problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
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7. If $A$ is the sum of the absolute values of all roots of the equation
$$
x=\sqrt{19}+\frac{91}{\sqrt{19}+\cdots \sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}
$$
Find $A^{2}$. | 7. The fraction to the right of the equation can be simplified to $\frac{a x+b}{c x+d}$ $(a, b, c, d$ are real numbers), then the equation is a quadratic equation, with at most two roots. H $x=\sqrt{19}+\frac{91}{x}$ has roots that are also $x=\frac{\sqrt{19}-\sqrt{38}}{2}$, which are the only two roots of the original... | 383 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. How many real numbers $a$ are there such that $x^{2}+a x+6 a$ $=0$ has only integer solutions?
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 at the end is not part of the translation but is provided to clarify that the format and structure of the original text have been maintained in the translation. | 8. Let $x^{2}+a x+6 a=$ O have integer solutions $m, n$ $(m \leqslant n)$. Then we have
$$
a=-(m+n), \quad 6 a=m n .
$$
Since $a$ must be an integer, we also have
$$
-6(m+n)=m n \text { . }
$$
H $(m+6)(n+6)=36$. Since $36=36 \times 1=18 \times 2=12 \times 3=6 \times 6$, the solutions satisfying $m \leqslant n$ are
$$... | 10 | Other | math-word-problem | Yes | Yes | cn_contest | false |
9. If $\sec x+\tan x=\frac{22}{7}, \csc x+\cot x$ $=\frac{m}{n}$, where $\frac{m}{n}$ is a reduced fraction, find $m+n$.
| $$
\begin{array}{l}
\text { 9. } \because \sec ^{2} x-\tan ^{2} x=1 \text {, let } p=\frac{22}{7}, \\
\therefore \sec x-\tan x=\frac{1}{p}, \sec x+\tan x=p .
\end{array}
$$
Adding these two equations, we get $2 \sec x=p+\frac{1}{p}$. Subtracting them, we get
$$
\begin{aligned}
2 \tan x & =p-\frac{1}{p} . \\
& \therefo... | 44 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. For the string of three letters $\hat{\mathrm{i}}$ "aaa" and "bbb", they are transmitted through a circuit, with each string being sent one letter at a time. Due to issues with the line, each of the 6 letters has a $\frac{1}{3}$ chance of being transmitted incorrectly (an $a$ is received as a $b$, or a $b$ is recei... | $x_{1}=a, y_{1}=b, i=1,2,3$.
Let $S_{1} \propto S_{2}$ represent that $S_{1}$ is lexicographically before $S_{2}$, and $P\left(S_{1} \propto S_{2}\right)$ represent the probability that $S_{1}$ is lexicographically before $S_{2}$. Due to the independence of the correct transmission of letters, we have
$$
\begin{array}... | 532 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. As shown in the figure, 12 congruent disks are placed on the circumference of a circle $C$ with radius 1, such that they cover $C$. No two disks overlap, and adjacent disks are tangent to each other. The sum of the areas of these disks can be written as
$$
\pi(a-b \sqrt{c}) \text {, where }
$$
$a, b, c$ are positiv... | 11. By the people's knowledge, 6-certainly passes through the tangency points of each circle, and each tangent line passes through the center $O$ of circle $C$ (as shown in the figure).
$P$ is the center of a circle,
$$
\begin{array}{l}
\angle P B O=90^{\circ}, \\
\angle B O A=\frac{1}{12} 2 \pi=\frac{\pi}{6}, \\
\angl... | 135 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. Rhombus $P Q R S$ is inscribed in rectangle $A B C D$, such that $P, Q, R, S$ are interior points on $\overline{A B}, \overline{B C}, \overline{C D}, \overline{D A}$. Given that $P B=15, B Q=20, P R=30, Q S$ $=40$. If the reduced fraction $\frac{m}{n}$ is the perimeter of rectangle $A B C D$, find $m+n$. | 12. As shown in the figure, let $x, y$ represent the lengths of $Q C, R C$ respectively. By symmetry, the lengths of $SA, PA$ are also $x, y$.
The diagonals of the rectangle intersect at the center H of the rectangle, and they are perpendicular to each other, thus forming eight right triangles. Six of these right trian... | 677 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. In a drawer, there are red and blue socks, no more than 1991 in total. If two socks are drawn without replacement, the probability that they are the same color is $\frac{1}{2}$. How many red socks can there be at most in this case? | 13. Let the number of red and black balls be $x$ and $y$ respectively. Given that the probability of randomly picking two balls of different colors is $\frac{1}{2}$, we have $\frac{x y}{C_{x+y}^{2}}=\frac{1}{2}$.
Therefore, $(x+y)(x+y-1)=4 x y$,
Thus, $(x-y)^{2}=x+y$.
Hence, the total number of balls is a perfect squa... | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. For a positive integer $n$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value. Where $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ are positive integers, and their sum is 17. There is a unique value of $n$ for which $S_{n}$ is an integer, find $n$. | 15. Consider each $l_{i}=\sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the hypotenuse of a right-angled triangle, with the two legs being $2 k-1$ and $a_{b}$. When these right-angled triangles are placed together to form a ladder, let $A, B$ be the starting and ending points, respectively.
The distance from $A$ to $B$ is
$$
\begin{... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Let $x$ be a cube root of 1 different from 1, find the value of $x^{\text {D}}$ $+x^{2}$. $(n \in N)$ | Solve: From $x^{3}=1$, i.e., $x^{3}-1=0$, which is also $(x-1)\left(x^{2}+x+1\right)=0$.
Since $x \neq 1$, it follows that $x^{2}+x=-1$.
When $n=3k$,
$$
x^{\mathrm{n}}+x^{2 n}=\left(x^{3}\right)^{k}+\left(x^{3}\right)^{2 k}=2 \text {; }
$$
When $n=3 k+1$,
$$
\begin{array}{l}
x^{\mathrm{n}}+x^{2 \mathrm{n}}=\left(x^{3}... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If two real-coefficient quadratic equations in $x$, $x^{2}+x+a=0$ and $x^{2}+a x+1=0$, have at least one common real root, then $a=$ $\qquad$ | 1. $-2 ;$ | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, there are 1990 points distributed on a straight line. We will mark the midpoints of all possible line segments with these points as endpoints. Try to find the minimum number of distinct midpoints that can be obtained.
untranslated text:
直线上外布着1990个点, 我们来标出以这些点为端点的一切可能的线段的 中点. 试求至少可以得出多少个互不重合的中点.
translated te... | Four, Solution: Let the two points that are farthest apart be denoted as $A, B$. Consider the 1988 line segments formed by $A$ and the other 1988 points excluding $B$. The midpoints of these 1988 line segments are all distinct, and their distances to point $A$ are less than $\frac{1}{2} A B$.
Similarly, for point $B$,... | 3977 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. For $\triangle A B C=$ with sides $a, b, c$, construct squares outward on each side, with areas sequentially $S_{a}, S_{b}, S_{c}$. If $a+b+c=18$, find the minimum value of $S_{\mathrm{a}}+S_{\mathrm{b}}+S_{\mathrm{c}}$. | Given $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$ $\geqslant 0$, which means $2\left(a^{2}+b^{2}+c^{2}\right)=2(a b+b c+c a)$. Therefore, $3\left(a^{2}+b^{2}+c^{2}\right) \geqslant a^{2}+b^{2}+c^{2}+2 a b$ $+2 a c+2 b c=(a+b+c)^{2}$. Thus, $S_{1}+S_{v}+S_{r} \geqslant \frac{18^{2}}{3}=108$.
When $a=b=c$, the equality holds, so th... | 108 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. The teacher distributed a stack of books among A, B, C, L, E. $\frac{1}{4}$ of the books were given to $B$, then $\frac{1}{3}$ of the remaining books were given to $C$, and the rest were split equally between $D$ and $E$. If $E$ received 6 books, then the teacher originally had $\qquad$ books. | 2. 48 | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. (Zu Chongzhi Cup Mathematics Competition) The sum of all digits of the natural numbers $1, 2, 3, \ldots, 9999$ is $\qquad$ | Solve $0+9999, 1+9998, \cdots, 4999+5000$
all sum to 9999 without any carry, so the sum of digits is more:
$$
4 \times 9 \times 5000 \div 2-5=89995 \text {. }
$$
This is called the method of integer value combination. | 89995 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Six, in all integers that start and end with 1 and alternate between 1 and 0 (i.e., $101$, $10101$, $1010101$, etc.), how many of them are prime numbers? | Six, Proof that there is only one prime number 101.
If $n \geqslant 2$, then $A=10^{2 n}+10^{2 n-2}+\cdots+10^{2}$
$$
+1=\frac{\left(10^{\mathrm{n}+1}-1\right)\left(10^{\mathrm{n}+1}+1\right)}{99} \text {. }
$$
When $n=2 m+1$, $\frac{10^{2 m+2}-1}{99}=10^{2 \mathrm{~m}}$ $+\cdots+10^{2}+1, A$ is a composite number;
Wh... | 101 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $[x]$ denote the greatest integer not exceeding $x$. If
$$
\begin{array}{l}
f=[1 \mathrm{~g} 1]+[1 \mathrm{~g} 2]+[1 \mathrm{~g} 3]+\cdots+[1 \mathrm{~g} 1989] \\
+[1 \mathrm{~g} 1990], \text { then } f=
\end{array}
$$ | 3. $4863 ;$ | 4863 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. (Shanghai Mathematics, 1987)
$$
\log _{3}\left[(3+1)\left(3^{2}+1\right)\left(3^{4}+1\right) \cdot \ldots\right.
$$
- $\left.\left(3^{84}+1\right)+\frac{1}{2}\right]+\log _{3} 2$ The value is ( ).
(A) 32.
( B) 64.
(C) 128.
(D) None of the above. | $$
\begin{aligned}
\text { Original expression }= & \log _{3}(3-1)(3+1)\left(3^{2}+1\right) \\
& \cdot \cdots \cdot\left(3^{84}+1\right) \\
= & \log _{3} 3^{128}=128 .
\end{aligned}
$$
Expressing "3-1" as 2 in the second line is a kind of simplification. This can be referred to as the method of adding factors. | 128 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Two cars start from the same location at the same time, driving in the same direction at the same speed. Each car can carry a maximum of 24 barrels of gasoline. They cannot use any other fuel during the journey. Each barrel of gasoline can make a car travel 60 kilometers. Both cars must return to the starting point,... | If so, A gives B $(24-2 x)$ barrels of gasoline. B continues to move forward, carrying $(24-2 x)+(24-x)=48-3 x$ barrels of gasoline. According to the problem, we should have
$$
48-3 x \leqslant 24 \text {. That is, } x \geqslant 8 \text {. }
$$
After A and B separate, the distance B continues to travel is
$$
\begin{ar... | 1920 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
3. There is a sequence of numbers which are $1, 5, 11, 19, 29$, $A, 55$, where $A=$ $\qquad$ . | 3. 41; | 41 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Find the positive integer solutions to the equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$. | 3. Solution Since $x, y, z$ are positive integers and
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{6} \leqslant 1 .
$$
Assume without loss of generality that $1<x \leqslant y \leqslant z$, then $\frac{1}{x} \geqslant \frac{1}{y} \geqslant \frac{1}{z}$.
Thus, $\frac{1}{x}<\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslan... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Among all possible four-digit numbers formed using the digits $1,9,9,0$, for each such four-digit number and a natural number $n$, their sum when divided by 7 does not leave a remainder of 1. List all such natural numbers $n$ in descending order.
$$
n_{1}<n_{2}<n_{3}<n_{4}<\cdots \cdots,
$$
Find: the value of $n_{1... | 3. Solution
$1,0,9,0$ digits can form the following four-digit numbers: 1099, 1909, 1990, 9019, 9091, 9109, $9190,9901,9910$ for a total of nine. The remainders when these nine numbers are divided by 7 are $0,5,2,3,5,2,6$, 3 , 5 . Since $n$ and their sum cannot be divisible by 7 with a remainder of 1, $n$ cannot have a... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-c-a}{b}$ $=3$, and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0$. Then $x-a-b-c=$ | 5. 0 ;
The above text has been translated into English, maintaining the original text's line breaks and format. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, A and B start from locations $A, B$ respectively, heading towards each other. They meet at point $C$ on the way, after which A takes 5 hours to reach $B$, and B takes $3 \frac{1}{5}$ hours to reach $A$; it is known that A walks 1 kilometer less per hour than B. Find the distance between $A$ and $B$. | Three, Hint: Let $A C=S_{1}, C B=S_{2}$, the time required for both to meet is /small hour, then we get $t=4$ hours. From the problem, we solve to get $S_{1}=16, S_{2}=20$. Therefore, $A B=35$ km. | 35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. In $\triangle A B C$, $A B=A C=2, B C$ side has 100 different points $P_{1}, P_{2}, \cdots, P_{1} 00$. Let $m_{1}=A P_{1}^{2}+B P_{1} \cdot P_{1} C(i=1,2, \cdots, 100)$, then $m_{1}+m_{2}+\cdots+m_{100}$ equals what? (1990 National Junior High School Mathematics League) | Solve $\because A B$ $=A C$, according to Corollary 1 we have
$$
\begin{array}{c}
A P_{1}^{2}=A B^{2} \\
-B P_{\mathrm{i}} \cdot P_{1} C, \\
\therefore A P_{1}^{2} \\
+B P_{1} \cdot P_{1} C \\
=A B^{2}, \\
\therefore m_{\mathrm{i}}=A P_{1}^{2}+B P_{1} \cdot P_{\mathrm{i}} C \\
=A B^{2}=2^{2}=4,
\end{array}
$$
that is
... | 400 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { Four, Prove: } 1991^{1992}+1993^{1994} \\
+1995^{1996}+1997^{1998}+1999^{2000}
\end{array}
$$
is divisible by 5. | Four, Hint: By discussing the regularity of the powers of the unit digits, we know that the unit digits of the terms in the sum are $1, 9, 5, 9, 1$, and their sum is 25. Therefore, the sum can be divisible by 5. | 25 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 3. In $\triangle A B C$, $D$ is a point on side $B C$. It is known that $A B=13, A D=12, A C=15, B D=5$. What is $D C$? (3rd Zu Chongzhi Cup Junior High School Mathematics Invitational Competition) | According to Stewart's Theorem, we have
$$
A D^{2}=A B^{2} \cdot \frac{C D}{B C}+A C^{2} \cdot \frac{B D}{B C}-B D \cdot D C \text {. }
$$
Let $D C=x$,
then $B C=5+x$.
Substituting the known data
into the above equation, we get
$$
\begin{array}{l}
12^{2}=13^{2} \\
\cdot \frac{x}{5+x}+15^{2} \cdot \frac{5}{5+x}-5 x,
\e... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, given that $a$ is an integer, the equation $x^{2}+(2 a+1) x$ $+a^{2}=0$ has integer roots $x_{1}, x_{2}, x_{1}>x_{2}$. Try to find the value of $\sqrt[4]{x_{1}^{2}}-\sqrt[4]{x_{2}^{2}}$. | Three, Solution: From the problem, we know that the discriminant $4a + 1$ is a perfect square, so $a \geqslant 0$. Since $4a + 1$ is odd, we can set $(2k + 1)^2 = 4a + 1$, solving for $a = k(k + 1)$.
$$
x_{1,2} = \frac{1}{2} \left[ \left( -2k^2 - 2k - 1 \right) \pm \sqrt{(2k + 1)^2} \right].
$$
When $k \geqslant 0$, $... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. As shown in the figure, points $A, B, C, D$ lie on the same circle, and $BC=DC=4, AE=6$. The lengths of segments $BE$ and $DE$ are both positive integers. What is the length of $BD$? (1988
National Junior High School Mathematics Competition) | Solve in $\triangle B C D$, $B C=D C$, H
Inference 1 gives
$$
C E^{2}=B C^{2}-B E \cdot D E,
$$
$\because A, B, C, D$ are concyclic,
$$
\begin{array}{l}
\therefore B E \cdot D E=A E \cdot C E, \\
\therefore C E^{2}=B C^{2}-A E \cdot C E=4^{2}-6 \times C E,
\end{array}
$$
which is $C E^{2}+6 \cdot C E-16=0$.
Solving gi... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Factorize: $a^{3}+2 a^{2}-12 a+15$ $=$ $\qquad$ - If $a$ is a certain natural number, and the above expression represents a prime number, then this prime number is | 3. $\left(a^{2}-3 a+3\right)(a+5), 7$; | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. A hostess is waiting for 7 or 11 children to arrive, and she has prepared 77 marbles as gifts. She puts these marbles into $n$ bags so that each child (whether 7 or 11) can receive several bags of marbles, and the 77 marbles are evenly distributed among these children. Find the minimum value of $n$. | 10. Since these $n$ bags of sand can be divided into 7 portions, each containing 11 marbles, represented by 7 vertices $x_{1}, x_{2}, \cdots, x_{7}$. These $n$ bags of marbles can also be divided into 11 portions, each containing 7 marbles, represented by vertices $y_{1}, y_{2}, \cdots, y_{11}$. If a certain bag of mar... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. Place several points on the unit sphere such that the distance between any two points is (1) at least $\sqrt{2}$; (2) greater than $\sqrt{2}$. Determine the maximum number of points and prove your conclusion. | 13. (1) The maximum number of points is 6. If $A$ is one of these points, let's assume $A$ is at the North Pole, then the remaining points must all be in the Southern Hemisphere (including the equator). If there is only one point $B$ at the South Pole, then the rest of the points are all on the equator, in which case t... | 6 | Geometry | proof | Yes | Yes | cn_contest | false |
3. (China) Let $S=\{1,2,3, \ldots, 280\}$. Find the smallest natural number $n$ such that every subset of $S$ with $n$ elements contains 5 pairwise coprime numbers. | Let $A_{1}=\{S \mapsto$ all natural numbers divisible by $i\}, i=2,3,5,7$. Let $A=A_{2} \cup A_{3} \cup A \cup \cup A_{7}$. Using the principle of inclusion-exclusion, it is easy to calculate that the number of elements in $A$ is 216. Since any 5 numbers chosen from $A$ must have at least two numbers in the same $A_i$,... | 217 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
21. Let the weight of the counterfeit coin be $a$, and the weight of the genuine coin be $b$ $(a \neq b)$. There are two piles of three coins each, and it is known that each pile contains exactly one counterfeit coin. How many times at least must a precise scale (not a balance) be used to find these two counterfeit coi... | 21. (1) At least 3 weighings are required.
First, take one coin from each of the two piles, (1) if the weight is $2a$, then the real coins have been found; (2) if the weight is $2b$, then take one coin from the remaining two in each pile and weigh them separately, and the two fake coins can be found; (3) if the weight... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
1. A cube with an edge length of 3 is composed of 27 unit cubes. How many lines pass through the centers of 3 unit cubes? How many lines pass through the centers of 2 unit cubes? | There are 49 lines passing through the centers of 3 unit cubes; there are
$$
C_{27}^{2}-3 \times 49=204 \text { (lines). }
$$ | 204 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. It is known that a parliament has 30 members, where any two are either friends or enemies, and each member has exactly 6 enemies. Any three members form a committee. Find the total number of such committees where the three members are either all friends or all enemies. | 3. Let the set of all three-member committees that satisfy the requirements of the problem be denoted as $X$, and the number of elements in set $X$ be denoted as $x$. The number of other three-member committees is denoted as $y$, thus
$$
x+y=C_{30}^{s}=4060 \text {. }
$$
For any senator $a$, we denote the set of commi... | 1990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Find the maximum value of the expression
$$
|| \cdots|| x_{1}-x_{2}\left|-x_{3}\right|-\cdots \mid-x_{1000}
$$
where $x_{1} ， x_{2}, \cdots, x_{1000}$ are different natural numbers from 1 to 1990. | 7. Noting that when $x \geqslant 0, y \geqslant 0$, the following formulas hold:
$$
\begin{array}{l}
|x-y| \leqslant \max \{x, y\}, \\
\max \{\max \{x, y\}, z\}=\max \{x, y, z\},
\end{array}
$$
we can directly obtain
$$
\begin{array}{l}
\left.|| \cdots|| x_{1}-x_{2}\left|-x_{3}\right|-\cdots \mid-x_{\mathrm{n}}\right\... | 1989 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. There are two fleas at the two endpoints of the line segment $[0, 1]$. Some points are marked within the line segment. Each flea can jump over the marked points such that the positions before and after the jump are symmetric about the marked point, and they must not jump out of the range of the segment $[0, 1]$. Eac... | 4. The segment [0, 1] is divided into smaller segments by certain fixed points, with lengths $\frac{17}{23}$ and $\frac{19}{23}$. Therefore, the two fleas cannot land on the same segment after each taking one step (see the following supplement). From this, we can conclude that the minimum number of steps required must ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
5 . Find the integer solution to the equation
$$
\left[\frac{x}{1!}\right]+\left[\begin{array}{c}
x \\
2!
\end{array}\right]+\cdots+\left[\frac{x}{10!}\right]=1001
$$ | 5. From the old equation, we know that $x$ is a positive integer not exceeding 1001, so $x<6$!. Therefore, the last five terms on the left side of the equation can be removed. Each positive integer $x<6$! can be uniquely expressed as
$$
x=a \cdot 5!+b \cdot 4!+c \cdot 3!+d \cdot 2!+e,
$$
where $a$, $b$, $c$, $d$, $e$ ... | 584 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
```
11. (16th All-Russian Mathematics Competition) Replace the “*” in the following division problem with appropriate digits
$$
\begin{array}{l}
\text { * } 8 \text { * } \\
* * \sqrt{* * * * *} \\
-) * * * * * * \\
\frac{-) * *}{* *} \\
\frac{-) * * *}{0} \\
\end{array}
$$
``` | From the fact that the divisor multiplied by 8 yields a two-digit number, we know the divisor is no greater than 12. Furthermore, since the divisor multiplied by the first or last digit of the quotient results in a three-digit number, we know the divisor is no less than 12.
Therefore, the divisor is 12.
And since 12 mu... | 11868 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
13. (19th Austrian Mathematical Competition) Find $N=$ $19^{88}-1$ for all divisors $d=2^{2} \cdot 3^{b}$ (where $a, b$ are natural numbers) and determine the sum of these divisors $d$. | $$
\begin{aligned}
N & =(20-1)^{88}-1 \\
& =(1-4 \times 5)^{88}-1 \\
& =-C_{88}^{1} 4 \times 5+C_{88}^{2} 4^{2} \times 5^{2} \\
& -C_{88}^{8} 4^{8} \times 5^{3}+\cdots \\
& -C_{88}^{87} 4^{87} \times 5^{87}+C_{88}^{88} 4^{88} \times 5^{88} \\
& =-2^{5} \times 55+2^{8} M \\
& =2^{5}(-55+2 M),
\end{aligned}
$$
where $M$... | 744 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
15. (16th All-Russian Mathematics Competition) A desert is shaped like a half-plane, which is divided into many small squares of size $1 \times 1$. In the desert, 15 squares away from the boundary, there is a robot with energy $E=59$. The "energy consumption" of each small square is a natural number not greater than 5,... | Solve As shown in the figure, use routes with arrows to represent five different stubborn movement routes of the robot, * indicates the position of the robot.
The total energy consumption of the small squares passed by these five routes does not exceed $3 \times 88 + 2 \times 10 + 2 \times 5 = 294$ (the above routes p... | 59 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
18. (16th All-Russian Mathematics Competition) In the school football championship, each team must play a match against every other team. Each match awards 2 points to the winning team, 1 point to each team in the event of a draw, and 0 points to the losing team. It is known that one team has the highest score, but it ... | Let the team with the highest score, denoted as team $A$, be the champion. Suppose team $A$ wins $n$ matches and draws $m$ matches, then the total score of team $A$ is $2n + m$ points.
From the given conditions, every other team must win at least $n + 1$ matches, meaning their score is no less than $2(n + 1)$ points. ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. The figure formed by the diagonals of squares with side length 1 is called a cross. Prove: In a circle with a radius of 100, the number of non-overlapping crosses that can be placed does not exceed 80000. | Prove that for each cross, consider a circle with the center of the line segment as the center and the half of the cross diagonal (diagonal length... half) $\frac{1}{2} \cdot \frac{\sqrt{2}}{2}=\frac{1}{2 \sqrt{2}}$ as the radius.
Then, in the figure, the intersection $O_{1} O_{2} \leqslant r_{1}+r_{2}=2 r$ $=\frac{1}... | 80000 | Geometry | proof | Yes | Yes | cn_contest | false |
Three, given any 5 points on a plane, where no three points are collinear and no four points are concyclic. If a circle passes through three of these points, and the other two points are respectively inside and outside the circle, then it is called a "good circle".
Let the number of good circles be $n$, find all possib... | Three, among: 5 points, take any two points $A, B$ and draw the line through $A$, $B$. If the other three points $C, D, E$ are on the same side of line $A B$, then consider $\angle A C B, \angle A D B, \angle A E B$. Without loss of generality, if $\angle A C B < 180^{\circ}$, then circle $A D B$ is the unique good cir... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Five, try to select 1999 integers from the set $\{1,2,3, \cdots, 3000\}$ to form a subset $A$, such that if $x \in A$, then $2 x \notin A$. | Five, the selection method is as follows:
1) Select 1500 integers from 1501 to 3000 to form $A_{1}$.
2) From 1 to 1500, remove numbers 751 to 1500 (these numbers multiplied by 2 are in $A_{1}$), and select 376 to 750, a total of 375 integers to form $A_{2}$.
3) From 1 to 375, remove 188 to 375 (these numbers multiplied... | 1999 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For $a \geqslant 1$, calculate the sum of the infinite series
$$
\begin{array}{l}
\frac{a}{a+1}+\frac{a^{2}}{(a+1)\left(a^{2}+1\right)} \\
+\frac{a^{4}}{\left.(a+1) a^{2}+1\right)\left(a^{4}+1\right)} \\
+\frac{a^{8}}{(a+1)\left(a^{2}+1\right)\left(a^{4}+1\right)\left(a^{8}+1\right)}+\cdots
\end{array}
$$ | $\begin{array}{c}\text { I. When } a=1 \text {, } \Sigma=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots \\ =\frac{\frac{1}{2}}{1-\frac{1}{2}}=1 \text {. When } a>1 \text {, we have } \Sigma_{1}=\frac{a^{2}-a}{a^{2}-1}, \\ \Sigma_{2}=\frac{a^{4}-a}{a^{4}-1}, \Sigma_{3}=\frac{a^{8}-a}{a^{8}-1}, \cdots, \Sigma_{n}=\frac{a^{2}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, let $A B C$ be an equilateral triangle, and $P$ a point on its incircle. Prove that $P A^{2}+P B^{2}+P C^{2}$ is a constant. | Four, as shown in the figure, establish
a rectangular coordinate system,
the coordinates of point $P$ are
$$
\begin{array}{l}
\left(\frac{\sqrt{3}}{3} \cos \theta,\right. \\
\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3} \\
\cdot \sin \theta) .
\end{array}
$$
Let $\Sigma=P A^{2}+P B^{2}+P C^{2}$, then
$$
\begin{array}{l}
\Sigm... | 5 | Geometry | proof | Yes | Yes | cn_contest | false |
Let $S=\{1,2,3,4\}, a_{1}, a_{2}, \cdots$, be any permutation ending with 1, i.e., for any permutation $\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ of the four numbers in $S$ that does not end with 1, $\left(b_{4} \neq 1\right)$, there exist $i_{1}, i_{2}, i_{3}, i_{4}$, such that $1 \leqslant i_{1}<i_{2}<i_{3}<i_{4} \leq... | II. Hint: First consider the minimum value of the number of terms $k$ in any sequence containing a permutation of $S$.
1. For $S=\{1,2,3\}$, it can be proven that a sequence with only 6 terms cannot contain any permutation of $\{1,2,3\}$.
2. For $S=\{1,2,3,4\}$, prove that a sequence with only 11 terms cannot contain a... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Find the minimum value of the function $f(x)=\max \left\{x^{2}+1, \cos x, 2 x\right\} \quad(x \in R)$. | $$
\begin{array}{l}
\text { Sol } \because\left(x^{2}+1\right)-\cos x \\
=x^{2}+(1-\cos x) \geqslant 0, \\
\left(x^{2}+1\right)-2 x=(x-1)^{2} \geqslant 0, \\
\therefore \quad x^{2}+1 \geqslant \cos x, x^{2}+1 \geqslant 2 x \\
\text { Therefore, } f(x)=\max \left\{x^{2}+1, \cos x, 2 x\right\} \\
=x^{2}+1 \geqslant 1,
\... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The average score of the participants in the Zu Chongzhi Cup Mathematics Invitational at a certain school is 75. Among them, the number of male participants is $80\%$ more than that of female participants, and the average score of female participants is $20\%$ higher than that of male participants. Therefore, the av... | 1. 84 | 84 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. $A, B$ are two fixed points on a plane, find a point $C$ on the plane such that $\triangle A B C$ forms an isosceles triangle. There are $\qquad$ such points $C$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The blank space represented by $\qquad$ in the original text is kept as is in the translation. | 4.6.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 6 | Logic and Puzzles | other | Yes | Yes | cn_contest | false |
Three, (1) Prove: $\sqrt{a^{2}+\frac{1}{b^{2}}+\frac{a^{2}}{(a b+1)^{2}}}$ $=\left|a+\frac{1}{b}-\frac{a}{a b+1}\right|$
(2) Using or not using (1), calculate
$$
\sqrt{1+1990^{2}+\frac{1990^{2}}{1991^{2}}}-\frac{1}{1991}
$$ | Three, Hint: Square both sides of the equation, noting that
$$
\frac{a}{b}-\frac{a^{2}}{a b+1}-\frac{a}{b(a b+1)}=0,
$$
and then take the square root of both sides.
Note: In fact, the following equation also holds:
$$
\sqrt{a^{2}+b^{2}+\frac{a^{2} b^{2}}{(a+b)^{2}}}=\left|a+b-\frac{a b}{a+b}\right|
$$
Using equation ... | 1990 | Algebra | proof | Yes | Yes | cn_contest | false |
Example 6. Given the equation $x^{2}+(a-6) x+a=0$ ( $a$ $\neq 0$ ) with both roots being integers. Try to find the integer $a$. (1989, Sichuan Province Junior High School Mathematics Competition) | Let the two integer roots of the equation be \( y_{1} x_{1}, x_{2} \), and \( x_{1} \geqslant x_{2} \). By Vieta's formulas, we have
\[
\left\{\begin{array}{l}
x_{1}+x_{2}=6-a \\
x_{1} x_{2}=a
\end{array}\right.
\]
From (1) and (2), we get \( x_{1} x_{2} + x_{1} + x_{2} = 6 \), which simplifies to \( \left(x_{1}+1\rig... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Sure, here is the translation:
---
One, starting from the natural number 1, write down in sequence to form the following series of numbers: $12345678910111213 \cdots$. With each digit occupying one position, determine the digit at the 1992nd position. | In the following sequence of digits, there are 9 single-digit numbers, $2 \times 90$ two-digit numbers, and $3 \times 900$ three-digit numbers. From $(1992-9-2 \times 90) \div 3$ $=601$, we know that the 1992nd position is in the 601st three-digit number starting from 100, which is the unit digit of the natural number ... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. The lengths of the three sides of $\triangle ABC$ are $BC=17$, $CA=18$, and $AB=19$. Through a point $P$ inside $\triangle ABC$, perpendiculars $PD$, $PE$, and $PF$ are drawn to the three sides of $\triangle ABC$ ($D$, $E$, and $F$ are the feet of the perpendiculars), and $BD + CE + AF = 27$. Find the length... | Let $B D=x, B F=y$. Then $C D=17-x$, $A F=19-2$. Also, $B D+C E+A F=27$, so
$$
\begin{array}{l}
C E=27-B D-A F=8-x+y, \\
A B=18-C E=10+x-y .
\end{array}
$$
By Theorem 1 conclusion (2), we have
$$
\begin{array}{l}
x^{2}+(8-x+y)^{2}+(19-y)^{2} \\
=(17-x)^{2}+(10+x-y)^{2}+y^{2} .
\end{array}
$$
Simplifying, we get $2 x+... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. As shown in Figure 3, through an internal point $P$ of $\triangle ABC$, three lines parallel to the three sides are drawn, resulting in three triangles $t_{1}, t_{2}$, and $t_{3}$ with areas 4, 9, and 49, respectively. Find the area of $\triangle ABC$.
(2nd American
Mathematical Invitational) | This example can directly apply the conclusion (4) of Theorem 2 to obtain
$$
S \triangle A B C=144
$$ | 144 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Given $-\frac{x-b}{c}+\frac{x-b-c}{a}$ $+\frac{x-c-a}{b}=3$,
and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0$.
Then $x-a-b-c=$ $\qquad$ (8th Jincheng
Mathematics Competition) | From the known, we get $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(x-a-b-c)=0$, so we should fill in “0”. This is transforming the known to the unknown, the other is transforming the unknown to the known.
| 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. If $x^{3}-x^{2}+x-2=0$. Then $x^{4}+2 x^{3}-2 x^{2}+x-1=$ $\qquad$ . (1991, Hubei Huanggang Region Mathematics Competition). | Left= $\begin{aligned} & \left(x^{4}-x^{3}+x^{2}-2 x\right) \\ & +\left(3 x^{3}-3 x^{2}+3 x-6\right)+5 \\ = & x \cdot 0+3 \cdot 0+5=5 .\end{aligned}$ | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Calculate: $\left(1^{2}+3^{2}+5^{2}+\cdots+99^{2}\right)-\left(2^{2}\right.$ $\left.+4^{2}+6^{2}+\cdots+100^{2}\right)$ | $\begin{array}{l}\text { 2. Hint: Original expression }=\left(1^{2}-2^{2}\right)+\left(3^{2}-4^{2}\right) \\ +\left(5^{2}-6^{2}\right)+\cdots+\left(99^{2}-100^{2}\right)=-5050 .\end{array}$ | -5050 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. When a surveyor is measuring a piece of land, they first walk 100 meters from point $A$ at a bearing of $35^{\circ}$ north of east to point $B$, then from point $B$ at a bearing of $55^{\circ}$ north of west for 80 meters to point $C$, and from point $C$ at a bearing of $35^{\circ}$ south of west for 60 meters to po... | 3. Brief solution: ( 1 ) As shown in the figure.
( 2 ) This piece of land is a trapezoid.
$$
\begin{array}{c}
\text { ( } 3 \text { ) } S= \\
\frac{1}{2}(60+100) \times 80 . \\
=6400 \text { (square meters). }
\end{array}
$$ | 6400 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $a, b$ are integers, $a$ divided by 7 leaves a remainder of 3, and $b$ divided by 7 leaves a remainder of 5. When $a^{2}>4 b$, find the remainder when $a^{2}-4 b$ is divided by 7. | 4. Slightly explained: Let $a=$
$$
7 m+3, b=7 n+5 \text {, }
$$
where $m, n$ are integers, then
$$
\begin{aligned}
a^{2}-4 b & =(7 m+3)^{2}-4(7 n+5) \\
& =7\left(7 m^{2}+6 m-4 n-2\right)+3 .
\end{aligned}
$$
Since $7 m^{2}+6 m-4 n-2$ is an integer,
thus $a^{2}-4 b$ leaves a remainder of 3 when divided by 7. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. There are two specifications of steel bars, A and B, and C. It is known that 2 bars of type A, 1 bar of type B, and 3 bars of type C are a total of 23 meters long; 1 bar of type A, 4 bars of type B, and 5 bars of type C are a total of 36 meters long. How long are 1 bar of type A, 2 bars of type B, and 3 bars of type... | 6. Slightly explained: Let the lengths of copper strips of specifications A, B, and C be $x, y, z$ meters, respectively, then
$$
\left\{\begin{array}{l}
2 x+y+3 z=23, \\
x+4 y+5 z=36
\end{array}\right.
$$
From (1) - (2) $\times 2$, we get $y=7-z$.
Substituting $y$ into (2), we get $x=8-z$.
Substituting $x, y$ into $x+... | 22 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. There are several warriors, forming a rectangular formation that is exactly eight columns wide. If 120 more people are added to or 120 people are removed from the formation, a square formation can be formed in both cases. How many warriors are there in the original rectangular formation? | 10. Solution: Let the original number of soldiers be $8x$ people,
From the given, $8x+120$ and $8x-120$ are both perfect squares, then we have
$$
\left\{\begin{array}{l}
8 x+120=m^{2} \\
8 x-120=n^{3}
\end{array}\right.
$$
( $m, n$ are positive integers).
By (1) - (2) we get $m^{2}-n^{2}=240$,
which is $(m+n)(m-n)=240... | 904 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, in a $3 \times 3$ square grid, fill in nine different natural numbers such that the product of the three numbers in each row and the product of the three numbers in each column are all equal (we denote this product by $P$).
(1) Prove that this arrangement of numbers is possible.
(2) Determine which of the six ... | Three, (1) Proof It is easy to prove the following filling method 1 As shown in the figure, fill in 1, 2, 3, 4, 5, 6, $8, 15, 20$ nine different natural numbers, then the product of the three numbers in each row and the product of the three numbers in each column are all equal to 120.
Therefore, the filling method req... | 120 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that a certain four-digit number is exactly equal to the fourth power of the sum of its digits, then this four-digit number is $\qquad$ | 9. 2401 .
The above text has been translated into English, maintaining the original text's line breaks and format. | 2401 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
In isosceles $\triangle A B C$, it is known that $A B=A C$ $=k B C$, where $k$ is a natural number greater than 1. Points $D$ and $E$ are on $A B$ and $A C$ respectively, and $D B=B C=C E$. $C D$ and $B E$ intersect at $O$.
Find the smallest natural number $k$ such that $\frac{O C}{B C}$ is a rational number. | II. Solution: As shown, connect $D E$, it is easy to know that $B C E D$ is an isosceles trapezoid. Also, from the given conditions, we have
$$
\angle 2=\angle 1=\angle 3 \text {, }
$$
thus $\triangle O B C \cos \triangle B C D$,
which means $O C \cdot C D=B C^{2}$.
Also, $\frac{C O}{O D}={ }_{D E}^{B C}=\frac{A B}{A ... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four, find all such three-digit numbers. If a two-digit number itself increases by 3, then, the sum of the digits of the resulting number is equal to one third of the sum of the digits of the original three-digit number.
Translate the above text into English, please retain the original text's line breaks and format, a... | Let the three-digit number $abc$ satisfy the given conditions. Then
(1) $c \geqslant 7$. In fact, if $c \leqslant b$, then the sum of the digits of $\overline{abc}+3$ will increase by 3, which does not meet the conditions.
(2) $b \neq 9$. This is because if $c \geqslant 7, b=9, a=9$, then
$$
\overline{abc}+3=1000+(c+3... | 432 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. The expression $\frac{\left(2^{4}+\frac{1}{4}\right)\left(4^{4}+\frac{1}{4}\right)\left(6^{4}+\frac{1}{4}\right)}{\left(1^{4}+\frac{1}{4}\right)\left(3^{4}+\frac{1}{4}\right)\left(5^{4}+\frac{1}{4}\right)}$ $\times \frac{\left(8^{4}+\frac{1}{4}\right)\left(10^{4}+\frac{1}{4}\right)}{\left(7^{4}+\frac{1}{4}\right)\le... | $221$ | 221 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The denominator of a reduced fraction is 30, find the sum of all such positive rational numbers less than 10.
untranslated text remains unchanged:
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。 | 1. Xiao Guang 10 points: If a fraction can be simplified to the form $\frac{30 n+r}{30}$, where $n$ and $r$ are integers satisfying $0 \leqslant n \leqslant 9$ and $0 \leqslant r<30$.
When and only when $r$ is coprime with 30, $\frac{30 n+r}{30}$ is a simplified fraction, thus, $r \in\{1,7,11,13,17,19,23,29\}$.
Since... | 400 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. A decimal positive integer of at least two digits, where each digit is smaller than the digit to its right, is called an "ascending number." How many ascending numbers are there?
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。 | 2. An "ascending number" must have at least two non-zero digits. It is a subset of the set $S=\{1,2,3,4,5,6,7, 8, 9\}$ consisting of two or more elements. The set determined by all elements is unique. Therefore, the number of ascending numbers is the number of subsets of $S$ consisting of two or more elements. Since $S... | 502 | Number Theory | proof | Yes | Yes | cn_contest | false |
5. $S$ is the set of rational numbers $r$ where $0<r<1$, and $r$ has a repeating decimal expansion of the form $0 . a b c a b c a b c \cdots$ $=0 . \dot{a} b \dot{c}, a, b, c$ are not necessarily distinct. Among the elements of $S$ that can be written as a fraction in simplest form, how many different numerators are th... | 5. Since $0 . \dot{a} b \dot{c}=\frac{a b c}{999}$, and $999=3^{8} \cdot 37$. If $a b c$ is neither divisible by 3 nor by 37, then the fraction is in its simplest form. By the principle of inclusion-exclusion, we have
$$
999-\left(\frac{999}{3}+\frac{999}{37}\right)+\frac{999}{3 \cdot 37}=648
$$
such numbers.
In addit... | 660 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. In $\{1000,1001, \cdots, 2000\}$, how many pairs of consecutive integers can be added without carrying over a digit? | 6 . Let $n$ have the decimal representation $1 a b c$. If one of $a, b$, $c$ is $5, 6, 7$ or 8, then adding $n$ and $n+1$ requires carrying 3. If $b=9, c \neq 9$ or $a=9, b$ and $c$ are not 9, then adding $n$ and $n+1$ requires carrying.
If $n$ is not as described above, then it must be of one of the following forms:
... | 156 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. In trapezoid $A B C D$, $A B \| C D$, and $A B$ $=92, B C=50, C D=19, A D=70$. A circle with center $P$ on $A B$ is tangent to sides $B C$ and $A D$. If $A P=\frac{m}{n}$, where $m, n$ are coprime positive integers. Find $m+n_{0} \quad$ | 9. As shown in the figure, extend $A D$ and $B C$ to intersect at point $Q$. Since the distances from $P$ to $A Q$ and $B Q$ are equal, $P$ lies on the angle bisector of $\angle A Q B$, thus we have
$$
\frac{A P}{B P}=\frac{A Q}{B Q}
$$
Since $A B \parallel C D$, we have
$$
\frac{A Q}{B Q}=\frac{A D}{B C}=\frac{7}{5}.... | 164 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. Lines $l_{1}, l_{2}$ both pass through the origin, and in the first quadrant, they form angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ with the positive $x$-axis, respectively. Define $R(l)$ as the line obtained by reflecting $l$ about $l_{1}$ and then reflecting the result about $l_{2}$. Let $R^{(1)}(l)=R(l), R^{... | A line passing through the origin and forming angles $\theta_{0}, \theta, \lambda$ with the positive x-axis. The symmetric line $\lambda^{\prime}$ about $\lambda$ forms an angle with the x-axis of $\theta_{0} + (\theta_{0} - \theta) = 2 \theta_{0} - \theta$.
Therefore, the symmetric line $\lambda_{1}$ about $l_{1}$ pa... | 945 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. In a game, two players take turns to "eat squares" from a $5 \times 7$ grid chessboard. To "eat a square," a player selects an uneaten square and moves the piece to that square, then all the squares in the quadrant formed (along the left edge of the square upwards, and along the bottom edge of the square to the rig... | 12. In this game, the height of the table formed by the un-eaten squares from left to right is not constant (as shown in the right figure). It is easy to prove that this situation is both sufficient and necessary. Moreover, each shape can be completely described by a 12-step zigzag line. This zigzag line starts from th... | 792 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
13. In $\triangle A B C$, it is known that $A B=9, B C: C A$ $=40: 41$. Find the maximum value of the area of $\triangle A B C$. | 13. Let $AB = c$, $AC = br$, $BC = ar$, $(a < b)$. Establish a Cartesian coordinate system with point $A$ as the origin and the line $AB$ as the x-axis (as shown in the figure), set $A(0,0)$, $B(c, 0)$, $C(x, y)$.
From $\frac{BC}{AC} = \frac{a}{b}$, we get
$$
\frac{\sqrt{(x-c)^{2} + y^{2}}}{\sqrt{x^{2} + y^{2}}} = \fra... | 820 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
14. In $\triangle A B C$, points $A^{\prime}, B^{\prime}, C^{\prime}$ are on sides $B C, A C, A B$ respectively, $A A^{\prime}, B B^{\prime}, C C^{\prime}$ intersect at point $O$, and $\frac{A O}{O A^{\prime}}+\frac{B O}{O B^{\prime}}+\frac{C O}{O C^{\prime}}=92$. Find the value of $\frac{A O}{O A^{\prime}} \cdot \frac... | 14. As shown in the figure, since $\triangle A O B$ and $\triangle A^{\prime} O B$ have the same height, therefore,
$$
\begin{aligned}
\frac{A O}{O A^{\prime}} & =\frac{S_{\triangle A O B}}{S_{\triangle A^{\prime} O B}} \\
& =\frac{S_{\triangle C O A}}{S_{\triangle C O A^{\prime}}}+S_{\triangle C O A} \\
& =\frac{S_{\t... | 94 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. If there exists a positive integer $m$ such that $m!$ ends with exactly $n$ zeros, then the positive integer $n$ is called a "factorial tail number." How many non-"factorial tail number" positive integers are there less than 1992? | 15. Let $f(m)$ be the number of trailing zeros in the number $m!$. Clearly, $f(m)$ is a non-decreasing function of $m$, and when $m$ is a multiple of 5, we have
$$
\begin{aligned}
f(m) & =f(m+1)=f(m+2) \\
& =f(m+3)=f(m+4) \\
& 4 \times 1991=7964. \text{ Using (2), it is easy to get }
$$
$$
f(7965)=1988, f(7975)=1991 .
... | 396 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. A lottery ticket has 50 spaces arranged in sequence, and each participant should fill in the numbers 1 to 50 without repetition on their ticket, with the order determined by themselves. The host also fills in these 50 numbers on a ticket as the answer key. If a participant's number in any space matches the number in... | 4. 26 tickets. To ensure a win, for example, 26 tickets can be filled out as follows:
$$
\begin{array}{l}
1,2,3, \cdots, 25,26,27, \cdots, 50 \\
2,3,4, \cdots, 26,1,27, \cdots, 50 \\
3,4,5, \cdots, 1,2,27, \cdots, 50 \\
\cdots \cdots
\end{array}
$$
$$
\begin{array}{l}
25,26,1, \cdots, 23,24,27, \cdots, 50 \\
26,1,2, \c... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. On the planet, there are 100 mutually hostile countries. To maintain peace, they decide to form several alliances, with the requirement that each alliance includes no more than 50 countries, and any two countries must be in at least one alliance. How many alliances are needed at a minimum?
(a) What is the minimum nu... | 6. 6 alliances. Each country should join no less than 3 alliances, so the number of alliances is at least 6.
(a) Only by dividing 100 countries into 4 groups, each with 25 countries, and then through combinations, we get $C{ }_{4}^{2}=6$ alliances.
(b) Only by dividing 100 countries into 10 groups $a_{1}, a_{2}, a_{8},... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For every $A \subset S$, let
$$
S_{\mathrm{A}}=\left\{\begin{array}{ll}
(-)^{\mid \mathrm{A}} \mid \sum_{\mathbf{a} \in \mathrm{A}} a, & A \neq \varnothing, \\
0, & A=\varnothing .
\end{array}\right.
$$
Find $\sum_{\mathrm{A} \subset \mathrm{S}} S_{\mathrm{A}}$. | Solve: There are $2^{n-1}$ sets $A$ satisfying $n \in A \subset S$, and there are also $2^{n-1}$ sets $B$ satisfying $n \notin B \subset S$. Note that $B$ covers all subsets of $\{1,2, \cdots, n-1\}$, and each $A$ satisfying $n \in A \subset S$ can be obtained by adding the element $n$ to some $B$. For each $B \subset ... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Question 5 Let the set $S=\{1,2, \cdots, 1000\}$. Now for any non-empty subset $A$ of $S$, let $\alpha_{\mathrm{A}}$ denote the sum of the largest and smallest numbers in $A$, then, the arithmetic mean of all such $\alpha_{\Delta}$ is $\qquad$ | The answer is 1001.
Let $S=\{1,2, \cdots, n\}, n \in N$. Let $m_{A}, M_{A}$ represent the minimum and maximum elements of a non-empty subset $A$ of $S$, respectively. Then the arithmetic mean of all $\alpha_{A}$ is
$$
\frac{1}{2^{n}-1} \sum_{\substack{A \subset S \\ A \neq \varnothing}}\left(m_{A}+M_{A}\right).
$$
Sin... | 1001 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
19. For $0<x<\frac{\pi}{2}$, let $\frac{\operatorname{ctg}^{3} x}{\operatorname{ctg} 3 x}$ take all real numbers except those in the open interval $(a, b)$. Find $a+b$.
---
Note: The function $\operatorname{ctg} x$ is the cotangent function, often denoted as $\cot x$ in English. | 19. Notice that $\operatorname{ctg} 3 x=\frac{\operatorname{ctg}^{3} x-3 \operatorname{ctg} x}{3 \operatorname{ctg}^{2} x-1}$. Let $y=\frac{\operatorname{ctg}^{3} x}{\operatorname{ctg} 3 x}, t=\operatorname{ctg}^{2} x$, then
$$
3 t^{2}-(y+1)t+3 y=0 \text {. }
$$
Thus this equation has real roots, hence
$$
\Delta=(y+1)... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
21. Let when $0 \leqslant x \leqslant 1$, there exists a positive number $q$ such that $\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{\mathrm{t}}}{q}$ holds, find the smallest positive number $t$ that makes the above inequality true. For this smallest $t$ value, what is the minimum value of $q$ that makes the above inequa... | 21. Since $y=\sqrt{x}$ is strictly concave, for $0<x<1$, we have
$$
q \geqslant 2-(\sqrt{1+x}+\sqrt{1-x}).
$$
After two constant transformations, we get
$$
\begin{array}{l}
q \geqslant x^{1-2}\left(1+\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) \\
\cdot\left(1+\sqrt{1-x^{2}}\right).
\end{array}
$$
If $t<2$, then as $x$ app... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Test Question $A-1$. How many evil numbers have a 10-adic representation that is exactly 0,1 alternating and ends with 1? | Let $P_{\mathrm{k}}=1010 \cdots 101$ ( $k$ zeros) be a prime number in decimal notation, $k \geqslant 1$. Clearly, $P_{1}=101$ is a prime number. When $k>2$, if $k$ is odd, then 101 clearly divides $P_{\times 3}$. If $k$ is even, then
$$
\begin{array}{l}
11 P_{k}=11 \cdots 1(2 k+2 \text { ones}) \stackrel{\text { denot... | 101 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Test $B-5$. Let $T$ be the inscribed trapezoid $ABCD$ (counterclockwise) in the unit circle $O$, $ABCDI, AB=s_{1}$, $CD=s_{2}, OE=d, E$ is the intersection point of the heights. When $d \neq 0$, determine the minimum upper bound of $\frac{s_{1}-s_{2}}{d}$. If the minimum upper bound can be achieved, determine all such ... | Solve: As shown in the figure, establish a rectangular coordinate system with $O$ as the origin, $AB \perp x$-axis, and the coordinates of $E$ are $(d, 0)$. The equation of $BD$ can be set as $x - d = k y$, where $k^{-1}$ is the slope of line $BD$. Since $B$ and $D$ are on the unit circle, we have
$$
(d + k y)^{2} + y^... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. Try to find the four-digit number $\overline{x x y y}$, such that it is a perfect square. | Solution: Since the required square number $\overline{x x y y}=11 \cdot(100 x + y)$, therefore, $100 x+y$ can be divided by 11, i.e., $11 | (x + y)$. Noting that $x$ and $y$ are both digits, thus $x+y=11$. After verification, only $x=7, y=4$ satisfies the condition, hence the required four-digit square number is 7744. | 7744 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. Five monkeys found Wang Yiguan's peaches. How can they be fairly divided when one monkey secretly gets up, eats one peach, and the remaining peaches are exactly divided into five equal parts, after which it hides its share and goes back to sleep. The second monkey gets up, also eats one peach, and the remain... | This problem has many solutions, for example, it can first be transformed into an indeterminate equation, and then its positive integer solutions can be sought. However, there is a simpler solution: Imagine that when Sun Wukong, the Great Sage Equal to Heaven, found his subordinates facing such a difficult problem, he ... | 3121 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. The numbers 1447, 1005, and 1231 have something in common. That is, each number is a four-digit number starting with $\mathrm{i}$, and in each four-digit number, exactly two digits are the same. How many such four-digit numbers are there? (1st AIME) | Solve for the four-digit numbers that must contain one 1 or two 1s.
(i) The case with two $\hat{\mathrm{i}} 1$s
Choose two numbers from the remaining 9 digits (excluding 1), which has $C_{9}^{2}$ ways. Then, form any permutation of a three-digit number with these two numbers and one 1, which has $P_{3}^{3}$ ways. Thus,... | 432 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. Use 6 white beads, 8 blue beads, and 1 red bead to form a string. How many different ways can this be done?
保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a repetition of the instruction and should not be part of the translation. Here is the corrected version:
Example 2. Use 6 white beads, 8 blue beads... | This is a circular permutation problem. If we fix the red bead, it turns into a linear permutation problem. By the principle of the whole, apart from the red bead, the remaining 14 beads have $N_{1}=\frac{P_{14}^{14}}{P_{6}^{6} \cdot P_{8}^{8}}=3003$ ways to be strung.
Below, we discuss two cases based on whether the ... | 1519 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
List 3. From $1,2,3, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, such that $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways are there to select the numbers that meet the above requirements? (1389, National High School League) | Solution: Obviously, $a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+(14-$ $\left.a_{3}\right)=14$, where $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3,14-$ $a_{3} \geqslant 0$.
Transform the above equation to
$$
\left(a_{1}-1\right)+\left(a_{2}-a_{1}-3\right)+\left(a_{3}-a_{2}-3\right)+... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
List 5. Arrange 8 cards $A A B B C D E F$ in a row, the number of arrangements where the same letter cards are allowed to be adjacent is how many?
Arrange 8 cards $A A B B C D E F$ in a row, with the same letter cards allowed to be adjacent. How many such arrangements are there? | Let the number of ways to arrange 8 cards in a row be $S$, then $|S|=\frac{P_{8}^{8}}{P_{2}^{2} \times \bar{P}_{2}^{2}}$; let the situation where the two cards with the letter $\Lambda$ are adjacent be $A_{1}$, and the situation where the two cards with the letter $B$ are adjacent be $A_{2}$, then $\left|A_{1}\right|:=... | 5760 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. A person wrote 6 letters to 6 different people and prepared 6 envelopes with the recipients' addresses written on them. How many ways are there to place the letters into the envelopes so that no letter matches the recipient on the envelope? (Polish Competition Question) | Let's denote the set of all ways to place 6 letters into 5 different envelopes (i, $2, \cdots, 6$) such that one letter is left out, and the remaining letters are placed into the 5 envelopes in any order, as $\Lambda_{i}$. Then, $\left|A_{i}\right|=P_{5}^{5}=5!$.
Similarly,
$$
\begin{array}{l}
\left|A_{i} \cap A_{j}\ri... | 265 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. There are exactly 35 consecutive natural numbers whose integer parts of the arithmetic square roots are the same. Then, what is this identical integer?
---
The translation maintains the original text's format and line breaks as requested. | Analyzing to determine the integer part of a real number $a$ actually means finding two adjacent integers $n$ and $n+1$, such that $n \leqslant a < n+1$, at this point the integer part of $a$ is $n$. Using the property $2^{\circ}$ above, the answer to this question is 17. | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. The number of natural numbers $n$ that make $n^{2}-19 n+91$ a perfect square is?
Will the above text be translated into English, please keep the original text's line breaks and format, and output the translation result directly. | Notice that $n^{2}-19 n+91=(n-9)^{2}+(10-$ $n$ ). When $n>10$,
$$
(n-10)^{2}<(n-9)^{2}+(10-n)<(n-9)^{2} .
$$
The integers between two consecutive perfect squares cannot be perfect squares, therefore the natural numbers $\pi$ that make $n^{2}-10 n+91$ a perfect square can only be from the ten numbers $1,2,3, \cdots, 9,... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The graph of a quadratic function passes through $(1,0),(5, 0)$, the axis of symmetry is parallel to the $y$-axis, but does not pass through points above the line $y=2x$. Then the product of the maximum and minimum values of the coordinates of its vertex is $\qquad$ . | 1. 4.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Toss a coin, if it lands heads, point $P$ moves +1 on the number line, if it lands tails, it does not move. The coin is tossed no more than 12 times, and if point $P$ reaches coordinate +10, no more tossing occurs. Then the total number of different ways for point $P$ to reach coordinate +10 is . | 3. 66.
Obviously, the last coin toss must be heads.
(i) To get 10 points with 10 heads, there is 1 way.
(ii) For 11 tosses, since the last one is heads, the first 10 must include 1 tails, which has $C_{10}^{1}=10$ ways.
(iii) For 12 tosses, since the last one is heads, the first 11 must include 2 tails, which has $C_{... | 66 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $f(x)=(\sin x+4 \sin \theta+4)^{2}+(\cos x$ $-5 \cos \theta)^{2}$, the minimum value of $f(x)$ is $g(\theta)$. Then the maximum value of $g(\theta)$ is | 6. 49.
For $f(x)$, expanding and rearranging yields
$$
\begin{aligned}
f(x) & =8(1+\sin \theta) \sin x-10 \cos \theta \cdot \cos x \\
& -9 \sin ^{2} \theta+32 \sin \theta+42 \\
& =\sqrt{64(1+\sin \theta)^{2}+100 \cos ^{2} \theta} \\
& \cdot \sin (x+\varphi)-9 \sin ^{2} \theta+32 \sin \theta \\
& +42 . \\
g(\theta) & =... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
In the letter摘登 of the sixth issue of "Middle School Mathematics" in 1992, Comrade Yang Xuezhi from Fuzhou No. 24 High School mentioned a conjecture by a senior high school student: "Given $\triangle P_{1} P_{2} P_{3}$ and a point $P$ inside it. The lines $P_{1} P, P_{2} P, P_{3} P$ intersect the opposite sides at $Q_{... | First, it should be noted that the above "conjecture" has long been proposed as a correct proposition and has been proven. Now, we introduce a proof method as follows (figure omitted).
Proof: Let the areas of $\triangle P_{2} P P_{3}, \triangle P_{3} P P_{1}, \triangle P_{1} P P_{2}$ be $S_{1}, S_{2}, S_{3}$, respecti... | 6 | Inequalities | proof | Yes | Yes | cn_contest | false |
Five, 10 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 most purchased book? Explain your reasoning.
(Na Chengzhang provided the question) | Solution 1: Let's assume that each person buys at most one copy of the same book. Thus, by condition (1), 10 people buy a total of 30 books.
Let the minimum number of people who buy the most popular book be $n$. Assume $A$ buys books 甲, 乙, and 丙.
Clearly, $n > 3$. Because if $n \leqslant 3$, then at most two more peo... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
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