problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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7. In a lottery with 100000000 tickets, each ticket number consists of eight digits. A ticket number is called "lucky" if and only if the sum of its first four digits equals the sum of its last four digits. Then the sum of all lucky ticket numbers, when divided by 101, leaves a remainder of $\qquad$ | 7.0.
If the eight-digit number $x=\overline{a b c d e f g h}$ is lucky, then $y=99999999-x$ is also lucky, and $x \neq y$.
$$
\begin{array}{l}
\text { and } x+y=99999999=9999 \times 10001 \\
=99 \times 101 \times 10001,
\end{array}
$$
Therefore, the sum of all lucky numbers must be a multiple of 101. | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 If a natural number $K$ can be expressed as
$$
\begin{aligned}
K & =V(a, b, c) \\
& =a^{3}+b^{3}+c^{3}-3 a b c \quad (a, b, c \in \mathbf{N})
\end{aligned}
$$
then $K$ is called a "Water Cube Number".
Now, arrange all different Water Cube Numbers in ascending order to form a sequence $\left\{K_{n}\right\}$, ... | 【Analysis】Let $M=\{3 x \mid x \in \mathbf{N},(3, x)=1\}$.
Lemma A natural number $k$ is a water cube number if and only if $k \in \mathbf{N} \backslash M$.
Proof Note that,
$V(a, b, c)=a^{3}+b^{3}+c^{3}-3 a b c$
$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$=(a+b+c)^{3}-3(a+b+c)(a b+b c+c a)$.
Then when $3 \mid... | 2614032 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 The sequence $\left\{a_{n}\right\}$:
$1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots$, is called a "fractal sequence", and its construction method is as follows:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, to get $a_{2}=1, a_{3}=2$;
Then copy all the previous items $1, 1, 2$, and add... | Solving the construction method of the sequence $\left\{a_{n}\right\}$, we easily know
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \cdots
$$
Generally, $a_{2^{n-1}}=n$, which means the number $n$ first appears at the $2^{n}-1$ term, and if
$$
m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right),
$$
then $a_{m}=... | 3950 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the function be
$$
f(x)=\sin ^{4} \frac{k x}{10}+\cos ^{4} \frac{k x}{10}\left(k \in \mathbf{Z}_{+}\right) .
$$
If for any real number $a$, we have
$$
\{f(x) \mid a<x<a+1\}=\{f(x) \mid x \in \mathbf{R}\} \text {, }
$$
then the minimum value of $k$ is $\qquad$ | 6. 16 .
From the given conditions, we have
$$
\begin{array}{l}
f(x)=\left(\sin ^{2} \frac{k x}{10}+\cos ^{2} \frac{k x}{10}\right)^{2}-2 \sin ^{2} \frac{k x}{10} \cdot \cos ^{2} \frac{k x}{10} \\
=1-\frac{1}{2} \sin ^{2} \frac{k x}{5}=\frac{1}{4} \cos \frac{2 k x}{5}+\frac{3}{4},
\end{array}
$$
The function $f(x)$ re... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $a_{1}, a_{2}, a_{3}, a_{4}$ be four distinct numbers from $1, 2, \cdots, 100$, satisfying
$$
\begin{array}{l}
\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right) \\
=\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}\right)^{2} .
\end{array}
$$
Then the number of such ordered quadruples... | 8. 40 .
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right) \\
\geqslant\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}\right)^{2},
\end{array}
$$
Equality holds if and only if
$$
\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\frac{a_{... | 40 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. | Three, using these ten points as vertices and the connected line segments as edges, we get a simple graph $G$ of order 10.
The following proves: The number of edges in graph $G$ does not exceed 15.
Let the vertices of graph $G$ be $v_{1}, v_{2}, \cdots, v_{10}$, with a total of $k$ edges, and use $\operatorname{deg}\le... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Several pairwise non-overlapping isosceles right triangles with leg lengths of 1 are placed on a $100 \times 100$ grid paper. It is known that the hypotenuse of any right triangle is the diagonal of some unit square; each side of a unit square is the leg of a unique right triangle. A unit square whose diagonals are ... | 3. For a general $2 n \times 2 n$ grid paper, the maximum number of empty cells is $n(n-1)$.
In fact, the $2 n \times 2 n$ grid paper is enclosed by $2 n+1$ horizontal lines and $2 n+1$ vertical lines:
$$
\begin{array}{l}
\{(x, y) \mid x=k, 0 \leqslant k \leqslant 2 n, k \in \mathbf{Z}\}, \\
\{(x, y) \mid y=k, 0 \leqs... | 2450 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sets
$$
A=\left\{n^{2}+1 \mid n \in \mathbf{Z}_{+}\right\}, B=\left\{n^{3}+1 \mid n \in \mathbf{Z}_{+}\right\} \text {. }
$$
Arrange all elements in $A \cap B$ in ascending order to form the sequence $a_{1}, a_{2}, \cdots$. Then the units digit of $a_{99}$ is | 2. 2 .
From the given, we know that $A \cap B=\left\{n^{6}+1 \mid n \in \mathbf{Z}_{+}\right\}$.
Therefore, $a_{99}=99^{6}+1$.
Thus, its unit digit is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: \frac{x^{2}}{4}-\frac{y^{2}}{5}=1$, respectively. Point $P$ is on the right branch of the hyperbola $C$, and the excenter of $\triangle P F_{1} F_{2}$ opposite to $\angle P F_{1} F_{2}$ is $I$. The line $P I$ intersects the $x$-axis at point $Q$... | 6. 4 .
Since $I F_{1}$ is the angle bisector of $\angle P F_{1} Q$, we have
$$
\frac{|P Q|}{|P I|}=1+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|} \text {. }
$$
Let $P\left(x_{0}, y_{0}\right)$. Then, $\left|P F_{1}\right|=\frac{3}{2} x_{0}+2$.
By the optical property of the hyperbola, the tangent line to the hyp... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. Answer Questions (Total 56 points)
9. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, a_{3}=4, \\
a_{n}=a_{n-1}+a_{n-2}-a_{n-3}+1(n \geqslant 4) .
\end{array}
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Prove: $\frac{1}{a_{1... | (1) For $n \geqslant 4$, summing up we get
$$
a_{n}-a_{n-2}=a_{3}-a_{1}+n-3=n \text {. }
$$
When $n=2 m\left(m \in \mathbf{Z}_{+}\right)$,
$$
\begin{aligned}
a_{n} & =a_{2}+\sum_{k=1}^{m-1}\left(a_{2 k+2}-a_{2 k}\right) \\
& =2+\sum_{k=1}^{m-1}(2 k+2)=\frac{1}{4} n(n+2) ;
\end{aligned}
$$
When $n=2 m+1\left(m \in \ma... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Roll a die six times, let the number obtained on the $i$-th roll be $a_{i}$. If there exists a positive integer $k$, such that $\sum_{i=1}^{k} a_{i}=6$ has a probability $p=\frac{n}{m}$, where $m$ and $n$ are coprime positive integers. Then
$$
\log _{6} m-\log _{7} n=
$$ | 2. 1.
When $k=1$, the probability is $\frac{1}{6}$;
When $k=2$,
$$
6=1+5=2+4=3+3 \text {, }
$$
the probability is $5\left(\frac{1}{6}\right)^{2}$;
When $k=3$,
$$
6=1+1+4=1+2+3=2+2+2 \text {, }
$$
the probability is $(3+6+1)\left(\frac{1}{6}\right)^{3}=10\left(\frac{1}{6}\right)^{3}$;
When $k=4$,
$$
6=1+1+1+3=1+1+2+2... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the set
$$
S=\{1,2, \cdots, 12\}, A=\left\{a_{1}, a_{2}, a_{3}\right\}
$$
satisfy $a_{1}<a_{2}<a_{3}, a_{3}-a_{2} \leqslant 5, A \subseteq S$. Then the number of sets $A$ that satisfy the conditions is | 5. 185 .
Notice that, the number of all three-element subsets of set $S$ is
$$
\mathrm{C}_{12}^{3}=220 \text {. }
$$
The number of subsets satisfying $1 \leqslant a_{1}<a_{2}<a_{3}-5 \leqslant 7$ is $\mathrm{C}_{7}^{3}=35$, so the number of sets $A$ that meet the conditions of the problem is $220-35=185$. | 185 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. If a positive integer can be expressed as the difference of cubes of two consecutive odd numbers, then the positive integer is called a "harmonious number" (for example, $2=1^{3}-(-1)^{3}, 26=3^{3}-1^{3}, 2, 26$ are both harmonious numbers). Among the positive integers not exceeding 2016, the sum of all harmonious n... | 3. B.
Notice that,
$$
(2 k+1)^{3}-(2 k-1)^{3}=2\left(12 k^{2}+1\right) \text {. }
$$
From $2\left(12 k^{2}+1\right) \leqslant 2016 \Rightarrow|k|<10$.
Taking $k=0,1, \cdots, 9$, we get all the harmonious numbers not exceeding 2016, and their sum is
$$
\begin{array}{l}
{\left[1^{3}-(-1)^{3}\right]+\left(3^{3}-1^{3}\ri... | 6860 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
Three. (25 points) Given positive real numbers $x, y, z$ satisfying
$$
\begin{array}{l}
x y + y z + z x \neq 1, \\
\frac{\left(x^{2}-1\right)\left(y^{2}-1\right)}{x y} + \frac{\left(y^{2}-1\right)\left(z^{2}-1\right)}{y z} + \\
\frac{\left(z^{2}-1\right)\left(x^{2}-1\right)}{z x} = 4 .
\end{array}
$$
(1) Find the value... | (1) From the given equation, we have
$$
\begin{array}{l}
z\left(x^{2}-1\right)\left(y^{2}-1\right)+x\left(y^{2}-1\right)\left(z^{2}-1\right)+ \\
y\left(z^{2}-1\right)\left(x^{2}-1\right)=4 x y z \\
\Rightarrow \quad x y z(x y+y z+z x)-(x+y+z)(x y+y z+z x)+ \\
\quad(x+y+z)-x y z=0 \\
\Rightarrow \quad[x y z-(x+y+z)](x y... | 1 | Inequalities | proof | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
a+b+c=5, a^{2}+b^{2}+c^{2}=15, \\
a^{3}+b^{3}+c^{3}=47 . \\
\text { Find }\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)
\end{array}
$$ | Given $a+b+c=5, a^{2}+b^{2}+c^{2}=15$, we know
$$
\begin{array}{l}
2(a b+b c+c a) \\
=(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=10 \\
\Rightarrow a b+b c+c a=5 .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) .
\end{array}
$$
Then $47-3... | 625 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) As shown in Figure 3, in isosceles $\triangle ABC$, $AB = AC = \sqrt{5}$, $D$ is a point on side $BC$ other than the midpoint, the symmetric point of $C$ with respect to line $AD$ is $E$, and the extension of $EB$ intersects the extension of $AD$ at point $F$. Find the value of $AD \cdot AF$. | Three, connect $A E$, $E D$, $C F$. From the given conditions, we know
$$
\angle A B C=\angle A C B=\angle A E D \text {. }
$$
Then $A$, $E$, $B$, $D$ are concyclic
$$
\Rightarrow \angle B E D=\angle B A D \text {. }
$$
Since points $C$, $E$ are symmetric with respect to line $A D$
$$
\begin{array}{l}
\Rightarrow \an... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=0,5 a_{n+1}=4 a_{n}+3 \sqrt{1-a_{n}^{2}}(n \in \mathbf{N}) \text {. }
$$
Let $S_{n}=\sum_{i=0}^{n} a_{i}$. Then $S_{51}=$ $\qquad$ | 2. 48 .
Let $a_{n}=\sin \alpha_{n}$.
Suppose $\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$.
Then $30^{\circ}0 ; \\
\sin \left(\alpha_{n}-\theta\right), \cos \alpha_{n}<0 .
\end{array}\right.
$
Thus, $a_{n}=\left\{\begin{array}{l}\sin n \theta, n=0,1 ; \\ \sin 2 \theta, n=2 k\left(k \in \mathbf{Z}_{+}\right) ; \\... | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $AB$ is a diameter of the smallest circle with center $C(0,1)$ that has common points with the graph of the function $y=\frac{1}{|x|-1}$, and $O$ is the origin. Then $\overrightarrow{O A} \cdot \overrightarrow{O B}$ $=$ | 4. -2 .
For any point $P(x, y)$ on the function $y=\frac{1}{|x|-1}(x>1)$, we have
$$
R^{2}=x^{2}+(y-1)^{2}=x^{2}+\left(\frac{1}{x-1}-1\right)^{2} \text {. }
$$
Let $t=x-1$. Then $t>0$,
$$
\begin{array}{l}
R^{2}=(t+1)^{2}+\left(\frac{1}{t}-1\right)^{2} \\
=t^{2}+\frac{1}{t^{2}}+2\left(t-\frac{1}{t}\right)+2 .
\end{arr... | -2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. A drawer contains red and blue socks, with a total number not exceeding 2016. If two socks are randomly drawn, the probability that they are the same color is $\frac{1}{2}$. Then the maximum number of red socks in the drawer is . $\qquad$ | 5.990.
Let $x$ and $y$ be the number of red and blue socks in the drawer, respectively. Then
$$
\frac{x y}{\mathrm{C}_{x+y}^{2}}=\frac{1}{2} \Rightarrow (x-y)^{2}=x+y.
$$
Thus, the total number of socks is a perfect square.
Let $n=x-y$, i.e., $n^{2}=x+y$. Therefore, $x=\frac{n^{2}+n}{2}$.
Since $x+y \leqslant 2016$, ... | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. If the function $f(x)$ is an odd function with a period of 3, and when $x \in[0,1)$, $f(x)=3^{x}-1$, then $f\left(\log _{\frac{1}{3}} 54\right)=$ $\qquad$ . | 2. -1 .
Notice that,
$$
\log _{\frac{1}{3}} 54=\log _{\frac{1}{3}} 27+\log _{\frac{1}{3}} 2=-3-\log _{3} 2 \text {, }
$$
and $f(x)$ has a period of 3.
Therefore, $f\left(\log _{\frac{1}{3}} 54\right)=f\left(-\log _{3} 2\right)$.
Furthermore, since $f(x)$ is an odd function and $\log _{3} 2 \in[0,1)$, we have
$$
\begin... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the set $I=\{1,2, \cdots, n\}(n \geqslant 3)$. If two non-empty proper subsets $A$ and $B$ of $I$ satisfy $A \cap B=\varnothing, A \cup$ $B=I$, then $A$ and $B$ are called a partition of $I$. If for any partition $A, B$ of the set $I$, there exist two numbers in $A$ or $B$ such that their sum is a perfect square... | 6. 15 .
When $n=14$, take
$A=\{1,2,4,6,9,11,13\}$,
$B=\{3,5,7,8,10,12,14\}$.
It is easy to see that the sum of any two numbers in $A$ and $B$ is not a perfect square.
Thus, $n=14$ does not meet the requirement.
Therefore, $n<14$ also does not meet the requirement.
Now consider $n=15$.
Use proof by contradiction. Assum... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $x, y>0$, and $x+2 y=2$. Then the minimum value of $\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}$ is . $\qquad$ | 8. 2 .
$$
\begin{array}{l}
\text { Let } \boldsymbol{a}=\left(\sqrt{\frac{x^{2}}{2 y}}, \sqrt{\frac{4 y^{2}}{x}}\right), \boldsymbol{b}=(\sqrt{2 y}, \sqrt{x}) . \\
\text { Then }|\boldsymbol{a}|=\sqrt{\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}},|\boldsymbol{b}|=\sqrt{2 y+x}=\sqrt{2} \\
\boldsymbol{a} \cdot \boldsymbol{b}=x+2 ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then the set
$$
\{[x]+[2 x]+[3 x] \mid x \in \mathbf{R}\} \cap\{1,2, \cdots, 100\}
$$
has elements. | 4. 67 .
Let $f(x)=[x]+[2 x]+[3 x]$.
Then $f(x+1)=f(x)+6$.
When $0 \leqslant x<1$, all possible values of $f(x)$ are 0, 1, 2, 3.
Therefore, the range of $f(x)$ is
$$
S=\{6 k, 6 k+1,6 k+2,6 k+3 । k \in \mathbf{Z}\} \text {. }
$$
Thus, $S \cap\{1,2, \cdots, 100\}$ has $4 \times 17-1=67$ elements. | 67 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the sequence $\left\{\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}\right\}$ have the sum of its first $n$ terms as $S_{n}$. Then the number of rational terms in the first 2016 terms of the sequence $\left\{S_{n}\right\}$ is | - 1.43.
$$
\begin{array}{l}
\text { Given } \frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1} \\
\Rightarrow S_{n}=1-\frac{\sqrt{n+1}}{n+1} .
\end{array}
$$
Since $44<\sqrt{2016}<45$, the number of values for which $S$ is rational is 43. | 43 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the curve $C_{1}: y=\sqrt{-x^{2}+10 x-9}$ and point $A(1,0)$. If there exist two distinct points $B$ and $C$ on curve $C_{1}$, whose distances to the line $l: 3 x+1=0$ are $|A B|$ and $|A C|$, respectively, then $|A B|+|A C|=$ | 4.8.
Let points $B\left(x_{B}, y_{B}\right), C\left(x_{C}, y_{C}\right)$, the locus of points whose distance to point $A(1,0)$ is equal to the distance to the line $l: x=-\frac{1}{3}$ is given by the equation $y^{2}=\frac{8}{3} x-\frac{8}{9}$.
Solving the system of equations
$$
\begin{array}{l}
\left\{\begin{array}{l}... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. If the function $f(x)=\frac{a+\sin x}{2+\cos x}+b \tan x$ has a sum of its maximum and minimum values equal to 4, then $a+b=$ $\qquad$ | 5.3.
Given $f(x)=\frac{a+\sin x}{2+\cos x}+b \tan x$ has a maximum or minimum value, we know $b=0$.
Then $y=\frac{a+\sin x}{2+\cos x}$
$$
\begin{array}{l}
\Rightarrow \sin x-y \cos x=a-2 y \\
\Rightarrow \sin (x+\alpha)=\frac{a-2 y}{\sqrt{1+y^{2}}} \\
\Rightarrow|a-2 y| \leqslant \sqrt{1+y^{2}} .
\end{array}
$$
By Vi... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 One evening, 21 people made $n$ phone calls to each other. It is known that among them, there are $m$ ($m$ is an odd number) people $a_{1}, a_{2}$, $\cdots, a_{m}$ such that $a_{i}$ and $a_{i+1}\left(i=1,2, \cdots, m ; a_{m+1}=a_{1}\right)$ made phone calls. If no three people among these 21 people made phone... | Solve: Represent 21 people with 21 points.
If two people communicate by phone, connect the corresponding two points with an edge; otherwise, do not connect an edge, resulting in a graph $G$. It is known that graph $G$ contains an odd cycle, and there are no triangles in graph $G$. The task is to find the maximum number... | 101 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Connecting the intersection points of $x^{2}+y^{2}=10$ and $y=\frac{4}{x}$ in sequence, a convex quadrilateral is formed. The area of this quadrilateral is $\qquad$ | 2. 12 .
Let $A\left(x_{0}, y_{0}\right)\left(x_{0}>0, y_{0}>0\right)$ be the intersection point of the two curves in the first quadrant.
Since the two curves are symmetric with respect to the origin and the line $y=x$, the coordinates of the other three intersection points are
$$
B\left(y_{0}, x_{0}\right), C\left(-x... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with the sum of the first $n$ terms denoted as $S_{n}$. If $S_{6}=26, a_{7}=2$, then the maximum value of $n S_{n}$ is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result direct... | 3. 338 .
$$
\begin{array}{l}
\text { Given } S_{6}=26, a_{7}=2 \\
\Rightarrow a_{1}=6, d=-\frac{2}{3} \\
\Rightarrow S_{n}=-\frac{1}{3} n^{2}+\frac{19}{3} n \\
\Rightarrow n S_{n}=-\frac{1}{3} n^{3}+\frac{19}{3} n^{2} . \\
\text { Let } f(x)=-\frac{1}{3} x^{3}+\frac{19}{3} x^{2}(x \in \mathbf{R}) .
\end{array}
$$
Then... | 338 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
4. If the complex coefficient equation with respect to $x$
$$
(1+2 \mathrm{i}) x^{2}+m x+1-2 \mathrm{i}=0
$$
has real roots, then the minimum value of the modulus of the complex number $m$ is $\qquad$ | 4.2.
Let $\alpha$ be a real root of the original equation, $m=p+q$ i.
$$
\begin{array}{l}
\text { Then }(1+2 \mathrm{i}) \alpha^{2}+(p+q \mathrm{i}) \alpha+1-2 \mathrm{i}=0 \\
\Rightarrow\left\{\begin{array}{l}
\alpha^{2}+p \alpha+1=0, \\
2 \alpha^{2}+q \alpha-2=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) When $x \in [1,2017]$, find
$$
f(x)=\sum_{i=1}^{2017} i|x-i|
$$
the minimum value. | When $k \leqslant x \leqslant k+1(1 \leqslant k \leqslant 2016)$,
$$
\begin{array}{l}
f(x)=\sum_{i=1}^{k} i(x-i)+\sum_{i=k+1}^{2017} i(i-x) \\
=\left(k^{2}+k-2017 \times 1009\right) x+ \\
\frac{2017 \times 2018 \times 4035}{6}-\frac{k(k+1)(2 k+1)}{3}
\end{array}
$$
is a linear function, and its minimum value is attain... | 801730806 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (40 points) Given a $2016 \times 2016$ grid. Find the smallest positive integer $M$, such that it is possible to draw $M$ rectangles (with their sides on the grid lines) in the grid, and each small square in the grid has its sides included in the sides of one of the $M$ rectangles. | Second, $M=2017$.
First, we prove: All grid lines of a $2016 \times 2016$ square grid can be covered by the edges of 2017 rectangles.
In fact, all horizontal grid lines can be covered by the edges of 1008 $1 \times 2016$ rectangles and one $2016 \times 2016$ rectangle; all vertical grid lines can be covered by the edg... | 2017 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Title: A quadruple of positive integers $(p, a, b, c)$ that satisfies the following conditions is called a "Leyden quadruple":
(i) $p$ is an odd prime;
(ii) $a, b, c$ are distinct;
(iii) $p \mid (ab + 1), p \mid (bc + 1), p \mid (ca + 1)$.
(1) Prove that for each Leyden quadruple $(p, a, b, c)$, we have $p + 2 \leq \fr... | From the form, the Leiden quadruples that satisfy the conditions have a certain rotational symmetry. To facilitate expression and discovery of its inherent laws, it is extended as follows:
Definition If the $(k+1)$-tuple of positive integers $\left(p, a_{1}, a_{2}, \cdots, a_{k}\right)$ satisfies the following three p... | 5 | Number Theory | proof | Yes | Yes | cn_contest | false |
Given that $a$, $b$, and $c$ are three distinct real numbers. If the quadratic equations:
$$
\begin{array}{l}
x^{2} + a x + b = 0, \\
x^{2} + b x + c = 0, \\
x^{2} + c x + a = 0
\end{array}
$$
each have exactly one common root with any other, find the value of $a^{2} + b^{2} + c^{2}$.
(The 4th Chen Shengshen Cup Natio... | From [1] we know
$$
x_{1}=\frac{a-b}{a-c}, x_{2}=\frac{b-c}{b-a}, x_{3}=\frac{c-a}{c-b},
$$
where $x_{1}$ is the common root of equations (1) and (3), $x_{2}$ is the common root of equations (1) and (2), and $x_{3}$ is the common root of equations (2) and (3).
Since $x_{1}$ and $x_{2}$ are the two roots of equation (1... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $\sqrt{24-t^{2}}-\sqrt{8-t^{2}}=2$, then
$$
\sqrt{24-t^{2}}+\sqrt{8-t^{2}}=
$$
$\qquad$ | II. 1.8.
From the known equation and using the formula $a+b=\frac{a^{2}-b^{2}}{a-b}$, we get $\sqrt{24-t^{2}}+\sqrt{8-t^{2}}=\frac{24-8}{2}=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. A certain unit distributes a year-end bonus of 1 million yuan, with first prize at 15,000 yuan per person, second prize at 10,000 yuan per person, and third prize at 5,000 yuan per person. If the difference in the number of people between third prize and first prize is no less than 93 but less than 96, then the tota... | 3. 147.
Let the number of first prize winners be $x$, the number of second prize winners be $y$, and the number of third prize winners be $z$.
Then $1.5 x + y + 0.5 z = 100$
$$
\begin{array}{l}
\Rightarrow (x + y + z) + 0.5 x - 0.5 z = 100 \\
\Rightarrow x + y + z = 100 + 0.5(z - x). \\
\text{Given } 93 \leqslant z - ... | 147 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the curve
$$
(x-20)^{2}+(y-16)^{2}=r^{2} \text { and } y=\sin x
$$
have exactly one common point $P\left(x_{0}, y_{0}\right)$. Then
$$
\frac{1}{2} \sin 2 x_{0}-16 \cos x_{0}+x_{0}=
$$
$\qquad$ | 2. 20 .
From the given information, there is a common tangent line at point $P\left(x_{0}, y_{0}\right)$.
$$
\begin{array}{l}
\text { Therefore, } \frac{\sin x_{0}-16}{x_{0}-20} \cos x_{0}=-1 \\
\Rightarrow \frac{1}{2} \sin 2 x_{0}-16 \cos x_{0}+x_{0}=20 .
\end{array}
$$ | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the set $M=\{(a, b) \mid a \leqslant-1, b \leqslant m\}$. If for any $(a, b) \in M$, it always holds that $a \cdot 2^{b}-b-3 a \geqslant 0$, then the maximum value of the real number $m$ is $\qquad$. | 3. 1 .
Notice that,
$$
a \cdot 2^{b}-b-3 a \geqslant 0 \Leftrightarrow\left(2^{b}-3\right) a-b \geqslant 0
$$
holds for any $a \leqslant-1$.
$$
\text { Then }\left\{\begin{array}{l}
2^{b}-3 \leqslant 0, \\
2^{b}+b \leqslant 3
\end{array} \Rightarrow b \leqslant 1\right. \text {. }
$$ | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. If $P$ is the circumcenter of $\triangle A B C$, and $\overrightarrow{P A}+\overrightarrow{P B}+\lambda \overrightarrow{P C}=\mathbf{0}, \angle C=120^{\circ}$. Then the value of the real number $\lambda$ is $\qquad$. | 5. -1 .
Let the circumradius of $\triangle ABC$ be $R$.
From the given information, we have
$$
\begin{array}{l}
|\overrightarrow{P A}+\overrightarrow{P B}|^{2}=\lambda^{2}|\overrightarrow{P C}|^{2} \\
=|\overrightarrow{P A}|^{2}+|\overrightarrow{P B}|^{2}-2|\overrightarrow{P A}||\overrightarrow{P B}| \cos \frac{C}{2} ... | -1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $\alpha, \beta$ satisfy the equations respectively
$$
\begin{array}{l}
\alpha^{3}-3 \alpha^{2}+5 \alpha-4=0, \\
\beta^{3}-3 \beta^{2}+5 \beta-2=0
\end{array}
$$
then $\alpha+\beta=$ $\qquad$ | 9. 2 .
We have
$$
\begin{array}{l}
(\alpha-1)^{3}+2(\alpha-1)-1=0, \\
(1-\beta)^{3}+2(1-\beta)-1=0,
\end{array}
$$
which means $\alpha-1$ and $1-\beta$ are solutions to the equation $x^{3}+2 x-1=0$.
Since $x^{3}+2 x-1=0$ has only one solution, then
$$
\alpha-1=1-\beta \Rightarrow \alpha+\beta=2
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. (20 points) Given sets $A$ and $B$ are both sets of positive integers, and $|A|=20,|B|=16$. Set $A$ satisfies the following condition: if $a, b, m, n \in A$, and $a+b=$ $m+n$, then $\{a, b\}=\{m, n\}$. Define
$$
A+B=\{a+b \mid a \in A, b \in B\} \text {. }
$$
Determine the minimum value of $|A+B|$. | 14. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{20}\right\}$,
$$
\begin{array}{l}
B=\left\{b_{1}, b_{2}, \cdots, b_{16}\right\}, \\
C_{j}=\left\{a_{i}+b_{j} \mid i=1,2, \cdots, 20\right\},
\end{array}
$$
where $j=1,2, \cdots, 16$.
Thus, $A+B=\bigcup_{j=1}^{16} C_{j}$.
We now prove: $\left|C_{m} \cap C_{n}\right| \leqslant ... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $n$ be a positive integer such that $\sqrt{3}$ lies between $\frac{n+3}{n}$ and $\frac{n+4}{n+1}$. Then $n=$ $\qquad$ | 3. 4 .
Notice that,
$$
\frac{n+4}{n+1}=1+\frac{3}{n+1}<1+\frac{3}{n}=\frac{n+3}{n} \text {. }
$$
Since $\sqrt{3}$ is an irrational number, then
$$
\begin{array}{l}
1+\frac{3}{n+1}<\sqrt{3}<1+\frac{3}{n} \\
\Rightarrow \frac{3}{n+1}<\sqrt{3}-1<\frac{3}{n} \\
\Rightarrow n<\frac{3}{\sqrt{3}-1}<n+1 .
\end{array}
$$
Thus... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. If the sum of 12 distinct positive integers is 2016, then the maximum value of the greatest common divisor of these positive integers is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. 24 .
Let the greatest common divisor be $d$, and the 12 numbers be $a_{1} d$, $a_{2} d, \cdots, a_{12} d\left(\left(a_{1}, a_{2}, \cdots, a_{12}\right)=1\right)$.
Let $S=\sum_{i=1}^{12} a_{i}$. Then, $2016=S d$.
To maximize $d$, $S$ should be minimized.
Since $a_{1}, a_{2}, \cdots, a_{12}$ are distinct, then
$$
S \... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
9. In the sequence $\left\{a_{n}\right\}$, $a_{4}=1, a_{11}=9$, and the sum of any three consecutive terms is 15. Then $a_{2016}=$ | Ni, 9.5.
According to the problem, for any $n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
a_{n}+a_{n+1}+a_{n+2}=a_{n+1}+a_{n+2}+a_{n+3}=15 \\
\Rightarrow a_{n+3}=a_{n} .
\end{array}
$$
Then $a_{1}=a_{4}=1, a_{2}=a_{11}=9$,
$$
a_{3}=15-a_{1}-a_{2}=5 \text {. }
$$
Therefore, $a_{2016}=a_{3 \times 672}=a_{3}=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $m$ and $n$ be positive integers, and satisfy $24 m=n^{4}$. Then the minimum value of $m$ is $\qquad$ | 10. 54 .
$$
\begin{array}{l}
\text { Given } n^{4}=24 m=2^{3} \times 3 m \text {, we know that } \\
m_{\min }=2 \times 3^{3}=54 \text {. }
\end{array}
$$ | 54 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
12. Let $x \in \mathbf{R}$. Then the function
$$
f(x)=|2 x-1|+|3 x-2|+|4 x-3|+|5 x-4|
$$
has a minimum value of $\qquad$ | 12. 1 .
Notice,
$$
\begin{array}{l}
f(x)=2\left|x-\frac{1}{2}\right|+3\left|x-\frac{2}{3}\right|+ \\
4\left|x-\frac{3}{4}\right|+5\left|x-\frac{4}{5}\right| \\
=2\left(\left|x-\frac{1}{2}\right|+\left|x-\frac{4}{5}\right|\right)+ \\
\quad 3\left(\left|x-\frac{2}{3}\right|+\left|x-\frac{4}{5}\right|\right)+4\left|x-\fr... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let four complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$ satisfy
$$
\begin{array}{l}
\left|z_{1}-z_{2}\right|=1,\left|z_{3}-z_{4}\right|=2, \\
\left|z_{1}-z_{4}\right|=3,\left|z_{2}-z_{3}\right|=4, \\
z=\left(z_{1}-z_{3}\right)\left(z_{2}-z_{4}\right) .
\end{array}
$$
Then the maximum value of $|z|$ is | 5.14 .
Notice,
$$
\begin{array}{l}
|z|=\left|\left(z_{1}-z_{3}\right)\left(z_{2}-z_{4}\right)\right| \\
=\left|z_{1} z_{2}-z_{1} z_{4}-z_{3} z_{2}+z_{3} z_{4}\right| \\
=\left|\left(z_{1}-z_{2}\right)\left(z_{3}-z_{4}\right)+\left(z_{1}-z_{4}\right)\left(z_{2}-z_{3}\right)\right| \\
\leqslant\left|\left(z_{1}-z_{2}\ri... | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Find the number of all positive integer solutions $(x, y, z)$ to the equation $\arctan \frac{1}{x}+\arctan \frac{1}{y}+\arctan \frac{1}{z}=\frac{\pi}{4}$. | 10. By symmetry, let $x \leqslant y \leqslant z$.
Taking the tangent of both sides of the given equation, we get
$\frac{\frac{1}{y}+\frac{1}{z}}{1-\frac{1}{y z}}=\frac{1-\frac{1}{x}}{1+\frac{1}{x}}$
$\Rightarrow \frac{y+z}{y z-1}=\frac{x-1}{x+1}=1-\frac{2}{x+1}$.
If $x \geqslant 5$, then
$1-\frac{2}{x+1} \geqslant 1-\... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of zeros of the function $f(x)=x^{2} \ln x+x^{2}-2$ is . $\qquad$ | 3.1 .
From the condition, we have
$$
f^{\prime}(x)=2 x \ln x+x+2 x=x(2 \ln x+3) \text {. }
$$
When $0\mathrm{e}^{-\frac{3}{2}}$, $f^{\prime}(x)>0$.
Thus, $f(x)$ is a decreasing function on the interval $\left(0, \mathrm{e}^{-\frac{3}{2}}\right)$ and an increasing function on the interval $\left(\mathrm{e}^{-\frac{3}{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $z \in \mathbf{C}$. If the equation $x^{2}-2 z x+\frac{3}{4}+\mathrm{i}=0$ (where $\mathrm{i}$ is the imaginary unit) has real roots, then the minimum value of $|z|$ is $\qquad$ . | 7.1 .
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R}), x=x_{0}$ be a real root of the original equation.
$$
\begin{array}{l}
\text { Then } x_{0}^{2}-2(a+b \mathrm{i}) x_{0}+\frac{3}{4}+\mathrm{i}=0 \\
\Rightarrow\left\{\begin{array}{l}
x_{0}^{2}-2 a x_{0}+\frac{3}{4}=0, \\
-2 b x_{0}+1=0
\end{array}\right. \\
\Rightarrow ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $f(x)$ be a function defined on $\mathbf{R}$, if $f(0)$ $=1008$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+4)-f(x) \leqslant 2(x+1), \\
f(x+12)-f(x) \geqslant 6(x+5) .
\end{array}
$$
Then $\frac{f(2016)}{2016}=$ $\qquad$ . | 9. 504 .
From the conditions, we have
$$
\begin{array}{l}
f(x+12)-f(x) \\
=(f(x+12)-f(x+8))+ \\
\quad(f(x+8)-f(x+4))+(f(x+4)-f(x)) \\
\leqslant 2((x+8)+1)+2((x+4)+1)+2(x+1) \\
=6 x+30=6(x+5) . \\
\text { Also, } f(x+12)-f(x) \geqslant 6(x+5), \text { thus, } \\
f(x+12)-f(x)=6(x+5) .
\end{array}
$$
Then, $f(2016)$
$$
... | 504 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $f(x)$ is a periodic function on $\mathbf{R}$ with the smallest positive period of 2, and when $0 \leqslant x<2$, $f(x)=x^{3}-x$. Then the number of intersections between the graph of the function $y=f(x)$ and the $x$-axis in the interval $[0,6]$ is $\qquad$ . | 2.7.
When $0 \leqslant x<2$, let $f(x)=x^{3}-x=0$, we get $x=0$ or 1.
According to the properties of periodic functions, given that the smallest period of $f(x)$ is 2, we know that $y=f(x)$ has six zeros in the interval $[0,6)$.
Also, $f(6)=f(3 \times 2)=f(0)=0$, so $f(x)$ has 7 intersection points with the $x$-axis ... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. It is known that Team A and Team B each have several people. If 90 people are transferred from Team A to Team B, then the total number of people in Team B will be twice that of Team A; if some people are transferred from Team B to Team A, then the total number of people in Team A will be 6 times that of Team B. Then... | 8. 153.
Let the original number of people in team A and team B be $a$ and $b$ respectively.
Then $2(a-90)=b+90$.
Suppose $c$ people are transferred from team B to team A. Then $a+c=6(b-c)$.
From equations (1) and (2), eliminating $b$ and simplifying, we get
$$
\begin{array}{l}
11 a-7 c=1620 \\
\Rightarrow c=\frac{11 a... | 153 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
518 On a circle, initially write 1 and 2 at opposite positions. Each operation involves writing the sum of two adjacent numbers between them, for example, the first operation writes two 3s, the second operation writes two 4s and two 5s. After each operation, the sum of all numbers becomes three times the previous sum. ... | Observe the pattern, and conjecture that after a sufficient number of operations, the $n$ numbers written equal $\varphi(n)$. After each operation, the property that adjacent numbers are coprime remains unchanged, and the new number written each time is the sum of its two neighbors. Therefore, the number we are looking... | 2016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. If $f(x)=\sum_{k=0}^{4034} a_{k} x^{k}$ is the expansion of $\left(x^{2}+x+2\right)^{2017}$, then $\sum_{k=0}^{1344}\left(2 a_{3 k}-a_{3 k+1}-a_{3 k+2}\right)=$ $\qquad$ | 8. 2 .
Let $x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$. Then
$$
\begin{array}{l}
x^{2}+x+2=1 \\
\Rightarrow \sum_{k=0}^{1344}\left(a_{3 k}+a_{3 k+1} \omega+a_{3 k+2} \omega^{2}\right)=1
\end{array}
$$
Taking the conjugate of the above equation, we get
$$
\sum_{k=0}^{1344}\left(a_{3 k}+a_{... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Define the sequence $\left\{a_{n}\right\}: a_{n}$ is the last digit of $1+2+\cdots+n$, and $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then $S_{2016}=$ $\qquad$ . | 6.7 066 .
From the problem, we know
$$
\begin{array}{l}
\frac{(n+20)(n+20+1)}{2}=\frac{n^{2}+41 n+420}{2} \\
=\frac{n(n+1)}{2}+20 n+210 .
\end{array}
$$
Then $\frac{(n+20)(n+21)}{2}$ and $\frac{n(n+1)}{2}$ have the same last digit, i.e., $a_{n+20}=a_{n}$.
$$
\begin{array}{l}
\text { Therefore, } S_{2016}=S_{16}+100 S... | 7066 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. From five positive integers $a, b, c, d, e$, any four are taken to find their sum, resulting in the set of sums $\{44,45,46,47\}$, then $a+b+c+d+e=$ $\qquad$ . | 2. 57 .
From five positive integers, if we take any four to find their sum, there are five possible ways, which should result in five sum values. Since the set $\{44,45, 46,47\}$ contains only four elements, there must be two sum values that are equal. Therefore,
$$
\begin{array}{l}
44+44+45+46+47 \\
\leqslant 4(a+b+c... | 57 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that $M N$ is a moving chord of the circumcircle of equilateral $\triangle A B C$ with side length $2 \sqrt{6}$, $M N=4$, and $P$ is a moving point on the sides of $\triangle A B C$. Then the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is | 9.4.
Let the circumcenter of $\triangle ABC$ be $O$.
It is easy to find that the radius of $\odot O$ is $r=2 \sqrt{2}$.
Also, $MN=4$, so $\triangle OMN$ is an isosceles right triangle, and
$$
\begin{aligned}
& \overrightarrow{O M} \cdot \overrightarrow{O N}=0,|\overrightarrow{O M}+\overrightarrow{O N}|=4 . \\
& \text ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange the numbers $2, 3, 4, 6, 8, 9, 12, 15$ in a row so that the greatest common divisor of any two adjacent numbers is greater than 1. The total number of possible arrangements is ( ) .
(A) 720
(B) 1014
(C) 576
(D) 1296 | 4. D.
First, divide the eight numbers into three groups:
I $(2,4,8)$, II $(3,9,15)$, III $(6,12)$.
Since the numbers in group I and group II have no common factors, the arrangement that satisfies the condition must be:
(1) After removing the numbers 6 and 12, the remaining numbers are divided into three parts and arra... | 1296 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
9. As shown in Figure 1, in $\triangle A B C$,
$$
\begin{array}{l}
\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}, \\
\overrightarrow{A H} \cdot \overrightarrow{B C}=0, \\
\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0 .
\end{array}
$$
Then the eccentricity of the hyperbola passing through point $C$ an... | 9. 2 .
Given $\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow(\overrightarrow{C B}-\overrightarrow{C A}) \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow A C=B C$.
From $\overrightarrow{A H} \cdot \overrightarrow{B C}=0 \Rightarrow A H \perp B C$.
Since $\cos \frac... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure $2, P$ is a point on the incircle of square $A B C D$, and let $\angle A P C=\alpha$, $\angle B P D=\beta$. Then $\tan ^{2} \alpha+\tan ^{2} \beta$ $=$ | 4. 8 .
As shown in Figure 5, establish a Cartesian coordinate system.
Let the equation of the circle be $x^{2}+y^{2}=r^{2}$.
Then the coordinates of the vertices of the square are
$$
A(-r,-r), B(r,-r), C(r, r), D(-r, r) \text {. }
$$
If $P(r \cos \theta, r \sin \theta)$, then the slopes of the lines $P A, P B, P C$, ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. If three numbers are taken simultaneously from the 14 integers $1,2, \cdots, 14$, such that the absolute difference between any two numbers is not less than 3, then the number of different ways to choose is $\qquad$ | 3. 120 .
Let the three integers taken out be $x, y, z (x<y<z)$.
$$
\begin{array}{l}
\text { Let } a=x, b=y-x-2, \\
c=z-y-2, d=15-z .
\end{array}
$$
Thus, $a, b, c, d \geqslant 1$.
If $a, b, c, d$ are determined, then $x, y, z$ are uniquely determined.
Since $a+b+c+d=11$, it is equivalent to dividing 11 identical ball... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { Find the largest positive integer } n, \text { such that for positive real } \\
\text { numbers } \alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}, \text { we have } \\
\quad \sum_{i=1}^{n} \frac{\alpha_{i}^{2}-\alpha_{i} \alpha_{i+1}}{\alpha_{i}^{2}+\alpha_{i+1}^{2}} \geqslant 0\left(\alpha_{n+1}... | Let $a_{i}=\frac{\alpha_{i+1}}{\alpha_{i}}(i=1,2, \cdots, n)$.
Then $\prod_{i=1}^{n} a_{i}=1$, and $a_{i}>0$.
The inequality to be proved becomes
$\sum_{i=1}^{n} \frac{1-a_{i}}{1+a_{i}^{2}} \geqslant 0$.
Let $x_{i}=\ln a_{i}(i=1,2, \cdots, n)$.
Then $\sum_{i=1}^{n} \frac{1-\mathrm{e}^{x_{i}}}{1+\left(\mathrm{e}^{x_{i}}... | 5 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Given ten points in space, where no four points lie on the same plane. Some points are connected by line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. ${ }^{[1]}$
(2016, National High School Mathematics Joint Compe... | Proof Let $v$ be a vertex in graph $G$, and $N_{i}(v)$ denote the set of points at a distance $i$ from $v$. For example,
$$
N_{0}(v)=\{v\},
$$
$N_{1}(v)=\{u \mid u$ is adjacent to $v\}$,
$N_{2}(v)=\{w \mid w$ is adjacent to $u$, not adjacent to $v$, and $u$ is adjacent to $v\}$.
First, we prove a lemma.
Lemma Let $G=(... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Arrange the numbers in the set $\left\{2^{x}+2^{y} \mid x 、 y \in \mathbf{N}, x<y\right\}$ in ascending order. Then the 60th number is $\qquad$ (answer with a number).
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3.2064.
It is known that the number of combinations $(x, y)$ satisfying $0 \leqslant x<y \leqslant n$ is $\mathrm{C}_{n+1}^{2}$.
Notice that, $\mathrm{C}_{11}^{2}=55<60<66=\mathrm{C}_{12}^{2}$.
Therefore, the 60th number satisfies $y=11, x=4$, which means the 60th number is $2^{11}+2^{4}=2064$. | 2064 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Find the maximum value of $n$ such that there exists an arithmetic sequence $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfying
$$
\sum_{i=1}^{n}\left|a_{i}\right|=\sum_{i=1}^{n}\left|a_{i}+1\right|=\sum_{i=1}^{n}\left|a_{i}-2\right|=507 .
$$
$(2005$, China Southeast Mathematical Olympiad) | 【Analysis】Let's set $a_{i}=a-i d(d>0, i=1,2$, $\cdots, n)$. Then the given system of equations becomes
$$
\left\{\begin{array}{l}
\sum_{i=1}^{n}|a-i d|=507, \\
\sum_{i=1}^{n}|a+1-i d|=507, \\
\sum_{i=1}^{n}|a-2-i d|=507 .
\end{array}\right.
$$
Thus, the absolute value sum function $f(x)=\sum_{i=1}^{n}|x-i d|$ has thre... | 26 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $f(x)=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$, then the maximum value of $f(x)$ is $\qquad$ | $-、 1.11$.
From the fact that $f(x)$ is defined, we know
$$
0 \leqslant x \leqslant 13 \text {. }
$$
Then $\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$
$$
\begin{array}{l}
=\sqrt{\left(6 \sqrt{\frac{x+27}{36}}+2 \sqrt{\frac{13-x}{4}}+3 \sqrt{\frac{x}{9}}\right)^{2}} \\
\leqslant \sqrt{(6+2+3)\left(6 \times \frac{x+27}{36}+2 \tim... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 There are 68 pairs of non-zero integers on the blackboard. For a positive integer $k$, at most one of the pairs $(k, k)$ and $(-k, -k)$ appears on the blackboard. A student erases some of these 136 numbers so that the sum of any two erased numbers is not 0. It is stipulated that if at least one number from a ... | Given that $(j, j)$ and $(-j, -j)$ can appear at most as one pair, we can assume that if $(j, j)$ appears, then $j > 0$ (otherwise, replace $j$ with $-j$). For a positive integer $k$, all $k$ or $-k$ can be deleted from the blackboard, but not both. For each $k > 0$, delete $k$ with probability $p$ and $-k$ with probab... | 43 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $a \in \mathbf{R}$, the equation ||$x-a|-a|=2$ has exactly three distinct roots. Then $a=$ $\qquad$ | 11. 2 .
The original equation can be transformed into $|x-a|=a \pm 2$.
For the equation to have exactly three distinct roots, then $a=2$. At this point, the equation has exactly three distinct roots:
$$
x_{1}=2, x_{2}=6, x_{3}=-2 \text {. }
$$
Thus, $a=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Given that $a$, $b$, and $c$ are distinct integers. Then
$$
4\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2}
$$
the minimum value is $\qquad$ | 15.8.
Notice,
$$
\begin{array}{l}
4\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2} \\
=(a-b)^{2}+(b-c)^{2}+(c-a)^{2}+a^{2}+b^{2}+c^{2},
\end{array}
$$
its minimum value is 8. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If three distinct real numbers $a$, $b$, and $c$ satisfy
$$
a^{3}+b^{3}+c^{3}=3 a b c \text {, }
$$
then $a+b+c=$ $\qquad$ | 2.0.
Notice,
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\
=\frac{1}{2}(a+b+c)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right) \\
=0 .
\end{array}
$$
Since $a, b, c$ are not all equal, we have
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \neq 0 \text {. }
$$
Therefore, $a+b+c=0... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a$ and $b$ are real numbers. If the quadratic function
$$
f(x)=x^{2}+a x+b
$$
satisfies $f(f(0))=f(f(1))=0$, and $f(0) \neq f(1)$, then the value of $f(2)$ is $\qquad$. | 3. 3 .
It is known that $f(0)=b, f(1)=1+a+b$ are both roots of the equation $f(x)=0$.
$$
\begin{array}{l}
\text { Then } x^{2}+a x+b \equiv(x-b)(x-(1+a+b)) \\
\Rightarrow a=-1-a-2 b, b=b(1+a+b) \\
\Rightarrow a=-\frac{1}{2}, b=0 \\
\Rightarrow f(2)=3 .
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) A function calculator has a display screen and two operation keys. If the first operation key is pressed once, the number on the display screen will change to $\left[\frac{x}{2}\right]$ (where $[x]$ represents the greatest integer not exceeding the real number $x$); if the second operation key is pre... | Three, (1) Impossible.
Convert the number to binary. Then pressing the first operation key means removing the last digit of the number on the display; pressing the second operation key means appending 01 to the number on the display.
When the initial number is 1, after performing the above two operations, the resultin... | 233 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Prove: The function
$$
f(x)=x^{x+2}-(x+2)^{x}-2 x(x+1)(x+2)+2
$$
has only one integer zero in the interval $[0,+\infty)$ | 9. In fact, it is only necessary to prove that there is only one positive integer satisfying $f(x)=0$.
When $x \in \mathbf{Z}_{+}, x \geqslant 5$,
$$
\begin{array}{l}
(x+3)^{x}=\left(\frac{x+3}{2}\right)^{4}(x+3)^{x-4} 4^{2} \\
& (x+3)^{x}-(x-2)^{x}-2 x(x+1)(x+2)+2 \\
> & (x+3)(x+2)^{x-1}-(x+2)^{x}- \\
& 2 x(x+1)(x+2)... | 3 | Algebra | proof | Yes | Yes | cn_contest | false |
Three, (50 points) There are $12k$ people attending a meeting, each of whom has shaken hands with exactly $3k+6$ others, and for any two of them, the number of people who have shaken hands with both is the same. How many people attended the meeting? Prove your conclusion. | Three, abstracting a person as a point and two people shaking hands as a line connecting two points, then $A$ shaking hands with $B$ and $C$ corresponds to $\angle BAC$ in the graph. Since each person has shaken hands with $3k+6$ people, there are $12k \mathrm{C}_{3k+6}^{2}$ such angles in the graph. The number of pair... | 36 | Combinatorics | proof | Yes | Yes | cn_contest | false |
2. A four-digit number divided by 433 has a quotient of $a$ and a remainder of $r$ $(a 、 r \in \mathbf{N})$. Then the maximum value of $a+r$ is $\qquad$ . | 2.454.
Let the four-digit number be $433 a+r(0 \leqslant r \leqslant 432)$. Since $433 \times 24=10392>9999$, then $a \leqslant 23$.
When $a=23$, $433 \times 23+40=9999$. At this point, $a+r=23+40=63$. When $a=22$, $433 \times 22+432=9958$. At this point, $a+r=22+432=454$. In summary, the maximum value of $a+r$ is 454... | 454 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $[x]$ denote the greatest integer not exceeding the real number $x$,
$$
\begin{array}{c}
S=\left[\frac{1}{1}\right]+\left[\frac{2}{1}\right]+\left[\frac{1}{2}\right]+\left[\frac{2}{2}\right]+\left[\frac{3}{2}\right]+ \\
{\left[\frac{4}{2}\right]+\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{3}{3}... | 6.1078.
$$
2+4+\cdots+2 \times 44=1980 \text {. }
$$
For any integer \( k \) satisfying \( 1 \leqslant k \leqslant 44 \), the sum includes \( 2k \) terms with the denominator \( k \): \(\left[\frac{1}{k}\right],\left[\frac{2}{k}\right], \cdots,\left[\frac{2 k}{k}\right]\), whose sum is \( k+2 \).
Also, \( 2016-1980=3... | 1078 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. If real numbers $a, b, c$ make the quadratic function $f(x) = a x^{2} + b x + c$ such that when $0 \leqslant x \leqslant 1$, always $|f(x)| \leqslant 1$. Then the maximum value of $|a| + |b| + |c|$ is $\qquad$ | 7. 17.
Take $x=0, \frac{1}{2}, 1$.
From the problem, we have
$$
\begin{array}{l}
|c| \leqslant 1,|a+2 b+4 c| \leqslant 4, \\
|a+b+c| \leqslant 1 .
\end{array}
$$
Let $m=a+2 b+4 c, n=a+b+c$.
Then $a=-m+2 n+2 c, b=m-n-3 c$
$$
\begin{aligned}
\Rightarrow & |a| \leqslant|m|+2|n|+2|c| \leqslant 8, \\
& |b| \leqslant|m|+|n... | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. | Let the ten points be $A_{1}, A_{2}, \cdots, A_{10}$. Using these ten points as vertices and the line segments connecting them as edges, we obtain a simple graph $G$ of order 10. Let the degree of point $A_{i} (i=1,2, \cdots, 10)$ be $d_{i}$. Then the total number of edges in graph $G$ is $\frac{1}{2} \sum_{i=1}^{10} d... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the sum of the digits of the natural number $x$ be $S(x)$. Then the solution set of the equation $x+S(x)+S(S(x))+S(S(S(x)))=2016$ is $\qquad$ | $-1 .\{1980\}$.
It is easy to see that $x<2016$.
Note that, the sum of the digits of natural numbers less than 2016 is at most 28, for example, $S(1999)=28$, which indicates,
$$
S(x) \leqslant 28 \text {. }
$$
Furthermore, $S(S(x)) \leqslant S(19)=10$.
Finally, $S(S(S(x))) \leqslant 9$.
From the equation we get
$$
\be... | 1980 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Set $A=\left[\frac{7}{8}\right]+\left[\frac{7^{2}}{8}\right]+\cdots+\left[\frac{7^{2016}}{8}\right]$. Then the remainder when $A$ is divided by 50 is $\qquad$ | 3. 42.
Since $\frac{7^{2 k-1}}{8}$ and $\frac{7^{2 k}}{8}$ are not integers, and
$$
\frac{7^{2 k-1}}{8}+\frac{7^{2 k}}{8}=7^{2 k-1},
$$
for any $k \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
{\left[\frac{7^{2 k-1}}{8}\right]+\left[\frac{7^{2 k}}{8}\right]=7^{2 k-1}-1} \\
\equiv 7(-1)^{k-1}-1(\bmod 50) .
\end{arr... | 42 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Rectangle $R$ is divided into 2016 small rectangles, with each small rectangle's sides parallel to the sides of rectangle $R$. The vertices of the small rectangles are called "nodes". For a line segment on the side of a small rectangle, if both endpoints are nodes and its interior does not contain any other n... | Consider a graph $G$ with all nodes as vertices and basic segments as edges. Let the number of vertices in graph $G$ be $v$, and the number of edges be $e$. Treat the external region of rectangle $R$ as one face (region). Then, the total number of faces in graph $G$ is $f=2017$.
By Euler's formula, we have $v+f-e=2$.
T... | 4122 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
13. In $\triangle A B C$, $\angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively. Let
$$
\begin{array}{l}
f(x)=\boldsymbol{m} \cdot \boldsymbol{n}, \boldsymbol{m}=(2 \cos x, 1), \\
\boldsymbol{n}=(\cos x, \sqrt{3} \sin 2 x), \\
f(A)=2, b=1, S_{\triangle A B C}=\frac{\sqrt{3}}{2} . \\
\text { Then }... | $=13.2$.
It is easy to know, $f(x)=1+2 \sin \left(2 x+\frac{\pi}{6}\right)$.
Combining the conditions, we get $\angle A=\frac{\pi}{3}, c=2, a=\sqrt{3}$. Therefore, $\frac{a}{\sin A}=\frac{b+c}{\sin B+\sin C}=2$. | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}+a_{n+1}=n(-1)^{\frac{a(a+1)}{2}} \text {, }
$$
the sum of the first $n$ terms is $S_{n}, m+S_{2015}=-1007, a_{1} m>0$. Then the minimum value of $\frac{1}{a_{1}}+\frac{4}{m}$ is $\qquad$ . | 15.9.
$$
\begin{array}{l}
\text { Given } S_{2015}=a_{1}+\sum_{k=1}^{1007}\left(a_{2 k}+a_{2 k+1}\right) \\
=a_{1}-1008 \\
\Rightarrow m+a_{1}-1008=-1007 \\
\Rightarrow m+a_{1}=1 . \\
\text { Also, } m a_{1}>0 \text {, so } m>0, a_{1}>0 . \\
\text { Therefore, } \frac{1}{a_{1}}+\frac{4}{m}=\left(m+a_{1}\right)\left(\fr... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Define the length of intervals $(m, n)$, $[m, n)$, $(m, n]$, and $[m, n]$ to be $n-m$ (where $n, m \in \mathbf{R}$, and $n > m$). Then the sum of the lengths of the intervals of real numbers $x$ that satisfy
$$
\frac{1}{x-20}+\frac{1}{x-17} \geqslant \frac{1}{512}
$$
is $\qquad$ . | $-1.1024$.
Let $a=20, b=17, c=\frac{1}{512}$.
Then $a>b>c>0$.
The original inequality is equivalent to $\frac{2 x-(a+b)}{(x-a)(x-b)} \geqslant c$.
When $x>a$ or $x0, f(a)=b-a<0$.
Let the two real roots of $f(x)=0$ be $x_{1}$ and $x_{2}$ $\left(x_{1}<x_{2}\right)$. Then the interval of $x$ that satisfies $f(x) \leqslant... | 1024 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
2. The equation
$$
\sqrt[3]{(x+7)(x+8)}-\sqrt[3]{(x+5)(x+10)}=2
$$
has $\qquad$ real solutions $x$ that are not equal. | 2.4.
Let $a=\sqrt[3]{(x+7)(x+8)}$,
$$
b=\sqrt[3]{(x+5)(x+10)} \text {. }
$$
Then $a-b=2, a^{3}-b^{3}=6$.
Eliminating $a$ gives
$$
\begin{array}{l}
3 b^{2}+6 b+1=0 \\
\Rightarrow b_{1}=\frac{-3+\sqrt{6}}{3}, b_{2}=\frac{-3-\sqrt{6}}{3} .
\end{array}
$$
If $b_{1}=\frac{-3+\sqrt{6}}{3}$, then
$$
x^{2}+15 x+50+\left(\fr... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the function
$$
\begin{aligned}
f(x)= & a \tan ^{2017} x+b x^{2017}+ \\
& c \ln \left(x+\sqrt{x^{2}+1}\right)+20,
\end{aligned}
$$
where $a$, $b$, and $c$ are real numbers. If $f\left(\ln \log _{5} 21\right)=17$, then $f\left(\ln \log _{21} 5\right)=$ $\qquad$ | 3. 23 .
Let $g(x)$
$$
=a \tan ^{2017} x+b x^{2017}+c \ln \left(x+\sqrt{x^{2}+1}\right) \text {. }
$$
Then $g(-x)=-g(x)$
$$
\begin{array}{l}
\Rightarrow f(-x)-20=-(f(x)-20) \\
\Rightarrow f(-x)=40-f(x) .
\end{array}
$$
Therefore, $f\left(\ln \log _{21} 5\right)=f\left(-\ln \log _{5} 21\right)$
$$
=40-f\left(\ln \log ... | 23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the complex number $z$ satisfies
$$
(a-2) z^{2018}+a z^{2017} \mathrm{i}+a z \mathrm{i}+2-a=0 \text {, }
$$
where, $a<1, \mathrm{i}=\sqrt{-1}$. Then $|z|=$ $\qquad$ | 6. 1 .
Notice,
$$
z^{2017}((a-2) z+a \mathrm{i})=a-2-a z \mathrm{i} \text {. }
$$
Thus, $|z|^{2017}|(a-2) z+a \mathrm{i}|=|a-2-a z \mathrm{i}|$.
Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$.
$$
\begin{array}{l}
\text { Then }|(a-2) z+a \mathrm{i}|^{2}-|a-2-a z \mathrm{i}|^{2} \\
=|(a-2) x+((a-2) y+a) \mathrm{i}|^{2}-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. For any positive integer $n$, define
$$
S(n)=\left[\frac{n}{10^{[\lg n]}}\right]+10\left(n-10^{[\lg n]}\left[\frac{n}{10^{[\lg n]}}\right]\right) \text {. }
$$
Then among the positive integers $1,2, \cdots, 5000$, the number of positive integers $n$ that satisfy $S(S(n))=n$ is $\qquad$ . | 7. 135.
Let $t=10^{[18 n]}$, then
$$
S(n)=\left[\frac{n}{t}\right]+10\left(n-t\left[\frac{n}{t}\right]\right)
$$
Notice that, $n-t\left[\frac{n}{t}\right]$ is the remainder of $n$ modulo $t$, and $\left[\frac{n}{t}\right]$ is the leading digit of $n$.
We will discuss the cases separately.
(1) If $n$ is a one-digit nu... | 135 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given the ellipse $\Gamma: \frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, a line passing through the left focus $F(-2,0)$ of the ellipse $\Gamma$ with a slope of $k_{1}\left(k_{1} \notin\{0\right.$, $\infty\})$ intersects the ellipse $\Gamma$ at points $A$ and $B$. Let point $R(1,0)$, and extend $A R$ and $B R$ to intersect th... | 9. 305 .
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, $C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$,
$l_{A R}: x=\frac{x_{1}-1}{y_{1}} y+1$.
Substitute into the equation of the ellipse $\Gamma$, eliminate $x$ to get
$$
\frac{5-x_{1}}{y_{1}^{2}} y^{2}+\frac{x_{1}-1}{y_{1}} y-4=0 \text {. }
$... | 305 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Find the maximum value of the positive integer $r$ such that: for any five 500-element subsets of the set $\{1,2, \cdots, 1000\}$, there exist two subsets that have at least $r$ elements in common. | Three, first explain $r \leqslant 200$.
Take $k \in\{1,2, \cdots, 10\}$. Let
$$
A_{k}=\{100 k-99,100 k-98, \cdots, 100 k\} \text {. }
$$
Consider the set
$$
\begin{array}{l}
A_{1} \cup A_{5} \cup A_{6} \cup A_{7} \cup A_{9}, A_{1} \cup A_{2} \cup A_{7} \cup A_{8} \cup A_{10}, \\
A_{2} \cup A_{3} \cup A_{6} \cup A_{8} ... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Find the number of integers in the set $\left\{\left.\frac{2015[a, b]}{a+b} \right\rvert\, a 、 b \in \mathbf{Z}_{+}\right\}$. | Let $d=(a, b)$, and $a=A d, b=B d$, where $A$ and $B$ are coprime positive integers.
Since $[a, b]=A B(a, b)$, then
$(a+b)|2015[a, b] \Leftrightarrow(A+B)| 2015 A B$.
Because $(A, B)=1$, we have
$(A+B, A)=1,(A+B, B)=1$,
$(A+B, A B)=1$.
Thus, $(A+B) \mid 2015$.
For a fixed divisor $k$ of 2015 greater than 1, and $k$ is ... | 1007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Red, blue, green, and white four dice, each die's six faces have numbers $1, 2, 3, 4, 5, 6$. Simultaneously roll these four dice so that the product of the numbers facing up on the four dice equals 36, there are $\qquad$ possible ways. | 4. 48 .
$$
\begin{array}{l}
36=6 \times 6 \times 1 \times 1=6 \times 3 \times 2 \times 1 \\
=4 \times 3 \times 3 \times 1=3 \times 3 \times 2 \times 2 .
\end{array}
$$
For each of the above cases, there are respectively
$$
\begin{array}{l}
\frac{4!}{(2!)(2!)}=6,4!=24, \\
\frac{4!}{2!}=12, \frac{4!}{(2!)(2!)}=6
\end{ar... | 48 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 On the plane, there are $n(n \geqslant 5)$ distinct points, each point is exactly at a distance of 1 from four other points. Find the minimum value of such $n$. [2] | Keep points $A, B, D, E, G$ in Figure 1, construct $\square E A B C, \square D A G F, \square B A G I$ to get points $C, F, I$, and construct $\square C I F H$ to get point $H$.
$$
\begin{array}{l}
\text { From } B I=A G=A E=B C, \\
\angle I B C=\angle A B C-\angle A B I \\
=\left(180^{\circ}-\angle B A E\right)-\left(... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Find all positive integers $n$, such that all positive divisors of $n$ can be placed in the cells of a rectangular grid, satisfying the following constraints:
(1) Each cell contains a different divisor;
(2) The sum of the numbers in each row of cells is equal;
(3) The sum of the numbers in each column of cells is eq... | 2. $n=1$.
Assume all positive divisors of $n$ can be placed in a $k \times l$ $(k \leqslant l)$ rectangular grid, and satisfy the conditions.
Let the sum of the numbers in each column of the grid be $s$.
Since $n$ is one of the numbers in a column of the grid, we have $s \geqslant n$, and the equality holds if and onl... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. There are $n(n \geqslant 2)$ cards, each with a real number written on it, and these $n$ numbers are all distinct. Now, these cards are arbitrarily divided into two piles (each pile has at least one card). It is always possible to take one card from the first pile and place it in the second pile, and then take one c... | 6. The maximum possible value of $n$ is 7.
If given seven cards, each written with $0, \pm 1, \pm 2, \pm 3$, it is easy to verify that they meet the requirements.
Below is the proof that the number of cards cannot be more.
Take any one card as the first pile, and the remaining cards as the second pile. After the oper... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $k$ be a positive integer. Suppose that all positive integers can be colored using $k$ colors, and there exists a function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, satisfying:
(1) For any positive integers $m, n$ of the same color (allowing $m = n$), we have $f(m+n)=f(m)+f(n)$;
(2) There exist positive in... | 2. The minimum value of $k$ is 3.
First, construct an example for $k=3$.
Let $f(n)=\left\{\begin{array}{ll}2 n, & n \equiv 0(\bmod 3) ; \\ n, & n \equiv 1,2(\bmod 3)\end{array}\right.$
Then $f(1)+f(2)=3 \neq f(3)$, satisfying condition (2).
At the same time, color the numbers that are congruent to $0, 1, 2 \pmod{3}$ wi... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given natural numbers $a, b, c$ whose sum is $S$, satisfying $a+b=1014, c-b=497, a>b$. Then the maximum value of $S$ is ( ).
(A) 1511
(B) 2015
(C) 22017
(D) 2018 | $\begin{array}{l}\text { I. 1.C. } \\ \text { Given } S=a+b+c=1014+b+497 \text {, and } a>b \\ \Rightarrow 1014=a+b \geqslant b+1+b \\ \Rightarrow b \leqslant 506.6 \Rightarrow b_{\max }=506 \\ \Rightarrow S_{\text {max }}=1014+506+497=2017 .\end{array}$ | 2017 | Algebra | MCQ | Yes | Yes | cn_contest | false |
II. (25 points) Several boxes are unloaded from a cargo ship, with a total weight of 10 tons, and the weight of each box does not exceed 1 ton. To ensure that these boxes can be transported away in one go, the question is: what is the minimum number of trucks with a carrying capacity of 3 tons needed? | II. First, notice that the weight of each box does not exceed 1 ton. Therefore, the weight of boxes that each vehicle can transport at once will not be less than 2 tons. Otherwise, another box can be added.
Let $n$ be the number of vehicles needed, and the weights of the boxes transported by each vehicle be $a_{1}, a_... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Let $A=\{0,1, \cdots, 2016\}$. If a surjective function $f: \mathbf{N} \rightarrow A$ satisfies: for any $i \in \mathbf{N}$,
$$
f(i+2017)=f(i),
$$
then $f$ is called a "harmonious function".
$$
\begin{array}{l}
\text { Let } f^{(1)}(x)=f(x), \\
f^{(k+1)}(x)=f\left(f^{(k)}(x)\right)\left(k \in \mathbf... | On the one hand, note that 2017 is a prime number.
Let $g$ be a primitive root modulo 2017, then the half-order of $g$ modulo 2017 is 1008.
$$
\text{Let } f(i) \equiv g(i-1)+1(\bmod 2017) \text{.}
$$
Since $(g, 2017)=1$, $g(i-1)+1$ runs through a complete residue system modulo 2017.
Thus, the mapping $f: \mathbf{N} \r... | 1008 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
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