problem
stringlengths 2
5.64k
| solution
stringlengths 2
13.5k
| answer
stringlengths 1
43
| problem_type
stringclasses 8
values | question_type
stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | source
stringclasses 6
values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
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 one
$$
S_{i_{0}} \geqslant \frac{1500500}{100}=15005,
$$
Therefore, $A \geqslant 15005$.
Below, we construct a specific permutation to show that the smallest positive number sought is $A=15005$.
Arrange all natural numbers from 1001 to 2000 as follows:
$$
\begin{array}{l}
2000,1001,1900,1101,1800,1201, \\
1700,1301,1600,1401 ;
\end{array}
$$
Solution: Let $b_{1}, b_{2}, \cdots, b_{1000}$ be any permutation of 1001,1002,
$\cdots, 2000$.
$$
\begin{array}{l}
1999,1002,1899,1102,1799,1202, \\
1699,1302,1599,1402 ; \\
\text {... } \\
1901,1100,1801,1200,1701,1300, \\
1601,1400,1501,1300 .
\end{array}
$$
The arrangement rule is: In columns 1, 3, 5, 7, 9, each column of 100 numbers forms an arithmetic sequence with a common difference of -1; in columns 2, 4, 6, 8, 10, each column of 100 numbers forms an arithmetic sequence with a common difference of 1. The first term of each arithmetic sequence is the number in the first row. Denote the above arrangement as
$$
a_{1}, a_{2}, \cdots, a_{2000} .
$$
Let $S_{1}=a_{i}+a_{i+1}+\cdots+a_{i+9}(i=1,2, \cdots$, 1901). Then $S_{1}=15005, S_{2}=15004$. It is easy to prove by induction: If $i$ is odd, then $S_{i}=15005$; if $i$ is even, then $S=15004$. This example shows that $A$ can be 15005.
In summary, $A=15005$.
(Provided by Yan Zhenjun and Su Chun, University of Science and Technology of China)
|
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 .
|
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 the region enclosed by the curve $y=x^{2} (x \geqslant 0)$ and the two lines $x=1, y=100$ (including the boundaries). Then $N(A \cup B)+N(A \cap B)=$
|
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 value of $a_{10}$ that satisfies the above requirement.
|
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,
$$
S_{k}=S_{k-1}+a_{k} \leqslant 2 S_{k-1}+1 .
$$
By the given conditions, $S_{1}=a_{1}=1$. Thus, by (1) and using induction, we can deduce that
$$
S_{k} \leqslant 2^{k}-1(1 \leqslant k \leqslant m) .
$$
From $1500 \leqslant S_{m} \leqslant 2^{m}-1$, we get $m=11$.
From (1), we have
$$
\frac{S_{k+1}-1}{2} \leqslant S_{k} \leqslant 2^{k}-1 \quad(k=1,2, \cdots, 10).
$$
Since $S_{k}$ is a natural number, we have
$$
\left(\frac{S_{k+1}-1}{2}\right) \leqslant S_{k} \leqslant 2^{k}-1 .
$$
Here $(x)$ denotes the smallest integer greater than or equal to $x$. In particular,
$$
S_{10} \geqslant\left(\frac{S_{11}-1}{2}\right) \geqslant\left(\frac{1500-1}{2}\right)=750 .
$$
Also, $S_{8} \leqslant 2^{8}-1=255$, so
$$
2 a_{10} \geqslant a_{9}+a_{10}=S_{10}-S_{8} \geqslant 495 .
$$
Therefore, $a_{10}=248$.
Next, we will use some of the above inequalities to construct a set $A$ that meets the requirements, and $a_{10}=248$. Specifically, we take
$$
\begin{array}{l}
A=\{1,2,4,8,16,32,64,128, \\
247,248,750\} .
\end{array}
$$
Since $255=2^{7}+2^{6}+\cdots+2+2^{0}$,
using binary representation, we can see that for $n \leqslant 255$, there exists $S \subset\{1,2,4, \cdots, 128\}$ such that $\sigma(S)=n$. Thus, for $n \leqslant 255+247=502$, there exists $S \subset\{1,2,4, \cdots, 128,247\}$ such that $\sigma(S)=n$; for $n \leqslant 502+248=750$, there exists $S \subset\{1,2, \cdots, 247,248\}$ such that $\sigma(S)=n$; for $n \leqslant 750+750=1500$, there exists $S \subset A$ such that $\sigma(S)=n$.
|
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 $\triangle O N D \backsim$
$\triangle O M C$, then
$$
\frac{O D}{O C}=\frac{D N}{C M} \text {, i.e. } \frac{6-y}{12}=\frac{y}{6} .
$$
Solving for $y$ gives $y=2$, i.e., the radius of $\odot D$ is 2.
|
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 } x_{1}^{2}+x_{2}^{2}+\cdots
$$
is maximized, and the maximum value is
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+\cdots+x_{50}^{2}+x_{57}^{2} \\
=56+44^{2}=1992 .
\end{array}
$$
|
1992
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (This question is worth 20 points) In the regular quadrilateral pyramid $P$ $A B C D$, $A B=3, O$ is the projection of $P$ on the base, $P O=6, Q$ is a moving point on $A O$, and the section passing through point $Q$ and parallel to $P A, B D$ is a pentagon $D F G H L$, with the area of the section being $S$. Find the maximum value of $S$.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
$$
\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 E}{B A}=\frac{D L}{D A}=\frac{H L}{P A} . \\
\text { Hence, } E F=H L . \\
\text { Also, } \because P O \perp \perp \\
A B C D, B D \perp A C, \\
\therefore P A \perp B D, \\
E F \perp E L . \\
\quad \text { Also, } \because F H / / \\
B D, P-A B C D \text { is a }
\end{array}
$$
regular quadrilateral pyramid.
$$
\therefore P H=P F,
$$
thus $\angle P F G \cong \triangle P I I G$.
Therefore, $G F=G H$, so the pentagonal section $E F G K L$ is composed of two congruent right trapezoids.
Since $E L / / B D$, $\triangle A E L$ is an isosceles right triangle. Let $E Q=x$. Then $Q L=x$.
So, $\frac{E F}{P A}=\frac{B E}{B A}=\frac{O Q}{O A}=\frac{\frac{3 \sqrt{2}}{2}-x}{\frac{3 \sqrt{2}}{2}}$.
$$
E F=\left(1-\frac{\sqrt{2}}{3} x\right) P A \text {. }
$$
Similarly, $Q G=\left(1-\frac{\sqrt{2}}{6} x\right) P A$.
Also, $P A=\sqrt{P O^{2}+O A^{2}}=\frac{9}{2} \sqrt{2}$.
Therefore, $S=(E F+Q G) E Q$
$$
=\left(2-\frac{\sqrt{2}}{2} x\right) \cdot \frac{\sqrt{2}}{2} x \text {. }
$$
Thus, $S=-\frac{9}{2} x^{2}+9 \sqrt{2} x$.
When and only when $x=\sqrt{2}$, $S$ reaches its maximum value of 9.
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
$$
|
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 all sets that satisfy the above condition.
Where the notation $\min (a, b, c)$ denotes the minimum of $a, b, c$, and $\max (a, b, c)$ denotes the maximum of $a, b, c$.
```
|
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 $z_{0}=0$, satisfy the given conditions for a total of 7 points.
If $n>7$, we prove that there must be two points $z^{\prime}, z^{\prime \prime}$ such that
$$
\left|z^{\prime}-z^{\prime \prime}\right|7$, then at least one of the sectors must contain two points $Z^{\prime}, Z^{\prime \prime}$. Without loss of generality, assume
$$
\left|z^{\prime}\right| \leqslant\left|z^{\prime \prime}\right|, \angle Z^{\prime} Z_{0} Z^{\prime \prime}=\varphi0$.
|
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
$$
C_{38}^{2}+C_{37}^{2}=\frac{38 \times 37}{2}+\frac{37 \times 36}{2}=1369 .
$$
|
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 $\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z} \Rightarrow \sqrt{3}$,
with equality holding when $x=y=z=\frac{1}{\sqrt{3}}$.
|
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 \( T(x, x), A(0,1), B\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right), C\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right) \), we have \( y = TA + TB + TC \). It is easy to verify that \( \triangle ABC \) is an equilateral triangle centered at \( O(0, 0) \) with side length 1. According to the Fermat point principle, when \( T \) coincides with \( O \), \( TA + TB + TC \) reaches its minimum value. Therefore, the minimum value of \( y \) is obtained when \( T(0,0) \), which is \( y(0) = 3 \).
|
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 table lists the initial cases:
Carefully observing, it is not difficult to find that the values of $T$ do not include even numbers (in fact, all even numbers are eliminated in the first round). When $n=1,2,4,8,16$, the values of $T$ are all 1. It is also not difficult to notice an interesting phenomenon: when $2^{k}<n<2^{k+1}$, the values of $T_{n}$ form an arithmetic sequence of odd numbers: 3, 5, 7, ... Thus, we can reasonably conjecture that when $m$ is a natural number less than $2^{k-1}$, $T_{n}=2m+1$.
(Conjectures (I) and (II) are not difficult to prove; please verify them yourself.)
When $n=1989$, since $1989=2^{10}+965$, we have $m=965$, thus $T_{n}=2m+1=1931$. Therefore, the last remaining number is 1931.
From these two examples, we can see that for complex problems, one can start with simpler cases to identify patterns, providing clues and a good start for solving more complex original problems.
|
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 $a=6 a_{1}+r_{1}\left(0 \leqslant r_{1} \leqslant 5\right)$,
$$
\begin{array}{l}
b=3 b_{1}+r_{2}\left(0 \leqslant r_{2} \leqslant 2\right), \\
c=2 c_{1}+r_{3}\left(0 \leqslant r_{3} \leqslant 1\right) .
\end{array}
$$
Equation (1) becomes
$$
\begin{array}{l}
6\left(a_{1}+b_{1}+c_{1}\right)+r_{1}+2 r_{2}+3 r_{3}=1994 . \\
\therefore \quad r_{1}+2 r_{2}+3 r_{3} \equiv 1994 \equiv 2(\bmod 6) .
\end{array}
$$
When $r_{3}=0$, $r_{2}=0, r_{1}=2$, we have
$$
\begin{array}{l}
a_{1}+b_{1}+c_{1}=332 . \\
r_{2}=1, r_{1}=0, a_{1}+b_{1}+c_{1}=332 . \\
r_{2}=2, r_{1}=4, a_{1}+b_{1}+c_{1}=331 .
\end{array}
$$
When $r_{3}=1$, $r_{2}=0, r_{1}=5$, we have
$$
\begin{array}{l}
a_{1}+b_{1}+c_{1}=331 . \\
r_{2}=1, r_{1}=3, a_{1}+b_{1}+c_{1}=331 . \\
r_{2}=2, r_{1}=1, a_{1}+b_{1}+c_{1}=331 .
\end{array}
$$
$\therefore$ Equation (1) has a total of $2 C_{334}^{2}+4 C_{333}^{2}=332334$ non-negative integer solutions.
From the above two examples, we can see that when $a_{i}$ is relatively large, the number of solutions to the congruence
(3) $\sum_{i=1}^{n} a_{i} r_{i} \equiv r_{m} \ (\bmod \ a) \left(0 \leqslant r_{m}<a\right)$ is quite large, making the calculation inconvenient.
$$
\text { Note that, } \begin{aligned}
\sum_{i=1}^{n} a_{i} r_{i} & \leqslant \sum_{i=1}^{n}\left[a_{i}\left(\frac{a}{a_{i}}-1\right)\right] \\
= & n a-\sum_{i=1}^{n} a_{i},
\end{aligned}
$$
Thus, (3) can be transformed into
$$
\begin{array}{c}
a_{1} r_{1}+a_{2} r_{2}+\cdots+a_{n} r_{n}=s a+r_{m}(0 \leqslant s \leqslant \\
{\left[\frac{n a-\sum_{i=1}^{n} a_{i}}{a}\right] \text { or } 0 \leqslant s \leqslant\left[\frac{n a-\sum_{i=1}^{n} a_{i}}{a}\right]-1 \text { and } s}
\end{array}
$$
$\in Z$ ).
For each $s$, let (3) have $t_{i}$ solutions. $x_{1}^{\prime}+x_{2}^{\prime}$ non-negative integer solutions.
Therefore, (1) has a total of $\sum_{i \in 1=1}\left[t_{i} C_{\frac{m-1}{a}-i+n-1}^{a}\right]$ non-negative integer solutions.
|
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
$$
\begin{array}{ll}
& 12\left(x_{1}+y_{1}+z_{1}\right)+r_{1}+3 r_{2}+4 r_{3}=665 . \\
& \therefore r_{1}+3 r_{2}+4 r_{3} \equiv 665 \equiv 5(\bmod 12) . \\
& \because 0 \leqslant r_{1}+3 r_{2}+4 r_{3} \leqslant 28, \\
& r_{1}+3 r_{2}+4 r_{3}=5 \\
\text { or } & r_{1}+3 r_{2}+4 r_{3}=17 .
\end{array}
$$
It is easy to find that (1) has 3 solutions, (2) has 8 solutions.
$\therefore$ The original equation has $3 C_{\frac{655 ~ 5}{2}-2}^{12}+8 C_{\frac{665-5}{2}-1+2}^{12}$ $=17108$ non-negative integer solutions.
|
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 squares are parallel or perpendicular to the sides of $R$;
(2) Including $T$, no two small squares have any common points;
(3) For any two small squares including $T$, there must be a straight line passing through them (including edges) that is parallel to one of the sides of $R$.
How many more small squares can be placed in $R$ besides $T$ while satisfying the requirements? Prove your conclusion.
|
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 placed on one of the 8 larger squares of side length 2, which are outside $T$ and bounded by dashed lines (such as $A B C D$).
Obviously, each larger square can contain at most one small cube that meets the requirements.
c) On any two larger squares that share a common vertex with $T$ (such as $A B C D$ and $C E F G$), two small cubes cannot be placed simultaneously, otherwise condition (3) would not be satisfied. The 8 larger squares can be divided into 4 pairs in this manner, and each pair of larger squares can contain at most one small cube that meets the requirements, meaning that the number of small cubes placed does not exceed 4.
d) As shown in the figure, placing 4 small cubes, it is easy to see that the conditions are met.
Therefore, the maximum number of small cubes that can be placed is 4.
(Maoyoude provided)
|
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 with the above process, we can reduce the target expression to a lower-degree algebraic expression in terms of known values.
$$
\begin{array}{l}
m^{4}+n^{4}=\left(m^{2}+n^{2}\right)^{2}-2 m^{2} n^{2} \\
=9-2 \times 1=7, \\
m^{5}+n^{5}=\left(m^{2}+n^{2}\right)\left(m^{3}+n^{3}\right)-m^{2} n^{2}(m+ \\
n)=11 .
\end{array}
$$
|
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 (as shown in the figure above), we have
$$
\frac{r_{2}+r_{1}}{r_{2}-r_{1}}=\frac{r_{3}+r_{2}}{r_{3}-r_{2}} \text {. }
$$
Using the property of the ratio of the sum and difference, we get $\frac{r_{2}}{r_{1}}=\frac{r_{3}}{r_{2}}$.
By analogy with the above process, it is easy to deduce the case for five circles.
If their radii are $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$ respectively, then we have
$$
\frac{r_{2}}{r_{1}}=\frac{r_{3}}{r_{2}}=\frac{r_{4}}{r_{3}}=\frac{r_{5}}{r_{4}}=k \text {, }
$$
which means $2 r_{i+1}=2 r_{1} k^{i}(i=1,2,3,4)$.
Therefore, $\left(2 r_{3}\right)^{2}=4 r_{1}^{2} k^{4}=2 r_{1} \cdot 2 r_{5}$.
Thus, $2 r_{3}=24$.
|
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
$$
x_{1}+x_{2}=-\frac{b}{a} \text {. }
$$
Therefore,
$$
\begin{array}{l}
a\left(x_{1}+x_{2}\right)^{2}+b\left(x_{1}+x_{2}\right) \\
=a\left(-\frac{b}{a}\right)^{2}+b\left(-\frac{b}{a}\right)=0 .
\end{array}
$$
|
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 $x_{1}+x_{2}+\cdots+x_{2 n}=0$,
and. $A_{i}=x_{i}+x_{i+1}+x_{i+2}$
$$
=\left\{\begin{array}{ll}
-3, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { are all odd, } \\
-1, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { exactly 2 are odd, } \\
+1, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { exactly 1 is odd, } \\
+3, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { are all even. }
\end{array}\right.
$$
where $x_{2 n+i}=x_{i}$.
$$
\text { By } \begin{aligned}
0 & =3\left(x_{1}+x_{2}+\cdots+x_{2 n}\right) \\
& =A_{1}+A_{2}+\cdots+A_{2 n} \\
& =(-3 a)-b+\cdots+3 d) \\
& =-3(a-c)-(b \cdots c)
\end{aligned}
$$
we get $\quad \frac{b-c}{a-d}=-3$.
|
-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 $x=\frac{\pi}{4}$; when $\frac{\pi}{2} \leqslant x \leqslant \pi$, the equation becomes $\sin x=-\cos x$, which has only one solution $x=\frac{3 \pi}{4}$. Therefore, $\sin x=|\cos x|$ has 2 roots in $[0,2 \pi]$, and thus $\sin |x|=|\cos x|$ has $10 \times 2=20$ solutions in $(-10 \pi, 10 \pi)$.
|
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,
$$
\begin{array}{l}
\sum_{k=1}^{9}[\sqrt{k}]=\sum_{i=1}^{9} i(2 i+1) \\
=2 \cdot \sum_{i=1}^{9} i^{2}+\sum_{i=1}^{9} i \\
=2 \times \frac{9 \times 10 \times 19}{6}+\frac{9 \times 10}{2}=615 .
\end{array}
$$
|
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 $\frac{a^{b}+b^{2 a}}{a^{-b}+b^{-a}}$. (1986, National Junior High School Competition)
|
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 $b$:
$$
\left(1-x_{0}\right) a^{2}+\left(x_{0}^{2}+2\right) a-\left(x_{0}^{2}+2 x_{0}\right)=0,
$$
$$
\left(1-x_{0}\right) b^{2}+\left(x_{0}^{2}+2\right) b-\left(x_{0}^{2}+2 x_{0}\right)=0 \text {. }
$$
It is evident that $a, b$ are the two distinct positive integer roots of the equation
$$
\left(1-x_{0}\right) y^{2}+\left(x_{0}^{2}+2\right) y-\left(x_{0}^{2}+2 x_{0}\right)=0
$$
By Vieta's formulas, we get
$$
a+b=\frac{x_{0}^{2}+2}{x_{0}-1}, a b=2+\frac{x_{0}^{2}+2}{x_{0}-1} .
$$
Thus, $a b=2+(a+b)$.
If $a>b>1$, then $b=1+\frac{2}{a}+\frac{b}{a}a>1$, leading to $a=2, b=4$. Therefore,
$$
\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}=a^{b} b^{a}=2^{4} \times 4^{2}=256 .
$$
|
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}
\begin{tabular}{l}
$5+1$ \\
$4+2$ \\
$3+3$
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
$4+1+1$ \\
$3+2+1$ \\
$2+2+2$
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
$3+1+1+1$ \\
$2+2+1+1$
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{c}
$2+1+1$ \\
$+1+1$
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{c}
$1+1+1$ \\
$+1+1+1$
\end{tabular}
\end{tabular} \\
\hline \begin{tabular}{l}
\begin{tabular}{c}
$P(6$, \\
$k)$
\end{tabular}
\end{tabular} & 1 & 3 & 3 & 2 & 1 & 1 \\
\hline
\end{tabular}
Thus, $P(6)=\sum_{i=1}^{6} P(6, i)=11$.
|
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
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}=20, \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}=k \text { (2) } \\
x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}=400, \\
x_{1} x_{2} x_{3} x_{4}=384 .
\end{array}\right.
$$
Without loss of generality, let $x_{1} x_{2}=24$.
From (4), we get $x_{3} x_{4}=16$.
Substituting (5) and (3) into (2) and (3), we get
$$
\begin{array}{l}
24+\left(x_{1}+x_{2}\right)\left(x_{3}+x_{4}\right)+16=k, \\
24\left(x_{3}+x_{4}\right)+16\left(x_{1}+x_{2}\right)=400 .
\end{array}
$$
Solving (1) and (8) simultaneously, we get $x_{3}+x_{4}=10, x_{1}+x_{2}=10$.
Substituting into (7), we get $k=24+10 \cdot 10+16=140$.
|
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 taken, so the parity of $\sum_{k=9}^{23} a_{k}$ is known. From the three parities, it is easy to determine the parity of $a_{1}+2 \cdot \sum_{k=2}^{23} a_{k}$. If it is odd, then $a_{1}$ is odd; otherwise, $a_{1}$ is even.
Clearly, asking only once is not enough. Asking only twice is also not enough.
It is easy to see that in the two questions, $a_{1}$ must be included exactly once. Suppose $a_{1}$ and $a_{2}$ are included in the first question. We then consider two cases:
(1) If $a_{2}$ is not included in the second question, then changing the parity of both $a_{1}$ and $a_{2}$ simultaneously does not affect the answer, so the parity of $a_{1}$ cannot be determined.
(2) If $a_{2}$ is included in the second question, since $a_{1}$ is not included in the second question, the second question must include some $a_{3}$, which does not appear in the first question. Then, changing the parity of $a_{1}$, $a_{2}$, and $a_{3}$ simultaneously does not affect the answer, so the parity of $a_{1}$ cannot be determined.
In summary, at least three questions are required.
|
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 $j=1,2, \cdots, 10$, the product
$$
\left(b_{1}+a_{j}\right)\left(b_{2}+a_{j}\right) \cdots\left(b_{10}+a_{j}\right)
$$
is equal to the same number, and find this number.
|
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 the theory of polynomials that there must be
$$
F(x)=\left(x-b_{1}\right)\left(x-b_{2}\right) \cdots\left(x-b_{10}\right) .
$$
Thus, from (2) we know
$$
\begin{aligned}
& \left(b_{1}+a_{j}\right)\left(b_{2}+a_{j}\right) \cdots\left(b_{10}+a_{j}\right) \\
= & F\left(-a_{j}\right) .
\end{aligned}
$$
Then, from (1) we know $F\left(-a_{j}\right)=-100$.
This proves that the constant sought is -100.
|
-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 $8 \cdot 9^{m-1}$. The sum of their reciprocals is less than $\frac{8 \cdot 9^{m}}{10^{m}}=8\left(\frac{9}{10}\right)^{m}$, so the sum of all such numbers is less than $\sum_{m=0}^{\infty} 8\left(\frac{9}{10}\right)^{m}=\frac{8}{1-\frac{9}{10}}=80$. They must be bounded.
|
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 & 3 & 4 \\
\hline 2 & 5 & 6 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 1 & 3 & 5 \\
\hline 2 & 4 & 6 \\
\hline
\end{tabular}
In summary, there are 5 ways to fill.
(Provided by Lai Guangfa, Hefei No. 46 Middle School, 230061)
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 3. Let } x>\frac{1}{4} \text {, simplify } \sqrt{x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}} \\ -\sqrt{x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1}}=\end{array}$
|
$\begin{array}{l}\text { 3. } \because \sqrt{x+\frac{1}{2} \pm \frac{1}{2} \sqrt{4 x+1}} \\ =\sqrt{\frac{2 x+1 \pm \sqrt{4 x+1}}{2}} \\ =\frac{1}{2} \sqrt{4 x+2 \pm 2 \sqrt{4 x+1}} \\ =\frac{1}{2} \sqrt{(\sqrt{4 x+1} \pm 1)^{2}} \\ =\frac{1}{2}|\sqrt{4 x+1} \pm 1|, \\ \therefore \text { when } x>\frac{1}{4} \text {, the original expression }=\frac{1}{2}(\sqrt{4 x+1}+1) \\ -\frac{1}{2}(\sqrt{4 x+1}-1)=1 . \\\end{array}$
|
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 $1992^{1992}$ is divided by 1993 is 1.
|
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 remained equal, and the number of desks in each classroom increased compared to before. How many classrooms are there now?
|
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 natural number, and 539 is an odd number,
$x-9$ must be an odd number.
Therefore, $x$ is an even number. Hence, the possible values of $x$ are even numbers greater than or equal to 18, and $\frac{1080}{x}$ and $\frac{539}{x-9}$ are natural numbers.
When $x=18$, $\frac{539}{18-9}$ is not a natural number, so it is discarded.
When $x=20$, $\frac{1080}{20}=54, \frac{539}{20-9}=49$.
Thus, $x=20$ is a solution that satisfies the conditions.
Since $1080=2^{3} \cdot 3^{3} \cdot 5$,
the even factors of 1080 greater than 20 and less than itself are 30, 40, 60, 120, 360, 540. And
$$
\begin{aligned}
30-9= & =21,40-9=31,60-9=51,120-9= \\
111,360-9 & =351,540-9=531
\end{aligned}
$$
None of these can divide 539, so these even numbers do not satisfy the conditions.
Therefore, $x=20$ is the only solution that satisfies the conditions.
Answer: The school now has 20 classrooms.
(Author: Taiyuan Education College, 030001,
|
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 dragon has $n$ legs, according to the problem, we get
$$
\left\{\begin{array}{l}
x+3 y=26, \\
n y+40 x=298 .
\end{array}\right.
$$
The system of equations has 3 unknowns and 2 equations, with two unknowns in each equation, and $x, y, n$ are all positive integers. For such equations (systems), there is the following definition
Definition If an equation (system) has multiple positive integer solutions (etc.), such an equation (system) is called an indeterminate equation (system).
Generally speaking, simple indeterminate equations are solved under certain conditions, usually seeking their integer solutions.
Below, we will introduce the solution methods for simple indeterminate equations through some examples.
|
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
$$
\begin{array}{l}
n y+40(26-3 y)=298, \\
n y=120 y-742 . \\
n=\frac{120 y-742}{y} .
\end{array}
$$
$n$ should be a positive integer, thus, $y$ should be a divisor of $120 y-742$.
When $y=1$, $n=\frac{120-742}{1}$ (discard);
When $y=2$, $n=\frac{120 \times 2-742}{2}$ (discard);
When $y=7$, $n=\frac{120 \times 7-742}{7}=14$;
When $y=8$, $n=\frac{120 \times 8-742}{8}$ (discard).
Substituting $y=7$ into (2), we get $x=5$.
Substituting $\left\{\begin{array}{l}x=5, \\ y=7\end{array}\right.$ into (2), we get $n=14$.
Therefore, the dragon has 14 feet.
In the process of solving this problem, analyzing that $y$ is a divisor of $120 y-742$ is the key, which is the origin of the "divisor analysis method".
|
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.
If there is a 4th point $D \in M$, then there must be a point $D' \in M$, such that $D'$, $A$, $D$ form an equilateral triangle, $\angle DAD' = 60^\circ$, so one of the points must be outside $\triangle ABC$. Therefore, without loss of generality, assume $D$ is within the segment $AC$, and $BD$ intersects the chord $AC$ at $K$. Also, without loss of generality, assume $D'$ is within the segment $BC$. Rotating $\triangle ABD$ by $60^\circ$ around $B$ to coincide with $\triangle CBD'$, then $K$ coincides with a point $K'$ on $BD'$. $\angle BCK' = \angle BAK = 60^\circ > 30^\circ$. Thus, $K'$ is not within the segment $BC$, and $D'$ is not within the segment $BC$, leading to a contradiction. Therefore, $M$ cannot contain a 4th point.
Therefore, $M$ contains exactly three points, which form an equilateral triangle.
|
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 $B^*$ and $C^*$ in the area notation might be a typographical error or a specific notation in the context of the problem. For clarity, it is assumed that $B^*$ and $C^*$ refer to the points $B'$ and $C'$ respectively. If this is not the case, please clarify the notation.
|
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, $\frac{S_{\triangle A A^{-} A^{\prime}}+S_{\triangle B B^{\circ} B^{\prime}}+S_{\triangle C C C C}}{S_{\triangle A B C}}$
$$
=3+\sum \frac{a(a+b+c)}{b c} .
$$
Thus, $\frac{S_{A^{\prime} \pi B^{\prime} C C^{\prime} A^{*}}}{S_{\triangle A B C}}=3+\sum \frac{a(a+b+c)}{b c}-2+3$
$$
\begin{array}{l}
=4+(a+b+c) \sum \frac{a}{b c} \\
\geqslant 4+3 \sqrt[3]{a b c} \cdot 3 \sqrt[3]{\frac{a}{b c} \cdot \frac{b}{c a} \cdot \frac{c}{a b}}=13 .
\end{array}
$$
|
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 $(m+1)!,(m+2)!, \cdots,(m+99999)!$ are the same as or one more than the previous one. Moreover (due to the increase in the fifth digit from the left), the first four digits must increase by 1 after at most 10 numbers. Thus, the first four digits of the 100000 numbers $m!,(m+1)!, \cdots,(m+99999)!$ cover all 10000 values, and among them, 1993 must appear.
|
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 $y$, then before $x$ and $y$ "meet", they are combined with $x$ and $y$. That is, we have to operate on two numbers of the form $\left(\left(\cdots\left(\left(x * a_{1}\right) * a_{2}\right) \cdots\right) * a_{1}\right)$ and $\left(\left(\cdots\left(\left(y * b_{1}\right) * b_{2}\right) \cdots\right) *\right.$ $\left.b_{n}\right)$, which are respectively equal to
$$
\begin{aligned}
X & =a_{1} a_{2} \cdots a_{k}\left(x+\frac{1}{a_{1}}+\frac{1}{a_{1} a_{2}}+\cdots+\frac{1}{a_{1} a_{2} \cdots a_{k}}\right) \\
& =a_{2} \cdots a_{k}\left(a_{1} x+1+u\right), \\
Y & =b_{1} b_{2} \cdots b_{n}\left(y+\frac{1}{b_{1}}+\frac{1}{b_{1} b_{2}}+\cdots+\frac{1}{b_{1} b_{2} \cdots b_{n}}\right) \\
& =b_{1} b_{2} \cdots b_{n}(y+v),
\end{aligned}
$$
where $u>0$ and $v>0$ represent the sum of the remaining numbers in the respective parentheses. We can assume $b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n}$, if there is $b_{j}>b_{j+1}$, then by swapping $b_{j}$ and $b_{j+1}$, we can increase $Y$.
Now, let's swap the positions of $a_{1}$ and $y$, obtaining
$$
\begin{array}{c}
X^{\prime}=a_{2} \cdots a_{k}(x y+1+u), \\
Y^{\prime}=b_{1} b_{2} \cdots b_{n}\left(a_{1}+v\right), \text { then we have } \\
\left(X^{\prime} Y^{\prime}-1\right)-(X Y-1)=X^{\prime} Y^{\prime}-X Y \\
=a_{2} \cdots a_{k} b_{1} b_{2} \cdots b_{n}\left[(x y+1+u)\left(a_{1}+v\right)\right. \\
\left.-\left(a_{1} x+1+u\right)(y+v)\right] \\
=c\left(a_{1}-y\right)(u+1-v x),
\end{array}
$$
where $c>0, a_{1}>y, v \leqslant \frac{1}{b_{1}}+\frac{1}{b_{1}^{2}}+\cdots \leqslant \frac{1}{b_{1}-1} \leqslant \frac{1}{x}$, i.e., $u+1-v x>0$, thus $X^{\prime} Y^{\prime}-X Y>0$. This means that by the above adjustment, we can increase the intermediate result, which contradicts our assumption about the optimality of the operation process.
Now, we can follow the optimal procedure we have proven to obtain the maximum possible value $A$, and it is easy to see that its last digit is 2.
|
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 there are at least $n$ teams participating, then $n=$ $\qquad$ .
|
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,
$2n + m > 2(n+1)$. This implies $m \geqslant 3$.
Thus, there exists a team that draws with team $A$ and scores no less than $2(n+1) + 1$, so we have
$2n + m > 2(n+1) + 1$, which implies $m \geqslant 4$.
Clearly, team $A$ must win at least one game, i.e., $n \geqslant 1$.
Therefore, team $A$ must play against at least 5 teams, meaning there are at least 6 teams participating in the competition. A tournament schedule for 6 teams that meets the conditions of the problem can be designed (omitted).
|
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
$$
\left\{\begin{array}{l}
k-96=2^{s}, \\
k+96=2^{m-t}=2^{t}, s<t .
\end{array}\right.
$$
Eliminating $k$ yields $2^{s}\left(2^{t-s}-1\right)=2^{6} \cdot 3$. Since $2^{t-s}-1$ is odd, we get $s=6,2^{t-s}-1=3$. Thus, $t=8, m=s+t=14$.
|
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 $\triangle ABC$, with $OD=\frac{1}{3} AD$.
Let $\angle PDO=\alpha\left(0<\alpha<\frac{\pi}{2}\right)$, $AB=a$, and $PD=x$. Then $OD=\frac{\sqrt{3}}{6} a$, $OD=x \cos \alpha$, and $a=2 \sqrt{3} x \cos \alpha$.
Also, $PO=PD \sin \alpha=x \sin \alpha$.
$$
\begin{aligned}
V & =\frac{1}{3} S_{\triangle ABC} \cdot PO \\
& =\frac{1}{3} \cdot \frac{\sqrt{3}}{4} a^{2} \cdot x \sin \alpha=\sqrt{3} x^{3} \cos ^{2} \alpha \sin \alpha .
\end{aligned}
$$
In $\triangle AGD$, by the cosine rule,
$$
AG^{2}=16=AD^{2}+DG^{2}-2 AD \cdot DG \cos \alpha,
$$
which is
$$
\begin{aligned}
16= & (3 x \cos \alpha)^{2}+\left(\frac{x}{3}\right)^{2} \\
& -2 \cdot 3 \cdot x \cos \alpha \cdot \frac{x}{3} \cdot \cos \alpha, \\
x^{2} & =\frac{144}{63 \cos ^{2} \alpha+1} . \\
V^{2} & =3\left(\frac{144}{63 \cos ^{2} \alpha+1}\right)^{3} \cos ^{4} \alpha \sin ^{2} \alpha \\
& =\frac{3 \times 144^{3}}{32^{2}}\left(\frac{1}{63 \cos ^{2} \alpha+1}\right)^{3} \\
& \cdot 32 \cos ^{2} \alpha \cdot 32 \cos ^{2} \alpha \cdot \sin ^{2} \alpha \\
& \leqslant \frac{4 \cdot 3^{7}}{\left(63 \cos ^{2} \alpha+1\right)^{3}}\left(\frac{32 \cos ^{2} \alpha+31 \cos ^{2} \alpha+1}{3}\right)^{3} \\
& =4 \times 3^{4},
\end{aligned}
$$
$$
V \leqslant 18 \text {. }
$$
The volume of the triangular pyramid reaches its maximum value of 18 when and only when $32 \cos ^{2} \alpha=\sin ^{2} \alpha$, i.e., $\alpha=\operatorname{arctg} 4 \sqrt{2}$.
|
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 translation result directly.
|
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
$$
A F=A D=A E,
$$
$$
\begin{array}{l}
\angle F A E=\angle A F G=\angle A E G=\angle F G E \\
=90^{\circ} .
\end{array}
$$
Therefore, quadrilateral $A F G E$ is a square.
Let $A F=A D=x$, then
$$
\begin{array}{l}
B G=x-3, C G=x-2 . \\
(x-3)^{2}+(x-2)^{2}=(2+3)^{2} .
\end{array}
$$
Solving for $x$ gives $x=6$ (negative value is discarded).
Thus, the area of $\triangle A B C$ is $=\frac{1}{2} \times 5 \times 6=15$.
|
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 $A P_{1}=A P_{2}$, $B P_{2}=B P_{3}, C P_{3}=C P_{4}, A P_{4}=A P_{5}, B P_{5}=$ $B P_{6}, \cdots$. Then $\left|P_{2} P_{1994}\right|=$ $\qquad$
|
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 point. We have
$$
\begin{array}{l}
P_{2}-A=e^{A i}\left(P_{1}-A\right), \\
P_{5}-A=e^{A i}\left(P_{4}-A\right) .
\end{array}
$$
Subtracting, $P_{5}-P_{2}=e^{A i}\left(P_{4}-P_{1}\right)$.
Similarly, $P_{6}-P_{3}=e^{R i}\left(P_{5}-P_{2}\right)$
$$
=e^{(A-B) i}\left(P_{4}-P_{1}\right),
$$
$$
\begin{aligned}
P_{7}-P_{4} & =e^{C_{i}}\left(P_{6}-P_{3}\right) \\
& =e^{(A+B+C) i}\left(P_{4}-P_{1}\right) \\
& =e^{\pi i}\left(P_{4}-P_{1}\right)=-\left(P_{4}-P_{1}\right),
\end{aligned}
$$
we get $P_{7}=P_{1}$:
Similarly, $P_{n+6}=P_{n}$.
Since $1994=6 \times 332+2$,
we know $P_{1994}=P_{2}$.
Thus, $\left|P_{2} P_{1994}\right|=0$.
|
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 yuan cheaper than the retail price for 60 pencils. Now, a class monitor, Xiao Wang, comes to buy pencils. If he buys one pencil for each student in the class, he must pay the retail price, which costs $m$ yuan ($m$ is a positive integer); but if he buys 10 more pencils, he can pay the wholesale price, which also costs exactly $m$ yuan. How many students are there in Xiao Wang's class?
|
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 $\frac{m}{x}$ yuan, and the wholesale price is $\frac{m}{x+10}$ yuan. According to the problem, we have the equation
$$
\frac{60 m}{x}-\frac{60 m}{x+10}=1,
$$
which is $x^{2}+10 x-600 m=0$,
where $40<x \leqslant 50$.
|
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
$$
\frac{70}{70+y}=\frac{4}{3} \cdot \frac{y}{70}, y=35 .
$$
From (2), we get $x=56$.
Therefore, $S_{\triangle A B C}=315$.
|
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 $4x$.
$$
\begin{aligned}
V & =\frac{1}{4} \cdot 4 x(80-2 x)(50-2 x) \\
& \leqslant \frac{1}{4} \cdot\left(\frac{4 x+80-2 x+50-2 x}{3}\right)^{3} \\
& =\frac{130^{3}}{108} .
\end{aligned}
$$
However, the condition for equality to hold should be $4 x=80 - 2 x=50-2 x$, which is impossible. Thus, this solution is incorrect. To achieve equality and find the extremum, the correct approach is
$$
\begin{array}{l}
V=\frac{1}{12} \cdot 6 x \cdot(80-2 x) \cdot 2(50-2 x) \\
\leqslant \frac{1}{12} \cdot\left[\frac{6 x+(80-2 x)+(100-4 x)}{3}\right]^{3} \\
=18000 .
\end{array}
$$
Equality holds when and only when $6 x=80-2 x=100-4 x$, i.e., $x=10$, at which $V$ reaches its maximum value.
This solution is concise, but why is it transformed this way?
This can be explained by introducing parameters into the inequality to obtain the answer.
$$
\begin{array}{l}
V=\frac{1}{\alpha \beta} \cdot \alpha x \cdot \beta(50-2 x) \cdot(80-2 x) \\
\leqslant \frac{1}{\alpha \beta}\left(\frac{\alpha x+50 \beta-2 \beta x+80-2 x}{3}\right)^{3} \\
=\frac{1}{\alpha \beta}\left[\frac{(50 \beta+80)+(\alpha-2 \beta-2) x}{3}\right]^{3} .
\end{array}
$$
To make (1) independent of $x$, it is necessary that
$$
\alpha-2 \beta-2=0 .
$$
For equality to hold, it must also be true that
$$
\alpha x=50 \beta-2 \beta x=80-2 x .
$$
From (2) and (3), we can obtain $\alpha=6, \beta=2$.
|
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 destination simultaneously?
|
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 reach the destination is at least 12 hours.
|
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 straight every time he reaches point $B$. The father runs 100 meters in 20 seconds, and the son runs 100 meters in 19 seconds. If they run at these speeds, how many times will they meet in 35 minutes from the start?
|
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
$$
\frac{250 N_{1}+x}{100 / 20}=\frac{400 N_{2}+x}{100 / 19} .
$$
This simplifies to $\frac{x}{200}=38 N_{2}-25 N_{1}$ being an integer. Since $0 \leqslant \frac{x}{200} \leqslant 1$, it can only be that $x=0$ or $x=200$. Therefore, their meeting points are at $A$ or $B$.
$\therefore 38 N_{2}-25 N_{1}-1$ or $38 N_{2}-25 N_{1}=1$.
Solving $38 N_{2}-25 N_{1}=0$ gives $N_{1}=38 k, N_{2}=$ $25 k .(k \in \mathbb{Z})$
Solving $38 N_{2}-25 N_{1}=1$ gives $N_{1}=38 k_{1}+3$, $N_{2}=25 k_{1}+2 .(k \in \mathbb{Z})$
Given that the father and son each run one lap in 50 and 76 seconds, respectively.
And $\left[\frac{2100}{50}\right)=42$ (laps), $\left[\frac{2100}{76}\right]=27$ (laps).
Their meeting at point $A$ can be determined by the solution to $38 N_{1}-25 N_{2}=0$.
Therefore, for $N_{1}=38 k, N_{2}=25 k$, we can take $k=$ 0,1 .
Similarly, for $N_{1}=38 k_{1}+3, N_{2}=25 k_{1}+2$, we can take $k_{1}=0,1$.
Thus, excluding the starting point, they meet 3 times. Answer: From the start of the race, they meet 3 times in 35 minutes.
|
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 $x^{4}+36 \leqslant 13 x^{2}$.
|
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,
$$
\begin{array}{l}
f\left(x_{1}\right)-f\left(x_{2}\right) \\
=\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{2}^{2}+x_{1} x_{2}-3\right)<0, \\
f\left(x_{1}\right)<f\left(x_{2}\right) .
\end{array}
$$
This shows that $f(x)$ is an increasing function on $[-3,-2]$ and $[2,3]$, and the maximum value is at the right endpoint of the interval. Therefore,
$$
\max f(x)=\max \{f(-2), f(3)\}=18
$$
|
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 issue. And prove it.
|
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 the following method, 100000 license plates can be issued while complying with the rule: the last 5 digits can be filled arbitrarily (there are $10^{5}$ ways to do this), and the first digit is filled with the unit digit of the sum of the last 5 digits, then if 2 numbers differ in only one of the last 5 digits, their first digit will also be different.
|
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 exists a circle that can cover all the points, and therefore, there must also exist a smallest circle that can cover all the points. Let this circle be $\odot C$, with radius $r$. By the problem statement, $r \geqslant 1$.
We now prove that $r \leqslant 1$.
If there is only one point or no points on the circumference of $\odot C$ among the eight points, then there must be a circle with a radius smaller than $r$ that can cover all the points. $\odot C$ would not be the smallest circle covering all the points, which is impossible. Therefore, the circumference of $\odot C$ must have at least two points.
Assume that $\odot C$ has exactly two points, then these two points must be at the ends of a diameter, otherwise $\odot C$ can still be reduced. Since the distance between the two points is no more than 2, the radius $r$ of $\odot C$ is $\leqslant 1$.
If $\odot C$ has more points, $a_{1}, a_{2}, \cdots, a_{k}(3 \leqslant k \leqslant n)$, then these $k$ points cannot all be on a minor arc, otherwise $\odot C$ can still be reduced. This means that at least three points are on a major arc, and the triangle formed by these three points does not contain an obtuse angle. $\odot C$ is the smallest circle containing these three points. Since a circle with a radius of 1 that covers these three points can completely cover $\odot C$, it follows that $r \leqslant 1$.
Since $r \geqslant 1, r \leqslant 1$, thus $r=1$.
|
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 who got $n$ questions correct & 7 & 8 & 10 & 21 & $\cdots$ & 15 & 6 & 3 & 1 \\
\hline
\end{tabular}
If it is also known that the students who got 4 questions and more than 4 questions correct each averaged 6 questions correct, and the students who got 10 questions and 10 questions or fewer correct each averaged 4 questions correct. How many people does this table at least include?
|
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 is $15+6+3+1=25$ (people), and the total number of questions they got correct is $15 \times 12+6 \times 13+3 \times 14+1$ $\times 15=315$ (questions).
Let $x_{0}, x_{1}, \cdots, x_{15}$ represent the number of people who got 0, 1, $\cdots, 15$ questions correct, respectively. From the problem, we have
$$
\frac{4 x_{4}+5 x_{5}+\cdots+15 x_{15}}{x_{4}+x_{5}+\cdots+x_{15}}=6,
$$
and $\frac{0 x_{0}+x_{1}+2 x_{2}+\cdots+10 x_{10}}{x_{0}+x_{1}+\cdots+x_{10}}=4$,
which means $4 x_{4}+5 x_{5}+\cdots+15 x_{15}=6\left(x_{4}+x_{5}+\cdots+x_{15}\right)$,
$$
0 x_{0}+x_{1}+2 x_{2}+\cdots+10 x_{10}=4\left(x_{0}+x_{2}+\cdots+x_{10}\right) \text {. }
$$
Subtracting the two equations, we get
$$
\begin{array}{l}
11 x_{11}+12 x_{12}+\cdots+15 x_{15}-\left(x_{1}+2 x_{2}+3 x_{3}\right) \\
= 6\left(x_{4}+x_{5}+\cdots+x_{15}\right)-4\left(x_{0}+x_{1}+\cdots+x_{10}\right) \\
= 6\left(x_{11}+x_{12}+\cdots+x_{15}\right)-4\left(x_{0}+x_{1}+x_{2}+x_{3}\right) \\
+2\left(x_{4}+x_{5}+\cdots+x_{10}\right) \\
= 4\left(x_{11}+x_{12}+\cdots+x_{15}\right)-6\left(x_{0}+x_{1}+x_{2}+x_{3}\right) \\
+2\left(x_{0}+x_{1}+\cdots+x_{15}\right) \\
= 4 x_{11}+4\left(x_{12}+x_{13}+x_{14}+x_{15}\right)-6\left(x_{0}+x_{1}+x_{2}\right. \\
\left.\quad+x_{3}\right)+2\left(x_{0}+x_{1}+\cdots+x_{15}\right) .
\end{array}
$$
And we have already found that $x_{0}+x_{1}+x_{2}+x_{3}=46$,
$$
\begin{array}{l}
x_{12}+x_{13}+x_{14}+x_{15}=25, \\
0 x_{0}+x_{1}+2 x_{2}+3 x_{3}=91, \\
12 x_{12}+13 x_{13}+14 x_{14}+15 x_{15}=315,
\end{array}
$$
Substituting these into the above equation, we get
$$
\begin{array}{l}
11 x_{11}+315-91 \\
=4 x_{11}+4 \times 25-6 \times 46+2\left(x_{0}+x_{1}+\cdots+x_{15}\right) .
\end{array}
$$
$$
\text { Therefore, } x_{0}+x_{1}+\cdots+x_{15}=200+3.5 x_{11},\left(x_{11} \geqslant 0\right)
$$
Thus, when $x_{11}=0$, the total number of people counted in the statistics $x_{0}+x_{1}+\cdots+$ $x_{15}$ is at least 200 people.
Solution 2 (The four sums are given the same as in Solution 1)
Let this table have counted a total of $y$ people, and the number of people who got 11 questions correct is $x$ people. From the problem, we can derive the functional relationship between $y$ and $x$
$$
6(y-46)-315-4(y-x-25)+11 x-91 \text {, }
$$
Simplifying, we get $y=200+3.5 x(x \geqslant 0)$,
Therefore, when $x=0$, $y=200$ is the minimum.
Thus, this table has counted at least 200 people.
|
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 from $B$ to $A . B$ can temporarily store several balls, but must follow the "last in, first out" rule. How many different arrangements of the balls are possible in $C$ after all the balls have been moved into $C$?
|
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) (2) (1) $\rightarrow B \rightarrow C$, (2) $\rightarrow B$, (3) $\rightarrow B \rightarrow C$, (2) $\rightarrow C$ )
(2) (1)(3) (1) $\rightarrow B$, (2) $\rightarrow B \rightarrow C$, (1) $\rightarrow C$, (3) $\rightarrow$ $B \rightarrow C$ )
(2) (3) (1) (1) $\rightarrow B$, (2) $\rightarrow B \rightarrow C$, (3) $\rightarrow B \rightarrow C$, (1) $\rightarrow C$ )
(3)(2)(1) (1) $\rightarrow B$, (2) $\rightarrow B$, (3) $\rightarrow B \rightarrow C$, (2) $\rightarrow$ $C,(1) \rightarrow C$ )
For 4 balls, when the 1st ball is at the bottom, there are 5 ways (first let (1) $\rightarrow B \rightarrow C$, then let (2), (3), (1) correspond to the 5 arrangements of (1), (2), (3) above). When the 2nd ball is at the bottom, there are also 5 ways (first let (1) $\rightarrow B$, (2) $\rightarrow B \rightarrow C$, then let (1), (3), (1) correspond to the 5 arrangements of (1), (2), (3) above, but remove the repeated operation of (1) $\rightarrow B$ in the first step). When the 3rd ball is at the bottom, there are 3 ways: (3) (2)(1)(4), (3)(2)(4)(1), (3)(4)(2)(1). When the 4th ball is at the bottom, there is only 1 way: (4)(3)(2)(1). Therefore, there are a total of 14 ways.
|
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 conditions of the problem.)
|
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 $P Q R S$ as their bases.)
|
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 minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$ is 36.
|
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 that any two numbers selected as side lengths uniquely determine a right-angled triangle. Try to find the maximum value of $k$.
|
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 $2 a_{i} \leqslant a_{j}$. Therefore, any two numbers $a_{i}, a_{j}$ can uniquely determine an isosceles triangle. Hence, $k \geqslant 11$.
(2) Suppose $a_{1}<a_{2}<a_{3}<\cdots<a_{12}$ are any 12 numbers selected from $A$, and any two of these numbers can uniquely determine an isosceles triangle. Then, $a_{1} \geqslant 1, a_{2} \geqslant 2, a_{3} \geqslant 4, a_{4} \geqslant 8, \cdots, a_{11} \geqslant 1024, a_{12} \geqslant 2048$. This contradicts $a_{12} \leqslant 1994$. This indicates that at most 11 numbers can be selected from $A$ such that any two of them can uniquely determine an isosceles triangle. Therefore, the maximum value of $k$ is 11.
(Tongtan Town, Huxian County, Sichuan, 646111, Answered by Zhong Dingji)
|
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 all the stones away?
|
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; second, take away $2^{9}=$ 512 stones from each pile that has enough; …… finally, take away the stones from the piles that have only $2^{0}=1$ stone left. This way, a total of 11 operations are used.
Since there are piles with only one stone, there must be an operation to take one stone; if the remaining operations all take more than 2 stones, then the piles with exactly two stones cannot be taken away. Therefore, 2 operations can take at most $1+2=3$ stones, ..., 10 operations can take at most $1+2+$ $2^{2}+\cdots+2^{9}=1023$ stones. Thus, 1950 piles of stones require at least 11 operations.
|
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 segments $A_{i} A_{i+1} (i=1,2, \cdots, k)$. When $n=2 k+1 (k \in N)$, the lines connecting vertex $A_{i} (1 \leqslant i \leqslant 2 k+1)$ to the midpoint of segment $A_{k+i} A_{k+i+1}$ are the axes of symmetry.
When $n=10$, if the vertices $A_{1}, A_{2}, A_{4}, A_{6}$, and $A$ are colored red, the regular decagon does not possess the property mentioned in the problem. Let the lines connecting $A_{i} A_{5+i} (i=1,2, \cdots, 5)$ be denoted as $l_{i}$. Let the perpendicular bisectors of the segments $A_{i} A_{i+1} (i=1,2, \cdots, 5)$ be denoted as $l_{i+\frac{1}{2}} (i=1,2, \cdots, 5)$. Thus, the complete set of axes of symmetry for the regular decagon are $l_{1}, l_{2}, l_{3}, l_{1}, l_{5}, l_{\frac{3}{2}}, l_{\frac{5}{2}}, l_{\frac{7}{2}}, l_{\frac{9}{2}}, l_{\frac{1}{2}}$, a total of 10.
When $i=1,2,4$, $l_{i}$ maps point $A_{i}$ to itself. Therefore, $l_{1}, l_{2}, l_{4}$ are not the axes of symmetry required by the problem. $l_{3}$ maps point $A_{2}$ to point $A_{4}$, $l_{5}$ maps point $A_{4}$ to point $A_{6}$, $l_{\frac{3}{2}}$ maps point $A_{1}$ to point $A_{2}$, $l_{\frac{5}{2}}$ maps point $A_{1}$ to point $A_{4}$, $l_{\frac{7}{2}}$ maps point $A_{1}$ to point $A_{6}$, $l_{\frac{9}{2}}$ maps point $A_{2}$ to point $A_{7}$, and $l_{\frac{11}{2}}$ maps point $A_{4}$ to point $A_{7}$. Therefore, none of the 10 axes of symmetry satisfy the property mentioned in the problem. Thus, the regular decagon does not possess the property mentioned in the problem.
Similarly, it can be proven that when $n=11,12,13$, if the points $A_{1}, A_{2}, A_{4}, A_{6}, A_{7}$ are all colored red, these regular $n$-sided polygons do not possess the property mentioned in the problem.
Next, we prove that a regular 14-sided polygon possesses the property mentioned in the problem.
For a regular 14-sided polygon $A_{1} \wedge \mathrm{i} 2 \cdots A_{14}$, there are 7 axes of symmetry that do not pass through the vertices. When $i$ is odd, the point $A_{i}$ is called an odd vertex, and when $i$ is even, the point $A_{i}$ is called an even vertex. Clearly, each axis of symmetry that does not pass through a vertex makes odd vertices and even vertices symmetric to each other. Suppose there are $m$ red odd vertices $(m=0,1, \cdots, 5)$, then there are $5-m$ red even vertices. The number of lines connecting red odd vertices to red even vertices is
$$
m(5-m) \leqslant\left(\frac{5}{2}\right)^{2}<7.
$$
Since $m(5-m)$ is an integer, then $m(5-m) \leqslant 6$. This indicates that the number of perpendicular bisectors of segments connecting red odd vertices to red even vertices is at most 6. Therefore, there is at least one axis of symmetry that does not pass through a vertex and ensures that the symmetric point of any red point is not red.
|
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.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Hence, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (10 points) Person A and Person B process a batch of parts. If A and B work together for 6 days, and then A continues alone for 5 more days, the task can be completed; if A and B work together for 7 days, and then B continues alone for 5 more days, the task can also be completed. Now, if A first processes 300 parts, and then the two work together for 8 days to complete the task. How many parts are there in total?
|
Four, suppose there are $s$ parts in total, person A completes $x$ parts per day, and person B completes $y$ parts per day. According to the problem, we have
$$
\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$ is 3982.
|
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
\[
p^{2}+q^{2}=r^{2} \text {. }
\]
where \(p, q, r\) are all positive integers, and \(p<r, q<r, p, q, r\) with at least two being prime numbers.
Further discussion leads to the conclusion that the minimum value of \(c\) is 169.
|
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 point 400 meters from $B$. Then the distance between $A$ and $B$ is meters.
|
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 $e \leqslant 6, f \geqslant 1$, it can only be that $f=1, e=6$.
Furthermore, since $3 \mid \overline{c d e}$, i.e., $3 \mid \overline{c 56}$, therefore, $c+5$ is divisible by 3.
And since $4 \mid \bar{a} \bar{a} c$, it follows that $c$ is even, so $c=4$. Further deducing, $b=2, a=3$.
Thus, $\overline{a b c d e f}=324561$.
|
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. But the right side has both the numerator and denominator positive, and the difference between the denominator and the numerator is $5\left(a^{2}+\right.$ $\left.b^{2}\right)-5>0$, making the fraction value less than 1. The equation cannot hold.
If $a^{2}+b^{2}<0$, similarly, it can be deduced that the equation does not hold.
Only when $a^{2}+b^{2}=1,|z|=1$.
|
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 $k$ gaps in a clockwise direction and color the endpoint reached red. Continue this process. How many red points can be obtained at most on the circumference? Prove your conclusion.
|
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 $2, 4, 6, \cdots, 2k, \cdots, 2n$ are renumbered as $1^{*}, 2^{*}, 3^{*}, \cdots, k^{*}, \cdots, n^{*}$, it is easy to see that the number of red points that can be dyed starting from the $2k$-th point on a circle with $2n$ points is the same as the number of red points that can be dyed starting from the $k$-th point on a circle with $n$ points, and they form a one-to-one correspondence.
(2) On a circle with $2n$ points, if the first red point is an odd-numbered point, then the second red point must be an even-numbered point. By (1), in this case, it can also correspond to a coloring method starting from the $k$-th point on a circle with $n$ points, and the number of red points dyed is exactly one more than the number of red points obtained by this coloring method on the circle with $n$ points.
(3) Let the maximum number of red points that can be dyed on a circle with $n$ points be $f(n)$, and suppose the coloring method starting from the $k$-th point achieves this maximum value. Then, on a circle with $2n$ points, the coloring method starting from the $k$-th point should achieve the maximum value $f(n)+1$ (since the second red point is the $2k$-th point). Therefore, $f(2n)=f(n)+1$.
(4) From (3), we have
$$
\begin{array}{l}
f(800)=f(400)+1=f(200)+2 \\
=f(100)+3=f(50)+4=f(25)+5 .
\end{array}
$$
(5) To calculate $f(25)$, note that on a circle with 25 points, if the number of the first red point is a multiple of 5, then the number of each red point dyed afterward will also be a multiple of 5. Therefore, the number of red points that can be dyed starting from such a point will not exceed 5.
If the number of the first red point is not a multiple of 5, then the number of each red point dyed afterward will also not be a multiple of 5. Therefore, starting from such a point, at most 20 red points can be dyed.
On the other hand, starting from a certain such point, we can actually dye 20 red points, for example, starting from the 1st point, we can sequentially dye 1, 2, 4, 8, 16, 7, 14, 3, 6, 12, 24, 23, 21, 17, 9, 18, 11, 22, 19, 13. Therefore, $f(25)=20$.
From (4) and (5), we get $f(800)=20+5=25$, i.e., the maximum number of red points that can be dyed is 25.
(Jiangxi Nanchang Vocational and Technical Normal University, 330013, Gou Ping Sheng's solution)
|
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 leaves a remainder of 7.
|
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 $y$.
|
$$
\begin{array}{l}
\text { Solve } x=(20-1)^{94}-1 \\
=20^{94}-C_{94}^{1} \cdot 20^{93}+\cdots-C_{94}^{93} \cdot 20 \\
=2^{3}\left(2 n_{1}-235\right) \\
=2^{3}\left[2\left(n_{1}-118\right)+1\right] .\left(n_{1} \in \mathbb{N}\right) \\
x=(18+1)^{94}-1 \\
=18^{94}+C_{94}^{1} \cdot 18^{93}+\cdots+C_{94}^{93} \cdot 18 \\
=3^{2}\left(3 n_{2}+188\right) \\
=3^{2}\left[3\left(n_{2}+62\right)+2\right] .\left\{n_{2} \in \mathbb{N}\right) \\
x=(20-1)^{94}-1 \\
=20^{94}-C_{94}^{1} \cdot 20^{93}+\cdots+C_{94}^{92} \cdot 20^{2} \\
-C_{94}^{93} \cdot 20 \\
=5\left(5 n_{3}+376\right) \\
=5\left[5\left(n_{3}+75\right)+1\right] .\left(n_{3} \in \mathbb{N}\right) \\
\end{array}
$$
$$
\therefore y=2^{m} \cdot 3^{n} \cdot 5^{l} \text { satisfies } 0 \leqslant m \leqslant 3,0 \leqslant n \leqslant
$$
$2,0 \leqslant l \leqslant 1$, and $m+n+l \neq 0$. Therefore,
$$
\begin{aligned}
S= & \left(2^{0}+2^{1}+2^{2}+2^{3}\right)\left(3^{0}+3^{1}+3^{2}\right) \\
& \cdot\left(5^{0}+5^{1}\right)-1 \\
= & 1169
\end{aligned}
$$
|
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 as level $k+1$. Let $r_{k}$ represent the number of main characters at level $k$, and $t_{k}$ represent the number of other men at level $k$. The number of fathers of level $k+1$ men does not exceed $\frac{1}{2} r_{k-1} + t_{k+1}$, because each main character has brothers. At the same time, the number of fathers of level $k-1$ men is no less than $r_{k}$, because each main character has a son. Therefore,
$$
r_{k} \leqslant \frac{1}{2} r_{k+1} + t_{k+1}, \quad k=0,1,2, \cdots.
$$
Additionally, $1 \leqslant \frac{1}{2} r_{0} + t_{0}$. Adding all the inequalities together, we get
$$
\frac{1}{2}\left(r_{0} + r_{1} + \cdots\right) + 1 \leqslant t_{0} + t_{1} + \cdots,
$$
Thus, $\left(r_{0} + r_{1} + \cdots\right) + \left(t_{0} + t_{1} + \cdots\right)$
$$
\geqslant \frac{3}{2}\left(r_{0} + r_{1} + \cdots\right) + 1 = \frac{3}{2} \cdot 10 + 1 = 16.
$$
Therefore, there are at least 16 men in the photo.
There are 10 main characters (numbered 1 to 10) and 16 men in the photo, as shown in the diagram. Horizontal lines connect brothers, while other lines (from top to bottom) connect fathers and sons.
The 10 photos are: $(3,1,2)$; $(5,2,1)$; $(7,3,4)$; $(9,4,3)$; $(11,5,6)$; $(12,6,5)$; $(13,7,8)$; $(14,8,7)$; $(15,9,10)$; $(16,10,9)$.
|
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 satisfied, as shown in the figure above.
|
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 $A$ (and not coinciding with $B$) or a row adjacent to $B$ (and not coinciding with $A$) as $C$.
Let $b$ be the number of the marked cell in row $B$. If $b \leqslant n-\left[\frac{n}{2}\right]$ or $b>\left[\frac{n}{2}\right]+1$, then in rows $A$ and $B$ there exists a rectangle with an area of at least $n$ that does not contain any marked cells, so,
$$
n-\left[\frac{n}{2}\right]7$, then the area of both rectangles is not less than $n$, but row $C$ has only one marked cell, meaning one of these rectangles does not contain a marked cell.
Thus, we have proved that $n \leqslant 7$. A $7 \times 7$ board that meets the conditions is shown above.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. $n$ is a natural number, $19 n+14$ and $10 n+3$ are both multiples of some natural number $d$ not equal to 1, then $d=$ $\qquad$ .
|
3. 83.
Since $\frac{19 n+14}{10 n+3}=1+\frac{9 n+11}{10 n+3}$,
and $\frac{10 n+3}{9 n+11}=1+\frac{n-8}{9 n+11}$,
also $\frac{9 n+11}{n-8}=9+\frac{83}{n-8}$,
since 83 is a prime number, if 83 and $n-8$ are coprime, then $n=91$.
When $n=91$,
$$
\begin{array}{l}
19 n+14=19 \times 91+14=83 \times 21, \\
10 n+3=10 \times 91+3=83 \times 11, \\
\therefore d=(19 n+14,10 n+3)=83 .
\end{array}
$$
|
83
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.