problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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2. Mark 10 points on a circle, how many different polygons can be formed using these points (all or part)? (Polygons are considered the same only if their vertices are exactly the same).
| $$
\begin{array}{l}
\left.+C_{10}^{2}+\cdots+C_{10}^{0}+C_{10}^{10}\right]-\left[C_{10}^{0}+C_{10}^{\overrightarrow{1}}\right. \\
\left.+C_{10}^{2}\right]=(1+1)^{10}-(1+10+45)=968 \text{ different convex polygons.} \\
\end{array}
$$
For $3 \leqslant k \leqslant 10$, every selection of $k$ points can form a convex $k$-... | 968 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3 . Let $n$ be a positive integer, $d$ be a digit among the 10 decimal digits, and suppose $\frac{n}{810}=0 . d 25 d 25 d 25 \cdots$, find $n$.
| $$
\begin{array}{c}
3 . \because \frac{n}{810}=0 . d 25 d 25 d 25 \cdots, \\
\therefore 1000 \cdot \frac{n}{810}=d 25 . d 25 d 25 \cdots \\
\therefore \frac{999 n}{810}=1000 \cdot \frac{n}{810}-\frac{n}{810} \\
=d 25=100 d+25, \\
\therefore \quad 999 n=810(100 d+25),
\end{array}
$$
And $(750,37)=1$, so $37 \mid (4d+1)... | 750 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. If $a<b<c<d<e$ are consecutive positive integers, $b+c+d$ is a perfect square, and $a+b+c+d+e$ is a perfect cube, what is the minimum value of $c$? | $$
\begin{aligned}
4 . & b+c+d=3 c \\
& a+b+c+a+e=5 c
\end{aligned}
$$
and $b+c+d=m^{2}, a+b+c+d+e=n^{3}$
$$
\begin{aligned}
\therefore \quad 3 c & =m^{2}, \\
5 c & =n^{3} .
\end{aligned}
$$
$$
\begin{array}{l}
\therefore 3^{3} \mid c_{0} \text { and } 5\left|n_{0} \quad \therefore 5^{2}\right| c, \quad \therefore 25 ... | 675 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Two Skaters
Allie and Bllie are located
at points $A$ and $B$ on
a smooth ice surface, with $A$ and $B$
being 100 meters apart. If Allie starts from $A$
and skates at 8 meters/second
along a line that forms a $60^{\circ}$
angle with $A B$, while Bllie starts from $B$ at 7 meters/second
along the shortest line to me... | 6. Let the two people meet at point $C$ after $t$ seconds, then the sides of $\triangle A B C$ are $A B=100, A C=8 t, B C=7 t$. By the cosine rule, we get
$$
\begin{aligned}
(7 t)^{2}= & 100^{2}+(8 t)^{2}-2(100)(8 t) \cos 60^{\circ} \\
\therefore \quad 0 & =3 t^{2}-160 t+2000 \\
& =(3 t-100)(t-20) .
\end{aligned}
$$
$\... | 160 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. If the integer $k$ is added to $36,300,596$, respectively, the results are the squares of three consecutive terms in an arithmetic sequence, find the value of $k$. | 7. Let $\left\{\begin{array}{l}36+k=(n-d)^{2} , \\ 300+k=n^{2} \\ 596+k=(n+d)^{2}\end{array}\right.$
(1)
From (3) - (1) we get $4 n d=560$,
$\therefore n d=140$.
From (2) $\times 2-(1)-$ (3) we get $2 d^{2}=32$,
$\therefore d= \pm 4$, then $n= \pm 35$.
From (2), $k=n^{2}-300=35^{2}-300=925$. | 925 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Euler's conjecture was refuted by American mathematicians in 1960, who proved that there exists a positive integer $n$ such that $133^{5}+110^{5}+84^{5}+27^{5}=n^{5}$. Find $n$ when it is satisfied. | 9. Clearly $n \geqslant 134$. Now we find the upper bound of $n$.
$$
\begin{aligned}
\because \quad n^{5}= & 133^{5}+110^{5}+84^{5}+27^{5}<133^{5} \\
& +110^{5}+(27+84)^{5}<3(133)^{5} \\
& <\frac{3125}{1024}(133)^{5}=\left(\frac{5}{4}\right)^{5}(133)^{5}, \\
\therefore \quad & n<\frac{5}{4}(133), \text { i.e., } n \leq... | 144 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. Given an integer array consisting of 121 integers, where each number is between 1 and 1000 (repetition is allowed), this array has a unique mode (i.e., the integer that appears most frequently). Let $D$ be the difference between this mode and the arithmetic mean of the array. When $D$ reaches its maximum value, wha... | 11. Let $M$ and $m$ be the "mode" and "arithmetic mean" respectively. Without loss of generality, assume $M \geqslant m$. To maximize $D$, then $M=1000$ (otherwise, if $M=1000-k$, then increasing $M$ by $k$ while $m$ increases by no more than $k$, $D$ is obviously non-decreasing). In the case of $M=1000$, we must make ... | 947 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. As shown in the figure, $ABCD$ is a tetrahedron, $AB=41$, $AC=7$, $AD=18$, $BC=36$, $BD=27$, $CD=13$. Let $d$ be the distance between the midpoints of $AB$ and $CD$. Find the value of $d^{2}$. | 12. By the median formula, we get $m_{s}^{2}=\frac{1}{4}\left(2 b^{2}+2 c^{2}\right.$ $\left.-a^{2}\right)$, so $P C^{2}=\frac{1}{4}\left[2(A C)^{2}+2(B C)^{2}\right.$ $\left.-(A B)^{2}\right]=\frac{1009}{4}$.
$$
\begin{array}{l}
P D^{2}\left.=\frac{1}{4}\left[2(A D)^{2}+2(B D)^{2}\right]-(A B)^{2}\right] \\
=\frac{42... | 137 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5 . There are two piles of stones. If 100 stones are taken from the first pile and placed in the second pile, then the number of stones in the second pile will be twice that in the first pile. Conversely, if some stones are taken from the second pile and placed in the first pile, then the number of stones in the first ... | Let the number of stones in the first and second piles be denoted as $x, y$. Let $z$ be the number of stones taken from the second pile and placed into the first pile. Thus, we obtain the equations
$$
\begin{array}{l}
2(x-100)=y+100, \\
x+z=6(y-z) .
\end{array}
$$
From (1), we get $y=2 x-300$, substituting this into (... | 170 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. The decimal representation of the natural number $A$ is $\overline{a_{n} a_{n}-\overline{1} \cdots a_{1} a_{0}}$, let
$f(A)=2^{\mathrm{n}} a_{0}+2^{\mathrm{n}-1} a_{1}+\cdots+2 a_{\mathrm{n}-1}^{\mathrm{n}}+a_{\mathrm{n}}$ denote $A_{1}=f(A), A_{\mathrm{i}+\mathrm{i}}=f\left(A_{\mathrm{i}}\right),(i=1,2, \cd... | Prove $f(A)=2^{n} a_{0}+2^{n-1} a_{1}^{\gamma}+\cdots+2 a_{n-1}$
$$
\begin{array}{l}
+a_{n} \leqslant\left(2^{n}+2^{n-1}+\cdots+2+1\right)_{9} \\
=9\left(2^{n+1}-1\right) .
\end{array}
$$
While $A \geqslant 10^{\text { }} a_{\text {n }}$
Next, we prove that when $n \geqslant 2$, $A>f(A)$.
By $a_{n} \cdot 10^{n} \geqsl... | 19 | Number Theory | proof | Yes | Yes | cn_contest | false |
To 9. On a plane, there is a fixed point $P$, consider all possible equilateral triangles $ABC$, where $AP=3, BP=2$. What is the maximum length of $CP$? (1961 Autumn Competition) | Solve for Ling and Hui
$$
\begin{array}{l}
\angle A P B=\alpha . \\
\angle B A P=\beta,
\end{array}
$$
Given $A B^{2}=3^{2}+2^{2}$
$$
\begin{aligned}
- & 12 \cos \alpha, \\
\cos \beta & =\frac{3^{2}+A B^{2}-2^{2}}{6 A B}, \sin \beta=\frac{2 \sin \alpha}{A B} .
\end{aligned}
$$
From this, we get
$$
\begin{aligned}
\co... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Given $\triangle A B C$, extend the three sides by 1, 2, 3 times respectively, to get $\triangle A^{\prime} B^{\prime} C^{\prime}$. Ask how many times the area of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is compared to the area of $\triangle A B C$. | Theorem: Let the area of $\triangle ABC$ be $S$. Extend the sides $AB, BC, CA$ of $\triangle ABC$ such that $BB'=\lambda_{1} AB$, $CC'=\lambda_{2} BC$, $AA'=\lambda_{3} CA$, and let $\triangle A' B' C'$ be the resulting triangle with area $S'$. Then,
$$
S'=1+\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{1} \lambda_{2}+\... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
17. In the plane, 7 points are given, connect them with some line segments so that
(1) among any three points, at least two are connected,
(2) the number of line segments is minimized.
How many line segments are there? Provide such a graph. | 17. The figure below shows that 9 line segments are sufficient.
Now prove that at least 9 line segments are needed.
If point $A$ is not an endpoint of any line segment, then due to (1), the other 6 points must connect at least $C_{6}^{2}>9$ line segments.
If point $A$ is the endpoint of only 1 line segment, then due... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Arrange the positive rational numbers in the following sequence,
$$
\begin{array}{l}
\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \\
\frac{2}{3}, \frac{1}{4}, \cdots,
\end{array}
$$
Then the position number of the number $\frac{1989}{1949}$ is $\qquad$ | 3. The number's position in the original
sequence is $(1+2+\cdots+3937)+$ $1949=7753902$, | 7753902 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. A hotel has 90 vacant rooms, each with a unique key. 100 guests arrive, and keys need to be distributed so that any 90 of them can stay in the 90 rooms, with each person getting one room (assuming there is no limit to the number of keys that can be issued for each room or the number of keys each person can receive).... | 8. At least 990 keys should be prepared. | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Six, there are 6000 points inside a circle, and no three points are collinear:
(1) Can this circle be divided into 2000 parts, each containing exactly three points? How to divide it?
(2) If the three points in each part satisfy: the distance between any two points is an integer and does not exceed 9, then using the thr... | (1) 6000 points inside a circle can determine $C_{6000}^{2}$ lines. Since $C_{6000}^{2}$ is a finite number, there must exist a tangent line to the circle that is not parallel to any of the $C_{5000}^{2}$ lines, denoted as $l$. Moving $l$ parallel within the circle, it is clear that the 6000 points will be crossed (if ... | 22 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Find the integer solution to the equation
$$
\left(1+\frac{1}{m}\right)^{m+1}=\left(1+\frac{1}{1988}\right)^{1088}
$$ | Answer $m=-1989$. | -1989 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. All vertices of a broken line lie on the faces of a cube with an edge length of 2, and each segment of the broken line is 3 units long. This broken line connects two farthest vertices of the cube. How many segments does such a broken line have at least?
untranslated part:
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:... | Solve: Consider the cube circumscribing a sphere with center $A$ and radius 3, intersecting the cube's faces at three arcs: $K \hat{L}, \widehat{L N}, \widehat{N K}$ (Figure 3). The points $K, L, N$ on the edges bisect these three edges. In fact, $A D_{1}=\sqrt{8}$, so $L D_{1}=\sqrt{A} L^{\overline{2}-\overline{D_{1}^... | 6 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
6. On the blackboard, there are numbers 1 and 2. It is stipulated that new numbers can be written according to the following method: If there are numbers $a$ and $b$ on the blackboard, then the number $a b + a + b$ can be written. Using this method, can the following numbers be obtained:
(a) Number 131213
(b) Number 12... | Let the new number $ab + a + b$ be $c$. This means $c + 1 = ab + a + b + 1 = (a + 1)(b + 1)$. This implies that if the number written on the blackboard is replaced by a number that is 1 greater, then each new number will be the product of two existing numbers. Starting with the numbers 2 and 3, after several multiplica... | 13121 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. A country has 21 cities, and some airlines can implement air transportation between these cities. Each airline connects pairs of cities with non-stop flights (and several airlines can operate flights between the same two cities at the same time). Every two cities are connected by at least one non-stop flight. How ma... | To make this country form an aviation network that meets the conditions required by the problem, there must be at least 21 airlines, because the total number of non-stop routes is no less than $20+19+\cdots+3+2+1=210$, and each airline provides $4+3+2+1=10$ non-stop routes. Figure 5 is an example of a service route map... | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. (Wenzhou Junior High School Math Competition Question, 1987) Let the natural number $n$ have the following property: from 1, 2, ..., $n$, any 51 different numbers chosen will definitely have two numbers whose sum is 101. The largest such $n$ is $\qquad$ | Consider $\{1,2, \cdots, n\}$ as the vertex set. When the sum of two numbers is 101, connect the corresponding two vertices to form a graph $G$. Clearly, the original problem is equivalent to finding the largest $n$ such that any selection of 51 vertices in $G$ must include two adjacent vertices.
i) When $51 \leqslant ... | 100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. (85 Provincial Six Autonomous Regions High School Mathematics Joint Competition Question) A football invitational tournament involves sixteen cities, each city sending Team A and Team B. According to the competition rules, each pair of teams plays at most one match, and teams from the same city do not play a... | Prove that if two teams have played against each other, then an edge is connected between the corresponding two vertices to form graph $G$. Let the team from City A be $v^{*}$, then the original problem is equivalent to: find the degree of the vertex corresponding to the other team from City A.
Obviously, $\max d\left(... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. (Hefei Mathematical Competition Question in 1983) A new station is opened, and several bus routes are planned to serve the community. Their wishes are: (1) to open as many routes as possible; (2) each route must have at least one bus stop; (3) ensure that each bus stop is served by at least two different rou... | Let $S$ be the number of lines that can be opened, and consider the lines as vertices to form a graph $K_{\mathrm{s}}$. Label the 1983 stations as $A_{1}, A_{2}, \cdots, A_{19}$. If two lines have a common station, color the edge between the corresponding two vertices with color $C$.
From (2), every edge of $K_{\mathrm... | 63 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 10. (19th All-Soviet Union Middle School Olympiad Problem) A quadratic trinomial $a x^{2}+b x+c$, where $a>100$. How many different integer points can there be at which the absolute value of its value does not exceed 50? | Assume there are three different integer points satisfying the given condition, then there must be two points located on the same side of the axis of symmetry $x=-\frac{b}{2 a}$ of the quadratic function $y=a x^{2} +b x+c$, or one point is located on this axis of symmetry. Without loss of generality, assume
$-\frac{b}{... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Column 1. A tour group is selecting visit locations from $A, B, C, D, E$ under the following constraints:
(1) If visiting $A$, must also visit $B$,
(2) At least one of $D, E$ must be visited,
(3) Only one of $B, C$ can be visited,
(4) Both $C, D$ must be visited or neither,
(5) If visiting $E$, both $A, D$ must be visi... | 堔 We adopt the貑 deduction method.
(1)
(4)
(2)
$1^{\circ}$ If going to $A \Rightarrow$ must go to $B \Rightarrow$ do not go to $C \Rightarrow$ do not go to $D \Rightarrow$ must not go to $A$. Going to $A, D$. This leads to the same contradiction as $1^{\circ}$, so do not go to $B$.
(5)
$3^{\circ}$ If going to $E \Righta... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. Find the largest integer $A$, such that for any permutation of all natural numbers from 1 to 100, there are 10 consecutive numbers whose sum is greater than or equal to $A$.
[22nd Polish Mathematical Olympiad] | This problem, although it is to find the maximum integer $A$, is actually an existence problem, i.e., there exists a maximum integer $A$ such that the sum of 10 consecutive numbers is not less than $A$.
Solution: Let $T=\left(a_{1}, a_{2}, \cdots, a_{100}\right)$ be a permutation of the natural numbers from 1 to 100.
... | 505 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
In 1988, the Chinese Junior High School Mathematics League had the following problem: If natural numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5}$, what is the maximum value of $x_{5}$? | This problem is novel and unique, and some students find it difficult to start. Below, I will first discuss the solution to this problem.
Algorithm 1: Given the symmetry of the equation, we can assume without loss of generality that \( x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5} \), and then ... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. The vertex of the parabola is at the origin, and the focus F is the center of the circle given by $x^{2}+y^{2}-4 x=0$. A line passing through point $F$ with a slope of 2 intersects the parabola at points $A$ and $D$, and intersects the circle at points $B$ and $C$.
Find $|A B|+|C D|$. | $$
\begin{array}{l}
\text { Given } F \text { as the pole, and the positive direction of the } x \text { axis as the polar axis, we establish the polar coordinate system. } \\
\left.\begin{array}{l}
\left.\therefore \text { The parabola equation is: } \rho=\frac{p}{1-\cos \theta}\right\} \Rightarrow \rho= \\
\because F... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
, Example 6. Find a four-digit number, which when multiplied by 4 is exactly equal to its reverse. | Assuming the four-digit number $x y z t(x \neq 0$, $1 \neq 0, y, z$ are integers) exists, at this point
$$
\text { 4. } x y z t=t z y x \text {. }
$$
Since $4 \cdot x y z t \geqslant 4 x \cdot 1000$,
$\overline{t z y x}7$. Thus, $t=8$ or $t=9$.
But $t=9$ is impossible, because if $t=9$, then the left-hand side of (*)... | 2178 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. (IMO-19-2) In a sequence of real numbers, the sum of any 7 consecutive terms is negative, while the sum of any 11 consecutive terms is positive. How many terms can such a sequence have at most?
保留源文本的换行和格式，翻译结果如下：
Example 7. (IMO-19-2) In a sequence of real numbers, the sum of any 7 consecutive terms is ne... | Given that a junior high school student is not clear about the concept of "sequence", we can simply change the term "sequence" to "numbers arranged according to a certain rule".
Let the sequence be $a_{1}, a_{2}, \cdots a_{2}, \cdots, a_{0}$, i.e., $n \geq 17$. According to the problem:
$$
\begin{array}{l}
a_{k}+a_{k+... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $m, n$ be positive integers, prove that there exists a constant $\alpha>1$ independent of $m, n$, such that when $\frac{m}{n}<\sqrt{7}$, we have $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\alpha}{n^{2}}$. What is the largest value of $\alpha$? | Solve $\alpha=3$.
$$
\begin{array}{l}
\quad 7-\frac{m^{2}}{n^{2}} \geqslant \frac{a}{n^{2}} \Leftrightarrow 7 n^{2}-m^{2} \geqslant \alpha_{0} \\
\because \quad m+7 \equiv m(\bmod 7) \\
\therefore \quad(m+7)^{2} \equiv m^{2}(\bmod 7)
\end{array}
$$
$\therefore\left\{m^{2}(\bmod 7)\right\}$ is a purely periodic sequence... | 3 | Inequalities | proof | Yes | Yes | cn_contest | false |
Six The terms of the sequence $x_{1}, x_{2}, x_{3}, \cdots$ are non-zero real numbers, and satisfy $x_{\mathrm{a}+2}=\frac{x_{\square} \cdot x_{\mathrm{n}+1}}{2 x_{\mathrm{a}}-x_{\mathrm{a}+1}} \quad(n=1,2, \cdots)$,
(1) Find the necessary and sufficient conditions for $x_{1}$ and $x_{2}$ such that all $x_{0}$ are inte... | Six, Solution (1) Hint: By mathematical induction, it is easy to prove
$$
x_{\mathrm{n}}=\frac{x_{1} x_{2}}{(n-1) x_{1}-(n-2) x_{2}},
$$
i.e., $x_{\mathrm{n}}=\frac{x_{1} x_{2}}{n\left(x_{1}-x_{2}\right)+\left(2 x_{2}-x_{1}\right)}$.
When $x_{1} \neq x_{2}$ and $n$ is sufficiently large, there must be $x_{0}<1$. When ... | 27 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given a set $A$ composed of functions, there are 10 odd functions, 8 increasing functions defined on the interval $(-\infty,+\infty)$, and 12 functions whose graphs pass through the origin. Then the set $A$ can have at most $\qquad$ elements. | 5. 14.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Insert “+” or “- -” between $1,2,3, \cdots, 1989$, what is the smallest non-negative number that the sum can achieve? | Except for 995, the numbers $1,2,3, \cdots, 1989$ can all be divided into 994 pairs: $(1,1989),(2,1988)$, $\cdots$, (994,996). Since the parity of the two numbers in each pair is the same, the result of the operation, regardless of how “+” or “-” signs are placed before each pair, can only be an even number. And 995 is... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Does a ten-digit number exist that is divisible by 11 and has all different digits on each of its positions?
翻译完成，保留了原文的换行和格式。 | Solve Write a three-digit number that can be divided by 11 and has different digits on each place, such as 275, 396, 418. Using these three numbers, it is not difficult to construct a ten-digit number that can be divided by 11. For example:
$$
\begin{array}{l}
2753964180 \\
=275 \cdot 10^{7}+396 \cdot 10^{4}+418 \cdot ... | 1427385960 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
*8. In decimal, find the smallest natural number: its square number starts with 19 and ends with 89. | The last digit of the required number can only be 3 or 7. We write out all two-digit numbers ending in 3 and 7, and we get $17, 33, 67, 83$. To ensure that the first two digits of the natural number $x$ are 19, the inequality $19 \leqslant x^{2} \cdot 10^{-N}<20$ must hold, where $N$ is a natural number. When $N = 2k$ ... | 1383 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Find the smallest natural number $x$, such that in decimal notation, the number $x^{3}$ has the first three digits and the last four digits as 1. | Solving the problem, the last four digits are only related to the last four digits of $x$.
Number $x^{3}$ ends with 1, so the last digit of number $x$ can only be 1. Cubing the numbers 01, 11, 21, ..., 91, we confirm that only $71^{3}=357911$. Therefore, the last two digits of number $x$ are 71. Cubing the numbers 071... | 1038471 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. If $p, q$ are both natural numbers, and the two roots of the equation $p x^{2}-$ $q x+1985=0$ are both prime numbers, then what is the value of $12 p^{2}+q$? (85 Beijing Mathematics Competition Question) | Since $1985=5 \times 397$, and the two roots $x_{1}, x_{2}$ are both prime numbers, and $x_{1} x_{2}=\frac{1985}{p}=\frac{5 \times 397}{p}$, thus $p=1$, then $x_{1}, x_{2}$ are 5 and $397, x_{1}+x_{2}=\frac{q}{p}$ $=q=402$, therefore $12 p^{2}+q=12+402=414$. | 414 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. For the equation $(1989 x)^{2}-1988 \cdot 1990 x$ $-1=0$, the larger root is $r$. For the equation $x^{2}+1989 x-1990=0$, the smaller root is $s$. Find the value of $r-s$. (Adapted from the 1984 Beijing Mathematics Competition) | Solve the equation $(1989 x)^{2}-1988 \cdot 1990 x-1$ $=0$, we get
$(1989 x)^{2}-(1989-1)(1989+1) x-1=0$.
Obviously, $1989^{2}-(1989-1)(1989+1)-1$ $=0$, so the equation has one root $x_{1}=1$, then the other root is $\boldsymbol{x}_{2}=-\frac{1}{1989^{2}}$, thus the larger root $r=1$.
Similarly, for the equation $x^{2... | 1991 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. If $k$ is an integer, the quadratic equation $(k-1) x^{2}-p x+k=0$ has two positive integer roots, find $k^{k 0}\left(p^{p}+k^{k}\right)$. (84 Beijing Mathematics Competition Question) | Let the two roots be $x_{1}, x_{2}$, then $x_{1}, x_{2}$ are both positive integers. By Vieta's formulas, $x_{1} x_{2}=\frac{k}{k-1}$ is also a positive integer. Since $k$ is a positive integer, if $k-1 \neq 1$, then $k-1$ and $k$ are coprime. In this case, $\frac{k}{k-1}$ cannot be an integer, so $k-1=1$, i.e., $k=2$ ... | 1984 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given the equation $x^{2}+(a-6) x+a=0$ $(a \neq 0)$ has two integer roots, find the integer value of $a$.
---
The equation $x^{2}+(a-6) x+a=0$ $(a \neq 0)$ is known to have two integer roots. Try to find the integer value of $a$. | Let the two integer roots of the equation be \(x_{1}, x_{2}\), and \(x_{1} \geqslant x_{2}\). By Vieta's formulas, we have:
\[
\left\{\begin{array}{l}
x_{1}+x_{2}=6-a, \\
x_{1} x_{2}=a .
\end{array}\right.
\]
(1)
\[
\begin{array}{l}
+ \text { (2): } x_{1} x_{2}+x_{1}+x_{2}=6, \\
\therefore \quad\left(x_{1}+1\right)\le... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $x=\sqrt{19-8 \sqrt{3}}$, then the value of the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{3}-7 x^{2}+5 x+15}$ is $\qquad$ | $5$ | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. A person is walking along the tram route, and a tram catches up from behind every 12 minutes, while a tram comes from the opposite direction every 4 minutes. Assuming both the person and the trams are moving at a constant speed, then the trams are dispatched from the starting station every $\qquad$ minutes. | $6$ | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, as shown in the figure, in $\triangle A B C$, $\angle A=90^{\circ}$, $A D \perp B C$ at $D, P$ is the midpoint of $A D$, $B P$ intersects $A C$ at $E, E F \perp B C$ at $F, A E=3, E C=12$. Find the length of $E F$. | Extend $F E, B A$ to intersect at point $H$,
$\left.\begin{array}{l}A D \| H F \\ P \text { is the midpoint of } A D\end{array}\right\} \Rightarrow H E=E F$.
Also, $\angle H A C=90^{\circ}$ ?
$\left.\begin{array}{l}\angle E F C=90^{\circ} \\ \text { and on the same side of } H C\end{array}\right\} \Rightarrow H, A, F, ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. Consider the following sequences:
$$
\begin{array}{l}
3,7,11,15,19,23,27,31, \cdots \\
2,5,8,11,14,17,20,23, \cdots
\end{array}
$$
The 20th pair of identical numbers in them is $\qquad$ | 10. 239 . | 239 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let the integer part of $\frac{\sqrt{13}+3}{\sqrt{13}-3}$ be $m$, and the decimal part be $n$, then the value of $198 m+9 n+n^{2}+1$ is $\qquad$ | 3. 1790 , | 1790 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Find the largest odd number that cannot be expressed as the sum of three distinct composite numbers.
untranslated text:
4.求一个不能用三个不相等的合数之和表示的最大奇数。 | The smallest sum of three unequal composite numbers is
$$
4+6+8=18 \text{.}
$$
We prove that 17 is the largest odd number that cannot be expressed as the sum of three unequal composite numbers.
In fact, it is only necessary to prove that any odd number $2k-1$ not less than 19 can always be expressed as the sum of thr... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, $E, F$ are
on the sides $B C$ and $C D$
of rectangle $A B C D$,
if the areas of $\triangle C E F$,
$\triangle A B E$, $\triangle A D F$
are 3,
4, 5 respectively. Find the area $S$ of $\triangle A E F$. | Let $AB = a, BC = b$, then $BE = \frac{8}{a}$,
$$
\begin{array}{l}
CE = b - \frac{8}{a}, DF = \frac{10}{b}, FC = a - \frac{10}{b}. \\
\left\{\begin{array}{l}
\frac{1}{2}\left(b - \frac{8}{a}\right) \times \left(a - \frac{10}{b}\right) = 3, \\
ab = 3 + 4 + 5 + S .
\end{array}\right. \\
S = \sqrt{144 - 80} = 8 .
\end{arr... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. When programming a computer to print out 10000 numbers, there is a bug, every time it prints the number 3, it prints “X” instead, how many numbers are printed incorrectly.
保留了源文本的换行和格式。 | 12. 3439 . | 3439 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four, as shown in the figure, $P$ is a point inside the square $ABCD$, $PA=5$, $PB=8$, $PC=13$. Find the area of square $ABCD$.
---
The translation maintains the original text's line breaks and format. | Four, Hint: Like tears, draw $P E \perp$ $A B, P F \perp B C$, and let the side length of the square be a, $\quad P_{E}=x, \quad P F$ $=y$, solve to get the area of square $A B C D$ is approximately 153. | 153 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. There is a car, the circumference of its front wheel is $5 \frac{5}{12}$ meters, and the circumference of its rear wheel is $6 \frac{1}{3}$ meters. Then it must travel $\qquad$ meters to make the number of revolutions of the front wheel 99 more than that of the rear wheel. | 2. 3705 ; | 3705 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. (Addition to the problem $1^{*}$ ) $n$ countries, each country's team of 3 representatives, form $m$ representative meetings $A_{0}(1), A_{0}(2), \cdots, A_{0}(m)$ called a meeting circle. If
1 each meeting has $n$ representatives, exactly 1,
() no two meetings have the same representatives;
(3) $A_{0}(i)$ and $A_{0... | We use $(m, n)$ to denote a conference circle with $n$ countries and $m$ meetings, and use $\left(A_{\mathrm{n}}(i), j\right)$ to denote a concept formed by all the representatives and the $j$-th representative of the $n+1$-th country in the meeting $A_{n}(i)$. If $A,(0), \cdots, A \sim(m)$ is an $(m, n)$ assembly, the... | 1990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. In a dark room, a drawer contains socks of two colors and two sizes, 4 pairs in total (one pair of each color and size). How many socks must be taken out to ensure that there are two pairs of different colors and different sizes? | New $A_{1}, A, B, B$ each two. Period out 6 as $A ., A_{1}, 1, B_{1}, B_{1}, B_{2}$, does not meet the question's requirements. But when any 7 are taken out, there must be one color (let's say A) with 4 pieces all taken out, and among the 3 pieces of another color, there must be two. | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. For the set $\{00,01, \cdots, 98,99\}$, a subset $X$ satisfies: in any infinite sequence of digits, there are two adjacent digits that form an element of $X$. What is the minimum number of elements that $X$ should contain? | 4. For any $i, j \in\{0,1, \cdots, 9\}, \lambda$ should include $i j$ or $i i$ (otherwise the sequence $i j i j i \ldots$ would satisfy the problem's requirements). There are $10+C_{10}^{2}=55$ such unordered pairs $(i, j)$, hence $|X| \geqslant 55$.
On the other hand, if we take $X=\{i j: 0 \leqslant i \leqslant j \l... | 55 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The street plan of a city is a $5 \times 5$ square grid, (as shown in the figure) and there is a sweeper at point $A$. i. Find the total length of all streets.
保留了原文的换行和格式，翻译结果如上。 | 4. This graph has 16 odd vertices (the four inner points on each boundary) and can be divided into eight pairs of adjacent ones. Therefore, the length of the shortest cleaning route is $60+8=68$. | 68 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. i will arrange $1 \times 1$ square tiles into a strip with a diagonal length of 100. Find the approximate minimum value of the total length of the cuts, not exceeding 2. | For $a>$. $a^{2}+b^{2}=100^{2}, ab=1$,
$$
\therefore c+b=\sqrt{100^{2}+2}>100 .
$$
The perimeter of the rectangle $2(a+b)$ is due to the original square's perimeter.
Given its two uses, $2 L+4 \geqslant 2(a+b)$, $L \geqslant a+b-2>98$.
On the other hand,
Any two parallelograms with the same base and height can be:
t... | 99 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. The sequence $2,3,5,6,7,10,11, \cdots$ contains all positive integers that are neither perfect squares nor perfect cubes, find the 500th term of this sequence. | 1. Answer: 528. | 528 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. $n$ is the smallest positive integer satisfying the following condition:
(2) $n$ has exactly 75 positive divisors (including 1 and itself). Find $\frac{n}{75}$. | 5. To ensure that $n$ is divisible by 75 and the obtained $n$ is the smallest, we can set $n=2^{\gamma_{13}} r_{2} \gamma_{3}$ and
$(r_{1}+1)(r_{2}+1)(r_{3}+1)=75$ $(r_{2} \geqslant 1, r_{3} \geqslant 2)$.
It is easy to prove that when $r_{1}=r_{2}=4, r_{3}=2$, $n$ has the minimum value. At this time, $\frac{n}{75}=2^... | 432 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A biologist wants to estimate the number of fish in a lake. On May 1, he randomly caught 60 fish and marked them, then released them back into the lake. On September 1, he found that some of the fish were no longer in the lake (due to death or migration), and on September 1, 40% of the fish in the lake were not ther... | 6. Let the number of fish in the lake on May 1 be $x$, and the number of fish in the lake on September 1 be $y$. According to the problem, we have
$$
y=0.75 x+0.40 y \text {. }
$$
On September 1, the number of tagged fish in the lake is $0.75 \times 60 = 45$. Assuming that the tagged fish on September 1 can represent ... | 1050 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. The vertices of a triangle are $P(-8$, 5), $Q(-15,-19), R(1,-7), \angle P$ bisector can be expressed as $a x+2 y+c=0$, find $a+c$,
| 7. Extend $P R$ to $T$, such that $P Q=P T$. Since $P Q=25, P R=15$, then the coordinates of $T$ are $(7,-15)$, and the angle bisector intersects $Q T$ at the midpoint $(-4,-17)$.
Thus, the slope of the angle bisector is $-\frac{11}{2}$, and the equation is
$$
\begin{array}{r}
11 x+2 y+78=0 . \\
\text { Hence } a+c=11+... | 89 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Find $i \div j$.
Translate the text above into English, keep the original text's line breaks and format, and output the translation result directly. | 9. Tossing a coin 10 times has $2^{10}$ outcomes. Let $A(n)$ be the number of outcomes where no heads appear consecutively. By enumeration, it is easy to see that $A(1)=2, A(2)=3, A(3)=5$. It can also be proven by induction that there is a recurrence relation:
$$
A(n+2)=A(n+1) + A(n) \text {. }
$$
This is because $A(n... | 73 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. We notice that $6!=8 \cdot 9 \cdot 10$. Try to find the largest positive integer $n$ such that $n!$ can be expressed as the product of $n-3$ consecutive natural numbers. | 11. If $n!$ can be expressed as the product of $(n-3)$ consecutive integers, then there exists an integer $k$, such that
$$
\begin{array}{l}
n!=(n+k)(n+k-1) \cdots(k+4) \\
=-(n+k)! \\
(k+3)!
\end{array}
$$
Originally,
$$
\begin{array}{l}
\frac{n+k}{k+3} \cdot \frac{n+k-1}{k+2} \cdot \ldots \cdot \frac{n+2}{5} \cdot \... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
The sum of all side lengths and diagonal lengths of a 12-sided polygon can be written in the form $a+b \sqrt{2}+c \sqrt{3}+d \sqrt{6}$, where $a, b, c$, $d$ are positive integers. Find $a+b+c+d$.
The text above is translated into English, preserving the original text's line breaks and format. | Coinciding with the origin, one vertex is at $(12,0)$, and the coordinates of the other vertices are $(12 \cos k x, 12 \sin k x)$, where $x=30^{\circ}$, $k=1,2, \cdots, 11$. The length of the line segment connecting $(12,0)$ and $(12 \cos k x, 12 \sin k x)$ is $24 \sin \frac{k x}{2}$. Therefore, from this,
$$
\begin{ar... | 720 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. Let $T=\left\{9^{4} \mid k\right.$ be an integer, $0 \leqslant k$ $\leqslant 4000\}$. It is known that $9^{1000}$ has 3817 digits, and its most significant digit is 9. How many elements in $T$ have 9 as their most significant digit? | 13. Notice that $9^{k}$ has one more digit than $9^{k-1}$, unless $9^{1}$ starts with the digit 9. In the latter case, by long division, $9^{k-1}$ starts with the digit 1, and $9^{k}$ has the same number of digits. Therefore, from $9^{0}$ to $9^{4000}$, there are 3816 digit increases, so there must be 184 times when th... | 184 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. $A B C D$ is a rectangle, $A B=12 \sqrt{3}, B C=$ $13 \sqrt{3}$, diagonals $A C, B D$ intersect at $P$. If the triangle $A B P$ is cut off and $A P, B P$ are joined, find its volume.
| 14. As shown in the figure, $N$ is the intersection of the altitude from $P$ to $BCD$.
It is easy to prove that $N$ is the circumcenter of $\angle BCD$, and
\[
\begin{array}{l}
r=\frac{n^{2}}{\sqrt{4 n^{2}-m^{2}}} . \\
\because P B=\frac{1}{2} \sqrt{m^{2}+n^{2}}, \text { by the Pythagorean theorem, we have } \\
P N=\f... | 594 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
$$
f_{u:-1}(k)=f_{1}\left(f_{u}(k)\right) .
$$
Find $f_{1991}\left(2^{1000}\right)$.
| Let $k$ be much smaller, and we can estimate the value range of $f_{1}(k)$ as follows:
Suppose the positive integer $a$ has $m$ digits, then by changing all the digits of $a$ to 9, we have
$$
f_{1}(a) \leqslant 9^{2} m^{2}=81 m^{2}.
$$
If $a \leqslant b$, then $m$ is no greater than the integer part of $\lg b$ plus 1... | 256 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $d_{1}, d_{2}, \cdots, d_{k}$ be all the divisors of the positive integer $n$, $1=d_{1}<d_{2}<d_{3}<\cdots<d_{k}=n$. Find all $n$ such that $k \geqslant 4$ and $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}=n$. | Number, but the sum of the squares of four odd numbers is even, which cannot equal $n$. Contradiction.
Thus, $n$ is even, $d_{2}=2$.
If $n$ is a multiple of 4, then $4 \in\left\{d_{3}, d_{4}\right\}$. Among the squares of the 4 factors, there are already two $\left(2^{2}\right.$ and $\left.4^{2}\right)$ that are multip... | 130 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. A natural number $N$ has exactly 12 divisors (including 1 and $N$), and these divisors are numbered in increasing order: $d_{1}<$ $d_{2}<\cdots<d_{12}$. The divisor with the index $d{ }_{4}-1$ equals $\left(d_{1}+d_{2}+d_{1}\right) \times d_{8}$, find $N$.
| 4. First, use the relationship
$$
d_{d_{4}-1}=\left(d_{1}+d_{2}+d_{4}\right) \times d_{8} \geqslant d_{5} \times d_{8}
$$
$=N$, we get $d_{4}=13, d_{5}=d_{2}+14, \quad N$ $=\left(d_{2}+14\right) \times d_{8}$. Then, using the enumeration method to take $d_{2}=2,3$, $5, 7, 11$ to get the only possible solution $d_{2}=3,... | 1989 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. If a $23 \times 23$ square is composed of 1000 squares of sizes $1 \times 1, 2 \times 2, 3 \times 3$, how many $1 \times 1$ squares are needed at minimum? | 4. At least 1 $1 \times 1$ square is needed. First, it needs to be shown that a $23 \times 23$ square can be composed of the following: $2 \times 2, 3 \times 3$ squares, and exactly 1 $1 \times 1$ square. Then, it also needs to be proven that a $1 \times 1$ square is indispensable. | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. As shown in the figure,
$2, P$ is a point inside the triangle,
where the areas of the four
small triangles are marked in the figure. Find the area of $A B C$. (Third A1ME)
Note: The translation keeps the original text's line breaks and format as requested. However, the mathematical notation and the figure r... | Solve $-\frac{S \triangle \triangle P D}{S \triangle C D}=\frac{B D}{C D}$,
X
$\frac{S \therefore \operatorname{LA} 1}{S \triangle \mathrm{CAD}}=\frac{B D}{C D}$.
Let $S \triangle \mathrm{BPF}=x, S: A P E=y$,
Then
$$
\begin{array}{l}
S \frac{\mathrm{BAD}}{S-\mathrm{CAD}}=-\frac{S P C D}{S \triangle \mathrm{CPD}}=\frac... | 315 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Calculate: $\sqrt{31 \cdot 30 \cdot 29 \cdot 28+1}$. (7th American Invitational Mathematics Examination) | ```
\begin{array}{l}
\because \sqrt{ }(x+1) x(x-1)(x-2)+1 \\
= \sqrt{ }\left(x^{2}-x-2\right)\left(x^{2}-x\right)+1 \\
= \sqrt{ }\left(x^{2}-x-1\right)^{2} \\
=\left|x^{2}-x-1\right| . \\
\therefore \text { when } x=30 \text {, the original expression }=\left|30^{2}-30-1\right| \\
=869 .
\end{array}
``` | 869 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. Take a point $P$ inside $\triangle ABC$, and draw three lines through $P$ parallel to the three sides of $\triangle ABC$,
thus forming three triangles
$t_{1}, t_{2}, t_{3}$ with areas
4, 9, 49 (as
shown in Figure 3).
Find the area of $\triangle ABC$.
(From the 2nd $\triangle I M E$) | $$
\begin{array}{l}
\text{Given that } \triangle t_{1}, \triangle t_{2}, \triangle t_{3} \text{ and } \triangle A B C \text{ are all similar triangles.} \\
\text{Let the areas of } \triangle A B C, \triangle t_{1}, \triangle t_{2}, \triangle t_{3} \text{ be } S, S_{1}, S_{2}, S_{3} \text{, respectively.} \\
\text{It is... | 144 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. A football invitational tournament has 16 cities participating, each city sending Team A and Team B. According to the competition rules, each pair of teams plays at most one match, and the two teams from the same city do not play against each other. If, after two days of the tournament, it is found that exce... | Solve the problem with a general discussion. Suppose there are $n$ cities participating in the competition, all of which meet the conditions of the problem. Let the number of matches played by Team B of City $A$ be $a_{a}$, obviously $a_{1}=0$.
In the case of $n$ cities, according to the competition rules, each team c... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 - What is the maximum number of knights that can be placed on an $8 \times 8$ chessboard so that no two knights attack each other (assuming there are enough knights)? | We will alternately color the chessboard in black and white, so there will be 32 black squares and 32 white squares. According to the knight's move (see Figure 1), a knight on a black square can only capture a knight on a white square. Therefore, placing knights on all black squares means they will not capture each oth... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. Given the sequence $\left\{x_{0}\right\}: x_{n+1}=$ $\frac{x_{\mathrm{n}}+(2-\sqrt{3})}{1-x_{n}(2-\sqrt{3})}$. Find the value of $x_{1001}-x_{401}$. | Given the shape of the recurrence relation, we can set $x_{\mathrm{n}}=\operatorname{tg} \alpha_{\mathrm{n}}$, and since $\operatorname{tg} \frac{\pi}{12}=2-\sqrt{3}$, we know that $x_{n+1}=\operatorname{tg} \alpha_{n+1}$
$$
=\frac{\operatorname{tg} \alpha_{\mathrm{a}}+\operatorname{tg} \frac{\pi}{12}}{1-\operatorname{... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. As shown in Figure 5, in $\triangle ABC$, $AB=2$, $AC=3$, I, II, and III respectively denote the squares constructed on $AB$, $BC$, and $CA$. What is the maximum value of the sum of the areas of the three shaded regions? (1988, National Junior High School League)
---
The translation preserves the original ... | Solve in $\triangle A D E$ and $\triangle A B C$ !! $\angle B A C + \angle D A E = 180^{\circ}$.
Also, $A C = A D, A B = A E$,
thus $S \triangle A D B = S \triangle A B C$.
Similarly, the areas of the other two shaded triangles are also equal to $S \triangle \triangle B C$.
$$
\begin{array}{r}
\therefore S \text { shad... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. A person earns a monthly salary of 200 yuan and receives a bonus of 20 yuan at the end of the month. After receiving the salary each month, they arrange the monthly expenses according to $\frac{4}{5}$ of the current savings. How much will be the remaining balance in the twelfth month? | Let the balance at the end of the $n$-th month be $a_{\mathrm{n}}$ yuan. According to the problem,
$$
\begin{array}{l}
a_{1}=200 \times\left(1-\frac{4}{5}\right)+20=60, \\
a_{2}=260 \times\left(1-\frac{4}{5}\right)+20=72 .
\end{array}
$$
In general, $a_{0}=\left(a_{0-1}+200\right) \times\left(1-\frac{4}{5}\right)+20$,... | 75 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. The lengths of the three sides $a, b, c$ of a triangle are all integers, and $a \leqslant b \leqslant c$. If $b=10$, then the number of such triangles is ( ).
(A) 10.
(B) 55.
(C) $10^{2}$.
(D) Infinitely many.
(1990, Suzhou High School Competition) | When $b=n$, take $a=k(1 \leqslant k \leqslant n)$, and $b \leqslant c<a+b$, then $n \leqslant c<n+k$, at this time the value of $c$ has exactly $k$ possibilities, i.e., $c=n, n+1, \cdots, n+k-1$. The table is as follows:
\begin{tabular}{c|c|c|c}
\hline$a$ & $b$ & $c$ & Number of triangles \\
\hline 1 & $n$ & $n$ & 1 \\... | 55 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
1. (1987, National Junior High School League) Among the triangles with three consecutive natural numbers as sides and a perimeter not exceeding 100, the number of acute triangles is $\qquad$ | (29)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 29 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. (1986, Shanghai) The three sides of a triangle are all positive integers, one of which has a length of 4, but it is not the shortest side. How many different triangles are there? ...
The translation is provided while retaining the original text's line breaks and format. | (8).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. A pile of toothpicks 1000 in number, two people take turns to pick any number from it, but the number of toothpicks taken each time must not exceed 7. The one who gets the last toothpick loses. How many toothpicks should the first player take on the first turn to ensure victory? (New York Math Competition) | From $1000=125 \times 8$, we know that one should first dare to take 7 moves, so that the latter can achieve a balanced state: $124 \times(7+1)+1$. | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. On a $4 \times n$ chessboard, if any cells are colored red, white, or blue, then the necessary and sufficient condition for the existence of a monochromatic rectangle is $n=19$. | If $n=19$, then in a $4 \times 19$ chessboard, there must be $\left[\frac{4 \times 19-1}{3}\right]+1=26$ small squares of the same color, let's assume they are red. At this point,
$$
\begin{array}{l}
m=4, n=19, a=26, q=1, r=7, \\
r C_{4+1}^{2}+(n-r) C_{4}^{2}=7, \\
C_{\mathrm{m}}^{2}=6,
\end{array}
$$
The conditions o... | 19 | Combinatorics | proof | Yes | Yes | cn_contest | false |
1. In $\triangle A B C$, $D, E, F$ are the midpoints of $B C$, $C A, A B$ respectively, and $G$ is the centroid. For each value of $\angle B A C$, how many non-similar $\triangle A B C$ are there such that $A E G F$ is a cyclic quadrilateral? | 1. From $A, E, G, F$ being concyclic, we get
$$
\begin{aligned}
\angle C G_{E} & =\angle B A C \\
= & \angle C E D .
\end{aligned}
$$
Let $C G$ intersect $D E$ at $M$. Then from (1),
it is easy to derive
$$
C M \times C G=C E^{2},
$$
i.e., $\frac{1}{2} m_{\mathrm{e}} \times \frac{2}{3} m_{\mathrm{c}}=\left(\frac{1}{2... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4.199J individuals are divided into several mutually disjoint subsets, such that
(a) in each subset, no one knows everyone else.
(b) in each subset, among any three people, at least two do not know each other.
(c) in each subset, for any two people who do not know each other, there is exactly one person in the subset w... | 4. (i) Consider only one group. Let $y_{1}$ and $y_{2}$ be unknown to each other within the same group. By (c), there exists $x$ who knows both $y_{1}$ and $y_{2}$. Suppose, apart from $x$ and themselves, $y_{1}$ knows $z_{11}, z_{12}, \cdots, z_{11}$. $y_{2}$ knows $z_{21}$, $z_{22}, \cdots, z_{2 \mathrm{k}}$.
By (b)... | 398 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 10: There are 100 points on a plane, where the distance between any two points is no less than 3. Now, connect every two points that are exactly 3 units apart with a line. Prove: the number of these line segments will not exceed 300. (1984, Beijing Junior High School Competition) | First, consider the maximum number of points whose distance from a certain point $A$ is: 3. Let $B_{1} A=B_{2} A=\cdots=B_{1} A$
$$
\begin{array}{l}
=3, \text { and } B_{1} B_{2} \geqslant 3, B_{2} B_{3} \geqslant 3, \cdots, B_{\mathrm{k}-1} B_{1} \\
\geqslant 3, B_{\mathrm{k}} B_{1} \geqslant 3 . \text { Then } \angle... | 300 | Combinatorics | proof | Yes | Yes | cn_contest | false |
3. The decimal representation of the natural number $A$ is $\overline{a_{n} a_{n-1} \cdots a_{0}}$. Let $f(A)=2^{\mathrm{n}} a_{0}+2^{\mathrm{n}-1} a_{1}+\cdots+2 a_{\mathrm{n}-1}+a_{\mathrm{n}}$, and denote $A_{1}=f(A), A_{1+1}=f\left(A_{1}\right) . \quad(i=1,2$, $\cdots, n$ ) Prove:
i) There must be a $k \in \mathbb{... | Proof (i) When $n=0$. For any $k$, we have $A_{\mathrm{s}}=A$.
$$
\begin{array}{l}
\text { When } n=1 \text {, } \\
A-f(A)-10 a_{1}+a_{0}-2 a_{0}-a_{1} \\
=9 a_{1}-a_{0} \geqslant 9 \times 1-9=0 . \\
\text { When } n \geqslant 2 \text {, } \\
A-f(A) \\
\geqslant 10^{\mathrm{n}}-\left(2^{\mathrm{n}}+2^{\mathrm{n}+1}+\cd... | 19 | Number Theory | proof | Yes | Yes | cn_contest | false |
In the May 1989 issue of this journal, there was a problem from the Zhejiang Province High School Mathematics Competition as follows:
Arrange the positive rational numbers: $\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}, \cdots$, then t... | The original answer is $(1+2+\cdots+3937)+1949$ $=7753902$.
In fact, the original sequence is formed by segments of fractions with the sum of the numerator and denominator being 2, 3, 4, ․ respectively. The fraction 1989/1949 has a sum of numerator and denominator of 3938, making it the 1949th number in the 3937th seg... | 7749965 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
On a plane, there is a convex quadrilateral $A B C D$.
(1) If there exists a point $P$ on the plane such that the areas of $\triangle A B P, \triangle B C P, \triangle C D P, \triangle D A P$ are all equal, what condition should the quadrilateral $A B C D$ satisfy?
(2) How many such points $P$ can there be on the plane... | . Solving (i) First consider the case where $P$ is inside the quadrilateral.
If points $A, P, C$ are collinear, and points $B, P, D$ are also collinear, then quadrilateral $ABCD$ is a parallelogram, and $P$ is the intersection of the two diagonals.
If points $A, P, C$ are not collinear, by the fact that $\triangle PAB... | 1 | Geometry | proof | Yes | Yes | cn_contest | false |
Three, there are 10 birds on the ground pecking at food, among which any 5 birds, at least 4 are on a circle. How many birds are there on the circle with the most birds, at minimum?
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。
---
Three, there are 10 birds on the ground pecking at food, among which any 5 birds, at least 4 are... | We use 10 points to represent 10 birds.
(1) Among the 10 points, there must be 5 points that are concyclic. If not, then any 5 points among the 10 points are not concyclic. The 10 known points can form $C_{10}^{5}=252$ groups of 5 points. According to the given, each group of 5 points has four points that are concyclic... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. If from the numbers $1,2, \cdots, 141$, we sequentially select $a_{1}, a_{2}, a_{2}$, such that $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$, then how many different ways of selection are there that meet the above requirements? | Let $S=\{1,2, \cdots, 14\}, S^{\prime}=\{1, 2, \cdots, 10\}$. Suppose $\left\{a_{1}, a_{2}, a_{3}\right\}$ is a subset of $S$ and satisfies $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-2, a_{3}^{\prime}=a_{3}-4$, i.e., the following mapping is established:
$$
\begin{array}{c}
\left(a_{1}, a_{2}, a_{3}\right) \rightarrow... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. (Shanghai $\cdot$ Senior High School) If positive integers $p, q, r$ make the quadratic equation $p x^{2}-{ }_{y} x+r=0$ have two distinct real roots in the open interval $(0,1)$, find the minimum value of $p$.
| Let $\alpha, \beta$ be the two roots of the quadratic equation $p x^{2}-q x+\gamma=0$, $\alpha \neq \beta, 0<\alpha, \beta<1$. Then
$$
\alpha \beta(1-\alpha)(1-\beta) = \frac{r}{p} \left[1 - \frac{q}{p} + \frac{r}{p}\right] = \frac{r}{p^{2}}(p-q+r).
$$
Therefore, $p^{2} > 16 r(p-q+r)$.
Since the quadratic term coeffic... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. (First State "Killing Hope Cup" - Grade 10) Let the function $f(n)=k$, where $n$ is a natural number, and $k$ is the $n$-th digit after the decimal point of the irrational number $\pi=3.1415926535 \cdots$. It is also defined that $f(0)=3$.
$$
\text { Let } F(n)=\frac{f(f(f(f \cdots(f(n)) \cdots)))}{10 \uparrow \bar{... | It can be easily proven that for non-negative integers $n$, $F(n)=1$ always holds.
Below $f(1990)+f(5)+f(13)$ and $f(1990) f(3)$
- $f(25)$ are all non-negative integers, so $F(f(1990)+f(5)$
$$
+f(13) \equiv 1 \equiv F[(1990) f(3) f(25)] .
$$ | 1 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 1. (14th All-Russian Mathematics Competition) Let
$$
\frac{1}{1+\frac{1}{1+\frac{1}{1+}}+\frac{1}{1}}=-\frac{m}{n},
$$
where $m$ and $n$ are coprime natural numbers, and the left side of the equation has 1988 fraction lines. Calculate $m^{2}+m n-n^{2}$. | Let the value of the complex fraction with $k$ layers of fraction lines be $\frac{m_{k}}{n_{k}},\left(m_{k}, n_{k}^{*}\right)=1$, then
$$
\frac{m_{\mathrm{k}+1}}{n_{\mathrm{k}+1}}=\frac{1}{1+\frac{m_{\mathrm{k}}}{n_{\mathrm{k}}}}=\frac{n_{\mathrm{k}}}{m_{\triangle}+n_{\mathrm{k}}},
$$
i.e., $m_{\mathrm{k}+1}=n_{\mathr... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
19. (Wuhu City)
In a square plate $ABCD$ with a side length of 8 units, there is an equilateral triangular colored plate $OEF$ (vertex $O$ is painted red, $E, F$ are white) placed as shown (E coincides with D, and $EF$ coincides with $AD$). Now, the triangular colored plate rotates to the left with $F$ as the pivot un... | The "left advance" rule for the triangular color board is:
?.
(1) Every 3 left advances, the red vertex is inside the disk;
(2) Every 16 left advances, the triangular board returns to its original starting position, but the red vertex moves to the point F on the AD side;
(3) Every 48 left advances, the triangular board... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. (Lokni Army) Let $A$ be a finite set, $N=\{1, 2, 3, \cdots\}$. If there exists a function $f: N \rightarrow A$ with the following property: if $|i-j|$ is a prime number, then $f(i) \neq f(j)$. Try to find the minimum number of elements in $A$.
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。 | Consider in $N$ taking $i=1,3,6,8$. The absolute value of the difference between any two of these numbers is a prime number, so their images $f(1), f(3), f(6), f(8)$ are the same, meaning that the set $A$ must have at least 4 elements.
Below, we prove that the minimum number of elements in set $A$ is 4. For this, we c... | 4 | Logic and Puzzles | other | Yes | Yes | cn_contest | false |
1. Given $x y+x+y=71, x^{2} y+x y^{2}$
$=830 . x, y$ are positive integers, find $x^{2}+y^{2}$. | 1. Answer: 146. (Hint: Let $a=x+y$, $b=x y$ ) | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Expand $(1+0.2)^{1000}$ using the binomial theorem, i.e. let $C_{1000}^{0}(0.2)^{0}+C_{1000}^{1}(0.2)^{1}+\cdots+$ $C_{1000}(0.2)^{1000}=A_{0}+A_{1}+\cdots+A_{1000}$. Let $A_{h}=C_{1000}^{k}(0.2)^{h}, k=0,1,2, \cdots$, 1000 . Find the value of $k$ for which $A_{k}$ is the largest. | 3. For $1 \leqslant k \leqslant 1000$,
$$
\begin{array}{l}
\frac{A_{\mathrm{s}}}{A_{\mathrm{k}-1}}=\frac{\frac{1000!}{k!(1000-k)!}(0.2)^{k}}{\frac{1000!}{(k-1)!(1001-k)!}(0.2)^{k}} \\
=\frac{1001-k}{k}(0.2) \text {. } \\
\end{array}
$$
When and only when $1001-k>5k$, this ratio is greater than 1. That is, when $k \leq... | 166 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. The number of real solutions to $\frac{1}{5} \log _{2} x=\sin (5 \pi x)$ is? | Consider only those $x$ that satisfy $\left|\frac{1}{5} \log _{2} x\right| \leqslant 1$. Solving this, we get $\frac{1}{32} \leqslant x \leqslant 32$.
First, consider $\frac{1}{32} \leqslant x<1$, in this case $-1 \leqslant \frac{1}{5} \log _{2} x <0$. When $\frac{1}{5} \leqslant x \leqslant \frac{2}{5}, \frac{3}{5} \... | 159 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
5. For a given rational number, represent it as a reduced fraction. Then find the product of its numerator and denominator. How many rational numbers between 0 and 1 have a product of their numerator and denominator equal to 20! ? | 5. 20! has 8 prime factors: $2, 3, 5, 7, 11, 13, 17, 19$. A prime factor appearing in the denominator cannot appear in the numerator (and vice versa). For each prime factor, it can be in the numerator or in the denominator, so there are $2^8 = 256$ choices. However, not all of them are less than 1. In fact, the numbers... | 128 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Find the value of $r$ such that
$$
\begin{array}{l}
{\left[r+\frac{19}{100}\right]+\left[r+\frac{20}{100}\right]+\cdots} \\
+\left[r+\frac{91}{100}\right]=546 .
\end{array}
$$
Find $[100 r]$. (Where $[x]$ denotes the greatest integer not greater than $x$) | 6. The left side of the equation has 73 terms, each being $[r]$ or $[r+1]$. This is because $\frac{19}{100}, \frac{20}{100}, \cdots, \frac{91}{100}$ are all less than 1, and $73 \times 7 < 546 < 73 \times 8$. To make the sum 546, $[r]$ must be 7.
Assume $\left[r+\frac{k}{100}\right]=7(19 \leqslant k \leqslant m),[r+$ ... | 743 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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