problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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4. Arrange the numbers $\{2 n+1\}$ in a cycle of: one, two, three, four terms: (3), $(5,7),(9,11,13)$, $(15,17,19,21),(23),(25,27),(29,31$, $33),(35,37,39,41),(43), \cdots$, then the sum of the numbers in the 100th parenthesis is $\qquad$. | 4. The 100th parenthesis
is the fine one in the
25th group,
the subsequence of parentheses in.
Accordingly, it forms
an arithmetic sequence
with the first term being 72 and the
common difference being 80. Therefore, its sum is $72+$ (25-
$$
\text { 1) } \times 80=1992 \text {. }
$$ | 1992 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Determine the smallest positive integer A with the following property: for any permutation of all positive integers from 1001 to 2000, it is possible to find 10 consecutive terms whose sum is greater than or equal to $A$.
| (Continued from page 28)
$S_{i}=b_{i}+b_{i+1}+\cdots+b_{i+9}(i=1,2, \cdots, 1991)$.
Then
$$
\begin{array}{l}
S_{1}+S_{11}+S_{21}+\cdots S_{1991} \\
=b_{1}+b_{2}+\cdots+b_{2000} \\
=\frac{(1001+2000) \times 1000}{2} \\
=1500500 .
\end{array}
$$
Thus, among $S_{1}, S_{11}, S_{21}, \cdots, S_{1991}$, there is at least on... | 15005 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $A, B, C$ be the three interior angles of $\triangle ABC$, then the imaginary part of the complex number
$$
\frac{(1+\cos 2B+i \sin 2 B)(1+\cos 2 C+i \sin 2 C)}{1+\cos 2 A-i \sin 2 A}
$$
is . $\qquad$ | 3. 0 .
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the text "3. 0 ." is already in a numerical and punctuation format that is universal and does not require translation, the output remains the same:
3. ... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $x, y$ be coprime natural numbers, and $xy=$ 1992. Then the number of different ordered pairs $(x, y)$ is $\qquad$ | 5. 8.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. In the parallelepiped $A E C D-A_{1} B_{1} C_{1} D_{1}$, it is known that the diagonals $A_{1} C=4, B_{1} D=2$. If a point $P$ in space satisfies $P A_{1}=3, P C=5$, then $P B_{1}^{2}+P D^{2}=$ | 13. 28 . | 28 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. For the right square prism $A B C D-A_{1} B_{1} C_{1} D_{1}$ with a base edge length of 1. If the dihedral angle $A-B D_{1}-C$ is $\frac{2 \pi}{3}$, then $A A_{1}=$ $\qquad$ | 15. 1.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
16. For a plane region $D$, let $N(D)$ denote the number of all integer points (i.e., points on the $X O Y$ plane where both coordinates $x, y$ are integers) within $D$. If $A$ represents the region enclosed by the curve $y=x^{2} (x \geqslant 0)$ and the two lines $x=10, y=1$ (including the boundaries); $B$ represents ... | 16. 1010 . | 1010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants. If $P(1)=10, P(2)$ $=20, P(3)=30$, then $P(10)+P(-6)=$ $\qquad$ . | 4.
$$
8104 .
$$ | 8104 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $\sigma(S)$ denote the sum of all elements in a non-empty set of integers $S$. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be a set of positive integers, and $a_{1}<a_{2}<\cdots<a_{n}$. For each positive integer $n \leqslant$ 1500, there exists a subset $S$ of $A$ such that $\sigma(S)=n$. Find the minimum... | Let $S_{k}=a_{1}+a_{2}+\cdots+a_{k} \quad(1 \leqslant k \leqslant 11)$. By the given conditions, there exists an index $m$ such that $S_{m-1}S_{k-1}+1$, and $a_{1}+\cdots+a_{k-1}=S_{k-1}$. Thus, there does not exist $S \simeq A$ such that $\sigma(S)=S_{k-1} +1$. Therefore,
$$
S_{k}=S_{k-1}+a_{k} \leqslant 2 S_{k-1}+1 .... | 248 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The chord $AB=18$ of the sector $OAB$, a circle $C$ with radius 6 is exactly tangent to $OA$, $OB$, and the arc $\widehat{AB}$. Another circle $D$ is tangent to circle $C$, $OA$, and $OB$ (as shown in the figure). Then the radius of circle $D$ is $\qquad$. | 4. Solution As shown in the figure, let the radii of $\odot O, \odot D$ be $x, y$ respectively. Then $\alpha C=x-6$.
Since $\triangle O L B \backsim \triangle O M C$.
Then
$$
\frac{O B}{O C}=\frac{L B}{C M},
$$
which means $\frac{x}{x-6}=\frac{9}{6}$.
Solving for $x$ gives $x=18, Q C=12$, $O D=$ $6-y$.
Also, since $\... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $M=2^{7} t=3^{5} s$, where $t$ is odd, and $s$ cannot be divided by 3. Then the sum of all divisors of $M$ in the form of $2^{P} 3^{q}$ is $\qquad$ . | 3. The sum of all divisors of $M$ is
$$
\begin{array}{l}
\left(1+2+\cdots+2^{7}\right)\left(1+3+\cdots+3^{5}\right) \\
=92820 .
\end{array}
$$ | 92820 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $x_{1}, x_{2}, \cdots, x_{57}$ are all positive integers, and $x_{1}+x_{2}+\cdots+x_{57}=100$. Then the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{57}^{2}$ is $\qquad$ | 5. Solution For natural numbers $x_{i}, x_{j}$, if $x_{i}x_{i}^{2}+x_{j}^{2} .
\end{array}
$$
Thus, when the sum $x_{i}+x_{j}$ remains constant, the value is maximized by setting one of the numbers to 1 and the other to the sum minus 1, leading to
$$
\text { to } x_{1}=x_{2}=\cdots=x_{56}=1, x_{57}=44 \text {, when } ... | 1992 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (This question is worth 20 points) In the regular quadrilateral pyramid $P$ $A B C D$, $A B=3, O$ is the projection of $P$ on the base, $P O=6, Q$ is a moving point on $A O$, and the section passing through point $Q$ and parallel to $P A, B D$ is a pentagon $D F G H L$, with the area of the section being $S$. Fi... | $$
\begin{array}{l}
\text { Three, Solution: As shown in the figure, } \because P A / / \text { plane } D F G H L, \\
\therefore P A / / E F, P A / / H L, P A / / Q G . \\
\text { Also, } \because B D / / \text { plane } D F G H L, \\
\therefore B D / / E L, B D / / F H . \\
\text { Therefore, } \frac{E F}{P A}=\frac{B... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Sure, here is the translated text:
```
II. (This question is worth 35 points) Let the set $Z=\left\{z_{1}\right.$, $\left.z_{2}, \cdots, z_{n}\right\}$ satisfy the inequality
$$
\min _{i \neq j}\left|z_{i}-z_{j}\right| \geqslant \max _{i}\left|z_{i}\right| \text {. }
$$
Find the largest $n$, and for this $n$, find al... | Let $\left|z_{m}\right|=\max \left|z_{i}\right|$.
Thus, all points $z_{i}$ on the plane are distributed within a circle of radius $\left|z_{m}\right|=R$ centered at $z_{0}=0$.
It can be seen that on the circumference $R$, the six vertices $z,(j=1,2,3,4,5,6)$ of any inscribed regular hexagon, along with $z_{0}=0$, sati... | 7 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
2. If $n$ is a natural number, and $n^{3}+2 n^{2}+9 n+8$ is the cube of some natural number, then $n=$ $\qquad$ . | 2. 7 .
Since $n \in N$,
$$
\begin{aligned}
n^{3} & <n^{3}+2 n^{2}+9 n+8 \\
& <(n+2)^{3}=n^{3}+6 n^{2}+12 n+8,
\end{aligned}
$$
thus, only
$$
\begin{array}{l}
n^{3}+2 n^{2}+9 n+8 \\
=(n+1)^{3}=n^{3}+3 n^{2}+3 n+1 .
\end{array}
$$
Solving for the positive root $n=7$. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. From the set $\{19,20,21, \cdots, 91,92,93\}$, the total number of ways to select two different numbers such that their sum is even is $\qquad$ | 4. 1369.
From 1993 - a total of 93-19+1=75 numbers. Among them, there are 38 odd numbers and 37 even numbers. The sum of the two selected numbers is even if and only if the two numbers have the same parity. Therefore, the total number of ways to select is
$$
C_{38}^{2}+C_{37}^{2}=\frac{38 \times 37}{2}+\frac{37 \times... | 1369 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. If $x>0, y>0, c>0$, and $x^{2}+y^{2}+$ $z^{2}=1$. Then the minimum value of the expression $\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z}$ is
| 6. 3 .
Let $\frac{yz}{x}=a, \frac{xz}{y}=b, \frac{xy}{z}=c$, then
$$
x^{2}+y^{2}+z^{2}=1 \Leftrightarrow ab+bc+ca=1 .
$$
Therefore, $a^{2}+b^{2}+c^{2} \geqslant ab+bc+ca=1$.
Thus, $(a+b+c)^{2}$
$$
=a^{2}+b^{2}+c^{2}+2(ab+bc+ca) \geqslant 3,
$$
which implies $\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z} \Rightarrow \sqrt{3... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{aligned}
y=\sqrt{2 x^{2}} & -2 x+1 \\
& +\sqrt{2 x^{2}-(\sqrt{3}-1) x+1} \\
& +\sqrt{2 x^{2}-(\sqrt{3}+1) x+1}
\end{aligned}
$$
Find the minimum value of the function above. | Obviously, since
\[
\begin{aligned}
y=\sqrt{x^{2}}+ & (x-1)^{2} \\
& +\sqrt{\left(x-\frac{\sqrt{3}}{2}\right)^{2}+\left(x+\frac{1}{2}\right)^{2}} \\
+ & \sqrt{\left(x-\frac{\sqrt{3}}{2}\right)^{2}+\left(x-\frac{1}{2}\right)^{2}},
\end{aligned}
\]
then for the points \( T(x, x), A(0,1), B\left(\frac{\sqrt{3}}{2}, -\frac... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. Arrange $1,2,3, \cdots, 1989$ in a circle, and starting from 1, cross out every other number. (That is, keep 1, cross out 2, keep 3, cross out $4, \cdots$), and repeat this process multiple times until only one number remains. What is this number? | Solving: Arranging so many numbers in a circle and performing deletions makes it difficult to see the pattern. Therefore, one should start with simpler cases. Suppose $n$ numbers are arranged in a circle, and the specified operations are performed. Let the last remaining number be denoted as $T_{n}$. The following tabl... | 1931 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. How many sets of positive integer solutions does the equation $x+2 y+3 z=2000$ have? | Solve: This problem is equivalent to $x-1+2(y-1)+3(z-1) = 2000-6$, i.e.,
$$
a+2 b+3 c=1994
$$
Find the number of non-negative integer solutions.
Let $a=6 a_{1}+r_{1}\left(0 \leqslant r_{1} \leqslant 5\right)$,
$$
\begin{array}{l}
b=3 b_{1}+r_{2}\left(0 \leqslant r_{2} \leqslant 2\right), \\
c=2 c_{1}+r_{3}\left(0 \leq... | 332334 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. How many non-negative integer solutions does the equation $x+3 y+4 z=665$ have?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Let $x=12 x_{1}+r_{1}\left(0 \leqslant r_{1} \leqslant 11\right)$,
$$
\begin{array}{l}
y=4 y_{1}+r_{2}\left(0 \leqslant r_{2} \leqslant 3\right), \\
z=3 z_{1}+r_{3}\left(0 \leqslant r_{3} \leqslant 2\right) .
\end{array}
$$
The equation becomes
$$
\begin{array}{ll}
& 12\left(x_{1}+y_{1}+z_{1}\right)+r_{1}+3 r_{2}+4 r... | 17108 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example. Divide 20 people into 4 groups, with each group having $4,5,5,6$ people, how many ways are there to do this? | Given $m=20, t=3, h_{1}=1, h_{2}=2, h_{3}=1$, hence
$$
\begin{aligned}
f_{20}^{4}= & \frac{20!}{4!\cdot 5!\cdot 5!\cdot 6!\cdot 1!\cdot 2!\cdot 1!} \\
& =4888643760 .
\end{aligned}
$$ | 4888643760 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four, as shown in the figure, $R$ is a square grid composed of 25 small squares with a side length of 1. Place a small square $T$ with a side length of 1 on the center square of $R$, and then place more small squares with a side length of 1 in $R$ according to the following requirements:
(1) The sides of these small sq... | Four, Answer: The maximum number of small cubes that can be placed is 4.
Proof As shown in the figure,
a) The small cubes that satisfy condition (3) can only be placed within the figure formed by removing a small square of side length 1 from each corner of $R$.
b) The small cubes that satisfy condition (3) must also be... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. If the equation $x^{2}+p x+q=0$ and the equation $x^{2}-\frac{1}{q} x+\frac{1}{q}=0$ both have equal real roots, and their roots are reciprocals of each other, then the value of $1992 p^{2}-12 q-6$ is . $\qquad$ | 3. 1983 | 1983 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$ (Math Competition) | By analyzing, we compare $E_{1}$ with $x^{2}-x-1=0$ and find that $m, n$ are the two distinct real roots of the equation. By Vieta's formulas, we get
$$
\begin{aligned}
m+n= & 1, m n=-1 \\
m^{2}+n^{2} & =(m+n)^{2}-2 m n=3 \\
m^{3}+n^{3} & =(m+n)\left(m^{2}+n^{2}\right)-m n(m+n) \\
& =4 .
\end{aligned}
$$
By analogy wi... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. Five circles are tangent to each other in sequence, and also tangent to lines $a$, $b$. If the diameters of the smallest and largest circles are 18 and 32, respectively, find the diameter of the circle in the middle. (34th American High School Mathematics Competition) | Analyzing the original problem involving the relationship between five circles, it is not easy to grasp. First, consider the case of three circles:
Let three circles $O_{1}, O_{2}, O_{3}$ be tangent to each other in sequence. If their radii are $r_{1}, r_{2}, r_{3}$ respectively, then by the similarity of triangles (a... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. For the quadratic function $y=a x^{2}+b x(a b \neq 0)$, when $x$ takes $x_{1}, x_{2}\left(x_{1} \neq x_{2}\right)$, the function values are equal. Then, when $x$ takes $x_{1}+x_{2}$, the function value is | 2. Let $x$ take $x_{1}, x_{2}$, the function values are $-c$, then we have
$$
\left\{\begin{array}{l}
a x_{1}^{2}+b x_{1}=-c, \\
a x_{2}^{2}+b x_{2}=-c .
\end{array}\right.
$$
That is, $x_{1}, x_{2}$ are the two distinct roots of the quadratic equation
$$
a x^{2}+b x+c=0
$$
and
$$
x_{1}+x_{2}=-\frac{b}{a} \text {. }
... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Place the natural numbers $1,2,3,4, \cdots, 2 n$ in any order on a circle. It is found that there are $a$ groups of three consecutive numbers that are all odd, $b$ groups where exactly two are odd, $c$ groups where exactly one is odd, and $d$ groups where none are odd. Then $\frac{b-c}{a-d}=$ $\qquad$ . | 4. -3 .
Let the numbers on a circle be recorded in reverse order as $x_{1}, x_{2}$, $\cdots, x_{2 n}$, and satisfy
$$
x_{i}=\left\{\begin{array}{l}
-1, \text { when } x_{i} \text { represents an odd number, } \\
+1, \text { when } x_{i} \text { represents an even number. }
\end{array}\right.
$$
Then $x_{1}+x_{2}+\cdo... | -3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The number of solutions to the equation $\sin |x|=|\cos x|$ in the interval $[-10 \pi, 10 \pi]$ is $\qquad$ | 1. E0.
Since $\sin |x|$ and $|\cos x|$ are both even functions, we only need to consider the number of solutions within $[0,10 \pi]$. In this case, the equation becomes $\sin x=|\cos x|$. When $0 \leqslant x \leqslant \frac{\pi}{2}$, the equation becomes $\sin x=\cos x$, which has only one solution $x=\frac{\pi}{4}$; ... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. $[x]$ represents the greatest integer not exceeding $x$, then
$$
\sum_{k=1}^{99}[\sqrt{k}]=
$$
$\qquad$ | 2. 615 .
Obviously, when $i^{2} \leqslant k \leqslant(i+1)^{2}-1=i^{2}+2 i(i=1,2,3$ $\cdots, 9)$, we have $[\sqrt{k}]=i$. Therefore,
$$
\begin{array}{l}
\sum_{k=1}^{9}[\sqrt{k}]=\sum_{i=1}^{9} i(2 i+1) \\
=2 \cdot \sum_{i=1}^{9} i^{2}+\sum_{i=1}^{9} i \\
=2 \times \frac{9 \times 10 \times 19}{6}+\frac{9 \times 10}{2}=... | 615 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. Two quadratic equations with unequal leading coefficients
$$
\begin{array}{l}
(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0, \quad (1) \\
(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 x\right)=0
\end{array}
$$
(Where $a, b$ are positive integers) have a common root. Find the value of $\frac{... | Analyzing, since the two equations have a common root (known number), let's denote it as $x_{0}$. Clearly, $x_{0} \neq 1$ (otherwise $a=b$), so the equation in terms of the unknown $x$ can be transformed into equations in terms of $a$ and $b$:
$$
\left(1-x_{0}\right) a^{2}+\left(x_{0}^{2}+2\right) a-\left(x_{0}^{2}+2 x... | 256 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. When $n=6$, find the value of $P(6)$.
| Solve: Make the partition table for 6
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline \begin{tabular}{l}
\begin{tabular}{l}
$k-$ \\
partition \\
\hline
\end{tabular}
\end{tabular} & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline \begin{tabular}{l}
partition \\
ways
\end{tabular} & \begin{tabular}{l}
6
\end{tabular} & \begin{tabular}{l}
\be... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The two roots of the equation $x^{2}+p x+q=0$ are non-zero integers, and $p+q=198$, then $p=$ $\qquad$ . | 2. -202
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | -202 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. $p, q$ are natural numbers, and there are $11 q-1 \geqslant 15 \psi, 10 p \geqslant 7 q$ $+1, \frac{7}{10}<\frac{q}{p}<\frac{11}{15}$. Then the smallest $q=$ $\qquad$ . | 2. $q=7$.
Solve $\left.\begin{array}{l}11 q-1 \geqslant 15 p, \\ 10 p \geqslant 7 q+1\end{array}\right\} \Rightarrow 11 q-1 \geqslant \frac{3}{2}(7 q+1) \Rightarrow q$
$\geqslant 5$. When $q=5,6$, check and find it does not meet the requirements, $q=7$ when, $p=5$. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
* 5. For the quartic equation $x^{4}-20 x^{3} + k x^{2}-400 x+384=0$, the product of two of its four roots is 24, then the value of $k$ is $\qquad$. | 5. 140 .
Let the four roots of the equation be $x_{1}, x_{2}, x_{3}, x_{4}$, then by Vieta's formulas, we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}=20, \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}=k \text { (2) } \\
x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2}... | 140 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
*Five, (20 points) 100 matchboxes, numbered 1 to 100. We can ask whether the total number of matches in any 15 boxes is odd or even. What is the minimum number of questions needed to determine the parity (odd or even) of the number of matches in box 1? | Let $a_{i}$ represent the number of matches in the $i$-th box.
The first time, boxes 1 to 15 are taken, so the parity of $\sum_{k=1}^{15} a_{k}$ is known;
The second time, boxes 2 to 8 and 16 to 23 are taken, so the parity of $\sum_{k=2}^{8} a_{k}+\sum_{k=16}^{23} a_{k}$ is known;
The third time, boxes 9 to 23 are ta... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
*Two. (35 points) Suppose $a_{1}$, $a_{2}, \cdots, a_{10}$ and $b_{1}, b_{2}, \cdots, b_{10}$
are sequences composed of unequal complex numbers. It is known that for $i=1,2$, $\cdots, 10$,
$$
\left(a_{1}+b_{i}\right)\left(a_{2}+b_{i}\right) \cdots\left(a_{10}+b_{i}\right)=100 .
$$
Prove that for any $j=1,2, \cdots, 10... | Let
$$
F(x)=\left(a_{1}+x\right)\left(a_{2}+x\right) \cdots\left(a_{10}+x\right)-100 .
$$
Then, by the given conditions, $b_{1}, b_{2}, \cdots, b_{10}$ are the 10 distinct complex roots of the polynomial $F(x)$. Since $F(x)$ is a 10th-degree polynomial in $x$ with the coefficient of $x^{10}$ being 1, it follows from t... | -100 | Algebra | proof | Yes | Yes | cn_contest | false |
Example 8. From the harmonic series $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\cdots$, remove all terms where the denominator's decimal representation contains the digit 9. The resulting series is necessarily bounded $[1]$. | Prove that in the interval $\left[10^{m}, 10^{m+1}-1\right]$, natural numbers all have $m$ digits, except that the first digit cannot be 0 or 9, while the other digits can be any of the 9 digits $0,1,2$, $\cdots, 8$. Therefore, the number of natural numbers in this interval that do not contain the digit 9 is $8 \cdot 9... | 80 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 6. The center of square $ABCD$ is $O$, and its area is $1989 \mathrm{~cm}^{2} . P$ is a point inside the square, and $\angle O P B=45^{\circ}, P A : P B=5 : 14$. Then $P B=$ $\qquad$ . (1989, National Junior High School League) | Analysis: The answer is $P B=42 \mathrm{~cm}$. How do we get it?
Connect $O A, O B$. It is easy to know that $O, P, A, B$ are concyclic, so $\angle A P B=\angle A O B=90^{\circ}$.
Therefore, $P A^{2}+P B^{2}=A B^{2}=1989$. Since $P A: P B=5: 14$, we can find $P B$. | 42 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) Fill in the right table with $1,2,3,4,5,6$ respectively, so that in each row, the number on the left is less than the number on the right, and in each column, the number on top is less than the number below. How many ways are there to fill the table? Provide an analysis process. | Five, as shown in the right figure, from the known information we can get: $a$ is the smallest, $f$ is the largest, so $a=1$, $f=6$. From $bd$, we need to discuss the following two cases:
(1) When $bd$, then $b=3, d=2, c=4$ or
5.
In this case, there are the following two ways to fill:
\begin{tabular}{|l|l|l|}
\hline 1 ... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 3. Let } x>\frac{1}{4} \text {, simplify } \sqrt{x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}} \\ -\sqrt{x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1}}=\end{array}$ | $\begin{array}{l}\text { 3. } \because \sqrt{x+\frac{1}{2} \pm \frac{1}{2} \sqrt{4 x+1}} \\ =\sqrt{\frac{2 x+1 \pm \sqrt{4 x+1}}{2}} \\ =\frac{1}{2} \sqrt{4 x+2 \pm 2 \sqrt{4 x+1}} \\ =\frac{1}{2} \sqrt{(\sqrt{4 x+1} \pm 1)^{2}} \\ =\frac{1}{2}|\sqrt{4 x+1} \pm 1|, \\ \therefore \text { when } x>\frac{1}{4} \text {, th... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$4.1992^{1002}$ when divided by 1993 leaves a remainder of | $$
\begin{aligned}
4 \cdot 1992^{1002} & =(1993-1)^{2 \cdot 2 \cdot 2 \cdot 3 \cdot 83} \\
& =\left(1993^{2}-2 \cdot 1993+1\right)^{2 \cdot 2 \cdot 3 \cdot 83}
\end{aligned}
$$
Continuing the calculation in a similar manner, the first term that appears is 1993, and the last term is 1. Therefore, the remainder when $19... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Five. (16 points) A certain middle school originally had several classrooms, each with an equal number of desks, totaling 539 desks. This year, the school added 9 new classrooms in a newly built teaching building, increasing the total number of desks to 1080. At this point, the number of desks in each classroom remaine... | Five, let there be $x$ classrooms now, then the original number of classrooms is $x-9$. According to the problem, we have
$\frac{1080}{x}>\frac{539}{x-9}$, which means $\frac{1080}{x}-\frac{539}{x-9}>0$,
or $\frac{1080 x-9720-539 x}{x(x-9)}>0, x>\frac{9720}{541}$.
Thus, $x>17$.
Also, since $\frac{539}{x-9}$ is a natura... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1.40 A centipede with 40 legs and a dragon with 3 heads are in the same cage, with a total of 26 heads and 298 legs. If the centipede has 1 head, then the dragon has $\qquad$ legs. (1988, Shanghai Junior High School Competition)
If the centipede has $x$ individuals, and the dragon has $y$ individuals, and each ... | From (1), we get $x=26-3 y$.
Since $0 \leqslant x \leqslant 26$, and $x$ takes integer values, we have $0 \leqslant y \leqslant 8 \frac{2}{3}$, and $y$ takes integer values, so $0 \leqslant y \leqslant 8$.
Substituting (3) into (2), we get
$$
\begin{array}{l}
n y+40(26-3 y)=298, \\
n y=120 y-742 . \\
n=\frac{120 y-742}... | 14 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If the inequality $|x-2|+|x-1| \geqslant a$ holds for all real numbers $x$, then the maximum value of $\boldsymbol{a}$ is $\qquad$ . | 3. 1.
Make $y=|x-2|+|x-1|$ $=\left\{\begin{array}{l}3-2 x, \quad x<1 \\ 1, \quad 1 \leqslant x \leqslant 2 \\ 2 x-3, \quad x>2\end{array}\right.$ Obviously, $|x-2|+|x-1| \geqslant 1$. Therefore, the maximum value of $a$ is 1. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
2. For a finite set of points $M$ in the plane, it has the property: for any two points $A, B$ in $M$, there must exist a third point $C \in M$ such that $\triangle A B C$ is an equilateral triangle. Find the maximum value of $|M|$. | 2. Let $A, B \in M$ and $AB$ be the longest, by the given condition there exists $C \in M$, such that $\triangle ABC$ is an equilateral triangle. Clearly, the point set $M$ is entirely within the union of the three segments $CAB$, $ABC$, $BCA$ with $A$, $B$, $C$ as centers and $AB$ as the radius, as shown in the figure... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. For $\triangle ABC$ with side lengths $AB=c, BC=a, CA=b$. Extend $AB$ to $A''$ such that $BA''=a$, and extend in the opposite direction to $B'$ such that $AB'=b$. Similarly, obtain $A', C', B'', C''$ as shown in the figure. Prove: $\frac{S_{A' B^* B^{\top} C C^* A^*}}{S_{ABC}} \geqslant 13$.
---
Note: The symbols ... | 4. Obviously, $\triangle A B^{\prime} C^{\prime \prime} \cong \triangle C^{\prime} B A^{\prime \prime} \cong \triangle B^{\prime \prime} C A^{\prime} \cong \triangle A B C$.
$$
\frac{S_{\triangle A A^{*} A}}{S_{\triangle A B C}}=\frac{(a+b)(a+c)}{b c}=1+\frac{a(a+b+c)}{b c},
$$
Therefore, $\frac{S_{\triangle A A^{-} A... | 13 | Geometry | proof | Yes | Yes | cn_contest | false |
4. Does there exist a natural number $n$, such that the first four digits of $n!$ are 1993? | 4. Existence. Let $m=1000100000$. When $k<99999$, if $(m+k)!=\overline{a b c d} \cdots$, then $(m+k+1)!=(m+k)!\times (m+k+1)=\overline{a b c d} \times 10001 \cdots=\overline{a b c x \cdots}$, where $x=d$ or $d+1$. Therefore, if $n!=\overline{a b c d \cdots}$, then the first four digits of $(m+1)!,(m+2)!, \cdots,(m+9999... | 1993 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. There are 128 ones written on the blackboard. In each step, you can erase any two numbers $a$ and $b$ on the blackboard, and write the number $a \cdot b + 1$. After doing this 127 times, only one number remains. Denote the maximum possible value of this remaining number as $A$. Find the last digit of $A$.
| 8. Let's prove that by operating on the smallest two numbers on the board at each step, we can achieve the maximum final number. For convenience, we use $a * b$ to denote the operation of removing $a$ and $b$, and immediately adding $a b + 1$. Suppose at some step we do not operate on the smallest two numbers $x$ and $... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. Find the minimum value of the function $f(u, v)=(u-v)^{2}+\left(\sqrt{2-u^{2}}\right.$ $\left.-\frac{9}{v}\right)^{2}$. (1983 Putnam Competition) | Analysis: The problem is equivalent to finding the shortest distance (squared) between points on two curves. A sketch reveals that $[f(u, v)]_{\min }=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
*5. In a football championship, each team is required to play a match against all other teams. Each match, the winning team gets 2 points, a draw results in 1 point for each team, and the losing team gets 0 points. It is known that one team has the highest score, but it has won fewer matches than any other team. If the... | 5. $n=6$. We call the team $A$ with the highest score the winning team. Suppose team $A$ wins $n$ games and draws $m$ games, then the total score of team $A$ is $2n + m$ points.
From the given conditions, each of the other teams must win at least $n+1$ games, i.e., score no less than $2(n+1)$ points. Therefore,
$2n + ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
* 6 . If $2^{10}+2^{13}+2^{m}$ is a perfect square, then the natural number $m=$ $\qquad$ | 6. $m=14$. Let $k^{2}=2^{10}+2^{13}+2^{m}$, then
$$
\begin{aligned}
2^{m} & =k^{2}-\left(2^{10}+2^{13}\right)=k^{2}-\left(2^{5} \cdot 3\right)^{2} \\
& =(k+96)(k-96) .
\end{aligned}
$$
Therefore, there exist two non-negative integers $s$ and $t$, such that
$$
\left\{\begin{array}{l}
k-96=2^{s}, \\
k+96=2^{m-t}=2^{t}, ... | 14 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (This question is worth 20 points) If the distance from a vertex of the base of a regular tetrahedron to the centroid of the opposite face is 4, find the maximum volume of this regular tetrahedron.
| Three, as shown in the figure, for the regular triangular pyramid $P-ABC$, $G$ is the centroid of the side face $\triangle PBC$, and $AG=4$.
Extend $PG$ to intersect $BC$ at $D$. Clearly, $D$ is the midpoint of $BC$.
Draw $PO \perp$ the base $ABC$ at $O$. Then $O$ lies on $AD$, and $O$ is the centroid of $\triangle A... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3.8 Arrange 8 people in a row, where A and B are two of them, the number of different arrangements where there are exactly three people between A and B is $\qquad$ (express the answer as a numerical value).
Translate the above text into English, please keep the original text's line breaks and format, and output the tr... | 3. 5760 | 5760 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. $A, B, C$ are three points on line $l$, and $A B=B C=5$, and $P$ is a point outside line $l$, $\angle A P B=\frac{\pi}{2}, \angle B P C=\frac{\pi}{4}$. Then the distance from $P$ to line $l$ is $\qquad$ | 3. 2
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. As shown in the figure, in $\triangle A B C$, $A D \perp B C$ at $D, \angle C A B=45^{\circ}, B D=3, C D=2$. Find the area of $\triangle A B C$. | Solve: Construct $\triangle A E B$ and $\triangle A C F$ symmetric to $\triangle A B D$ and $\triangle A D C$ with respect to $A B$ and $A C$ respectively. Extend $E B$ and $F C$ to intersect at point $G$. In quadrilateral $A F G E$, according to the transformation, we have
$$
A F=A D=A E,
$$
$$
\begin{array}{l}
\angle... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
* 2. On the three sides of $\triangle A B C$, take points $P_{1}, P_{2}$, $P_{3}, P_{4}, P_{5}, P_{6}, \cdots$, such that $P_{1}, P_{4}, P_{7}, \cdots$ are on $A C$, $P_{2}, P_{5}, P_{8}, \cdots$ are on $A B$, and $P_{3}, P_{6}, P_{9}, \cdots$ are on $B C$, and $A P_{1}=A P_{2}$, $B P_{2}=B P_{3}, C P_{3}=C P_{4}, A P_... | 2. 0
Construct the incircle of $\triangle ABC$. By the property of tangent segments, we can prove that $P_{n+6}=P_{6}$. Therefore, $P_{1994}=P_{2}$, which gives $P_{2} P_{1994}=$ 0. Now, let's handle this using complex numbers.
Set up the complex plane, and let each point's letter represent the complex number at that... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. There is a batch of parts, with the smallest diameter being $12 \mathrm{~mm}$ and the largest being $12.5 \mathrm{~mm}$. If $x$ parts are randomly selected, there will always be 2 parts with a diameter difference less than $0.01 \mathrm{~mm}$, then the minimum value of $x$ is | 6. 52
Divide a line segment between 12 to $12.5 \mathrm{~mm}$ into $n$ equal parts. If $n+1$ items are taken, then at least 2 items have a diameter difference that falls within the same equal part interval. From $\frac{12.5-12}{n} \leq 0.5$, we get the smallest $n=51$. | 52 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
II. (35 points) A stationery store that operates both wholesale and retail has stipulated: If a customer buys 51 pencils or more (including 51), they will be charged at the wholesale price; if they buy 50 pencils or fewer (including 50), they will be charged at the retail price. The wholesale price for 60 pencils is 1 ... | Get $x=-5+\sqrt{25+600 m}$,
$(25+600 m)$ is a perfect square.
From $40<-5+\sqrt{25+600 m} \leqslant 50$,
we get $3 \frac{1}{3}<m \leqslant 5$.
When $m=4$, $25+600 m$ is not a perfect square, discard it.
When $m=5$, $x=50$ is the solution.
Second, let the class have $x$ students, then the retail price of pencils is $\f... | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. As shown in the figure, $A D, B E, C F$ intersect at a point $P$ inside $\triangle A B C$, dividing $\triangle A B C$ into six smaller triangles, with the areas of four of these smaller triangles already given in the figure. Find the area of $\triangle A B C$. | Let the unknown areas of the two smaller triangles be $x$ and $y$, then
$$
\frac{B D}{D C}=\frac{40}{30}=\frac{84+x}{70+y},
$$
i.e., $\frac{84+x}{70+y}=\frac{4}{3}$.
Also, $\frac{A E}{E C}=\frac{70}{y}=\frac{84+x}{40+30}$.
i.e., $\frac{84+x}{70}=\frac{70}{y}$.
Dividing (1) by (2), we get
$$
\frac{70}{70+y}=\frac{4}{3... | 315 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 8. There is a rectangular sheet of size $80 \times 50$. Now, we need to cut off a square of the same size from each corner and then make it into an open box. What should be the side length $y$ of the square to be cut off so that the volume of this open box is maximized? | Let the side length of the cut-out square be $x$, then the volume of the box made is
$$
V=x(80-2 x)(50-2 x) .
$$
To find the maximum value of $V$, a common approach is to try to make the product a constant, which involves converting $x$ into $4x$.
$$
\begin{aligned}
V & =\frac{1}{4} \cdot 4 x(80-2 x)(50-2 x) \\
& \leq... | 10 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. 1000 teachers and students of a school are to visit a place 100 km away from the school. There are five cars available, each capable of carrying 50 people, with a speed of 25 km/h, while the walking speed of a person is 5 km/h. How much time is required for all teachers and students to arrive at the destinat... | Let's assume they walk $3 x$ kilometers throughout the journey. According to the problem, we have
$$
\frac{100-3 x+100-4 x}{25}=\frac{x}{5} .
$$
Thus, $\frac{200-7 x}{5}=x$, which means $3 x=50$.
The total time spent walking and riding is
$$
\frac{100-3 x}{25}+\frac{3 x}{5}=12 \text{. }
$$
Answer: The total time to r... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. In the figure, the large circle is a 400-meter track, and the track from $A$ to $B$ is 200 meters long. The straight-line distance is 50 meters. A father and son start running counterclockwise from point $A$ along the track for a long-distance run. The son runs the large circle, while the father runs straigh... | Let the number of laps the father and son run when they meet be $N_{1}, N_{2}\left(N_{1}, N_{2}\right.$ are positive integers), and let the distance from point $A$ to the meeting point on the left half-circle be $x$ meters $(0 \leqslant x \leqslant 200)$. According to the problem, we have
$$
\frac{250 N_{1}+x}{100 / 20... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. (47th Putnam Competition) Find and prove the maximum value of $f(x)=x^{3}-3 x$, where $x$ is a real number satisfying $x^{4}+36 \leqslant 13 x^{2}$.
保留源文本的换行和格式，翻译结果如下：
Example 1. (47th Putnam Competition) Find and prove the maximum value of $f(x)=x^{3}-3 x$, where $x$ is a real number satisfying $x^{4}+36... | From $x^{4}+36 \leqslant 13 x^{2}$, we can get
$$
-3 \leqslant x \leqslant-2,2 \leqslant x \leqslant 3 \text {. }
$$
Let $x_{i}3$. Then,
$$
\begin{array}{l}
f\left(x_{1}\right)-f\left(x_{2}\right) \\
=\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{2}^{2}+x_{1} x_{2}-3\right)<0, \\
f\left(x_{1}\right)<f\left(x_{2}\right) .... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The license plate numbers issued by a city consist of 6 digits (from 0 to 9), but it is required that any 2 license plates must have at least 2 different digits (for example, license plates 038471 and 030471 cannot be used simultaneously). Try to find the maximum number of different license plates that the city can ... | 5. The answer is 100000. If 100001 license plates are issued, then by the pigeonhole principle, at least 10001 numbers will have the same first digit, and similarly, at least 1001 numbers will have the same second digit, $\cdots$, at least 2 numbers will have the same fifth digit, which would violate the rule. Using th... | 100000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. There are $n$ points on a plane, where any three points can be covered by a circle of radius 1, but there are always three points that cannot be covered by any circle of radius less than 1. Find the minimum radius of a circle that can cover all $n$ points. | We prove that the radius of the smallest circle covering $n$ points is 1.
Since any three points can be covered by a circle with a radius of 1, the distance between any two points is no more than 2.
Thus, a circle with a radius of 2 centered at any one of the points can cover all the points. This implies that there e... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) A certain mathematics competition had a total of 15 questions. The table below shows the statistics for the number of people who got $n (n=0,1,2, \cdots, 15)$ questions correct.
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
\hline$n$ & 0 & 1 & 2 & 3 & $\cdots$ & 12 & 13 & 14 & 15 \\
\hline Number of people wh... | Three, Solution 1 From the statistical table, we know: The total number of people who got $0-3$ questions correct is $7+8+10+21=40$ people, and the total number of questions they got correct is $7 \times 0+8 \times 1+10 \times 2+21 \times 3=91$ (questions); The total number of people who got $12-15$ questions correct i... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
*Three. (20 points) 4 small balls with a radius of $r$ are placed in cylinder $A$, numbered from top to bottom as $1,2,3,4 . A$'s base radius is slightly larger than $r . B, C$ are cylinders identical to $A$. The balls in $A$ are moved unidirectionally to $C$ via $B$, meaning no balls can be moved from $C$ to $B$, or f... | Three, first consider the case of three small balls. At this time, there are a total of 5 arrangement methods, each method (from bottom to top) and the operation process is represented as follows:
(1)(2)(3) (1) $\rightarrow B \rightarrow C$, (2) $\rightarrow B \rightarrow C$, (3) $\rightarrow B \rightarrow$ C)
(1) (3) ... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. In trapezoid $A B C D$, $A D / / B C, \angle B=$ $30^{\circ}, \angle C=60^{\circ}, E, M, F, N$ are the midpoints of $A B, B C$. $C D, D A$ respectively. Given $B C=7, M N=3$. Then $E F$ | 4. 4(Extend $B A, C D$ to meet at $H$, then $\triangle B H C$ is a right triangle.) | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the equation $a x^{2}+b x+c=0(a \neq 0)$, the sum of the two roots is $S_{1}$, the sum of the squares of the two roots is $S_{2}$, and the sum of the cubes of the two roots is $S_{3}$. Then the value of $a S_{1}+b S_{2}+c S_{3}$ is $\qquad$ | 2. 0
(Hint: Let the two roots of the equation be $x_{1}, x_{2}$. It is easy to see that $a x_{1}^{3}+b x_{1}^{2}+c x_{1}=$ $\left.0, a x_{2}^{3}+b x_{2}^{2}+c x_{2}=0.\right)$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $x=1-\sqrt{3}$. Then $x^{5}-2 x^{4}-2 x^{3}$ $+x^{2}-2 x-1$ is
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. 1 (Hint: Original expression $=x^{3}\left(x^{2}-\right.$
$$
\left.2 x-2)+\left(x^{2}-2 x-2\right)+1\right)
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let there be $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying: $\left|x_{1}\right|$ $<1(i=1,2, \cdots, n)$, and $\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|$ $=19+\left|x_{1}+x_{2}+\cdots+x_{n}\right|$. Then the minimum value of $n$ is | 5. 20 (Hint:
From $\left|x_{i}\right|<1(i=1,2, \cdots, n)$, it is easy to know that $19 \leqslant\left|x_{1}\right|+\left|x_{2}\right|+\cdots$ $+\left|x_{n}\right|<n$. When $n=20$, we have
$$
x_{i}=\left\{\begin{array}{l}
0.95(i=1,3, \cdots, 19), \\
-0.95(i=2,4, \cdots, 20)
\end{array}\right.
$$
satisfying the condit... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. As shown in the figure, in isosceles $\triangle ABC$, $AB=AC$, $\angle A=120^{\circ}$, point $D$ is on side $BC$, and $BD=1$, $DC=2$, then $AD=$ | 5. 1 (Hint: Draw $D N \perp A B$, and draw the median $A M$ of side $B C$. Thus, we can prove $\triangle A D N \cong \triangle A D M$, getting $A D=$ $2 D N=1$. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. $M$ is a point inside the convex quadrilateral $A B C D$, and the points symmetric to $M$ with respect to the midpoints of the sides are $P, Q, R, S$. If the area of quadrilateral $A B C D$ is 1, then the area of quadrilateral $P Q R S$ is equal to | 6. 2 (Hint: Connect the midpoints of each pair of adjacent sides of quadrilateral $A B C D$, to get a parallelogram, whose area is easily known to be $\frac{1}{2}$. On the other hand, these four sides are precisely the midlines of four triangles that share $M$ as a common vertex and have the four sides of quadrilateral... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. If real numbers $x, y, z$ satisfy the equation
$$
\sqrt{x+5+\sqrt{x-4}}+\frac{|x+y-z|}{4}=3 \text {, }
$$
then the last digit of $(5 x+3 y-3 z)^{1994}$ is | 1. 4 | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $x, y, z>0$ and $x+y+z=1$. Then $\frac{1}{x}$ $+4+9$ has the minimum value of $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. 36
$$
\begin{aligned}
\text { Given } & \left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)(x+y+z) \\
& \geqslant(1+2+3)^{2}=36 \\
\quad & \frac{1}{x}+\frac{4}{y}+\frac{9}{z} \geqslant \frac{36}{x+y+z}=36 .
\end{aligned}
$$
Equality holds if and only if $x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}$. Therefore, the min... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
* 5. For a positive integer $m$, its unit digit is denoted by $f(m)$, and let $a_{n}=f\left(2^{n+1}-1\right)(n=1,2, \cdots)$. Then $a_{1994}=$ $\qquad$. | 5. 7
$$
\begin{aligned}
1994 & =4 \times 498+2, \\
a_{1994} & =f\left(2^{4 \times 198+9}-1\right)=f\left(2^{4 \times 198} \times 8-1\right) \\
& =f\left(16^{448} \times 8-1\right)=f(6 \times 8-1)=7 .
\end{aligned}
$$ | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Initially 17. M $1,2,3,4, \cdots, 1994$ are 1994 numbers from which $k$ numbers are arbitrarily selected, such that any two numbers selected as side lengths uniquely determine a right-angled triangle. Try to find the maximum value of $k$.
In the 1994 numbers $1,2,3,4, \cdots, 1994$, arbitrarily select $k$ numbers, so ... | Solve for the maximum value of $k$ being 11.
(1) From $A=\{1,2,3,4, \cdots, 1994\}$, select $\{1,2,4,8,16,32,64,128,256,512,1024\}$, a total of 11 numbers, using the recursive formula: $a_{1}=1, a_{n+1}=2 a_{n}$. Any two numbers $a_{i}, a_{j}$ $(i<j)$ in this set satisfy $a_{i}<a_{j}$, and $2 a_{i} \leqslant a_{j}$. Th... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
The 24th All-Union Mathematical Olympiad has a problem:
There are 1990 piles of stones, with the number of stones being $1, 2, \cdots$, 1990. The operation is as follows: each time, you can choose any number of piles and take the same number of stones from each of them. How many operations are needed at least to take ... | Solve the confusion of
$$
\begin{array}{l}
1990=: 2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+0 \cdot 2^{5} \\
+0 \cdot 2^{4}+0 \cdot 2^{3}+2^{2}+2^{1}+0 \cdot 2^{0},
\end{array}
$$
and write $1,2, \cdots, 1989$ in binary form. The operation is as follows:
First, take away $2^{10}=1024$ stones from each pile that has enough; seco... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Find the smallest natural number $n$ with the following property: when any five vertices of a regular $n$-gon $S$ are colored red, there is always a line of symmetry $l$ of $S$ such that the reflection of each red point across $l$ is not a red point. (Supplied by Hu Chengzhang, Nankai University) | When $n \leqslant 9$, a regular $n$-sided polygon clearly does not possess the property mentioned in the problem.
For a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, when $n=2 k (k \in N)$, there are $2 k$ axes of symmetry: the lines $A_{i} A_{k+i} (i=1,2, \cdots, k)$ and the perpendicular bisectors of the seg... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}-0.25^{2} \div\left(-\frac{1}{2}\right)^{4} \times(-2)^{3}+\left(1 \frac{3}{8}+\right. \\ \left.2 \frac{1}{3}-3.75\right) \times 24\end{array}$ | One, 7.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Ten, (10 points) If $a>0, b>0$, and $\sqrt{a}(\sqrt{a} + \sqrt{b}) = 3 \sqrt{b}(\sqrt{a} + 5 \sqrt{b})$. Find the value of $\frac{2a + 3b + \sqrt{ab}}{a - b + \sqrt{ab}}$. | $$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (10 points) Person A and Person B process a batch of parts. If A and B work together for 6 days, and then A continues alone for 5 more days, the task can be completed; if A and B work together for 7 days, and then B continues alone for 5 more days, the task can also be completed. Now, if A first processes 300 par... | Four, suppose there are $s$ parts in total, person A completes $x$ parts per day, and person B completes $y$ parts per day. According to the problem, we have
$$
\left(\begin{array}{l}
6 x+6 y+5 x=s, \\
7 x+7 y+5 y=s, \\
300+8(x+y)=s .
\end{array}\right.
$$
Solving this, we get $s=2700$ ( $\uparrow$ ). | 2700 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If natural numbers $m, n$ satisfy $m+n>m n$, then the value of $m$ $+n-m n$ is $\qquad$ | 2. 1. From the given, we have $\frac{1}{n}+\frac{1}{m}>1$. If $n, m \geqslant 2$, then $\frac{1}{n}$ $+\frac{1}{m} \leqslant \frac{1}{2}$, which is a contradiction. Therefore, at least one of $m, n$ must be 1. Without loss of generality, let $m=1$, then $m+n-m n=1+n-1 \cdot n=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. In $\triangle A B C$, $A B \leqslant A C \leqslant B C$, and the smallest interior angle is not less than $59^{\circ}$. Then the maximum value of the largest interior angle is $\qquad$ degrees. | 2. According to the problem, we have $\angle C \leqslant \angle B \leqslant \angle A$. From $59^{\circ} \leqslant \angle C \Rightarrow$ $59^{\circ} \leqslant \angle B$, then $\angle B+\angle C \geqslant 118^{\circ}$. Therefore, $\angle A \leqslant 62^{\circ}$. | 62 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. $a, b, c$ are all positive integers, and satisfy $a b+b c=$ $3984, a c+b c=1993$. Then the maximum value of $a b c$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 6. Since 1993 is a prime number, from $c(a+b)=1993$, we get $c=1, a+b=1993$. Substituting $b=1993-a$ into the other equation, we get $a^{2}-1992 a+1991=0$. Solving this, we get $a_{1}=1, a_{2}=1991$, then $b_{1}=1992, b_{2}=2$. Therefore, $(a, b, c)$ has two sets $(1,1992,1),(1991,2,1)$. Thus, the maximum value of $abc... | 3982 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. The roots $x_{1}, x_{2}$ of the equation $x^{2}-a x-a=0$ satisfy the relation $x_{1}{ }^{3}+x_{2}{ }^{3}+x_{1}{ }^{3} x_{2}{ }^{3}=75$. Then $1993+5 a^{2}$ $+9 a^{4}=$ $\qquad$ | 1. According to Vieta's formulas, we have $x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3}=75$. Thus, $a^{2}=25, a^{4}=625$. Therefore, $1993+5 a^{2}+9 a^{4}=7743$. | 7743 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a-b=5+\sqrt{6}, b-c=5-$ $\sqrt{6}$. Then $a^{2}+b^{2}+c^{2}-a b-b c-c a=$. | 2. Given, we have $a-c=10$. Then
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=\underline{81} .
\end{array}
$$ | 81 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. $a, b, c$ are all natural numbers greater than 20, one of them has an odd number of positive divisors, the other two each have exactly three positive divisors, and $a+b=c$. The smallest value of $c$ that satisfies the above conditions is . $\qquad$ | 4. Note first that a natural number has an odd number of positive divisors if and only if it is a perfect square. In particular, if a natural number has exactly 3 positive divisors, this number must be the square of a prime.
Let \(a=p^{2}, b=q^{2}, c=r^{2}\). From \(a+b=c\), we get
\[
p^{2}+q^{2}=r^{2} \text {. }
\]
w... | 169 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Person A and Person B start walking towards each other at a constant speed from points $A$ and $B$ respectively, and they meet for the first time at a point 700 meters from $A$; then they continue to walk, with A reaching $B$ and B reaching $A$, and both immediately turning back, meeting for the second time at a poi... | 5. Let the distance between places $A$ and $B$ be $x$ meters, the time for the first meeting be $t$, and the time for the second meeting be $2t$. According to the problem,
$$
\frac{700}{t}=\frac{x-700+400}{2 t} \text{, solving for } x=1700 \text{. }
$$ | 1700 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$二 、$ (16 points) $1,2,3,4,5,6$ each is used once to form a six-digit number $\overline{a b c d e f}$, such that the three-digit numbers $\overline{a b c}$, $\tilde{b} c \bar{d}, \bar{c} d e, \overline{d e f}$ can be successively divisible by $4,5,3,11$. Find this six-digit number. | II. Since $5 \mid \overline{b c d}$, therefore, $d=5$.
Also, since $11 \mid \overline{d e f}$, therefore, $d+f-e$ is a multiple of 11.
But $3 \leqslant d+f \leqslant 5+6=11, 1 \leqslant e \leqslant 6$,
so $-3 \leqslant d+f-e \leqslant 10$.
Therefore, it can only be that $d+f-e=0$, i.e., $5+f=e$.
Also, since $e \leqslan... | 324561 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Divide the sequence $2,6,10,14, \cdots$ into groups in order, the first group has 2 terms $(2,6)$, the second group has 6 terms $(10,14, \cdots, 30)$, $\cdots$ the $k$-th group has $4 k-2$ terms. Then 1994 belongs to the $\qquad$ group. | 2.16
1994 is the 499th term. The first $k$ groups have a total of $2+6+\cdots+(4 k-2)=2 k^{2}$ terms. From $2 k^{2}<499$, we have $k \leqslant 15$. Therefore, 1994 is in the 16th group. | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. If the complex number $z$ satisfies $3 z^{6}+2 i \cdot z^{5}-2 z-3 i=$ 0 . Then $|z|=$ $\qquad$ . | 6. $|z|=1$.
The equation is $z^{5}=\frac{2 z+3 i}{3 z+2 i}$. Let $z=a+b i$.
$$
\left|z^{5}\right|=\frac{|2 z+3 i|}{|3 z+2 i|}=\sqrt{\frac{4\left(a^{2}+b^{2}\right)+12 b+9}{9\left(a^{2}+b^{2}\right)+12 b+4}} \text {. }
$$
If $a^{2}+b^{2}>1$, then the left side of (1) $=|z|^{5}=\left(\sqrt{a^{2}+b^{2}}\right)^{5}>$ 1. B... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Initially $20^{\circ}$. There are 800 points on the circumference, labeled in a clockwise direction as $1,2, \cdots, 800$. They divide the circumference into 800 gaps. Now, choose one point and color it red, then follow the rule to color other points red one by one: if the $k$-th point has been colored red, then move $... | Proof Consider a circle with $2n$ points in general.
(1) On a circle with $2n$ points, if the first red point is an even-numbered point, for example, the $2k$-th point, then according to the coloring rule, each red point dyed afterward will also be an even-numbered point. At this time, if the points numbered $2, 4, 6, ... | 25 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 11. Let $a_{n}=6^{n}-8^{n}$. Find the remainder when $a_{94}$ is divided by 49. (Adapted from the first American Mathematical Invitational Competition) | $$
\begin{aligned}
a_{94} & =6^{94}-8^{94}=(7-1)^{94}-(7+1)^{94} \\
= & -2\left(C_{94}^{1} \cdot 7^{93}+C_{94}^{3} \cdot 7^{91}+\cdots\right. \\
& \left.+C_{94}^{91} \cdot 7^{3}+C_{94}^{93} \cdot 7\right) \\
= & 49 k-2 \cdot 94 \cdot 7 \\
= & 49(k-27)+7 .(k \in Z)
\end{aligned}
$$
$\therefore a_{94}$ when divided by 49... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 12. Given $x=19^{94}-1, y=2^{m} \cdot 3^{n} \cdot 5^{l}$ $(m, n, l$ are non-negative integers, and $m+n+l \neq 0)$. Find the sum $S$ of all divisors of $x$ that are of the form $y$. | $$
\begin{array}{l}
\text { Solve } x=(20-1)^{94}-1 \\
=20^{94}-C_{94}^{1} \cdot 20^{93}+\cdots-C_{94}^{93} \cdot 20 \\
=2^{3}\left(2 n_{1}-235\right) \\
=2^{3}\left[2\left(n_{1}-118\right)+1\right] .\left(n_{1} \in \mathbb{N}\right) \\
x=(18+1)^{94}-1 \\
=18^{94}+C_{94}^{1} \cdot 18^{93}+\cdots+C_{94}^{93} \cdot 18 \\... | 1169 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. 4. In a family photo album, there are 10 photos. Each photo has 3 people, with the man standing in the middle, his son standing on his left, and his brother on his right. If it is known that the 10 men standing in the middle are all different, then what is the minimum number of different people in these photos? | 9. 4. 16.
We refer to the 10 men standing in the middle of the photo as the main characters. We then classify all the men in the photo into levels: men who do not have a father in any photo are classified as level 0; when $k=0,1,2$, $\cdots$, men who have a father classified as level $k$ in some photo are classified a... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9.7. Place an even number of pawns on each row, each column, and each diagonal of a chessboard. Question: For this arrangement, what is the maximum number of pawns that can be placed? | 9. 7. 48.
Firstly, note that on a chessboard, there are 16 diagonals containing an odd number of squares, and these diagonals do not share any squares. Therefore, the number of pieces will not exceed \(64-16=48\). If a piece is placed in each square except those on the two main diagonals, the condition will be satisfi... | 48 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.7. A square wooden board is divided into $n^{2}$ unit squares by horizontal and vertical lines. Mark $n$ squares so that any rectangle with an area of at least $n$ and whose sides lie along the grid lines contains at least one marked square. Find the largest $n$ that satisfies this condition. | 10.7. 7 .
Obviously, if $n$ marked cells satisfy the conditions of the problem, then in each row and each column there is exactly one marked cell. Let $n \geqslant 3$ (obviously, $n=2$ is not the maximum), take the first row with a marked cell as $A$, the row adjacent to $A$ as $B$, and take either a row adjacent to $... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. $n$ is a natural number, $19 n+14$ and $10 n+3$ are both multiples of some natural number $d$ not equal to 1, then $d=$ $\qquad$ . | 3. 83.
Since $\frac{19 n+14}{10 n+3}=1+\frac{9 n+11}{10 n+3}$,
and $\frac{10 n+3}{9 n+11}=1+\frac{n-8}{9 n+11}$,
also $\frac{9 n+11}{n-8}=9+\frac{83}{n-8}$,
since 83 is a prime number, if 83 and $n-8$ are coprime, then $n=91$.
When $n=91$,
$$
\begin{array}{l}
19 n+14=19 \times 91+14=83 \times 21, \\
10 n+3=10 \times 9... | 83 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
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