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Example 1 Find the positive integer root of the equation
$$
\frac{2}{n}+\frac{3}{n+1}+\frac{4}{n+2}=\frac{133}{60}
$$
(1990, Shanghai Junior High School Mathematics Competition) | Solution: Since $n$ is a natural number, by the properties of natural numbers and
$$
\begin{array}{l}
\frac{1}{n}>\frac{1}{n+1}>\frac{1}{n+2} \\
\therefore \frac{2+3+4}{n+2}<\frac{2}{n}+\frac{3}{n+1}+\frac{4}{n+2} \\
<\frac{2+3+4}{n} .
\end{array}
$$
That is, $\frac{9}{n+2}<\frac{133}{60}<\frac{9}{n}$.
Solving this, ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, when arranging student dormitories, it is stipulated that two students live in one room. The school's preference order for various combinations of students from grades $A, B, C, D, E$ is
$$
A A, A B, A C, A D, A E, B B, B C, B D,
$$
$B E, C C, C D, C E, D D, D E, E E$.
That is, two students from grade $A$ living... | Three, Solution: According to the problem, we have
$$
a a>a b>a c>a d>a e>b b>b c>b d>b e>c c>c d
$$
$>c e>d d>d e>e e$, which means
$$
\left\{\begin{array}{l}
a>b>c>d>e, \\
a>\frac{b^{2}}{e}, \\
b>\frac{c^{2}}{e}, \\
c>\frac{d^{2}}{e} .
\end{array}\right.
$$
From (2), (3), and (4), we know that for a given $t$, to ma... | 69 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
I. (25 points) The Xiguang Factory glasses workshop has received a batch of tasks, requiring the processing of 6000 Type A parts and 2000 Type B parts. This workshop has 214 workers, and each of them can process 3 Type B parts in the time it takes to process 5 Type A parts. These workers are to be divided into two grou... | One, Solution: Let the number of people in the group that includes those with the surname $A$ be $x$, and in a unit of time, the number of $\mathrm{C} . \mathrm{A}$ parts processed by one person is $5 k$, then the number of the other parts is $3 k$.
The time required to process type $A$ parts is $t_{A}(x)=\frac{6000}{5... | 137 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (35 points) The real number sequence $a_{1}, a_{2} \cdots, a_{1997}$ satisfies:
$$
\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\left|a_{1996}-a_{1997}\right|=
$$
1997. If the sequence $\left\{b_{n}\right\}$ satisfies:
$$
b_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}(k=1,2, \cdots, 1997),
$$
find the max... | $$
\begin{array}{l}
\text { III. Solution: From } \left.\left|b_{k}-b_{k+1}\right|=\frac{1}{k(k+1)} \right\rvert\,(k+1)\left(a_{1}+a_{2}\right. \\
\left.+\cdots+a_{k}\right)-k\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) \mid \\
=\frac{1}{k(k+1)}\left|a_{1}+a_{2}+\cdots+a_{k}-k a_{k+1}\right| \\
\left.=\frac{1}{k(k+1)} \rig... | 1996 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 The first 100 natural numbers are arranged in a certain order, then the sum of every three consecutive numbers is calculated, resulting in 98 sums, of which the maximum number of sums that can be odd is how many?
(21st Russian Mathematical Olympiad) | Proof: First, we prove that it is impossible for all 98 sums to be odd. We use proof by contradiction.
If all the sums are odd, then the arrangement of these 100 natural numbers can only be one of the following four cases:
(1) odd odd odd odd odd odd ...; (2) odd even even odd even even ...;
(3) even odd even even odd... | 97 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Solve the equation
\[
\begin{array}{l}
\sqrt{x^{2}+5 x-14}+\sqrt{x+7}+\sqrt{2-x}+x-5 \\
=0 .
\end{array}
\] | Solution: By the properties of quadratic radicals, we get
$$
\left\{\begin{array}{l}
x^{2}+5 x-14 \geqslant 0, \\
x+7 \geqslant 0, \\
2-x \geqslant 0 .
\end{array}\right.
$$
Solving, we get $-7 \leqslant x \leqslant-7$ or $2 \leqslant x \leqslant 2$.
That is, $x=-7$ or $x=2$.
Upon verification, $x=2$ is a root of the ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 For a finite set $A$, function $f: N \rightarrow A$ has only the following property: if $i, j \in N, |H| i-j |$ is a prime number, then $f(i) \neq f(j)$. How many elements does set $A$ have at least? | Solution: Let $|A|$ denote the number of elements in the finite set $A$, and estimate the lower bound of $|A|$.
Since the absolute value of the difference between any two numbers among $1, 3, 6, 8$ is a prime number, by the problem's condition: $f(1)$, $f(3)$, $f(6)$, $f(8)$ are four distinct elements in $A$. Therefor... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Five, let $A=\{1,2,3, \cdots, 17\}$. For any function $f: A \rightarrow A$, denote
$$
f^{[1]}(x)=f(x), f^{[k+1]}(x)=f\left(f^{[k]}(x)\right)
$$
$(k \in \mathbb{N})$. Find the natural number $M$ such that:
$$
\begin{array}{l}
\text { (1) When } m<M, 1 \leqslant i \leqslant 16, \text { we have } \\
f^{[m]}(i+1)-f^{[m]}(i... | Five, the required $M_{0}=8$.
First, prove $M_{0} \geqslant 8$.
In fact, we can define the mapping $f(i) \equiv 3 i-2(\bmod 17)$, where $i \in A, f(i) \in A$.
If $f(i) \equiv f(j)(\bmod 17)$,
then $3 i-2 \equiv 3 j-2(\bmod 17)$,
we have $i \equiv j(\bmod 17)$,
$$
\therefore i=j \text {. }
$$
The mapping $f$ is a mappi... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. Solve the equation
$$
\begin{array}{l}
\sqrt{x-1}+\sqrt{2 x-3}+\sqrt{3 x-5} \\
+\sqrt{4 x-7}=5 x-6 .
\end{array}
$$
(First Yangtze Cup Correspondence Competition for Junior High School Students) | Solution: By applying the method of completing the square, the original equation can be transformed into
$$
\begin{array}{l}
(\sqrt{x-1}-1)^{2}+(\sqrt{2 x-3}-1)^{2} \\
+(\sqrt{3 x-5}-1)^{2}+(\sqrt{4 x-7}-1)^{2}=0 . \\
\therefore \sqrt{x-1}-1=\sqrt{2 x-3}-1 \\
\quad=\sqrt{3 x-5}-1=\sqrt{4 x-7}-1=0 .
\end{array}
$$
Solv... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
24. (15 points) Given the parabola $y^{2}=\operatorname{tar}(0<a<1)$ with focus $\vec{F}$, a semicircle is constructed above the $x$-axis with center $A(a+4,0)$ and radius $|A F|$, intersecting the parabola at two distinct points $M$ and $N$. Let $P$ be the midpoint of segment $M N$.
(1) Find the value of $|M F|+|N F|$... | 24. (1) From the given, we have $F(a, 0)$, and the semicircle is
$$
[x-(a+4)]^{2}+y^{2}=16(y \geqslant 0) \text {. }
$$
Substituting $y^{2}=4 a x$ into the equation, we get
$$
x^{2}-2(4-a) x+a^{2}+8 a=0 \text {. }
$$
Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. Then, by the definition of the parabola... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (15 points) The function $f(k)$ is defined on $N$ and takes values in $N$, being a strictly increasing function (if for any $x_{1}, x_{2} \in A$, when $x_{1}<x_{2}$, we have $f\left(x_{1}\right)<f\left(x_{2}\right)$, then $f(x)$ is called a strictly increasing function on $A$), and satisfies the condition $f(f(k))=... | For $k \in N, f(f(k))=3_{i}^{2}$.
(1)
$\therefore f[f(f(\hat{i})]:=\hat{j}(3 k)$
Also, $f[f(f(i))]=3 f(k)$,
$\therefore f(3 k)=3 f(k)$.
If $f(1)=1$, substituting into (1) gives $f(1)=3$, a contradiction.
$\therefore f(1)=a>1$, but $f(f(1))=f(a)=3$.
By the strict monotonicity of $f(k)$, i.e., $1<a \Rightarrow f(1)<f(a)=... | 197 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Does there exist a six-digit number $A$, such that the last six digits of any number in $A, 2A, 3A, \cdots, 500000A$ are not all the same? | 5. Let $A$ be a six-digit number. If $A$ is even, then $500000A$ ends with six zeros; if $A$ is divisible by 5, then $200000A$ ends with six zeros. Therefore, we can assume that the last digit of $A$ is $1, 3, 5, 7,$ or $9$. For any odd digit $d$, there exists a one-digit number $b$ such that the last digit of $Ab$ is ... | 888889 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Find $A^{2}$, where $A$ is the sum of the absolute values of all roots of the following equation:
$$
x=\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}
$$
(9th American Invitational Mathematics Examination) | Solution: According to the structural characteristics of the original equation, set
$$
\begin{array}{ll}
x=\sqrt{19}+\frac{91}{y}, & y=\sqrt{19}+\frac{91}{z}, \\
z=\sqrt{19}+\frac{91}{u}, & u=\sqrt{19}+\frac{91}{v}, \\
v=\sqrt{19}+\frac{91}{x} . &
\end{array}
$$
Assume $x>y$ and $x<y$ respectively, then
$$
\begin{arra... | 383 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 If the real numbers $x, y, m$ satisfy the relation
$$
\begin{array}{l}
\sqrt{3 x+5 y-2-m}+\sqrt{2 x+3 y-m} \\
=\sqrt{x-199+y} \cdot \sqrt{199-x-y},
\end{array}
$$
try to determine the value of $m$.
(1994, Beijing Junior High School Mathematics Competition) | Analysis: An equation with one unknown cannot be solved by brute force, but only by intelligent methods. Observing the characteristics of the equation, all are square roots. By definition, the right side of the equation has
$$
\left\{\begin{array}{l}
x-199+y \geqslant 0, \\
199-x-y \geqslant 0,
\end{array}\right.
$$
w... | 201 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 2, the areas of $\triangle A_{1} B_{2} B_{3}$, $\triangle P B_{3} A_{2}$, and $\triangle P A_{3} B_{2}$ are $5, 8, 3$ respectively. Find the area of $\triangle A_{1} A_{2} A_{3}$. (Letters have been changed, the same below)
(1994, Japan Mathematical Olympiad Preliminary Problem) | $$
\begin{array}{c}
\frac{\lambda_{1}}{\mu_{1}}=\frac{\lambda_{0}}{v_{1}}=\frac{S_{\triangle A_{1} A_{2} A_{3}}}{S_{\triangle A_{1} B_{2} B_{3}}} \\
=\frac{\lambda_{1}+\mu_{1}+5+8+3}{5} . \\
\text { Also } \frac{8}{\lambda_{1}}=\frac{\mu_{1}}{3}, \\
\therefore \lambda_{1}=12, \mu_{1}=2, \\
S_{\triangle A_{1} A_{2} A_{3... | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 In the border desert area, patrol vehicles travel 200 kilometers per day, and each vehicle can carry enough gasoline to travel for 14 days. There are 5 patrol vehicles that set out from base $A$ simultaneously, complete their tasks, and then return along the same route to the base. To allow 3 of them to patro... | Analysis: The key is to list the relationship formula for the distance traveled by the car, and then solve it by analyzing the characteristics of the formula.
Let 5 cars travel to point $B$ in $x$ days, and 3 cars continue to travel for $y$ days, with the distance $s=200(x+y)$, which is to find the maximum value of $s... | 1800 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given that $x, y, z$ are two non-negative rational numbers, and satisfy $3x+2y+z=5, x+y-z=2$. If $S=2x+y-z$, then what is the sum of the maximum and minimum values of $S$?
(1996, Tianjin City Junior High School Mathematics Competition) | Analysis: By representing the algebraic expression $S$ with one letter and determining the range of this letter, we can find the maximum or minimum value of $S$.
From the given, we solve $y=\frac{7-4 x}{3}, z=\frac{1-x}{3}$, and.
$$
\left\{\begin{array}{l}
\frac{7-4 x}{3} \geqslant 0, \\
\frac{1-x}{3} \geqslant 0, \\
x... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given that $n$ is a positive integer, and $n^{2}-71$ is divisible by $7 n+55$. Try to find the value of $n$.
Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
Example 7 Given that $n$ is a positive integer, and $n^{2}-71$ is divisible by $7... | Analysis: Let $\frac{n^{2}-71}{7 n+55}=k$ ( $k$ is an integer), then $n^{2}$ $-7 k n-(55 k+71)=0$, and $\Delta=49 k^{2}+4$ (5jk $+71)=49 k^{2}+220 k+284$ should be a perfect square.
$$
\begin{array}{l}
\text { and }(7 k+15)^{?}=49 k^{2}+210 k+225 \\
<49 k^{2}+220 k+284 \\
<49 k^{2}+238 k+289 \\
=(7 k+17)^{2}, \\
\there... | 57 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
$(1996$, National Junior High School Mathematics Leagu... | $$
\begin{array}{l}
\text{Analysis: Let } x_{1} \text{ and } x_{2} \text{ be the x-coordinates of the points where the parabola intersects the x-axis, then } x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}. \text{ It is easy to know that } -\frac{b}{a}<0, \text{ and } -116 c(a-b+c). \\
\because a-b+c>0, c \geqslant 1... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Arrange the natural numbers $1,2,3, \cdots, 999$. 1.5 into the number $N=1234 \cdots 998999$. Then, the sum of the digits of $N$ is $\qquad$ | II. 1.13500.
Since $1.2, \cdots .999$ contains 999 numbers. |1. $1+998=2+997$ 999 , for sure, hence the sum of each pair is 27. There are 500 pairs. Therefore, the sum of the digits of $N$ is $500 \times 27=13500$. | 13500 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Natural numbers $m, n$ satisfy $8 m+10 n>9 m n$. Then
$$
m^{2}+n^{2}-m^{2} n^{2}+m^{4}+n^{4}-m^{4} n^{4}=
$$
$\qquad$ | 3. 2 .
It is known that $\frac{8}{n}+\frac{10}{m}>9$. It is easy to see that one of the two natural numbers $m, n$ must be 1 (otherwise $\frac{8}{m}+\frac{10}{m} \leqslant \frac{8}{2}+\frac{10}{2}=9$ which is a contradiction). Without loss of generality, let $m=1$, then
$$
\begin{array}{l}
m^{2}+n^{2}-m^{2} n^{2}+m^{4... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $(-1)^{n} 2^{n} \equiv b_{n}(\bmod 9), 0 \leq b_{n} \leq 8$. Then the range of values for $b_{n}$ is \{ $\qquad$ \}. | $\begin{array}{l}\text { II.1.1. } \\ \text { \| }(-1)^{n} 2^{3 n}=(-8)^{n}=(1-9)^{n} \\ \quad=C_{n}^{n}-C_{n}^{1} 9+C_{n}^{2} 9^{2}+\cdots+(-1)^{n} C_{n}^{n} 9^{n}, \\ \text { know }(-1)^{n} 2^{3 n} \equiv 1(\bmod 9) .\end{array}$ | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
The struggle of the land, by 1996, the forest coverage rate of the whole county had reached 30% (becoming an oasis). From 1997 onwards, each year will see such changes: 16% of the original desert area will be converted, and 4% of it will be eroded, turning back into desert.
(1) Assuming the total area of the county is ... | Let the accumulated area be $b_{n+1}$. Then
$$
a_{1}+b_{1}=1, \quad a_{n}+b_{n}=1 .
$$
According to the problem, $a_{n+1}$ consists of two parts: one part is the remaining area of the original oasis $a_{n}$ after being eroded by $\frac{4}{100} \cdot a_{n}$, which is
$$
a_{n}-\frac{4}{100} a_{n}=\frac{96}{100} a_{n} .
... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, color $a$ points on a line red, $b$ points only yellow, and $c$ points blue, such that no two adjacent points have the same color. The number of such coloring methods is denoted as $F(a, b, c)$. Obviously, we have:
(1) Prove: For natural numbers $a, b, c$ not all equal to 1, we have
$$
\begin{array}{l}
F(a, b, c... | The number of ways to color the 1st point yellow and blue.
$$
F(a, b, c)=\sum_{i=1}^{3} F_{i}(a, b, c) .
$$
And
$$
\begin{aligned}
& F_{1}(a, b, c)=F_{2}(a-1, b, c)+F_{3}(a-1, b, c) \\
= & F_{1}(a-1, b-1, c)+F_{3}(a-1, b-1, c) \\
& +F_{1}(a-1, b, c-1)+F_{2}(a-1, b, c-1) \\
= & F_{1}(a-1, b-1, c)+F_{3}(a-1, b-1, c) \\... | 10160640 | Combinatorics | proof | Yes | Yes | cn_contest | false |
The sum and product of the first 51.1995 integers are both equal to F $\cdot$ 1996. What is the sum of the absolute values of these 1995 integers? Please prove your conclusion.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Let these 1995 integers be denoted as $x_{1}, x_{2}, \cdots, x_{1995}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{1995}=1996, \\
x_{1} x_{2} \cdots \cdots x_{1995}=1996=2^{2} \times 499 .
\end{array}\right.
$$
(1) From (1), we know that among these 1995 numbers, there are an even number of odd num... | 2496 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Initial 52. Given real numbers $a, b, x, y$ satisfy $a x^{n}+b y^{n}=1+2^{n+1}$ for any natural number $n$. Find the value of $x^{a}+y^{b}$. | Let $S_{n}=a x^{\prime \prime}+b y^{\prime \prime}$. Then
$$
S_{1}=a x+b y=5 \text {, }
$$
$$
\begin{array}{l}
S_{0}=a x^{2}+b y^{2}=9 . \\
S=a x^{3}+b y^{3}=17 . \\
S_{1}=a x^{4}+b y^{\prime}=33 .
\end{array}
$$
Let $x+y=A, x y=B$. Then $x, y$ are the roots of the quadratic equation $t^{2}-A t+B=0$. Therefore,
$$
\be... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{c}
\text { Example } 1 \text { Calculate } \frac{a}{a^{3}+a^{2} b+a b^{2}+b^{3}}+ \\
\frac{b}{a^{3}-a^{2} b+a b^{2}-b^{3}}+\frac{1}{a^{2}-b^{2}}-\frac{1}{b^{2}+a^{2}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} .
\end{array}
$$
(1995, Tianjin City, Grade 7 "Mathematics Competition) | $\begin{array}{l}\text { Solution: Original expression }=\frac{a}{(a+b)\left(a^{2}+b^{2}\right)} \\ +\frac{b}{(a-b)\left(a^{2}+b^{2}\right)}+\frac{2 b^{2}}{a^{4}-b^{4}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} \\ =\frac{a^{2}+b^{2}}{\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)}-\frac{a^{2}+b^{2}}{a^{4}-b^{4}} \\ =\frac{1}... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 If $a+b+c=a b c \neq 0$, find the value of $\frac{\left(1-a^{2}\right)\left(1-b^{2}\right)}{a b}+\frac{\left(1-b^{2}\right)\left(1-c^{2}\right)}{b c}+$ $\frac{\left(1-c^{2}\right)\left(1-a^{2}\right)}{a c}$.
(1990, Wuhan City Mathematics Competition) | $\begin{array}{l}\text { Solution: Original expression }=\frac{1-a^{2}-b^{2}+a^{2} b^{2}}{a b} \\ +\frac{1-b^{2}-c^{2}+b^{2} c^{2}}{b c}+\frac{1-a^{2}-c^{2}+a^{2} c^{2}}{a c} \\ =\frac{1}{a b}-\frac{a}{b}-\frac{b}{a}+a b+\frac{1}{b c}-\frac{b}{c}-\frac{c}{b}+b c \\ +\frac{1}{a c}-\frac{a}{c}-\frac{c}{a}+a c \\ =\left(\... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Given that $\alpha$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$. Find the value of $\frac{\alpha^{3}-1}{\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}}$.
(1995, National Junior High School Mathematics League) | Solution: From the given, we have $\alpha^{2}+\alpha=\frac{1}{4}$.
To form the required expression containing $\alpha^{2}+\alpha$, we have
$$
\frac{\alpha^{3}-1}{\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}}=\frac{(\alpha-1)\left(\alpha^{2}+\alpha+1\right)}{(\alpha-1)\left(\alpha^{2}+\alpha\right)^{2}}=20 .
$$ | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. $1000,1001,1002, \cdots, 1996$ These 997 natural numbers appear in all the digits whose sum is
The sum of all the digits that appear in the 997 natural numbers $1000,1001,1002, \cdots, 1996$ is | 9. 14419 | 14419 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Full marks 20 points) Given the equation $x^{2}+m x-m+1=0$ (where $m$ is an integer) has two distinct positive integer roots. Find the value of $m$.
---
The translation maintains the original format and line breaks as requested. | Three, let the two distinct positive integer roots be $\alpha, \beta(\alpha<\beta)$. By Vieta's formulas, we have $\left\{\begin{array}{l}\alpha+\beta=-m, \\ \alpha \beta=-m+1 .\end{array}\right.$ Eliminating $m$, we get $\alpha \beta-\alpha-\beta=1$.
That is, $(\alpha-1)(\beta-1)=2$.
Then $\left\{\begin{array}{l}\alph... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (Full marks 20 points, (1) 10 points, (2) 10 points) A total of 240 students participating in a large group performance stand in a row facing the coach, and report numbers in sequence from left to right as $1,2,3,4,5, \cdots$. The coach asks all students to remember their reported numbers and perform the followin... | (1) Because all students who report numbers that are multiples of $15$, $21$, and $35$ but not multiples of $105$ turn twice, still facing the coach. Students who report numbers that are multiples of $3$ but not multiples of $15$ or $21$, multiples of $5$ but not multiples of $15$ or $35$, and multiples of $7$ but not ... | 118 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a, b, c$ are distinct. Find the value of $\frac{2 a-b-c}{(a-b)(a-c)}$ $+\frac{2 b-c-a}{(b-c)(b-a)}+\frac{2 c-a-b}{(c-a)(c-b)}$. | (Answer: 0) | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Real numbers $a, b, c$ satisfy $a+b+c=0, a b c>0$. If $x=$
$$
\begin{array}{l}
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}, y=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}\right. \\
\left.+\frac{1}{b}\right) \text {. Then, } x+2 y+3 x y=
\end{array}
$$
. | 1. 2
The above text has been translated into English, maintaining the original text's line breaks and format. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Six, (Full marks 12 points) On a circle, there are
12 points, one of which is painted red, and another is painted blue, with the remaining 10 points unpainted. Among the convex polygons formed by these points, those whose vertices include both the red and blue points are called bicolored polygons; those that include on... | For any bicolored $n(n \geqslant 5)$-gon, naturally, removing the red and blue vertices results in a colorless $(n-2)$-gon. Different bicolored $n$-gons, after removing the red and blue vertices, yield different colorless $(n-2)$-gons; conversely, for any colorless polygon, adding red and blue vertices can always resul... | 55 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Anita, Kieran, and Mitchell form a team to participate in a mixed cycling and running race, which takes place on a 3-kilometer circular track. It is known that each person's running speed is 5 kilometers per hour, and their cycling speed is 20 kilometers per hour. The starting point of the track is marked by a flag,... | 2. (a) It is easy to see that each person takes 12 minutes to run 1 km and 3 minutes to cycle 1 km. At the start, let Anita and Kemai cycle, while Mitchell runs. When they reach the 1 km mark, Kemai gives the bike to Mitchell and starts running. When they reach the 2 km mark, Anita gives the bike to Kemai and starts ru... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
3. If three integers $a, b, c (a \neq 0)$ make the equation $a x^{2}$ $+b x+c=0$ have two roots $a$ and $b$, then $a+b+c$ equals. $\qquad$ | 3. 18 .
From the relationship between roots and coefficients, we get
$$
\left\{\begin{array}{l}
a+b=-\frac{b}{a}, \\
a b=\frac{c}{a} .
\end{array}\right.
$$
From (1), we have $a^{2}+a b+b=0, (a+1) b=-a^{2}$.
Since $a \neq 0$, then $a+1 \neq 0$, $b \neq 0$. Thus,
$$
b=-\frac{a^{2}}{a+1}=-\frac{(a+1-1)^{2}}{a+1}=-a+1-\... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $X=\{-1,0,1\}, Y=\{-2,-1,0,1,2\}$, and for all elements $x$ in $X$, $x+f(x)$ is even. Then the number of mappings $f$ from $X$ to $Y$ is ( ).
(A) 7
(B) 10
(C) 12
(D) 15. | 4. (C).
$x$ and $f(x)$ are both even, or both odd.
When $x=0$, there are three possibilities: $-2, 0, 2$;
When $x=-1, 1$, there are two possibilities each for -1 and 1, making 4 possibilities.
Therefore, the total is $3 \times 4=12$. | 12 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
6. A monkey is on a ladder with $n$ steps: climbing up and down, it either ascends 16 steps or descends 9 steps each time. If it can climb from the ground to the very top step, and then return to the ground, the minimum value of $n$ is ( ).
(A) 22
(B) 23
(C) 24
(D) greater than 24. | 6. (C).
Assume the monkey climbs as follows: 0 $\nearrow 16 \searrow 7 \nearrow 23 \searrow 14 \searrow 5$ $\nearrow 21 \searrow 12 \searrow 3 \nearrow 19 \searrow 10 \searrow 1, \nearrow 17 \searrow 8 \nearrow 24 \searrow 15 \searrow 6 \nearrow$ $22 \searrow 13 \searrow 4$ Л $20 \searrow 11 \searrow 2$ オ゙ $18 \searro... | 24 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 1. } \lim _{n \rightarrow \infty} \frac{1}{\sqrt[3]{n}}\left(\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+ \\ \frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}+ \\ \left.\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}}\right)=\end{array}$ | $$
\approx .1 .1 \text {. }
$$
Since
$$
\begin{array}{l}
\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+\cdots+ \\
\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}} \\
=\frac{\sqrt[3]{2}-1}{2-1}+\cdots+\frac{\sqrt[3]{n}-\sqrt[3]{n-1}}{n-(n-1)} . \\
=\sqrt[3]{n}-1,
\end{array}
$$
Therefore, the required limit is $\l... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
5. On each face of an opaque cube, a natural number is written. If several (one, two, or three) faces of the cube can be seen at the same time, then find the sum of the numbers on these faces. Using this method, the maximum number of different sums that can be obtained is _.
翻译结果如下：
5. On each face of an opaque cube,... | 5.26 .
There are 6 cases where a single face can be seen, 12 cases where two faces sharing a common edge can be seen simultaneously, and 8 cases where three faces sharing a common vertex can be seen simultaneously, thus yielding 26 sums. The numbers on the faces can be filled in such a way that all 26 sums are distinc... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the sequence of positive integers $a_{1} 、 a_{2} 、 a_{3} 、 a_{4}$ be a geometric sequence, with the common ratio $r$ not being an integer and $r>1$. The smallest value that $a_{4}$ can take in such a sequence is $\qquad$ . | 6. 27 .
According to the problem, $r$ is a rational number, so there exist coprime positive integers $p$ and $q (q > p \geqslant 2)$, such that $r=\frac{q}{p}$. Therefore, $a_{4}=a_{1} r^{3}=\frac{a_{1} q^{3}}{p^{3}}$.
Since $a_{4}$ is an integer, $a_{1}$ must be a multiple of $p^{3}$.
Thus, let $a_{1}=k p^{3}$ ($k$ i... | 27 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Simplify $\frac{(x+b)(x+c)}{(a-b)(a-c)}+\frac{(x+c)(x+a)}{(b-c)(b-a)}$ $+\frac{(x+a)(x+b)}{(c-a)(c-b)}$. | (Answer: 1).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $x=b y+c z, y=c z+a x, z=a x$ $+b y$. Find the value of $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}$. | Solution: $\frac{a}{a+1}=\frac{a x}{a x+x}=\frac{a x}{a x+b y+c z}$.
Similarly, $\frac{b}{b+1}=\frac{b y}{a x+b y+c z}$,
$$
\frac{c}{c+1}=\frac{c z}{a x+b y+c z} \text {. }
$$
Adding them up, the value of the desired expression is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 If $\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=$ $\frac{t}{x+y+z}$, let $f=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}$. Prove: $f$ is an integer.
(1990, Hungarian Mathematical Competition) | Proof: If $x+y+z+t \neq 0$, by the property of proportion, we have
$$
\begin{array}{l}
\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y} \\
=\frac{t}{x+y+z}=\frac{x+y+z+t}{3(x+y+z+t)}=\frac{1}{3} . \\
\text { Then we have }\left\{\begin{array}{l}
y+z+t=3 x, \\
z+t+x=3 y, \\
t+x+y=3 z, \\
x+y+z=3 t .
\end{array}\right.
\e... | -4 | Algebra | proof | Yes | Yes | cn_contest | false |
Example 7 Given $4 x-3 y-6 z=0, x+2 y-7 z$ $=0(x y z \neq 0)$. Find the value of $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}$.
(1992, Sichuan Province Junior High School Mathematics League Preliminary) | Solution: Let $y=k_{1} x, z=k_{2} x$. Then
$$
\left\{\begin{array} { l }
{ 4 - 3 k _ { 1 } - 6 k _ { 2 } = 0 , } \\
{ 1 + 2 k _ { 1 } - 7 k _ { 2 } = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
k_{1}=\frac{2}{3}, \\
k_{2}=\frac{1}{3} .
\end{array}\right.\right.
$$
Then $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given the theorem: "If three prime numbers greater than 3, $a, b$, and $c$, satisfy the equation $2a + 5b = c$, then $a + b + c$ is a deficient number of the integer $n$." What is the maximum possible value of the integer $n$ in the theorem? Prove your conclusion. | Three, the maximum possible value of $n$ is 9.
First, prove that $a+b+c$ can be divided by 3.
In fact, $a+b+c=a+b+2a+5b=3(a+2b)$, so $a+b+c$ is a multiple of 3.
Let the remainders when $a$ and $b$ are divided by 3 be $r_{a}$ and $r_{b}$, respectively, then $r_{a} \neq 0, r_{b} \neq 0$.
If $r_{a} \neq r_{b}$, then $r_{a... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false |
3. In an equilateral $\triangle ABC$, $P$ is a point on side $AB$, $Q$ is a point on side $AC$, and $AP = CQ$. It is measured that the distance between point $A$ and the midpoint $M$ of line segment $PQ$ is $19 \mathrm{~cm}$. Then the distance from point $P$ to point $C$ is $\qquad$ $\mathrm{cm}$. | 3. 38 . | 38 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (Full marks 15 points) A natural number $a$ is exactly equal to the square of another natural number $b$, then the natural number $a$ is called a perfect square (for example, $64=8^{2}$, so $64$ is a perfect square).
If $a=1995^{2}+1995^{2} \cdot 1996^{2}+1996^{2}$, prove that $a$ is a perfect square, and write dow... | Let $x=1995$, then $x+1=1996$.
$$
\begin{aligned}
a= & 1995^{2}+1995^{2} \cdot 1996^{2}+1996^{2} \\
= & x^{2}+x^{2}(x+1)^{2}+(x+1)^{2} \\
= & (x+1)^{2}-2 x(x+1)+x^{2}+2 x^{2}(x+1) \\
& +x^{2}(x+1)^{2} \\
= & (x+1-x)^{2}+2 x(x+1)+[x(x+1)]^{2} \\
= & 1^{2}+2 x(x+1)+[x(x+1)]^{2} \\
= & {[1+x(x+1)]^{2} } \\
= & (1+1995 \... | 3982021 | Number Theory | proof | Yes | Yes | cn_contest | false |
Five. (Full marks: 15 points) From the 91 natural numbers $1,2,3, \cdots, 90,91$, select $k$ numbers such that there must be two natural numbers $p, q$ satisfying $\frac{2}{3} \leqslant \frac{q}{p} \leqslant \frac{3}{2}$. Determine the minimum value of the natural number $k$, and explain your reasoning. | Five, divide the 91 natural numbers from 1 to 91 into nine groups, such that the ratio of any two natural numbers in each group is no less than $\frac{2}{3}$ and no more than $\frac{3}{2}$, and the division is as follows:
$$
\begin{array}{l}
A_{1}=\{1\}, A_{2}=\{2,3\}, A_{3}=\{4,5,6\}, \\
A_{4}=\{7,8,9,10\}, \\
A_{5}=\... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 If $x, y, z$ are real numbers, and
$$
\begin{aligned}
(y-z)^{2} & +(z-x)^{2}+(x-y)^{2} \\
= & (y+z-2 x)^{2}+(z+x-2 y)^{2} \\
& +(x+y-2 z)^{2},
\end{aligned}
$$
find the value of $M=\frac{(y z+1)(z x+1)(x y+1)}{\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right)}$. | Solution: The condition can be simplified to
$$
x^{2}+y^{2}+z^{2}-x y-y z-z x=0 .
$$
Then $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0$,
which implies $x=y=z$.
Therefore, $M=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 20 points) There is a 14-digit number, the digit in the units place is 3 less than the digit in the tens place, and the new four-digit number formed by reversing its digits differs from the original four-digit number by 8987. Find this four-digit number and write out the reasoning process.
---
Trans... | Five, let the thousands digit of this four-digit number be $a$, the hundreds digit be $b$, the tens digit be $c$, and the units digit be $c-3$.
This number is $1000a + 100b + 10c + (c-3)$.
The new four-digit number is $1000(c-3) + 100c + 10b + a$.
According to the problem, we have
$$
\begin{array}{l}
1001(a+c-3) + 110... | 1996 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Five, in a chess tournament, there are an odd number of participants, and each participant plays one game against every other participant. The scoring system is as follows: 1 point for a win, 0.5 points for a draw, and 0 points for a loss. It is known that two of the participants together scored 8 points, and the avera... | Five, suppose there are $(n+2)$ players in total, except for 2 people who get 8 points, $n$ people on average get $k$ points each ($k$ is an integer).
$\because$ Each person plays one match with everyone else, and there are $(n+2)$ people,
$\therefore$ A total of $\frac{(n+1)(n+2)}{2}$ matches are played.
Since each ma... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 As shown, in Rt $\triangle ABC$, the hypotenuse $AB=5, CD \perp AB$. It is known that $BC, AC$ are the two roots of the quadratic equation $x^{2}-(2 m-1) x+4(m-1)=0$. Then the value of $m$ is $\qquad$. | Solution: Let $A C=b$, $B C=a$. By Vieta's formulas, we get $a+b=2 m-$
$$
\begin{array}{l}
1, a b=4(m-1) . \\
\begin{aligned}
\therefore A B^{2} & =a^{2}+b^{2}=(a+b)^{2}-2 a b \\
& =(2 m-1)^{2}-2 \times 4(m-1)=5^{2},
\end{aligned}
\end{array}
$$
i.e., $m^{2}-3 m-4=0$.
$$
\therefore m=4 \text { or } m=-1 \text {. }
$$
... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The train is $400 \mathrm{~m}$ long, and it takes 10 minutes to pass through the tunnel (from the front of the train entering the tunnel to the rear of the train leaving the tunnel). If the speed increases by 0.1 kilometers per minute, then it will take 9 minutes, and the length of the tunnel is $\qquad$ $ـ$. | 5. 8600 meters.
Let the full length be $x$ meters, and the original speed of the train be $v$ meters per minute. According to
$$
\left\{\begin{array}{l}
\frac{x+400}{v}=10, \\
\frac{x+400}{v+100}=9 .
\end{array} \text { Solving, we get } v=900\right. \text { (meters/minute.) }
$$ | 8600 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (Full marks 14 points) Given that the roots of the equation $x^{2}+p x+q=0$ are 1997 and 1998, when $x$ takes the integer values $0,1,2, \cdots, 1999$, the corresponding values of the quadratic trinomial $y=x^{2}+p x+q$ are $y_{0}$. Find the number of these values that are divisible by 6. | Let $y=(x-1997)(x-1998)$ be divisible by 6. Since when $x$ takes integer values, all $y$ values can be divisible by 2, we only need to examine the cases where the factors are divisible by 3.
$$
\begin{array}{l}
\text { (1) When } x-1997=3 k \text {, then } x-1998=3 k-1 \text {. } \\
\because 0 \leqslant x \leqslant 199... | 1333 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 13 Given that $x_{1}, x_{2}$ are the two real roots of the equation $x^{2}-(k-2) x + (k^{2}+3 k+5)=0$ (where $k$ is a real number). Then the maximum value of $x_{1}^{2}+x_{2}^{2}$ is $\qquad$ | According to Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=k-2, \quad x_{1} x_{2}=k^{2}+3 k+5 . \\
\therefore \quad x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \\
=(k-2)^{2}-2\left(k^{2}+3 k+5\right) \\
=9-\left(k+5\right)^{2} .
\end{array}
$$
From the discriminant theorem, we get
$$
\be... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. As shown in the figure, in the right trapezoid $A B C D$, the base $A B=13$, $C D=8, A D \perp A B$, and $A D=$ 12. Then the distance from $A$ to the side $B C$ is ( ).
(A) 12
(B) 13
(C) 10
(D) $\frac{12 \times 21}{13}$ | 6. (A).
Draw $C C^{\vee} \perp A B$, then
$$
B C=13-8=5 \text {. }
$$
Let the distance from $A$ to $B C$ be $h$,
Connect AC.
$$
\begin{aligned}
& \because S_{\triangle A C D}+S_{\triangle A B C} \\
& =S_{\text {UEABCD }}, \\
\therefore & \frac{1}{2} \times 8 \times 12+\frac{1}{2} \times h \times 13=\frac{1}{2}(8+13) ... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false |
2. Given $|x| \leqslant 2$, the sum of the maximum and minimum values of the function $y=x-|1+x|$ is $\qquad$ . | 2. -4 .
$$
\begin{array}{l}
\because|x| \leqslant 2, \text { i.e., }-2 \leqslant x \leqslant 2, \\
\therefore y=x-|1+x| \\
\quad=\left\{\begin{array}{ll}
2 x+1, & \text { when }-2 \leqslant x<-1, \\
-1, & \text { when }-1 \leqslant x \leqslant 2.
\end{array}\right.
\end{array}
$$
As shown in the figure.
Indeed, when $... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In $\triangle A B C$, $\angle B C A=90^{\circ}$, a perpendicular $C D \perp A B$ is drawn from $C$ intersecting $A B$ at $D$. Suppose the side lengths of $\triangle A B C$ are all integers, and $B D=29^{3}, \cos B=\frac{m}{n}$, where $m$ and $n$ are coprime positive integers. Then $m+n=$ $\qquad$ | 4. 450 .
Let the three sides of $\triangle ABC$ be $a, b, c$. It is easy to prove that $BC^2 = BD \cdot BA$. Therefore, $a^2 = 29^3 c$.
Since 29 is a prime number, $29^2 \mid a$.
Let $a = 29^2 k$ ($k$ is a positive integer), then $c = 29 k^2$.
Since $b = \sqrt{c^2 - a^2} = 29 k \sqrt{k^2 - 29^2}$ is an integer, we kno... | 450 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $M=\cos 5^{\circ} \sin 15^{\circ} \sin 25^{\circ} \sin 35^{\circ}, N=$ $\sin 5^{\circ} \cos 15^{\circ} \cos 25^{\circ} \cos 35^{\circ}$. Then $\frac{M}{N}=$ $\qquad$ . | $=1.1$.
$$
\begin{aligned}
\frac{M}{N} & =\frac{\frac{1}{2}\left(\sin 20^{\circ}+\sin 10^{\circ}\right) \cdot \frac{1}{2}\left(\cos 10^{\circ}-\cos 60^{\circ}\right)}{\frac{1}{2}\left(\sin 20^{\circ}-\sin 10^{\circ}\right) \cdot \frac{1}{2}\left(\cos 10^{\circ}+\cos 60^{\circ}\right)} \\
& =\frac{\left(\sin 20^{\circ}+... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $a+\lg a=10, b+10^{b}=10$. Then $a+b$ | 4. 10.
Thought 1: From the given information,
$$
\begin{array}{l}
a=10^{10-a}, \\
10-b=10^{b} .
\end{array}
$$
Subtracting, we get $10-a-b=10^{b}-10^{10-a}$.
If $10-a-b>0$, then $10-a>b$. By the monotonicity of the exponential function, we get $10^{10-a}>10^{b}$. Substituting into (1), we get
$$
010-a-b=10^{b}-10^{10... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
On September 4, 1996, scientists used a supercomputer to find the 33rd Mersenne prime, which is the largest prime number known to humans so far. It is: $2^{125787}-1$ (378632 digits). Try to find the last two digits of this prime number. | Solution: First, find the unit digit.
$$
\begin{array}{l}
\because 2^{1257787}-1=2 \cdot 4^{628893}-1=8 \cdot 16^{31446}-1 \\
\quad \equiv 8 \times 6-1(\bmod 10) \equiv 7(\bmod 10),
\end{array}
$$
$\therefore 2^{1257787}-1$ has a unit digit of 7.
Next, find the tens digit.
$$
\begin{aligned}
\because & \frac{1}{10}\lef... | 27 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
In the plane: There are 212 points all located within or on the boundary of a unit circle. Connect any two points to form a line segment. Prove: The number of line segments with length not greater than 1 is at least 1996. | Proof: (1) First, prove that in a circle or on any $\epsilon$ points, there must be two points whose distance is not greater than:
Let these 6 points be $A_{1}, A_{2}, \cdots, A_{6}$, and the center be $O$. If one point is at the center, the proposition holds. Otherwise, connect $O A_{i},(i=1,2, \cdots, 6)$, to get 6 ... | 1996 | Combinatorics | proof | Yes | Yes | cn_contest | false |
7. Let $a, b$ be unequal real numbers, and $a^{2}+2 a-5$ $=0, b^{2}+2 b-5=0$. Then $a^{2} b+a b^{2}=$ $\qquad$ . | （Answer:10） | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For example, 610 people go to the bookstore to buy books, it is known that
(1) each person bought three books;
(2) any two people have at least one book in common.
How many people at most bought the book that was purchased by the fewest people? | Solution: Let the number of people who bought the most popular book be $x$. Among the 10 people, person A bought three books. Since the other 9 people each have at least one book in common with A, and $9 \div 3=3$, it follows that among A's three books, the most popular one must have been bought by at least 4 people, s... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given that $x$, $y$, $z$ are positive integers, and $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(y+z)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Explanation: This is a typical example of constructing a geometric figure to solve a problem. As shown in Figure 1, construct $\triangle ABC$, where the lengths of the three sides are
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z, \\
c=z+x .
\end{array}\right.
$$
Then its area is
$$
\begin{aligned}
\triangle & =\sqrt{p(p... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Color each vertex of a square pyramid with one color, and make the endpoints of the same edge have different colors. If only 5 colors are available, then the total number of different coloring methods is $\qquad$ | Solution: Vertex $S$ can be colored with any of the $m$ colors, and the color on $S$ cannot appear on the vertices of the polygon $A_{1} A_{2} \cdots A_{n}$. The problem is then reduced to coloring the vertices of the polygon with $m-1$ colors, ensuring that adjacent vertices have different colors. Let there be $a_{n}$... | 420 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Prove that there exists a unique function $f: N \rightarrow$ $N$ satisfying for all $m, n \in N$, $f[m+f(n)]=n$ $+f(m+95)$. Find the value of $\sum_{k=1}^{19} f(k)$. | Solution: Observe and verify that $f(n)=n+95$ meets the requirement, at this time, $\sum_{k=1}^{19} f(k)=1995$.
Now prove the uniqueness.
When $n_{1} \neq n_{2}$,
$$
\begin{array}{l}
f\left[m+f\left(n_{1}\right)\right]=n_{1}+f(m+95) \\
\neq n_{2}+f(m+95)=f\left[m+f\left(n_{2}\right)\right],
\end{array}
$$
i.e., $f\lef... | 1995 | Algebra | proof | Yes | Yes | cn_contest | false |
Example 7 The function $f(x)$ is defined on the set of real numbers, and for all real numbers $x$ it satisfies the equations: $f(2+x)=f(2-x)$ and $f(x+7)=f(7-x)$. Suppose $x=0$ is a root of $f(x)=0$, and let $N$ denote the number of roots of $f(x)=0$ in the interval $[-1000,1000]$. Find the minimum value of $N$. | $$
\begin{array}{l}
\text { Solution: } \because f(x+2)=f(2-x), \\
\begin{array}{l}
\therefore f(x+4)=f[(x+2)+2] \\
\quad=f[2-(x+2)]=f(-x) . \\
\because f(x+7)=f(7-x), \\
\therefore f(x+14)=f[(x+7)+7]
\end{array} \\
\quad=f[7-(x+7)]=f(-x) .
\end{array}
$$
Thus, $f(x+14)=f(4+x)$,
which means $f(x+10)=f(x)$.
Therefore, ... | 401 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$ . | Solution: Since $m \neq n$, by the definition of roots, $m, n$ are two distinct roots of the equation $x^{2}-x-1=0$. By Vieta's formulas, we have
$$
\begin{array}{l}
m+n=1, m n=-1 . \\
\because m^{2}+n^{2}=(m+n)^{2}-2 m n \\
\quad=1^{2}-2 \times(-1)=3, \\
\quad m^{3}+n^{3}=(m+n)^{3}-3 m n(m+n) \\
\quad=1^{3}-3 \times(-... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Given 10 points of the World Team, where 5 of these points lie on a straight line, and no three points lie on another straight line besides these, the number of distinct rays that can be drawn through any 2 of these 10 points is $\qquad$. | 13. 78 | 78 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 The roots $x_{1}, x_{2}$ of the equation $x^{2}-a x-a=0$ satisfy the relation $x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3}=75$. Then $1993+5 a^{2}+$ $9 a^{4}=$ $\qquad$ | Solution: By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=a, x_{1} x_{2}=-a . \\
\therefore x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3} \\
=\left(x_{1}+x_{2}\right)^{3}-3 x_{1} x_{2}\left(x_{1}+x_{2}\right)+\left(x_{1} x_{2}\right)^{3} \\
=a^{3}-3(-a) a+(-a)^{3}=3 a^{2} .
\end{array}
$$
According to the probl... | 7743 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
20. For a certain commodity, if you buy 100 (including 100) pieces or less, it is settled at the retail price; if you buy 101 (including 101) pieces or more, it is settled at the wholesale price. It is known that the wholesale price is 2 yuan lower per piece than the retail price. A person originally intended to buy a ... | $\begin{array}{l}20 \\ 840\end{array}$ | 840 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, let $S=\{1,2,3,4\}, n$ terms of the sequence: $a_{1}$, $a_{2}, \cdots, a_{n}$ have the following property, for any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted as $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
| Three, the minimum value of $n$ is 8.
First, prove that each number in $S$ appears at least 2 times in the sequence $a_{1}, a_{2}, \cdots, a_{n}$. This is because, if a number in $S$ appears only once in this sequence, since there are 3 two-element subsets containing this number, but in the sequence, the adjacent pairs... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The average score of the participants in a junior high school mathematics competition at a certain school is 75 points. Among them, the number of male participants is $80\%$ more than that of female participants, and the average score of female participants is $20\%$ higher than that of male participants. Therefore,... | II. 1.84.
Let the number of female participants be $x$, then the number of male participants is $1.8 x$; Let the average score of male participants be $y$ points, then the average score of female participants is $1.2 y$ points. According to the problem, we have $\frac{1.8 x y + 1.2 x y}{x + 1.8 x} = 75$, which simplifi... | 84 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $p+q=96$, and the quadratic equation $x^{2}+p x$ $+q=0$ has integer roots. Then its largest root is | 2. 98 .
Let the two roots be $x_{1}$ and $x_{2}$. Then $96=(x_{1}-1)(x_{2}-1)-1$, so $(x_{1}-1)(x_{2}-1)=97$. Since 97 is a prime number, then $x_{1}-1= \pm 1$, $x_{2}-1= \pm 97$, hence the largest root is 98. | 98 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in the figure, the side lengths of $\triangle A B C$ are $A B=14, B C$ $=16, A C=26, P$ is a point on the angle bisector $A D$ of $\angle A$, and $B P \perp A D, M$ is the midpoint of $B C$. Find the value of $P M$ $\qquad$ | 3. 6 .
From the figure, take $B^{\prime}$ on $A C$ such that $A B^{\prime}=A B=14$, then $B^{\prime} C=12$. Since $\triangle A B B^{\prime}$ is an isosceles triangle, we know that the intersection point of $B B^{\prime}$ and $A D$ is $P$ (concurrency of five lines), so $P$ is the midpoint of $B B^{\prime}$. | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find the smallest natural number with the following properties:
(1) Its decimal representation ends with the digit 6;
(2) If the last digit 6 is deleted and this digit 6 is written in front of the remaining digits, then the resulting number is 4 times the original number.
(4th IM) | Solution: From (1), we know that the last digit of the required smallest natural number is 6. Combining (2) with the multiplication table, it is not difficult to know that its tens digit is 4, and it is also the last digit of the obtained number.
$46 \times 4=184$.
Thus, the tens digit of the obtained number should be ... | 153846 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For example, $1 \overline{a b c d}$ is a four-digit natural number. It is known that $\overline{a b c d}-\overline{a b c}-\overline{a b}-a=1995$. Try to determine this four-digit number $\overline{a b c d}$
(1995, Beijing Middle School Advanced Mathematics Competition for Junior High School Students) | Solution: For easier analysis, convert the known horizontal expression into a vertical format, we get
$$
\begin{array}{r}
1000 a+100 b+10 c+d \\
-100 a-10 b-c \\
-10 a-b \\
\text { +) } \begin{array}{r}
-a
\end{array} \\
\hline 1000 \times 1+100 \times 9+10 \times 9+5
\end{array}
$$
By the subtraction rule and borrowi... | 2243 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If the digits of a four-digit number are reversed to form a new four-digit number, the new number is exactly four times the original number. Find the original number.
(1988, Nanjing Mathematical Olympiad Selection Contest) | Solution: Let the required four-digit number be $\overline{a b c d}$, then the new four-digit number is $\overline{d c b a}$. According to the problem, we have
$\overline{a b c d}+\overline{d c b a}=5 \overline{a b c d}$.
Obviously, the last digit of $a+d$ can only be 0 or 5.
$\because 4 a \leqslant d \leqslant 9$, and... | 2178 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 A four-digit number has the sum of the squares of its first and last digits equal to 13, and the sum of the squares of its middle two digits equal to 85. If 1089 is subtracted from this number, the result is the same four digits written in reverse order. What is the original four-digit number? $\qquad$
(1994,... | Solution: Let the original four-digit number be $\overline{a b c d}$. According to the problem, we have
$$
\begin{array}{l}
a^{2}+d^{2}=13, \\
b^{2}+c^{2}=85, \\
\overline{a b c d}-1089=\overline{d c b a} .
\end{array}
$$
Analyzing from the first digit, it is easy to know
$$
a-d=1 \text {. }
$$
From (1) and (4), we g... | 3762 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $x, y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1 \\
(y-1)^{3}+1997(y-1)=1 .
\end{array}\right.
$$
Then $x+y=$ $\qquad$ (Proposed by the Problem Group) | $=、 1.2$.
The original system of equations is transformed into
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1, \\
(1-y)^{3}+1997(1-y)=-1 .
\end{array}\right.
$$
Since $f(t)=t^{3}+1997 t$ is monotonically increasing on $(-\infty,+\infty)$, and $f(x-1)=f(1-y)$, it follows that $x-1=1-y$, i.e., $x+y=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$, a line $l$ intersects the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$ such that there are exactly 3 lines $l$, then $\lambda=$
(Proposed by the Problem Committee) | 2. 4 .
First, note the following conclusion: For a chord passing through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ and intersecting the right branch at two points, the chord attains its minimum length $\frac{2 b^{2}}{a}=4$ if and only if the chord is perpendicular to the $x$-axis. (In fact, the polar ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $A B C D E F$ be a regular hexagon. A frog starts at vertex $A$ and can randomly jump to one of the two adjacent vertices each time. If it reaches point $D$ within 5 jumps, it stops jumping; if it does not reach point $D$ within 5 jumps, it stops after 5 jumps. How many different jumping sequences can the frog h... | 5. 26 .
As shown in the figure, it is clear that the frog cannot reach point $D$ by jumping 1 time, 2 times, or 4 times. Therefore, the frog's jumping methods are only of the following two scenarios:
(1) The frog reaches point $D$ after 3 jumps, with 2 ways to do so;
(2) The frog stops after a total of 5 jumps. In thi... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 What is the smallest positive integer that can be expressed as the sum of 9 consecutive integers, the sum of 10 consecutive integers, and the sum of 11 consecutive integers?
(11th American Invitational Mathematics Examination (AIME))
| Solution: Let the required positive integer be $A$. According to the problem, $A$ can be expressed as (where $p$, $n$, $k$ are all integers)
$$
\begin{aligned}
A & =(p+1)+(p+2)+\cdots+(p+9) \\
& =9 p+45, \\
A & =(n+1)+(n+2)+\cdots+(n+10) \\
& =10 n+55, \\
A & =(k+1)+(k+2)+\cdots+(k+11) \\
& =11 k+66 .
\end{aligned}
$$
... | 495 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (This question is worth 50 points) In a $100 \times 25$ rectangular table, each cell is filled with a non-negative real number. The number in the $i$-th row and $j$-th column is denoted as $x_{i, j} (i=1,2, \cdots, 100 ; j=1,2, \cdots, 25)$ (as shown in Table 1). Then, the numbers in each column of Table 1 are r... | Three, the minimum value of $k$ is 97.
(1) Take $x_{i, j}=\left\{\begin{array}{ll}0, & 4(j-1)+1 \leqslant i \leqslant 4 j \\ \frac{1}{24}, & \text { otherwise } i .\end{array}\right.$ $(j=1,2, \cdots, 25)$
At this time, $\sum_{j=1}^{25} x_{i, j}=0+24 \times \frac{1}{24} .(i=1,2, \cdots, 100)$
It satisfies the conditio... | 97 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 A four-digit number has the following property: dividing this four-digit number by its last two digits yields a perfect square (if the tens digit is zero, then divide by the units digit), and this perfect square is exactly the square of the first two digits plus 1. For example, $4802 \div 2=2401=49^{2}=$ $(48... | Solution: Let the first two digits of the four-digit number be $c_{1}$, and the last two digits be $c_{2}$. Then $10 \leqslant c_{1} \leqslant 99,1 \leqslant c_{2} \leqslant 99$. The four-digit number can be represented as $100 c_{1}+c_{2}$. According to the problem, we have
$$
\begin{array}{l}
100 c_{1}+c_{2}=\left(c_... | 1805 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $x=\frac{-\sqrt[1]{17}+\sqrt{\sqrt{17}+4 \sqrt{15}}}{2 \sqrt{3}}$. Then, $3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5=$ | 2. 0 .
From the given, we have
$$
2 \sqrt{3} x+\sqrt[4]{17}=\sqrt{\sqrt{17}+4 \sqrt{15}} \text {. }
$$
Squaring and rearranging, we get
$$
\begin{array}{l}
\sqrt{3} x^{2}+\sqrt[4]{17} x-\sqrt{5}=0 . \\
\text { Also, } 3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5 \\
=\left(3 x^{4}-2 \sqrt{3} \cdot \sqrt{5} x^{2}-5\right)-1... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that the integer $n$ is not a multiple of 5. Then the remainder when $n^{4}+4$ is divided by 5 is $\qquad$ . | 4. 0 .
$$
\text { Given } \begin{aligned}
& n^{4}+4=\left(n^{4}-1\right)+5 \\
= & \left(n^{2}+1\right)\left(n^{2}-1\right)+5 \\
= & \left(n^{2}-4\right)\left(n^{2}-1\right)+5\left(n^{2}-1\right)+5 \\
= & (n+2)(n-2)(n+1)(n-1) \\
& +5\left(n^{2}-1\right)+5,
\end{aligned}
$$
and $n$ is not a multiple of 5, so among $n+2,... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. $a, b$ are two positive integers, their least common multiple is 465696. Then the number of such ordered pairs of positive integers $(a, b)$ is ( ) .
(A) 144
(B) 724
(C) 1008
(D) 1155 | 4. (D).
From $[a, b]=465696=2^{5} \cdot 3^{3} \cdot 7^{2} \cdot 11$, we know that
$$
a=2^{a_{1}} \cdot 3^{a_{2}} \cdot 7^{a_{3}} \cdot 11^{\alpha_{4}}, b=2^{\beta_{1}} \cdot 3^{\beta_{2}} \cdot 7^{\beta_{3}} \cdot 11^{\beta_{4}},
$$
where $\alpha_{i}$ and $\beta_{i}$ are non-negative integers, and
$$
\begin{array}{l}... | 1155 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
6. For a cube wooden block $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with an edge length of 1, take points $P, Q, R$ on the three edges passing through vertex $A_{1}$, such that $A_{1} P=A_{1} Q=A_{1} R$. After cutting off the tetrahedron $A_{1}-P Q R$, use the section $\triangle P Q R$ as the base to drill a triangular pri... | 6. (B).
Through point $R$, a line parallel to $A_{1} C$ intersects $A C$ at a point $R^{\prime}$. Through point $Q$, a line parallel to $A_{1} C$ intersects $B_{1} C$ at a point $Q^{\prime}$. A plane through $R Q$ and parallel to $A_{1} C$ intersects the side face of the triangular prism at a point $C_{0}$ on the edge... | 6 | Geometry | MCQ | Yes | Yes | cn_contest | false |
1. Let $n=\underbrace{111 \cdots 11}_{1999 \uparrow 1}, f(n)=90 n^{2000}+20 n+$ 1997. Then the remainder when $f(n)$ is divided by 3 is | Ni.1.1.
A natural number $a$ has the same remainder when divided by 3 as its digit sum $S(a)$ when divided by 3. Therefore, $n=\underbrace{11 \cdots 111}_{1999 \uparrow 1}$ has the same remainder when divided by 3 as 1999, which is 1.
$90 n^{2000}$ has a remainder of 0 when divided by 3, and $20 n$ has a remainder of 2... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $a$, $b$, and $c$ are the lengths of the three sides of a right triangle, and for a natural number $n$ greater than 2, the following holds:
$$
\left(a^{n}+b^{n}+c^{n}\right)^{2}=2\left(a^{2 n}+b^{2 n}+c^{2 n}\right) .
$$
Then $n=$ | 6. 4 .
Let $x=a^{\frac{n}{2}}, y=b^{\frac{n}{2}}, z=c^{\frac{\pi}{2}}$, then
$$
\begin{aligned}
0= & 2\left(a^{2 n}+b^{2 n}+c^{2 n}\right)-\left(a^{n}+b^{n}+c^{n}\right)^{2} \\
= & 2\left(x^{4}+y^{4}+z^{4}\right)-\left(x^{2}+y^{2}+z^{2}\right)^{2} \\
= & x^{4}+y^{4}+z^{4}-2 x^{2} y^{2}-2 x^{2} z^{2}-2 y^{2} z^{2} \\
=... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Six. (12 points) On the hypotenuse $AB$ of the right triangle $\triangle ABC$, color the points $P$ that satisfy $PC^{2} = PA \cdot PB$ in red. How many red points are there at least, and at most, on the hypotenuse? | Six, as shown, let $B P=x$, then $A P=c-x$. Draw the altitude $C H$ on the hypotenuse, then $C H=\frac{a b}{c}, B H$ $=\frac{a^{2}}{b}$. Therefore,
$$
\left.A P=\frac{a^{2}}{c}-x \right| \, \text { ( } P \text { can be between } A \text { and } Y \text {, also please consider } H \text { and }
$$
$B$ points.
In the rig... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
21. A sequence of integers $a_{0}, a_{1}, \cdots, a_{n}$ is called a 二次 sequence if for each $i \in\{1,2, \cdots, n\}$, the equation
$$
\left|a_{i}-a_{i-1}\right|=i^{2} \text {. }
$$
holds.
(a) Prove that for any integers $b$ and $c$, there exists a natural number $n$ and a 二次 sequence such that $a_{0}=b, a_{n}=c$. A 二... | Solution: (a) Since $a_{i}-a_{i-1}= \pm i^{2}, 1 \leqslant i \leqslant n$,
and $c-b=a_{n}-a_{0}=\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)=\sum_{i=1}^{n}\left( \pm i^{2}\right)$.
Let $d=c-b$. We only need to prove that for any integer $d$, there exists a natural number $n$, such that we can make:
“ $\square 1^{2} \square... | 19 | Number Theory | proof | Yes | Yes | cn_contest | false |
2. In the Cartesian coordinate system, points of the form $\left(m, n^{2}\right)$ are painted red (where $m, n$ are integers), referred to as red points, and their surrounding points are not colored. Then, the parabola $y=x^{2}-$ $196 x+9612$ has $\qquad$ red points. | 2. 2 .
Let $\left(m, n^{2}\right)$ be on the parabola $y=x^{2}-196 x+9612$, then $n^{2}=m^{2}-196 m+9612$.
Completing the square and factoring, we get
$$
\begin{array}{l}
(n+m-98)(n-m+98) \\
=8=2 \times 4=(-2) \times(-4) .
\end{array}
$$
$\because m, n$ are integers,
$\therefore n+m-98$ and $n-m+98$ are of the same pa... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given, as shown in the figure, a semicircle $O$ with a diameter of $20 \mathrm{~cm}$ has two points $P$ and $Q$, $P C \perp A B$ at $C, Q D$ $\perp A B$ at $D, Q E \perp$ $O P$ at $E, A C=4 \mathrm{~cm}$. Then $D E=$ | $3.8 \mathrm{~cm}$.
Take the midpoint $M$ of $O P$ and the midpoint $N$ of $O Q$, and connect $C M, D N$, and $E N$. Then
$$
\begin{array}{c}
M C=P E=\frac{1}{2} O P \\
=\frac{1}{2} O Q=E N=D N, \\
\angle P M C=\angle M C O+ \\
\angle M O C=2 \angle M O C .
\end{array}
$$
Since $O, D, Q, E$ are concyclic, with $N$ as t... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
一、(满分 20 分) Determine the smallest positive integer $n$, whose last digit is 7, such that if the last digit 7 is moved to the front, the resulting number is 5 times the original number. | Let $n=\overline{a_{k} a_{k-1} \cdots a_{2} a_{1}}, x=\overline{a_{k} a_{k-1} \cdots a_{3} a_{2}}$, where $a_{1}$ $=7, k$ is a natural number greater than 1. Then
$$
\begin{aligned}
n & =\overline{a_{k} a_{k-1} \cdots a_{2}} \times 10+7=10 x+7, \\
5 n & =\overline{7 a_{k} a_{k-1} \cdots a_{3} a_{2}} \\
& =7 \times 10^{... | 142857 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Full marks 25 points) Does there exist 100 different lines in the plane, such that their intersection points total 1998? | Three, there are 1998 intersection points.
First, divide 98 lines into two sets of parallel lines, one set has $k$ lines, and the other set has $(98-k)$ lines. The number of intersection points is $k(98-k)$. When the value of the positive integer $1998-k(98-k)$ is minimized, $k=26$ or 72.
When $k=26$ or 72, $k(98-k)=2... | 1998 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
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