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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6. A square piece of paper is divided into two parts by a straight line, one of which is then divided into two parts by another straight line, and then one of the three pieces is divided into two parts by another straight line, and so on. In the end, 19 96-sided polygons and some other polygons are obtained. Then the m... | \%. 1766 . After each division, the sum of the angles of the resulting polygons increases by $2 \pi$. Thus, after $k$ divisions, $k+1$ polygons are obtained, the sum of whose interior angles is $2 \pi(k+1)$. Since among these $k+1$ polygons, there are 19 ninety-six-sided polygons, the sum of their interior angles is $1... | 1766 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four, (1/3 points) There are 99 numbers $a_{1}, a_{2}, a_{3}, \cdots, a_{98}$, defined as: $S_{1} = a_{1}, S_{2} = a_{1} + a_{2}, \cdots, S_{k} = a_{1} + a_{2} + \cdots + a_{k} (1 \leqslant k \leqslant 99)$. There are 100 numbers $b_{1}, b_{2}, \cdots, b_{100}$, and $b_{1} = 1, b_{2} = 1 + a_{1}, b_{3} = 1 + a_{2}, \cd... | $\begin{array}{l}\text { Four, (1) } \because S_{k}^{\prime}=b_{1}+b_{2}+\cdots+b_{k} \\ ==?+\left(1+a_{1}\right)+\left(1+a_{2}\right)+\cdots+\left(1+a_{k-1}\right) \\ =k+\left(a_{0}+a_{2}+\cdots+a_{k-1}\right) \\ =k+S_{k-1} \text {. } \\ \therefore S^{\prime}{ }_{4}-S_{2-1}=k \text {. } \\ \text { (2) } \because A=\fr... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Initially 35. 4 distinct prime numbers, none of whose last digit is 1, have a sum of squares equal to \(4pq\) (where \(p, q\) are primes, \(p \neq q\)). \(A=2^n pq\) (where \(n\) is an integer greater than 6) is a 6-digit number. The sum of the two 3-digit numbers formed by the first 3 digits and the last 3 digits of \... | Let
$$
2^{\prime \prime} p q=10^{5} a+10^{4} b+10^{3} c+10^{2} d+10 e+f,
$$
where $a, b, \cdots, f$ are all integers, $1 \leqslant a \leqslant 9,0 \leqslant b, c, \cdots, f \leqslant$
9. Then (1) can be written as
$$
\begin{aligned}
2^{2} p q= & 199(1002+10 s+c)+100(a+d) \\
& +10(b+e)+c+f \\
= & 3^{3} \cdot 37 \cdot 2... | 1995 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Second Pump Find all real numbers $p$ such that the cubic equation
$$
5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p
$$
has three roots that are all natural numbers. | The original problem is equivalent to
$$
5 x^{2}-5 p x+66 p-1=0
$$
having both roots as natural numbers.
Method 1: Let $u, v$ be the roots of equation (2), then
$$
\left\{\begin{array}{l}
u+v=p, \\
u v=\frac{1}{5}(66 p-1) .
\end{array}\right.
$$
Eliminating the parameter $p$ yields
$$
\begin{array}{l}
5 u v=66 u+66 v... | 76 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $x+\frac{1}{x}=3$, then $x^{4}+3 x^{3}-16 x^{2}+3 x-1$ ? The value equals $\qquad$ . | 5. From $x^{2}+1=3 x$, we get $x^{4}=7 x^{2}-1, x^{3}=3 x^{2}-x$. Therefore, the original expression $=7 x^{2}-1+9 x^{2}-3 x-16 x^{2}+3 x-17$ $=-18$.
Another solution: From the given $x^{2}+1=3 x, x^{2}-3 x=-1$, so,
$$
\begin{array}{l}
x^{4}+3 x^{3}-16 x^{2}+3 x-17=\left(x^{4}+2 x^{2}+1\right)+3 x^{3}+3 x \\
-18 x^{2}-... | -18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $x_{1}, x_{2}, \cdots, x_{19}$ be integers, and satisfy $x_{1}+x_{2}+\cdots$ $+x_{19}=95$. Then the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{19}^{2}$ is $\qquad$. | 6. Let's assume $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant \cdots \leqslant x_{19}$. Note that when $1<x_{i}$ $\leqslant x_{j}$, $x_{i}^{2}+x_{j}^{2}<x_{i}^{2}+1-2 x_{i}+2 x_{j}+1+x_{j}^{2}=\left(x_{i}-\right.$ $1)^{2}+(x,+1 \cdot 1)^{2}$. Therefore, when $x_{1}=x_{2}=\cdots=x_{18}=1, x_{10}=$ 77, $x^{2} \cdot x_... | 5947 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (This question is worth 20 points)
In the border desert area, patrol vehicles travel 200 kilometers every day. Each vehicle departs from base $A$ at the same time, completes the mission, and then returns along the original route to the base (returning together). After vehicles A and B reach point $B$ on the way, t... | When the car reaches point $B$ on the way, it takes $x$ days, and from $B$ to the farthest point, it takes $y$ days, then we have
$$
2[3(x+y)+2 x]=14 \times 5,
$$
which simplifies to $5 x+3 y=35$.
According to the problem, we need $x>0, y>0$, and
$$
14 \times 5-(5+2) x \leqslant 14 \times 3 \text {, }
$$
which simpli... | 1800 | Other | math-word-problem | Yes | Yes | cn_contest | false |
Three, (11 points) There are four numbers. Each time, select three of them, calculate their average, and then add the other number. Using this method, four calculations were made, resulting in the following four numbers: 86, 92, 100, 106. (1) Find the four numbers, (2) Find the average of these four numbers. | Three, let these four numbers be $a, b, c, d$. We can get
$$
\left\{\begin{array}{l}
\frac{a+b+c}{3}+d=86, \\
\frac{b+c+d}{3}+a=92, \\
\frac{c+d+a}{3}+b=100, \\
\frac{d+a+b}{3}+c=106 .
\end{array}\right.
$$
Solving it, we get $a=42, b=54, c=63, d=33$.
$$
\frac{1}{4}(a+b+c+d)=48 \text{. }
$$
Answer: These four numbers... | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11. There are 100 points on a plane, where the distance between any two points is no less than 3. Now, connect every two points whose distance is exactly 3 with a line segment. Prove: such line segments will not exceed 300.
(1984, Beijing Junior High School Mathematics Competition) | Given that the central angle subtended by an arc between two points is less than $60^{\circ}$, it means the distance between these two points is less than 3, which contradicts the problem's condition. Therefore, with $P$ as the endpoint, the maximum number of line segments that can be drawn to $A_{1}, A_{2}, A_{3}, A_{... | 300 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 2. Find all natural numbers with the following property: its unit digit is 6, and when its unit digit is moved to the front of the number, the resulting new number is four times the original number. | Let the required number be $a_{n} a_{n-1} \cdots \cdots a_{2} a_{1} 6$. According to the problem, we have $\overline{6 a_{n} a_{n-1} \cdots a_{2} a_{1}}=4 \overline{a_{n} a_{n-1} \cdots a_{2} a_{1} 6}$, where $0 \leqslant a_{i} \leqslant 9, a_{i} \in N$.
That is 6
$$
\begin{array}{l}
6 \cdot 10^{n}+a_{n} \cdot 10^{n-1}... | 153846 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. Let $S$ be a subset of the set of numbers $\{1,2,3, \cdots, 1989\}$, and the difference between any two numbers in $S$ is not equal to 4 or 7. How many numbers can $S$ contain at most? | Analysis Notice that the sum of 4 and 7 is 11. We first observe the first 11 natural numbers $1,2, \cdots, 11$, and find that $1,4,6,7,9$ satisfy the conditions of the problem. Then we observe the 11 numbers $12,13, \cdots, 22$, and find that $1+11,4+11,6+11,7+11,9+11$ also satisfy the conditions of the problem. And $1... | 905 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the arithmetic sequence (1): $5,8,11, \cdots$, and the arithmetic sequence (2): $1,5,9, \cdots$, both having 1996 terms, then there are $\qquad$ numbers that appear in both sequences. | (Answer: 498) | 498 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. What is the largest even integer that cannot be written as the sum of two odd composite numbers? (If a positive integer can be divided by a positive integer other than 1 and itself, then this positive integer is called a composite number).
(2nd American Invitational Mathematics Examination) | Solution: Because the last digit of a positive even number is $0, 2, 4, 6, 8$.
When the last digit is 0, there are even integers such as $0,10,20,30,40, \cdots$, among which $30=15+15$.
When the last digit is 2, there are even integers such as $2,12,22,32,42,52, \cdots$, among which $42=15+27,52=25+27, \cdots$. Theref... | 38 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Six, in a game played on a $1 \times 1000$ board, initially there are $n$ chips placed. The game is played by two players taking turns. The first player places a chip in any unoccupied cell, and the second player removes any number of chips occupying consecutive cells. If the first player can place all $n$ chips on the... | Six, let the two players be $A$ and $B$.
1. On the board, arrange 96 chips in the pattern of 8 chips, 1 empty space, 8 chips, 1 empty space, $\cdots$. This is a winning position for $A$. Clearly, $A$ can achieve this winning position. After $B$ moves a certain number of consecutive chips, $A$ can return them to their o... | 98 | Combinatorics | proof | Yes | Yes | cn_contest | false |
II. Write the number $1234567802011 \cdots 19941995$ on the blackboard, forming the integer $N_{1}$. Erase the digits of $N_{1}$ that are in even positions, leaving the remaining digits to form the integer $N_{2}$. Remove the digits of $N_{2}$ that are in odd positions, leaving the remaining digits to form the integer ... | $$
=、 9 \times 1+90 \times 2+900 \times 3+996 \times 4=6873 \text {. }
$$
The integer $N_{1}$ is composed of 6873 digits, which are sequentially numbered as $1,2,3, \cdots, 6873$. We still examine the set of numbers $\{1,2, \cdots, 6873\}$. Removing the digits at even positions from $N_{1}$ is equivalent to removing $... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $a, b, c$ satisfy $a+b+c=0, a b c>0$, and $x=\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}, y=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+$ $c\left(\frac{1}{a}+\frac{1}{b}\right)$. Then the algebraic expression $x^{18}-96 x y+y^{3}=$ | $$
=\sqrt{ } .-316
$$
Given $-b c>0$, and $a+b+c=0$, we know that among $a, b, c$, there must be two negative numbers. Without loss of generality, assume $a>0$. Hence $x=-1$.
$$
\begin{array}{l}
\text { ini } \begin{array}{l}
y=\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b} \\
=\frac{b+c}{a}+... | -316 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the parabola $y=x^{2}+k x+4-k$ intersects the $x$-axis at integer points $A, B$, and intersects the $y$-axis at point $C$. Then $S_{\triangle A B C}=$ $\qquad$ . | 3.24.
Given the equation $x^{2}+k x+4-k=0$ has two distinct integer roots $x_{1}, x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-k, \\
x_{1} x_{2}=4-k .
\end{array}\right.
$$
(2) - (1) gives $x_{1}+x_{2}-x_{1} x_{2}+4=0$,
$$
\left(x_{1}-1\right)\left(x_{2}-1\right)=5 \text {. }
$$
Thus, $\left\{\begin{array}{l}... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that the inverse function of $y=f(x)$ is $\varphi(x)$, and $\varphi(x)$ . $=\log _{m a_{0}{ }^{2}}\left(\frac{1996}{x}-\sin ^{2} 0\right), 0 \in\left(0, \frac{\pi}{2}\right)$. Then the solution to the equation $f(x)=$ 1996 is | $=, 1 .-1$.
Since $f(x)$ and $\varphi(x)$ are inverse functions of each other, solving $f(x) = 1996$ is equivalent to finding the value of $\varphi(1996)$.
$$
\varphi(1996)=\log _{\sec ^{2} \theta}\left(1-\sin ^{2} \theta\right)=-1 .
$$ | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 10. As shown in Figure 6-1, it is a part of a city's street map, with five roads running both longitudinally and latitudinally. If one walks from point $A$ to point $B$ (only from north to south, and from west to east), how many different ways are there to do so? | Solution: Because the number of different ways from the top-left vertex of each small square to the endpoint equals the sum of the number of different ways from the two adjacent vertices to the endpoint, we can deduce the data at the top-left corner of each small square in Figure 6-4, which represents the number of dif... | 70 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
When the expression changes, the minimum value of the fraction $\frac{3 x^{2}+6 x-5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$. (1993, National Junior High School Competition) | Let $y=\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$.
(*)
Rearranging (*), we get
$$
(y-6) x^{2}+(2 y-12) x+2 y-10=0 \text {. }
$$
Since $x$ is a real number, we have
$$
\Delta=(2 y-12)^{2}-4(y-6)(2 y-10) \geqslant 0 \text {, }
$$
which simplifies to $y^{2}-10 y+24 \leqslant 0$.
Solving this, we get $4 \leqslant y \le... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. Let $x$ be a positive real number, then the minimum value of the function $y=x^{2}-x+\frac{1}{x}$ is $\qquad$ - (1995, National Junior High School Competition) | - $y=x^{2}-x+\frac{1}{x}$ $=(x-1)^{2}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+1$.
From the above equation, it is known that when $x-1=0$ and $\sqrt{x}-\frac{1}{\sqrt{x}}$ $=0$, i.e., $x=1$, $y$ takes the minimum value of 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 12. $a, b, c$ are all integers, and $a b c=1990$. Find the minimum value of $a b+b c+c a$. (1990, Jinyun Cup Junior Competition) | $$
\begin{array}{l}
\text { Sol } \because 1990=1 \times 1 \times 1990 \\
=(-1) \times(-1) \times 1990 \\
= 1 \times(-1) \times(-1990) \\
= 1 \times 2 \times 995 \\
=(-1) \times(-2) \times 995 \\
= \cdots=2 \times 5 \times 199=1 \times 10 \times 199,
\end{array}
$$
When $a=-1, b=-1, c=1990$ (letters can be swapped), ... | -3979 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 13. Let $a, b, c$ be distinct integers from 1 to 9. What is the largest possible value of $\frac{a+b+c}{a b c}$? (1992, 1st Dannevirke-Shanghai Friendship Correspondence Competition) | Let $P=\frac{a+b+c}{a b c}$.
In equation (1), let $a, b$ remain unchanged temporarily, and only let $c$ vary, where $c$ can take any integer from 1 to 9. Then, from $P=\frac{a+b+c}{a b c}=\frac{1}{a b}+\frac{a+b}{a b c}$, we know that when $c=1$, $P$ reaches its maximum value. Therefore, $c=1$.
Thus, $P=\frac{a+b+1}{a ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. There are 120 equally distributed spheres inside a regular tetrahedron $A-B C D$. How many spheres are placed at the bottom of this regular tetrahedron? | Analysis (1) The base of the regular tetrahedron $A-BCD$ is the equilateral $\triangle BCD$. Assuming the number of spheres closest to the edge $BC$ is $n$, then the number of rows of spheres tangent to the base $\triangle BCD$ is also $n$, with the number of spheres in each row being $n, n-1, \cdots, 3,2,1$. Thus, the... | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. If 20 points divide a circle into 20 equal parts, then the number of regular polygons that can be formed with vertices only among these 20 points is ( ) .
(A) 4
(B) 8
(C) 12
(D) 24 | 6. (C).
Let the regular $k$-sided polygon satisfy the condition, then the 20 $-k$ points other than the $k$ vertices are evenly distributed on the minor arcs opposite to the sides of the regular $k$-sided polygon.
Thus, $\frac{20-k}{k}=\frac{20}{k}-1$ is an integer, so $k \mid 20$.
But $k \geqslant 3$,
$\therefore k=4... | 12 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
Example 6. When $|x+1| \leqslant 6$, the maximum value of the function $y=x|x|$ $-2 x+1$ is $\qquad$. (1994, National Junior High School Competition) | Solve: From $|x+1| \leqslant 6$ we get $-7 \leqslant x \leqslant 5$.
When $0 \leqslant x \leqslant 5$,
$$
y=x^{2}-2 x+1=(x-1)^{2} \text {. }
$$
Since $x=1 \in[0,5]$, and $1>\frac{-7+0}{2}, \\
\therefore y_{\text {max }}=f(5)=(5-1)^{2}=16; \\
y_{\text {min }}=f(1)=(1-1)^{2}=0 .
$$
When $-7 \leqslant x < 0$,
$$
y=x^{2}... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Given that $a, b, c$ are positive integers, and the parabola $y=$ $a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A, B$. If the distances from $A, B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
Translate the above text into English, please retain the origin... | Three, let the coordinates of $A, B$ be $\left(x_{1}, 0\right),\left(x_{2}, 0\right)$ and $x_{1}<0, x_{2}>0$,
\[
\begin{array}{l}
\therefore x_{1} x_{2}=\frac{c}{a}<0, \\
\therefore a c<0, \text { i.e., } a>0, c<0 .
\end{array}
\]
\[
\begin{array}{l}
\because x_{1}+x_{2}=-\frac{b}{a}>0, \\
\therefore b<0 .
\end{array}
... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the product of four real numbers is 1, and the sum of any one of the numbers with the product of the other three is 1000. Then the sum of these four numbers is | $=、 1.2000$.
These four real numbers are all non-zero and not all equal, denoted as $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$. Taking any one of these numbers $x_{i} (i=1,2,3,4)$, the other three numbers are $x_{j}$, $x_{k}$, $x_{r}$, then we have
$$
\left\{\begin{array}{l}
x_{i}+x_{j} x_{k} x_{r}=1000, \\
x_{i} x_{j} x_{k} ... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) As shown in the figure, given that $AB, CD$ are perpendicular chords in a circle $\odot O$ with radius 5, intersecting at point $P$. $E$ is the midpoint of $AB$, $PD=AB$, and $OE=3$. Try to find the value of $CP + CE$.
---
The translation is provided as requested, maintaining the original text's form... | By the intersecting chords theorem, we have
$$
C P \cdot P D=A P \cdot P B \text {. }
$$
Also, $A E=E B=\frac{1}{2} A B$,
$$
\begin{aligned}
\because C P \cdot P D & =\left(\frac{1}{2} A B+E P\right)\left(\frac{1}{2} A B-E P\right) \\
& =\frac{1}{4} A B^{2}-E P^{2} .
\end{aligned}
$$
And $P D=A B$,
$$
\therefore \qua... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 8. If the polynomial $P=2a^{2}-8ab+17b^{2}$ $-16a-4b+1990$, what is the minimum value of $P$? (1990, Junior High School Correspondence Contest in Some Provinces and Cities) | $$
\begin{array}{r}
\text { Sol } P=2 a^{2}-8 a b+17 b^{2}-16 a-4 b \\
+1990=2(a-2 b-4)^{2}+9(b-2)^{2}+1922 .
\end{array}
$$
From this, for all real numbers $a, b, P \geqslant 1922$. The equality holds only when $a-2 b-4=0, b-2=0$, i.e., $a=$ $8, b=2$. Therefore, the minimum value of $P$ is $1922$. | 1922 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The Dao sequence $\left\{\begin{array}{l}x_{1}=x_{2}=1, \\ x_{n+2}=a x_{n+1}+b x_{n}(n \in N) .\end{array}\right.$
If $T=1996$ is the smallest natural number such that $x_{T+1}=x_{T+2}=1$, then $\sum_{i=1}^{1006} x_{i}=$ $\qquad$ . | 2. 0 .
From $T=1996$ being the smallest natural number that makes $x_{1+1}=x_{T+2}=1$, we know that $a+b \neq 1$. Otherwise, if $a+b=1$, then $x_{3}=a x_{2}+b x_{1}=a+b=1$. This would mean that when $T=1$, $x_{T+1}=x_{T+2}=1$, which contradicts the fact that $T=1996$ is the smallest value. Furthermore, we have
$$
\beg... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. In the plane $\alpha$ there is a $\triangle A B C, \angle A B C=105^{\circ}$, $A C=2(\sqrt{6}+\sqrt{2})$. On both sides of the plane $\alpha$, there are points $S, T$, satisfying $S A=S B=S C=\sqrt{41}, T A=T B=T C=$ 5. Then $S T=$ $\qquad$. | 3. 8 .
From Figure 4, since $S A=S B=S C$, we know that the projection $D$ of $S$ on plane $\alpha$ is the circumcenter of $\triangle A B C$. Similarly, the projection of $T$ on plane $\alpha$ is also the circumcenter of $\triangle A B C$. By the uniqueness of the circumcenter, we have $S T \perp a$.
Connect $A D$. By... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given functions $f(x)$ and $g(x)$ are defined on $R$, and $f(x-y)=f(x) g(y)-g(x) f(y)$. If $f(1)=f(2) \neq$ 0, then $g(1)+g(-1)=$ $\qquad$ . | 4. 1 .
For $x \in R$, let $x=u-v$, then we have
$$
\begin{array}{l}
f(-x)=f(v-u) \\
=f(v) g(u)-g(v) f(u) \\
=-[f(u) g(v)-g(u) f(v)] \\
=-f(u-v)=-f(x) .
\end{array}
$$
Therefore, $f(x)$ is an odd function, and we have
$$
\begin{array}{l}
f(2)=f[1-(-1)] \\
=f(1) g(-1)-g(1) f(-1) \\
=f(1) g(-1)+g(1) f(1) .
\end{array}
$... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ have side lengths $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$, respectively, and their areas be $S$ and $S^{\prime}$. If for all values of $x$, $a x^{2}+b x+c=3\left(a^{\prime} x^{2}+b^{\prime} x+c^{\prime}\right)$ always holds, then $\fra... | II. 1. From the given, $\left(a-3 a^{\prime}\right) x^{2}+\left(b-3 b^{\prime}\right) x+(c$ $\left.-3 c^{\prime}\right)=0$ is an identity, hence $\left(a-3 a^{\prime}\right)=0,\left(b-3 b^{\prime}\right)=0$, $\left(c-3 c^{\prime}\right)=0$, which gives $\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}=3$,... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $a_{1}, a_{2}, \cdots, a_{k}$ be $k$ distinct positive integers, and $a_{1}+a_{\varepsilon}+\cdots+a_{k}=1995$. Then, the maximum value of $k$ is $\qquad$ | 2. $\because a_{1}, a_{2}, \cdots, a_{4}$ are all distinct, without loss of generality, let $a_{1}<a_{2}<\cdots<a_{k}$. Since $a_{1}, a_{2}, \cdots, a_{k}$ are all positive integers, we have $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots, a_{k} \geqslant k$, thus $1+2+\cdots+k \leqslant a_{1}+a_{2}+\cdots+a_{k}=$ 1995, ... | 62 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. If $\sqrt{a^{2}-1996}$ is an integer, then the minimum value of the integer $a$ is | 5. Consider the case of natural number $a$: Let $\sqrt{a^{2}-1996} = m$, then $a^{2}-m^{2}=1996$, $(a+m)(a-m)=1996$. Since $a+m$ and $a-m$ are both even numbers, and $1996=2 \times 998$ is the only way to split it into the product of two even numbers, we have $a+m=998, a-m=2$, solving for $a=500, m=498$. Since the larg... | -500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 14 points) If the sum of the digits of a natural number and the product of its digits add up to exactly the natural number itself, we call it a "lucky number." Try to find the sum of all "lucky numbers."
---
Translate the above text into English, please retain the original text's line breaks and for... | Let the $n$-digit natural number be $N=\overline{a_{1} a_{2} \cdots a_{n}}$, then
\[ N=a_{1} \times 10^{n-1}+a_{2} \times 10^{n-2}+\cdots+a_{n-1} \times 10+a_{n} \]
(Where $a_{1}, a_{2}, \cdots, a_{n}$ are the digits of $N$).
If $N$ is a "lucky number", then
\[
\begin{array}{l}
a_{1} \times 10^{n-1}+a_{2} \times 10^{... | 531 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of diagonals in a convex nonagon is $\qquad$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The blank space represented by $\qquad$ is kept as is in the translation. | 3. 27. (Hint: Each vertex leads to $(9-3)=6$ diagonals, there are 9 vertices in total, and each diagonal passes through two vertices, so the total number of diagonals is $6 \times 9 \div 2=27$.) | 27 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
7. Given that the six interior angles of a regular hexagon are all $120^{\circ}$, and the lengths of four consecutive sides are $1, 9, 9, 5$ cm respectively. Then, the perimeter of this hexagon is $\qquad$ cm. | 7. 42. (Hint: Extend the three pairs of opposite sides of the hexagon, the result is an equilateral triangle, it is easy to know that the side length of the equilateral triangle is 19 cm, thus find the other two side lengths of the hexagon to be 13 cm and 5 cm.) | 42 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. In the interior angles of an $n(n \geqslant 3)$-sided polygon, the maximum number of acute angles is $\qquad$ . | II, 1.3.
Let a polygon with $n$ sides have $k$ acute angles, then
$$
(n-2) \times 180^{\circ}<k \times
$$
$90^{\circ}+(n-k) \times 180^{\circ}$.
Thus, $k<4$.
As shown in the figure, construct a $60^{\circ}$ sector $A_{1} A_{2} A_{1}$. When $n=3$, $\triangle A_{1} A_{2} A_{3}$ has three interior angles, all of which are... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. There are two roads $O M, O N$ intersecting at a $30^{\circ}$ angle. Along the direction of road $O M$, 80 meters from $A$ is a primary school. When a tractor travels along the direction of $O N$, areas within 50 meters on both sides of the road will be affected by noise. Given that the speed of the tractor is 18 ki... | 3. 12 .
As shown in the figure, when the tractor reaches point $B$, it starts to have an impact, and when it reaches point $C$, the impact ceases. Thus, $A B = A C = 50$ meters. Draw $A D \perp$ $O N$ from $A$. Given that $\angle A O D = 30^{\circ}$, we know that $A D = \frac{1}{2} O A = 40$ meters. Therefore, $B D = ... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let real numbers $x, y$ satisfy the equation $9 x^{2}+4 y^{2}-3 x+2 y=0$. Then the maximum value of $z=3 x+2 y$ is $\qquad$ . | From $9 x^{2}+4 y^{2}=3 x-2 y$ we know $3 x-2 y \geqslant 0$ and $\frac{9 x^{2}+4 y^{2}}{3 x-2 y}$ $=1$,
we have $z=3 x+2 y=\frac{(3 x)^{2}-(2 y)^{2}}{3 x-2 y} \leqslant \frac{(3 x)^{2}+(2 y)^{2}}{3 x-2 y}$ $=1$ (equality holds when $y=0$). | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 17. Given as shown, in quadrilateral $ABCD$, $AD=DC=1, \angle DAB=$ $\angle DCB=90^{\circ}, BC, AD$ extended intersect at $P$. Find the minimum value of $AB \cdot S_{\triangle PAB}$.
(1994, Sichuan Province Junior High School Mathematics League Competition) | Let $DP = x$, then $PC = \sqrt{x^2 - 1}$.
Since $\triangle PCD \sim \triangle PAB$,
$\therefore CD \perp AB = PC : PA$.
$\therefore AB = \frac{CD \cdot PA}{PC} = \frac{x + 1}{\sqrt{x^2 - 1}}$.
Let $y = AB \cdot S_{\triangle PAB}$, then
$$
y = \frac{1}{2} AB^2 \cdot PA = \frac{(x + 1)^3}{2(x^2 - 1)}.
$$
Eliminating the... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Write $(\sqrt{7}-\sqrt{6})^{6}$ in the form $\sqrt{N+1}-\sqrt{N}$, where $N$ is a natural number. Then $N=$ $\qquad$ | 5. $N=76545000$.
Let $(\sqrt{7}-\sqrt{6})^{6}=\sqrt{N+1}-\sqrt{N}$, then $(\sqrt{7}-\sqrt{6})^{6}+\sqrt{N}=\sqrt{N+1}$.
Square it to get
$$
\begin{array}{l}
(\sqrt{7}-\sqrt{6})^{12}+2 \sqrt{N}(\sqrt{7}-\sqrt{6})^{6}+N \\
= N+1 \\
2 \sqrt{N}=\frac{1-(\sqrt{7}-\sqrt{6})^{12}}{(\sqrt{7}-\sqrt{6})^{6}} \\
=\frac{(7-6)^{6}... | 76545000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
*6. From $1,2, \cdots, 1996$, select $k$ numbers such that the sum of any two numbers cannot be divisible by their difference. Then the maximum possible value of $k$ is $\qquad$ . | 6. 666 .
Classify $1,2, \cdots, 1996$ by their remainder when divided by 3:
$$
\begin{array}{l}
3 k+1: 1,4,7, \cdots, 1996, \text { a total of 666; } \\
3 k+2: 2,5,8, \cdots, 1994, \text { a total of 665; } \\
3 k: 3,6,9, \cdots, 1995, \text { a total of 665. }
\end{array}
$$
Select all $3 k+1$ numbers, a total of 666... | 666 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For any natural number $n$, connect the origin $O$ and the point $A_{n}(n, n+3)$. Let $f(n)$ denote the number of integer points on the line segment $O A_{n}$, excluding the endpoints. Try to find: $f(1)+f(2)+\cdots+f(1996)$. | The equation of the line segment $O A_{n}$ is $y=\frac{n+3}{n} x(0 \leqslant x \leqslant n)$. Therefore, $f(x)$ is the number of integer solutions of this equation in the interval $(0, n)$. When $n=3 k$ (where $k$ is a positive integer), the equation becomes
$$
y=\frac{k+1}{k} x(0<x<3 k) \text {. }
$$
It has two sets ... | 1330 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
In $\triangle A B C$, $A B=37, A C=58$. With $A$ as the center and $A B$ as the radius, an arc is drawn intersecting $B C$ at point $D$, and $D$ is between $B$ and $C$. If the lengths of $B D$ and $D C$ are both integers, find the length of $B C$. | Solve as shown in the figure, first prove
$$
\begin{array}{l}
B C \cdot C D \\
=A C^{2}-A B^{2} .
\end{array}
$$
Draw $A H \perp B C$ at $H$, then
$$
\begin{array}{l}
A C^{2}=A H^{2}+C H^{2}, \\
A B^{2}=A H^{2}+B H^{2},
\end{array}
$$
we have
$$
\begin{array}{l}
A C^{2}-A B^{2}=C H^{2}-B H^{2} \\
\quad=(C H+B H)(C H-... | 57 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. If $a=\sqrt{17}-1$, find the value of $\left(a^{5}+2 a^{4}-17 a^{3}\right.$ $\left.-a^{2}+18 a-17\right)^{1993}$.
(Adapted from the 1987 Chongqing Junior High School Mathematics Invitational Competition) | Given $a+1=\sqrt{17}$, squaring both sides yields $a^{2}+2 a+1=17$.
Then
$$
\begin{array}{l}
\left(a^{5}+2 a^{4}-17 a^{3}-a^{2}+18 a-17\right)^{1993} \\
=\left[a^{5}+2 a^{4}-\left(a^{2}+2 a+1\right) a^{3}-a^{2}+\left(a^{2}\right.\right. \\
\left.+2 a+1+1) a-\left(a^{2}+2 a+1\right)\right]^{1993} \\
=(-1)^{1993}=-1 \tex... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. If $m^{2}=m+1, n^{2}=n+1$, then what is the value of $m^{5}+n^{5}$?
(1989, Jiangsu Province High School Mathematics Competition) | Let $S_{k}=m^{k}+n^{k}$, construct the recurrence relation for $S_{k}$.
$$
\begin{array}{l}
\because m^{2}=m+1, n^{2}=n+1, \\
\therefore m^{k}=m^{k-1}+m^{k-2}, n^{k}=n^{k-1}+n^{k-2} .
\end{array}
$$
Adding the two equations gives $S_{k}=S_{k-1}+S_{k-2}$.
$$
\begin{aligned}
\therefore S_{5} & =S_{4}+S_{3}=\left(S_{3}+S... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. Let real numbers $a, b, x$ and $y$ satisfy $a x + b y = 3, a x^2 + b y^2 = 7, a x^3 + b y^3 = 16, a x^4 + b y^4 = 42$. Find the value of $a x^5 + b y^5$.
(Eighth American Invitational Mathematics Examination) | Let $S_{\mathrm{n}}=a x^{n}+b y^{\star}, A=x+y, B=x y$, then $x, y$ are the two roots of the equation $u^{2}-A u+B=0$. Therefore,
$$
\begin{array}{l}
x^{2}-A x+B=0, y^{2}-A y+B=0 . \\
\therefore a x^{n}-A\left(a x^{n-1}\right)+B\left(a x^{n-2}\right)=0, \\
b y^{n}-A\left(b y^{n-1}\right)+B b y^{n-2}=0 .
\end{array}
$$
... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. Find the integer part of $(\sqrt{3}+1)^{6}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text{Solve for the dual number } B = (\sqrt{3}-1)^{6} \text{ of the number } A = (\sqrt{3}+1)^{6}, \text{ and examine} \\
A+B=(\sqrt{3}+1)^{6}+(\sqrt{3}-1)^{6} \\
= \left[(\sqrt{3}+1)^{3}\right]^{2}+\left[(\sqrt{3}-1)^{3}\right]^{2} \\
= \left[(\sqrt{3}+1)^{3}-(\sqrt{3}-1)^{3}\right] \\
+2[(\sqrt{3... | 415 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 10. Given $a^{2}+2 a-5=0, b^{2}+2 b-5=0$, and $a \neq b$. Then the value of $a b^{2}+a^{2} b$ is what?
(1989, Sichuan Province Junior High School Mathematics Competition) | Given that $a, b$ are the roots of the quadratic equation $x^{2}+$ $2 x-5=0$, then $a+b=-2, ab=-5$. Therefore, $ab^{2}+a^{2}b=ab(a+b)=10$. Note: Here, the definition of the roots is used to construct the equation. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a+b+c=0, a^{3}+b^{3}+c^{3}=0$. Find the value of $a^{15}+b^{15}+c^{15}$. | 2. From $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right.$ $-a b-b c-a c)$, we have $a b c=0$. Then at least one of $a, b, c$ is zero, for example, $c=0$, then it is known that $a, b$ are opposites, i.e., $a^{15}+b^{15}+c^{15}$ $=0$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. Find a natural number $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square.
(2nd All-Russian High School Mathematics Olympiad) | Let $2^{4}=x$, then $2^{8}=x^{2}, 2^{11}=2^{7} \cdot x$. Therefore, $2^{8}+2^{11}+2^{n}=x^{2}+2^{7} x+2^{n}$.
According to the condition for a quadratic trinomial to be a perfect square, we get $\Delta=\left(2^{7}\right)^{2}-4 \times 2^{n}=0$.
Solving for $n$ yields $n=12$. | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $\frac{x}{m}+\frac{y}{n}+\frac{z}{p}=1, \frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0$. Calculate the value of $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}}$. | 6. From $\frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0$, we have $\frac{m y z+n x z+p x y}{x y z}=0$.
Since $x, y, z$ are not zero, then $m y z+n x z+p x y=0$. Also,
$$
\begin{aligned}
& \frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}} \\
= & \left(\frac{x}{m}+\frac{y}{n}+\frac{z}{p}\right)^{2}-2\left(\frac{x y}{m ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Find the minimum value of $|x-1|+|x-2|+|x-3|+\cdots$ $+|x-1996|$. | 8. When $998 \leqslant x \leqslant 999$, the original expression has a minimum value, the minimum value is:
$$
\begin{array}{l}
(1996-1)+(1995-2)+\cdots+(999-998) \\
=1995+1993+\cdots+1 \\
=\frac{(1+1995) \times 998}{2}=998^{2}=996004 .
\end{array}
$$ | 996004 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that $x, y, z$ are three non-negative rational numbers, and satisfy $3 x$ $+2 y+z=5, x+y-z=2$. If $S=2 x+y-z$, then what is the sum of the maximum and minimum values of $S$? | 9. $\left\{\begin{array}{l}3 x+y+z=5, \\ x+y-z=2, \\ 2 x+y-z=S .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}x=S-2, \\ y=\frac{15-4 S}{3}, \\ z=\frac{3-S}{3} .\end{array}\right.$ Since $x, y, z$ are all non-negative, we have
$$
\left\{\begin{array} { l }
{ S - 2 \geqslant 0 , } \\
{ \frac { 1 5 - 4 S } ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Among the students in the second year of junior high school, 32 students participated in the math competition, 27 students participated in the English competition, and 22 students participated in the Chinese competition. Among them, 12 students participated in both math and English, 14 students participated in both... | 10. Let the number of people who participate in all three subjects be $x$, and the total number of participants be $y$, then
$$
\begin{aligned}
y & =32+27+22-12-14-10+x \\
& =45+x .
\end{aligned}
$$
Given $2 \leqslant x \leqslant 10$, hence $47 \leqslant y \leqslant 55$.
Only when $y=48$, it can be divisible by 8, so ... | 30 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
II. (16 points) A plot of land can be covered by $n$ identical square tiles. If smaller identical square tiles are used, then $n+76$ such tiles are needed to cover the plot. It is known that $n$ and the side lengths of the tiles are integers. Find $n$.
| Let the side length of the larger tile be $x$, and the side length of the smaller tile be $r$. Then $x, y$ are both natural numbers, and
$$
n x^{2}=(n+76) y^{2} \text {. }
$$
If $(x, y)=d$, let $x_{1}=\frac{x}{d}, y_{1}=\frac{y}{d}$, then $x_{1}, y_{1}$ are both natural numbers, and $\left(x_{1}, y_{1}\right)=1$. Thus... | 324 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four points are such that no three are collinear, and no two are coincident. Through each point, draw the perpendiculars to the lines joining the other three points. If we do not count the original four points, how many different intersection points are there among these perpendiculars? Prove your conclusion. | For each point, draw perpendicular lines to the lines connecting it to the other three points. Since there are 3 such lines for each point, and there are 4 points, there are a total of $4 \times 3 = 12$ perpendicular lines.
If every two perpendicular lines intersect, there would be $\frac{12 \times 11}{2} = 66$ inters... | 40 | Geometry | proof | Yes | Yes | cn_contest | false |
3. In $\triangle A B C$, $G$ is the centroid, and $I$ is the intersection of the angle bisectors of $\angle B$ and $\angle C$. If $I G / / B C$, and $B C=5$, then $A B+A C$ $=$ . $\qquad$ | 3. Connect $A G, B G$ and $C G$, and extend $A G$ to intersect $B C$ at $M$, then $S_{\triangle G B C}: S_{\triangle A B C}=G M: A M=1: 3$.
Connect $A I$, then by $I G$ $/ / B C$, we know
$$
\begin{array}{l}
S_{\triangle I B C}=S_{\triangle G B C}=\frac{S_{\triangle A B C}}{3}, \\
\therefore S_{\triangle I A B}+S_{\tr... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $x_{1}, x_{2}, \cdots, x_{51}$ be natural numbers, $x_{1}<x_{2}$ $<\cdots<x_{51}$, and $x_{1}+x_{2}+\cdots+x_{51}=1995$. When $x_{26}$ reaches its maximum value, the maximum value that $x_{51}$ can take is | 5. To maximize $x_{26}$, then $x_{1}, x_{2}, \cdots, x_{25}$ must be minimized, thus we have
$$
\begin{array}{l}
x_{26}+x_{27}+\cdots+x_{51} \\
=1995-(1+2+\cdots+25)=1670 .
\end{array}
$$
To maximize $x_{51}$, then $x_{26}, x_{27}, \cdots, x_{50}$ must be minimized, thus we have
$$
\begin{array}{l}
x_{26}+\left(x_{26}... | 95 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. There is a four-digit number. Dividing it
in the middle into two parts, we get the front and back two $\uparrow=$ digit numbers. By adding a 0 to the end of the front two-digit number, and then adding the product of the front and back two-digit numbers, we get exactly the original four-digit number. It is also know... | 3. 1995 | 1995 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 20 points, Question (1) 8 points, Question (2) 12 points) As shown in the figure, there is a cube-shaped wire frame, and the midpoints of its sides $I, J, K, L$ are also connected with wire.
(1) There is an ant that wants to crawl along the wire from point A to point G. How many shortest routes are th... | (1) "The shortest path" means that the ant can neither move left nor down, otherwise it would be taking a "detour" rather than the "shortest" path. The ant can crawl on face $ABCGFE$ or on $ADCGHE$. The number of "shortest paths" on these two faces is the same, and the possible paths can be represented by the following... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. Given $x, y, z \in R^{+}$, and $x+y+z=1$. Prove: $\frac{1}{x}+\frac{4}{y}+\frac{9}{z} \geqslant 36$.
(1990, Japan IMO Team Selection Test) | Prove that for $x=\sin ^{2} \alpha \cos ^{2} \beta, y=\cos ^{2} \alpha \cos ^{2} \beta$, $z=\sin ^{2} \beta$, where $\alpha, \beta$ are acute angles, we have
$$
\begin{array}{l}
\frac{1}{x}+\frac{4}{y}+\frac{9}{z} \\
=\left(1+\operatorname{ctg}^{2} \alpha\right)\left(1+\operatorname{tg}^{2} \beta\right)+4\left(1+\opera... | 36 | Inequalities | proof | Yes | Yes | cn_contest | false |
3. Let $a_{1}, a_{2}, \cdots, a_{n}$ represent any permutation of the integers $1,2, \cdots, n$. Let $f(n)$ be the number of such permutations such that
(i) $a_{1}=1$;
(ii) $\left|a_{i}-a_{i+1}\right| \leqslant 2 . i=1,2, \cdots, n-1$.
Determine whether $f(1996)$ is divisible by 3. | 3. Verify that $f(1)=f(2)=1$ and $f(3)=2$.
Let $n \geqslant 4$. Then it must be that $a_{1}=1, a_{2}=2$ or 3.
For $a_{2}=2$, the number of permutations is $f(n-1)$, because by deleting the first term and reducing all subsequent terms by 1, we can establish a one-to-one correspondence of sequences.
If $a_{2}=3$, then ... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. If $360^{x}=3, 360^{y}=5$, then $72^{\frac{1-2 x-y}{3(1-y)}}=$ | \begin{array}{l}=1.2. \\ \because 72=\frac{360}{5}=\frac{360}{360^{y}}=360^{1-y}, \\ \therefore \text { original expression }=\left(360^{1-y}\right)^{\frac{1}{3(2 x-y)}} \\ =(360)^{\frac{1}{3}(1-2 x-y)}-\left(360^{1-2 x-y}\right)^{\frac{1}{3}} \\ =\left(\frac{360}{\left(360^{x}\right)^{2} \cdot 360^{5}}\right)^{\frac{1... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If $x=\frac{1}{2}-\frac{1}{4 x}$, then $1-2 x+2^{2} x^{2}-2^{3} x^{3}+2^{4} x^{4}$ $-\cdots-2^{1995} x^{1995}$ is. $\qquad$. | 2. 1 .
From $x=\frac{1}{2}-\frac{1}{4 x}$ we can get $1-2 x+(2 x)^{2}=0$.
$$
\begin{array}{c}
\therefore \text { the original expression }=1-2 x\left[1-2 x+(2 x)^{2}\right]+(2 x)^{4}[1-2 x \\
\left.+(2 x)^{2}\right]-\cdots-(2 x)^{1993}\left[1-2 x+(2 x)^{2}\right]=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. From .1 to Э these nine natural numbers, select two, separately as the logarithm's number and base, to get different logarithmic values ( ).
(A) 52
(B) 53
(C) 57
(D) 72 | 2. (B).
Since the base is not 1, the logarithm and the base cannot take the same value, so only 64 different logarithmic forms can be taken. Among them, there are 8 cases where the true value is 1, all taking the value 0. In addition,
$$
\begin{array}{ll}
\log _{2} 3=\log _{4} 9 & \log _{2} 4=\log _{3} 9, \\
\log _{3}... | 53 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
4. Let $P=\{$ natural numbers no less than 3 $\}$. Define the function $f$ on $P$ as follows: if $n \in P, f(n)$ represents the smallest natural number that is not a divisor of $n$, then $f(360360)=$ $\qquad$ . | 4. 16 .
Since $360360=2^{3} \times 3^{2} \times 5 \times 7 \times 11 \times 13$, we know that the divisors of 360360 in ascending order are $1,2,3,4,5,6,7,8,9,10$, $11,12,13,14,15,18,20, \cdots$, the smallest natural number that is not a divisor of 360360 is 16. Therefore, $f(360360)=16$. | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$5 . n$ is a positive integer not exceeding 1996. If there is a 0 such that $(\sin \theta+i \cos \theta)^{n}=\sin n \theta+i \cos n 0$ holds, then the number of $n$ values satisfying the above condition is $\qquad$. | 5. 498 .
$$
\begin{array}{l}
\because(\sin \theta+i \cos \theta)^{n}=[i(\cos \theta-i \sin \theta)]^{n} \\
\quad=i^{n}(\cos n \theta-\sin n \theta)=i^{n-1}(\sin n \theta+i \cos n \theta), \\
\text { and }(\sin \theta+i \cos \theta)^{n}=\sin n \theta+i \cos n \theta, \\
\therefore i^{n-1}(\sin n \theta+i \cos n \theta)=... | 498 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. In the sequence of natural numbers starting from 1, certain numbers are colored red according to the following rules. First, color 1; then color two even numbers 2, 4; then color the three consecutive odd numbers closest to 4, which are $5, 7, 9$; then color the four consecutive even numbers closest to 9, which are ... | 6. 3929 .
The first time, color one number red: $1,1=1^{2}$;
The second time, color 2 numbers red: $2,4,4=2^{2}$;
The third time, color 3 numbers red: $5,7,9,9=3^{2}$;
Guessing, the last number colored red in the $k$-th time is $k^{2}$. Then the $k+1$ numbers colored red in the $(k+1)$-th time are:
$$
k^{2}+1, k^{2}+3... | 3929 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. If $x=\sqrt{19-8 \sqrt{3}}$, then the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15}=$ $\qquad$ | Solve: From $x=\sqrt{19-2 \sqrt{48}}=4-\sqrt{3}$, we get $x-4=-\sqrt{3}$. Squaring both sides and rearranging, we obtain
$$
x^{2}-8 x+13=0 \text {. }
$$
Therefore,
$$
\begin{aligned}
\text { Original expression } & =\frac{\left(x^{2}-8 x+13\right)\left(x^{2}+2 x+1\right)+10}{\left(x^{2}-8 x+13\right)+2} . \\
& =5 .
\e... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a, b$ be positive integers, and $a+b \sqrt{2}$ $=(1+\sqrt{2})^{100}$. Then the units digit of $a b$ is $\qquad$ | 6.4.
By the binomial theorem, we have
$$
a-b \sqrt{2}=(1-\sqrt{2})^{100} \text {. }
$$
Therefore, $a=\frac{1}{2}\left((1+\sqrt{2})^{100}+(1-\sqrt{2})^{100}\right)$,
$$
\begin{array}{l}
b=\frac{1}{2 \sqrt{2}}\left((1+\sqrt{2})^{100}-(1-\sqrt{2})^{100}\right) . \\
\text { Hence } \left.a b=\frac{1}{4 \sqrt{2}}(1+\sqrt{... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
46. Given that $\overline{a b}$ is a two-digit number, and $\overline{a b c d e}$ is a five-digit number. If $(\overline{a b})^{3}=\overline{a b c d e}$, find $\overline{a b c d e}$. | Solve $(\overline{a b})^{3}=\overline{a b c d e}=1000 \cdot \overline{a b}+\overline{c d e}$.
We have $\left[(\overline{a b})^{2}-1000\right] \cdot \overline{a b}=\overline{c d e}$.
Thus, $(\overline{a b})^{2}>1000$, which means $\overline{a b}>\sqrt{1000}>31$.
Therefore, $\bar{a} \bar{v} \geqslant 32$.
Also, $(\overli... | 32768 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 13. Given $4 x-3 y-6 z=0, x+2 y-7 z$ $=0, x y z \neq 0$. Then the value of $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}$ is equal to | Solve the system of equations by treating $\approx$ as a constant,
$$
\left\{\begin{array}{l}
4 x-3 y=6 z, \\
x-2 y=7 z,
\end{array}\right.
$$
we get
$$
\begin{array}{l}
x=3 z, y=2 z . \\
\text { Therefore, the original expression }=\frac{2 \cdot(3 z)^{2}+3 \cdot(2 z)^{2}+6 z^{2}}{(3 z)^{2}+5 \cdot(2 z)^{2}+7 z^{2}}=1... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 14. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq$ $n$, then $m^{5}+n^{5}=$ $\qquad$ | $$
\begin{array}{l}
\text { Sol } \because(a-b)\left(a^{n}+b^{n-1}\right) \\
=a^{n}+b^{n}+a b\left(a^{n-2}+b^{n-2}\right), \\
\therefore a^{n}+b^{n}=(a+b)\left(a^{n-1}+b^{n-1}\right) \\
-a b\left(a^{n-2}+b^{n-2}\right) \text {. } \\
\end{array}
$$
Let $S_{n}=a^{n}-b^{n}$, we get the recursive formula
$$
S_{n}=(a+b) S_... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the equation $a x^{2}+b x+c=0(a \neq 0)$, the sum of the roots is $s_{1}$, the sum of the squares of the roots is $s_{2}$, and the sum of the cubes of the roots is $s_{3}$. Then the value of $a s_{3}+$ $\left\langle s_{2}\right.$ $+c s_{1}$ is . $\qquad$ | (Tip: Let the two roots of the equation be $x_{1}, x_{2}$.
Then by definition, we have $a x_{1}^{2}+b x_{1}+c=0, a x_{2}^{2}+b x_{2}+c=0$. The original expression $=0$.) | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. If $a$ is a root of $x^{2}-3 x+1=0$, try to find the value of $\frac{2 a^{5}-5 a^{4}+2 a^{3}-8 a^{2}}{a^{2}+1}$. | From the definition of the root, we know that $a^{2}-3 a+1=0$.
Thus, $a^{2}+1=3 a$ or $a^{2}-3 a=-1$ or
$$
\begin{array}{l}
a^{3}=3 a^{2}-a \\
\therefore \text { the original expression }=\frac{a\left[2 a^{2}\left(a^{2}+1\right)-5 a^{3}-8 a\right]}{3 a} \\
=\frac{1}{3}\left(a^{3}-8 a\right)=\frac{1}{3}\left(3 a^{2}-9 a... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 8. Let } x>\frac{1}{4} \text {. Simplify } \sqrt{x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}} \\ -\sqrt{x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1}}=\end{array}$ | Let $a$
$$
\begin{array}{l}
=x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}, b=x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1} . \text { Original } \\
\text { expression }=1 .)
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given
$$
\left\{\begin{array}{l}
1988(x-y)+1989(y-z)+1990(z-x)=0, \\
1988^{2}(x-y)+1989^{2}(y-z)+1990^{2}(z-x)=1989 .
\end{array}\right.
$$
Find the value of $y-z$. | $y-z=-1989$ | -1989 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. The value of $x$ that satisfies the following equations is
$$
\begin{array}{l}
(123456789) x+9=987654321, \\
(12345678) x+8=98765432 . \\
(1234567) x+7=9876543 . \\
\cdots \cdots .
\end{array}
$$ | Observe the numerical changes on both sides of each equation, it is easy to know that the last equation should be $x+1=9$, i.e., $x=8$. Upon verification, $x=8$ is the solution. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. Let $G$ be a simple graph of order 10 and does not contain a cycle of length 4, $C_{4}$. Then the maximum number of edges in $G$ is 16. | Proof: Let $f(n)$ be the maximum number of edges in an $n$-order simple graph without $C_{4}$. Clearly, $f(n)=4$. Now we prove that $f(5)=6$. First, if $G$ is two triangles sharing exactly one common vertex, then $G$ is a 5-order graph with 6 edges and no $C_{4}$. Second, if a 5-order graph $G$ has 7 edges, then there ... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. Let $a^{2}+2 a-1=0, b^{4}-2 b^{2}-1=0$ and $1-a b^{2} \neq 0$. Then the value of $\left(\frac{a b^{2}+b^{2}+1}{a}\right)^{1990}$ is $\qquad$. | From the given, we have $\left(\frac{1}{a}\right)^{2}-\frac{2}{a}-1=0$, $\left(b^{2}\right)^{2}-2 b^{2}-1=0$. By the definition of roots, $\frac{1}{a}, b^{2}$ are the roots of the equation $x^{2}-2 x-1=0$. Then
$$
\begin{aligned}
\frac{1}{a}+b^{2} & =2, \frac{1}{a} \cdot b^{2}=-1 . \\
\therefore \text { the original ex... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Glue the bases of two given congruent regular tetrahedra together, precisely to form a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$ | 4. 3.
As shown in the figure, construct $CE \perp AD$, connect $EF$, it is easy to prove that $EF \perp AD$. Therefore, $\angle CEF$ is the plane angle of the dihedral angle formed by plane $ADF$ and plane $ACD$.
Let $G$ be the midpoint of $CD$. Similarly, $\angle AGB$ is the plane angle of the dihedral angle formed ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Choose several colors from the given six different colors to color the six faces of a cube, with each face being colored with exactly one color, and any two faces sharing a common edge must be colored differently. Then, the number of different coloring schemes is $\qquad$. (Note: If we color two identical cubes and ... | 5. 230 ways.
(1) Using 6 colors, the division is $\frac{6}{5 \times} \times \frac{4 \times 3 \times 2 \times 1}{6 \times 4}=$ 30 ways;
(3) Using 4 colors, the division is $C_{6}^{4} \cdot C_{4}^{2} \cdot \frac{2 \times 1}{2}=90$ ways;
(4) Using 3 colors, the painting method is $C_{6}^{3}=20$ ways; therefore, the total ... | 230 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. In the Cartesian coordinate plane, the number of integer points (i.e., points with both coordinates as integers) on the circumference of a circle centered at $(199,0)$ with a radius of 199 is $\qquad$ . | 6. 4 .
Let $A(x, y)$ be an integer point on circle $O$. As shown in the figure, the equation of circle $O$ is $y^{2}+(x-199)^{2}=$ $199^{2}$.
$$
\begin{array}{l}
\text { Clearly, } x=0, y=0 ; \\
x=199, y=199 ; \\
x=199, y=-199 ; x=389, y=0
\end{array}
$$
These are 4 solutions to the equation. However, when $y \neq 0,... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 8. Calculate
$$
\begin{array}{l}
\sqrt{3633 \times 3635 \times 3639 \times 3641+36} \\
-3636 \times 3638=
\end{array}
$$ | Let $3637=a$. Then
the original expression $=$
$$
\begin{array}{l}
\sqrt{(a-4)(a-2)(a+2)(a+4)+36} \\
-(a+1)(a-1) \\
=\sqrt{\left(a^{2}-10\right)^{2}}-\left(a^{2}-1\right) \\
=a^{2}-10-a^{2}+1=-9 . \\
\end{array}
$$ | -9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. (Ireland) A conference is attended by $12 k$ people, each of whom has greeted exactly $3 k+6$ others. For any two people, the number of people who have greeted both of them is the same. How many people attended the conference? | For any two people, let the number of other people who have greeted these two people be $n$. By the problem, $n$ is fixed. For a specific person $a$, let $\mathrm{B}$ be the set of all people who have greeted $a$, and $\mathrm{C}$ be the set of people who have not greeted $a$. Then, $\mathbf{B}$ contains $3 k+6$ people... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. (Flowing Orchid) For $x \geqslant 1$, let $p(x)$ be the smallest prime number that does not divide $x$, and $q(x)$ be the product of all primes less than $p(x)$. Specifically, $p(1)=2$. If some $x$ makes $p(x)=2$, then define $q(x)=1$. The sequence $x_{0}, x_{1}, x_{2}, \cdots$ is defined by the following formula, w... | Obviously, from the definitions of $p(x)$ and $q(x)$, it can be derived that for any $x$, $q(x)$ divides $x$. Therefore,
$$
x_{n=1}=\frac{x_{i}}{q\left(x_{x}\right)} \cdot p(x ;)
$$
Moreover, it is easy to prove by induction that for all $n$, $x_{n}$ has no square factors. Thus, a unique encoding can be assigned to $x... | 142 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. (India) Let $N$ be the set of all positive integers. Prove that there exists a unique function $f: \mathrm{N} \rightarrow \mathrm{N}$, satisfying
$$
f(m+f(n))=n+f(m+95),
$$
where $m, n$ are any elements of $N$. And calculate the value of $\sum_{k=1}^{19} f(k)$. | Prove that for all $n \geqslant 1$, let $F(n)=f(n)-95$. By substituting $k$ for $m+95$, the given condition becomes
$$
F(k+F(n))=n+F(k),
$$
where $n \geqslant 1, k \geqslant 96$. In (1), substitute $m$ for $k$, then add $k$ to both sides, and apply the function $F$, we get $F(k+n+F(m))=F(k+F(m+F(n)))$. Using (1) again... | 1995 | Algebra | proof | Yes | Yes | cn_contest | false |
1. $\sqrt{1991 \cdot 1993 \cdot 1995 \cdot 1997+16}=$ | Let $x=1994$, then
$$
\begin{array}{l}
\sqrt{1991 \cdot 1993 \cdot 1995 \cdot 1997+16} \\
=\sqrt{(x-3)(x-1)(x+1)(x+3)+16} \\
=\sqrt{\left(x^{2}-1\right)\left(x^{2}-9\right)+16}=\sqrt{x^{4}-10 x^{2}+25} \\
=\sqrt{\left(x^{2}-5\right)^{2}}=1994^{2}-5 \\
=3976031 .
\end{array}
$$ | 3976031 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $a, b$ be unequal real numbers, and $a^{2}+2 a=$ $5 \cdot b^{2}+2 b=5$. Then $a^{2} b+a b^{2}=$ | 4. (10).
It is known that $a, b$ are two unequal real numbers which are exactly the two real roots of the equation $x^{2}+2 x-5=0$. Therefore, $a+b=-2, ab=-5$.
Thus, $a^{2} b+ab^{2}=ab(a+b)=(-2) \cdot(-5)=$
10. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. If $m=1996^{3}-1995^{3}+1994^{3}-1993^{3}$ $+\cdots+4^{3}-3^{3}+2^{3}-1^{3}$, then the last digit of $m$ is | 5. (0).
$$
2^{3}-1^{3}=7,4^{3}-3^{3}=37,6^{3}-5^{3}=91,8^{3}-7^{3}=
$$
$169,10^{3}-9^{3}=271$. Therefore, the sum of the last digits of the first 10 numbers $10^{3}-9^{3}+8^{3}-7^{3}$ $+6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}$ is $1+9+1+$ $7+7=25$, and the last digit of 25 is 5. The last digit of the algebraic sum of every... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (Full marks 20 points) Given that positive integers $p, q$ are both prime numbers, and $7 p+q$ and $p q+11$ are also prime numbers. Calculate the value of $\left(p^{2}+q^{p}\right)\left(q^{2}+p^{q}\right)$. | Given that $p q+11$ is a prime number, we know that $p q+11$ must be odd. Therefore, $p q$ is even. So, at least one of $p, q$ must be even. But since $p, q$ are both primes, one of $p, q$ must be 2. If $p=q=2$, then $p q+11=15$ is not a prime number. Therefore, $p, q$ cannot both be 2. So, one and only one of $p, q$ i... | 221 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>0$. To make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is $\qquad$
(1993, National Competition) | Solution: According to the problem, all logarithms are positive. Therefore, from (*) we get
$$
\begin{aligned}
\text { LHS } & =\frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}} \\
& \geqslant \frac{(1+1+1)^{2}}{\log _{1993}\left(\frac{x_{0}... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Three. (This question is worth 16 points) Given the set $\{1,2,3,4,5, 6,7,8,9,10\}$. Find the number of subsets of this set that have the following property: each subset contains at least 2 elements, and the absolute difference between any two elements in each subset is greater than 1.
| Let $a_{n}$ be the number of subsets of the set $\{1,2, \cdots, n\}$ that have the given property.
The set $\{1,2, \cdots, n, n+1, n+2\}$ has subsets with the given property divided into two categories: the first category of subsets contains the element $n+2$, and there are $a_{n}+n$ such subsets (i.e., the union of e... | 133 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.3. $x, y$ are natural numbers. $k$ is a natural number of size 1. Find all natural numbers $n$ that satisfy: $3^{n}=x^{k}+y^{k}$, and provide a proof. | 10. 3. Answer: $n=2$.
At the time: Let $3^{n}=x^{k}+y^{k}$, where $x$ and $y$ are prime (assuming $x > y$), $k>1$, and $n$ is a natural number. Of course, neither $x$ nor $y$ can be divisible by 3.
If $k$ is even, then $x^{k}$ and $y^{k}$ leave a remainder of 1 when divided by 3. Thus, the sum of $x^{k}$ and $y^{k}$ ... | 2 | Number Theory | proof | Yes | Yes | cn_contest | false |
Three, (25 points) On the first day of operation, the factory's production does not exceed 20 units, and the daily production increases thereafter, but the increase in production each time does not exceed 20 units. When the daily production reaches 1996 units, find the minimum value of the total production of the facto... | Three, suppose the production on the $n$-th day after the start of work reaches 1996 pieces, and the production on the first day is $a_{1}$ pieces, with the increase in production on the $i$-th day being $a_{i}$ pieces $(i=2,3, \cdots, n)$, then we have
$$
a_{1}+a_{2}+a_{3}+\cdots+a_{n}=1996,0<a_{i} \leqslant 20 .
$$
... | 100600 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. The smallest natural number $a$ that makes the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$ $<a-1995 \frac{1}{3}$ hold for all natural numbers $n$ is $\qquad$. | 6. 1997.
Let $\dot{f}(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$, then $f(n+1)-f(n)=\frac{-1}{(2 n+3)(2 n+2)}<0$, which means $f(n)$ is a decreasing function when $n \geq 0$. Therefore, $f(1)$ is the maximum value. To ensure the inequality holds for all $n \in N$, the maximum value of $f(n)$ must be less t... | 1997 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
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