, n$ are positive integers greater than 1). First, he gives the eldest child 1 gold coin and $\frac{1}{7}$ of the remainder; then, from the remaining coins, he gives the second child 2 gold coins and $\frac{1}{7}$ of the remainde... | Let the number of gold coins left after giving to the $k$-th child be $a_{k}$, then $a_{0}=m, a_{n-1}=n$,
$$
a_{k}=a_{k-1}-\left(k+\frac{a_{k-1}-k}{7}\right)=\frac{6}{7}\left(a_{k-1}-k\right) \text {. }
$$
Thus, $a_{k}+6 k-36=\frac{6}{7}\left[a_{k-1}+6(k-1)-36\right]$.
This indicates that the sequence $b_{k}=a_{k}+6 k... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $\left|x_{i}\right|<1(i=1,2, \cdots, n)$, let $\sum_{i=1}^{n}\left|x_{1}\right|=19+\left|\sum_{k=1}^{n} x_{k}\right|$. Find the minimum value of $n$. | (Given: $\left|\sum_{i=1}^{n} x_{i}\right| \geqslant 0$, then $\sum_{i=1}^{n}\left|x_{i}\right| \geqslant 19$, since $\left|x_{i}\right|<1$, then $n \geqslant 20$. Taking $x_{i}=\frac{19}{20}(i=1,2, \cdots, 10), x_{j}=-\frac{19}{20}(j=11,12, \cdots, 20)$, we have $n=20$, hence $n_{\min }=20$.) | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If $a_{1}=1, a_{k}=k+a_{k-1}(2 \leqslant k \leqslant$ $n)$, then $a_{1}, a_{2}, \cdots, a_{n}$ is called a regular number. Question: How many numbers in the set $\{1,2, \cdots, 2001\}$ are either a regular number itself or the sum of several different regular numbers? | Solution: It is easy to know that $a_{k}=\frac{k(k+1)}{2}=C_{k+1}^{2}$, these regular numbers form the sequence $1,3,6,10,15,21,28,36,45,55$, ... First, we check the numbers that can be represented by regular numbers (i.e., can be written as the sum of several different regular numbers):
Among the positive integers le... | 1995 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For example, 42000 people are divided into $n$ groups, satisfying:
(i) No one in each group knows everyone in the same group;
(ii) In any three people in each group, at least two of them do not know each other;
(iii) For any two people who do not know each other in each group, there is exactly one person in the same gr... | Proof: Representing people as points, connect a line between points representing people who do not know each other within the same group, and do not connect a line between points representing people who know each other. The original conditions can be restated as:
(i) There are no isolated points in any group;
(ii) Amon... | 400 | Combinatorics | proof | Yes | Yes | cn_contest | false |
3. The three sides are three consecutive positive integers, and its perimeter is less than or equal to 100 of the acute triangle has $\qquad$.
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 is not part of the translation but is provided to clarify the instruction. The actual translation is above. | 3. 29 | 29 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
Four, among the 10-digit numbers where each digit is different, how many are multiples of 11111? Prove your conclusion.
In the 10-digit numbers where each digit is different, how many are multiples of 11111? Prove your conclusion. | Let $n=\overline{a b c d e f g h i j}$ satisfy the conditions, then $11111 \mid n$, and $a, b, \cdots, j$ are a permutation of $0,1,2, \cdots, 9$ $(a \neq 0)$.
Since $a+b+\cdots+j=0+1+\cdots+9=45$,
it follows that $9 \mid(a+b+\cdots+j)$. Therefore, $9 \mid n$.
Since 11111 and 9 are coprime, then $99999 \mid n$.
Let $x=... | 3456 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$1 . m$ is what integer when the equation
$$
\left(m^{2}-1\right) x^{2}-6(3 m-1) x+72=0
$$
has two distinct positive integer roots? | (Tip: $m^{2}-1 \neq 0, m \neq \pm 1$. Since $\Delta=36(m-3)^{2}>$ 0, hence $m \neq 3$. Using the quadratic formula, we get $x_{1}=\frac{6}{m-1}, x_{2}=$ $\frac{12}{m+1}$. Therefore, $(m-1)|6,(m+1)| 12$. So $m=2$, at this point, $x_{1}=6, x_{2}=4$.) | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. From 3 boys and $n$ girls, any 3 people are selected to participate in a competition, given that the probability of having at least 1 girl among the 3 people is $\frac{34}{35}$. Then $n=$ $\qquad$ . | 13.4. From the condition, $1-\frac{\mathrm{C}_{3}^{3}}{\mathrm{C}_{n+3}^{3}}=\frac{34}{35}$, solving for $n$ yields $n=4$. | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
14. There are 10 table tennis players participating in a round-robin tournament. The match results show that there are no draws, and among any 5 players, there is 1 player who wins against the other 4, and 1 player who loses to the other 4. Then the number of players who won exactly two matches is $\qquad$. | 14.1.
It can be proven that under the given conditions, no two players have the same number of wins. Therefore, the number of wins for 10 players are 10 different numbers: $0,1, \cdots, 9$. Hence, the number of players who win exactly two games is 1.
If not, suppose there exist $A$ and $B$ with the same number of win... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
16. (12 points) A clothing workshop has 4 teams, A, B, C, and D, with their daily production capacities for shirts and pants as shown in Table 1. Now, it is required to produce matching sets of shirts and pants (one shirt and one pair of pants make a set). How many sets can these 4 teams produce in 7 days? | 16. The ratio of shirts to skirts produced daily by groups $A$, $B$, $C$, and $D$ are $\frac{8}{10}$, $\frac{9}{12}$, $\frac{7}{11}$, and $\frac{6}{7}$, respectively, and
$$
\frac{6}{7}>\frac{8}{10}>\frac{9}{12}>\frac{7}{11} \text {. }
$$
Only by having the group with the highest efficiency in producing shirts make sh... | 125 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
18. (16 points) In $\triangle A B C$ with a fixed perimeter, it is known that $|A B|=6$, and when vertex $C$ is at a fixed point $P$, $\cos C$ has a minimum value of $\frac{7}{25}$.
(1) Establish an appropriate coordinate system and find the equation of the locus of vertex $C$;
(2) Draw a line through point $A$ that in... | 18. (1) Establish a Cartesian coordinate system with the line $AB$ as the $x$-axis and the perpendicular bisector of $AB$ as the $y$-axis. Let $|CA| + |CB| = 2a (a > 3)$ be a constant, then the locus of point $C$ is an ellipse with foci at $A$ and $B$. Therefore, the focal distance is $2c = |AB| = 6$.
Since $\cos C = \... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $m$ be a non-zero integer, and the quadratic equation in $x$, $m x^{2}-(m-1) x+1=0$, has rational roots. Find the value of $m$.
| (Given: Let $\Delta=(m-1)^{2}-4 m=n^{2}, n$ be a non-negative integer, then $(m-3)^{2}-n^{2}=8$, i.e., $(m-3-n)(m-3+n)=$ 8. Following Example 2, we get $\left\{\begin{array}{l}m=6, \\ n=1\end{array}\right.$ or $\left\{\begin{array}{l}m=0, \\ n=1\end{array}\right.$ (discard). Therefore, $m=6$, and the two roots of the e... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $\log _{a} x=24, \log _{b} x=40, \log _{\text {abc }} x=$
12. Then, $\log _{c} x=$ $\qquad$ . | \begin{array}{l}\text { II.7.60. } \\ \log _{x} c=\log _{x} a b c-\log _{x} a-\log _{x} b \\ =\frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60} .\end{array} | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $f(x)=\prod_{i=1}^{4}\left(x^{2}-8 x+c_{i}\right), M=|x|$ $f(x)=0\}$. It is known that $M=\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}\right.$, $\left.x_{7}, x_{8}\right\} \subseteq \mathbf{N}$. Then, $\max \left\{c_{1}, c_{2}, c_{3}, c_{4}\right\}-$ $\min \left\{c_{1}, c_{2}, c_{3}, c_{4}\right\}=$ $\qquad$ | 8.15.
Let the two roots of $x^{2}-8 x+c=0$ be $\alpha, \beta$, then $\alpha+\beta=8$. The unequal non-negative integer values of $(\alpha, \beta)$ are only $(0,8),(1,7),(2,6)$, $(3,5)$. Therefore, $\left\{c_{1}, c_{2}, c_{3}, c_{4}\right\}=\{0,7,12,15\}$. | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6.4. There is a 36-digit number, in which the digits $1,2, \cdots, 9$ each appear 4 times, and except for 9, all other digits are less than the digit that follows them. It is known that the first digit of the number is 9. What is the last digit of the number? Please provide all possible answers and prove that there are... | 6.4. The last digit of this number is 8.
According to the problem, only 9 can follow 8. If all four 8s are located within the number, then each of them is followed by one 9, and adding the one 9 at the beginning, there are a total of 5 nines. | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7.2. Person A and Person B perform division with remainder on the same number. A divides it by 8, and B divides it by 9. It is known that the sum of the quotient obtained by A and the remainder obtained by B equals 13. Try to find the remainder obtained by A.
| 7.2. Let $a$ and $b$ represent the quotient and remainder obtained by Jia, and let $c$ and $d$ represent the quotient and remainder obtained by Yi. Thus, we have
$$
8 a+b=9 c+d, a+d=13 \text{. }
$$
Substituting $d=13-a$ into the first equation, we get $9(a-c)=$ $13-b$, so $13-b$ is divisible by 9. Since $b$ can only b... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7.4. There are three piles of stones, and it is allowed to add stones to any pile, with the number of stones added must equal the sum of the number of stones in the other two piles at that time; it is also possible to remove stones from any pile when possible, with the number of stones removed equal to the sum of the n... | 7.4. Cannot.
Let the method of adding stones be $A$, and the method of reducing stones be $B$. For any given three piles of stones, let the $i$-th operation be $c_{i}(i=1,2, \cdots)$. To obtain the pile with the fewest stones, since $A$ operations are limited, there must be a case where “$c_{i}=A, c_{i+1}=B$”. In this... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
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