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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 $\left\{K_{n}\right\}$, called the Water Cube Number sequence, and let $K_{0}=0$.
(1) Find $n$ such that $K_{n}=2016$;
(2) Determine $K_{2016}$ and $S_{2016}=\sum^{2016} K_{n}$.
|
【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 $3 \mid (a+b+c)$, $3^{2} \mid V(a, b, c)$;
while when $3 \nmid (a+b+c)$, $3 \nmid V(a, b, c)$.
Therefore, each number in the set $M$ is not a water cube number.
Next, we prove: each natural number outside the set $M$ is a water cube number.
When $k \in \mathbf{N} \backslash M$:
If $k=0$, then $V(1,1,1)=0$;
If $k>0$, then
$$
k \in\{3 n+1,3 n+2,9(n+1) \mid k \in \mathbf{N}\},
$$
For each natural number $n$, we have
$$
\begin{array}{l}
V(n, n, n+1)=3 n+1, \\
V(n, n+1, n+1)=3 n+2, \\
V(n, n+1, n+2)=9(n+1),
\end{array}
$$
Hence $k$ is a water cube number.
The lemma is proved.
According to the lemma, we know that a positive integer $k$ is not a water cube number if and only if $k$ has the form $9 n+3$ or $9 n+6$, where $n$ is any natural number, i.e., among every nine consecutive positive integers, exactly seven are water cube numbers, and the other two are not water cube numbers.
Now list all positive integers in sequence, and divide them into segments of nine numbers each, from left to right:
$$
\begin{array}{l}
1,2, \cdots, 9 ; 10,11, \cdots, 18 ; \cdots ; \\
9 m-8,9 m-7, \cdots, 9 m ; \cdots .
\end{array}
$$
In the first $m$ segments, each segment has exactly seven water cube numbers, and $9 m$ is the largest water cube number among them, then
$$
K_{7 m}=9 m \text {. }
$$
(1) According to equation (1),
$$
\begin{array}{l}
2016=9 \times 224=K_{7 \times 224}=K_{1568} \\
\Rightarrow n=1568 .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence } S_{7 m}=\sum_{n=1}^{7 m} K_{n} \\
=\sum_{n=1}^{9 m} n-\sum_{n=0}^{m-1}(9 n+3)-\sum_{n=0}^{m-1}(9 n+6) \\
=\sum_{n=1}^{9 m} n-\sum_{n=0}^{m-1}(18 n+9) \\
=\frac{9 m(9 m+1)}{2}-9 \sum_{n=0}^{m-1}(2 n+1) \\
=\frac{9 m(9 m+1)}{2}-9 m^{2} \\
=\frac{9 m(7 m+1)}{2} .
\end{array}
$$
(2) Since 2016 is a multiple of both 7 and 9, then $m=288$.
Hence $K_{2016}=K_{7 \times 288}=9 \times 288=2592$,
$$
S_{2016}=\frac{9 m(7 m+1)}{2}=2614032 .
$$
|
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 the successor number 3 of 2, to get $a_{4}=1, a_{5}=1, a_{6}=2, a_{7}=3$;
Next, copy all the previous items $1,1,2, 1, 1, 2, 3$, and add the successor number 4 of 3, to get the first 15 items as $1,1,2,1,1,2,3,1,1,2,1,1,2,3,4$; and so on.
Try to find $a_{2000}$ and the sum of the first 2000 terms of the sequence $S_{2000}$. (2000, Nanchang City High School Mathematics Competition)
|
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 $a_{m}=a_{k}$.
Given $2000=2^{10}-1+977$,
$977=2^{9}-1+466,466=2^{8}-1+211$,
$211=2^{7}-1+84,84=2^{6}-1+21$,
$21=2^{4}-1+6$,
thus $a_{2000}=a_{977}=a_{466}=a_{211}$
$$
=a_{84}=a_{21}=a_{6}=2 \text {. }
$$
To find $S_{2000}$, we first calculate $S_{2^{n}-1}$.
From the construction method of the sequence $\left\{a_{n}\right\}$, we know that in the first $2^{n}-1$ terms of the sequence, there is exactly 1 $n$, 2 $n-1$s, $2^{2}$ $n-2$s, $\cdots, 2^{k}$ $n-k$s, $\cdots, 2^{n-1}$ 1s.
Thus, $S_{2^{n-1}}=n+2(n-1)+2^{2}(n-2)+$
$$
\begin{array}{c}
\cdots+2^{n-2} \times 2+2^{n-1} \times 1, \\
2 S_{2^{n-1}}=2 n+2^{2}(n-1)+2^{3}(n-2)+ \\
\cdots+2^{n-1} \times 2+2^{n} .
\end{array}
$$
Subtracting the above two equations, we get
$$
\begin{array}{l}
S_{2^{n}-1}=-n+\left(2+2^{2}+\cdots+2^{n-1}+2^{n}\right) \\
=2^{n+1}-(n+2) .
\end{array}
$$
When $m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$,
$$
\begin{array}{l}
S_{m}=S_{2^{n}-1}+\sum_{i=1}^{k} a_{\left(2^{n}-1\right)+i} \\
=S_{2^{n}-1}+\left(a_{1}+a_{2}+\cdots+a_{k}\right)=S_{2^{n}-1}+S_{k} .
\end{array}
$$
Thus, $S_{2000}=S_{2^{10}-1}+S_{977}, S_{977}=S_{2^{9}-1}+S_{466}$,
$$
\begin{array}{l}
S_{466}=S_{2^{8}-1}+S_{211}, S_{211}=S_{2^{7}-1}+S_{84}, \\
S_{84}=S_{2^{6}-1}+S_{21}, S_{21}=S_{2^{4}-1}+S_{6}, S_{6}=8 .
\end{array}
$$
Substituting into equation (1), we get
$$
\begin{aligned}
S_{2000}= & \left(2^{11}-12\right)+\left(2^{10}-11\right)+\left(2^{9}-10\right)+ \\
& \left(2^{8}-9\right)+\left(2^{7}-8\right)+\left(2^{5}-6\right)+6 \\
= & 3950
\end{aligned}
$$
|
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 $f(x)$ reaches its maximum value if and only if $x=\frac{5 m \pi}{k}(m \in \mathbf{Z})$.
According to the conditions, we know that any open interval of length 1, $(a, a+1)$, contains at least one maximum point. Therefore,
$$
\frac{5 \pi}{k}5 \pi \text {. }
$$
Conversely, when $k>5 \pi$, any open interval $(a, a+1)$ contains a complete period of $f(x)$, at which point, $\{f(x) \mid a<x<a+1\}=\{f(x) \mid x \in \mathbf{R}\}$.
In summary, the minimum value of $k$ is $[5 \pi]+1=16$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
|
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 $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ is
|
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
$$
\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\frac{a_{3}}{a_{4}},
$$
which means \(a_{1}, a_{2}, a_{3}, a_{4}\) form a geometric sequence.
Thus, the problem is equivalent to counting the number of geometric sequences \(\{a_{1}, a_{2}, a_{3}, a_{4}\}\) such that
$$
\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\} \subseteq\{1,2, \cdots, 100\}.
$$
Let the common ratio \(q \neq 1\) and \(q \in \mathbf{Q}\). Denote \(q=\frac{n}{m}\), where \(m\) and \(n\) are coprime positive integers, and \(m \neq n\).
First, consider the case \(n > m\).
In this case, \(a_{4}=a_{1}\left(\frac{n}{m}\right)^{3}=\frac{a_{1} n^{3}}{m^{3}}\).
Since \(m^{3}\) and \(n^{3}\) are coprime, \(l=\frac{a_{1}}{m^{3}} \in \mathbf{Z}_{+}\). Accordingly, \(a_{1}, a_{2}, a_{3}, a_{4}\) are \(m^{3} l, m^{2} n l, m n^{2} l, n^{3} l\), all of which are positive integers. This indicates that for any given \(q=\frac{n}{m}>1\), the number of sequences \(\{a_{1}, a_{2}, a_{3}, a_{4}\}\) that satisfy the conditions and have \(q\) as the common ratio is the number of positive integers \(l\) that satisfy the inequality \(n^{3} l \leqslant 100\), which is \(\left[\frac{100}{n^{3}}\right]\).
Since \(5^{3} > 100\), we only need to consider
$$
q=2,3, \frac{3}{2}, 4, \frac{4}{3}
$$
The total number of such geometric sequences is
$$
\begin{array}{l}
{\left[\frac{100}{8}\right]+\left[\frac{100}{27}\right]+\left[\frac{100}{27}\right]+\left[\frac{100}{64}\right]+\left[\frac{100}{64}\right]} \\
=12+3+3+1+1=20 .
\end{array}
$$
When \(n < m\), by symmetry, there are also 20 sequences \(\{a_{1}, a_{2}, a_{3}, a_{4}\}\) that satisfy the conditions.
In summary, there are 40 ordered tuples \(\left(a_{1}, a_{2}, a_{3}, a_{4}\right)\) that satisfy the conditions.
|
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 $\operatorname{deg}\left(v_{i}\right)$ to denote the degree of vertex $v_{i}$.
If $\operatorname{deg}\left(v_{i}\right) \leqslant 3$ for $i=1,2, \cdots, 10$, then $k=\frac{1}{2} \sum_{i=1}^{10} \operatorname{deg}\left(v_{i}\right) \leqslant \frac{1}{2} \times 10 \times 3=15$.
Assume there exists a vertex $v_{i}$ such that $\operatorname{deg}\left(v_{i}\right) \geqslant 4$. Without loss of generality, let $\operatorname{deg}\left(v_{1}\right)=n \geqslant 4$, and $v_{1}$ is adjacent to $v_{2}, v_{3}, \cdots, v_{n+1}$. Thus, there are no edges between $v_{2}, v_{3}, \cdots, v_{n+1}$, otherwise, a triangle would be formed. Therefore, there are exactly $n$ edges between $v_{1}, v_{2}, \cdots, v_{n+1}$.
For each $j(n+2 \leqslant j \leqslant 10)$, $v_{j}$ can be adjacent to at most one of $v_{2}, v_{3}$, $\cdots, v_{n+1}$ (otherwise, if $v_{j}$ is adjacent to $v_{s}$ and $v_{t}(2 \leqslant s<t \leqslant n+1)$, then $v_{1}, v_{s}, v_{j}, v_{t}$ would correspond to the four vertices of a spatial quadrilateral, which contradicts the given condition). Therefore, the number of edges between $v_{2}, v_{3}, \cdots, v_{n+1}$ and $v_{n+2}, v_{n+3}$, $\cdots, v_{10}$ is at most
$$
10-(n+1)=9-n \text {. }
$$
Among the $9-n$ vertices $v_{n+2}, v_{n+3}, \cdots, v_{n}$, since there are no triangles, by Turán's theorem, there are at most $\left[\frac{(9-n)^{2}}{4}\right]$ edges. Therefore, the number of edges in graph $G$ is
$$
\begin{array}{l}
k \leqslant n+(9-n)+\left[\frac{(9-n)^{2}}{4}\right] \\
=9+\left[\frac{(9-n)^{2}}{4}\right] \leqslant 9+\left[\frac{25}{4}\right]=15 .
\end{array}
$$
The graph in Figure 8 has 15 edges and meets the requirements. In summary, the maximum number of edges is 15.
|
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 not the hypotenuse of any right triangle is called a "blank square." Find the maximum number of blank squares.
|
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:
$$
\begin{array}{l}
\{(x, y) \mid x=k, 0 \leqslant k \leqslant 2 n, k \in \mathbf{Z}\}, \\
\{(x, y) \mid y=k, 0 \leqslant k \leqslant 2 n, k \in \mathbf{Z}\} .
\end{array}
$$
For an isosceles right triangle, if its right-angle vertex is at the bottom-left (top-left, bottom-right, top-right) corner of its cell, the isosceles triangle is called bottom-left (top-left, bottom-right, top-right). Bottom-left (top-left) and bottom-right (top-right) triangles are collectively referred to as bottom (top) triangles.
Let $u_{k}(0 \leqslant k \leqslant 2 n)$ denote the number of bottom right triangles with one leg on the line $y=k$, and $d_{k}(0 \leqslant k \leqslant 2 n)$ denote the number of top right triangles with one leg on the line $y=k$.
$$
\begin{array}{l}
\text { Then } u_{k}+d_{k}=2 n, u_{0}=d_{2 n}=2 n, \\
u_{k}+d_{k+1}=2 n+1 .
\end{array}
$$
Solving this, we get $u_{k}=2 n-k, d_{k}=k$. There are $u_{k}$ bottom triangles and $d_{k+1}$ top triangles between the lines $y=k$ and $y=k+1$. Therefore, the number of empty cells in this row is no more than
$$
2 n-\max \left\{u_{k}, d_{k+1}\right\} \text {. }
$$
Thus, the total number of empty cells is no more than
$$
2[0+1+\cdots+(n-1)]=n(n-1) \text {. }
$$
Below is an example with $n(n-1)$ empty cells.
The set of coordinates for the right-angle vertices of bottom-left triangles:
$$
\begin{array}{l}
\{(x, y) \mid x 、 y \in \mathbf{Z}, x 、 y \geqslant 0, x+y \leqslant n-1\} \cup \\
\{(x+k-1, n+k-x) \mid x 、 k=1,2, \cdots, n\},
\end{array}
$$
The set of coordinates for the right-angle vertices of top-right triangles:
$$
\begin{array}{l}
\{(2 n-x, 2 n-y) \mid x 、 y \in \mathbf{Z}, x 、 y \geqslant 0, \\
x+y \leqslant n-1\} \cup \\
\{(x+k-1, n+k-x) \mid x 、 k=1,2, \cdots, n\},
\end{array}
$$
The set of coordinates for the right-angle vertices of top-left triangles:
$$
\begin{array}{l}
\{(x, y) \mid x 、 y \in \mathbf{Z}, x=k, n+1+k \leqslant y \leqslant \\
2 n, k=0,1, \cdots, n-1\},
\end{array}
$$
The set of coordinates for the right-angle vertices of bottom-right triangles:
$$
\begin{array}{l}
\{(2 n-x, 2 n-y) \mid x, y \in \mathbf{Z}, x=k, \\
n+1+k \leqslant y \leqslant 2 n, k=0,1, \cdots, n-1\} .
\end{array}
$$
Thus, the required result is 2450.
|
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 $Q$. Then
$$
\frac{|P Q|}{|P I|}+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|}=
$$
$\qquad$
|
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 hyperbola at point $P$ is perpendicular to line $P Q$.
Thus, $l_{P Q}: y=\frac{-4 y_{0}}{5 x_{0}}\left(x-x_{0}\right)+y_{0}$.
Let $y=0$, we get $Q\left(\frac{9}{4} x_{0}, 0\right)$.
Therefore, $\frac{|P Q|}{|P I|}+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|}=1+2 \frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|}$
$$
=1+2 \times \frac{\frac{9}{4} x_{0}+3}{\frac{3}{2} x_{0}+2}=4
$$
|
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: $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2016}}<3$.
|
(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 $n=2 m+1\left(m \in \mathbf{Z}_{+}\right)$,
$$
a_{n}=a_{1}+\sum_{k=1}^{m}\left(a_{2 k+1}-a_{2 k-1}\right)=\frac{1}{4}(n+1)^{2} \text {. }
$$
Thus, $a_{n}=\left\{\begin{array}{ll}\frac{1}{4}(n+1)^{2}, & n \text { is odd; } \\ \frac{1}{4} n(n+2), & n \text { is even. }\end{array}\right.$
(2) When $n$ is odd,
$$
\begin{array}{l}
\frac{1}{a_{n}}=\frac{4}{(n+1)^{2}}<\frac{4}{(n-1)(n+1)} \\
=2\left(\frac{1}{n-1}-\frac{1}{n+1}\right) ;
\end{array}
$$
When $n$ is even,
$$
\begin{array}{l}
\frac{1}{a_{n}}=\frac{4}{n(n+2)}=2\left(\frac{1}{n}-\frac{1}{n+2}\right) . \\
\text { Therefore, } \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2016}} \\
=\left(\frac{1}{a_{1}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{2015}}\right)+ \\
\quad\left(\frac{1}{a_{2}}+\frac{1}{a_{4}}+\cdots+\frac{1}{a_{2016}}\right) \\
<\left[1+2\left(\frac{1}{2}-\frac{1}{2016}\right)\right]+2\left(\frac{1}{2}-\frac{1}{2018}\right)<3 .
\end{array}
$$
|
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 $k=4$,
$$
6=1+1+1+3=1+1+2+2 \text {, }
$$
the probability is $(4+6)\left(\frac{1}{6}\right)^{4}=10\left(\frac{1}{6}\right)^{4}$;
When $k=5$,
$$
6=1+1+1+1+2 \text {, }
$$
the probability is $5\left(\frac{1}{6}\right)^{5}$;
When $k=6$, the probability is $\left(\frac{1}{6}\right)^{6}$.
$$
\begin{array}{l}
\text { Hence } p=\frac{1}{6}+5\left(\frac{1}{6}\right)^{2}+10\left(\frac{1}{6}\right)^{3}+10\left(\frac{1}{6}\right)^{4}+ \\
\quad 5\left(\frac{1}{6}\right)^{5}+\left(\frac{1}{6}\right)^{6} \\
=\frac{1}{6}\left(1+\frac{1}{6}\right)^{5}=\frac{7^{5}}{6^{6}} \\
\Rightarrow n=7^{5}, m=6^{6} \\
\Rightarrow \log _{6} m-\log _{7} n=1 .
\end{array}
$$
|
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 numbers is $(\quad$.
(A) 6858
(B) 6860
(C) 9260
(D) 9262
|
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
$$
\begin{array}{l}
{\left[1^{3}-(-1)^{3}\right]+\left(3^{3}-1^{3}\right)+\left(5^{3}-6^{3}\right)+} \\
\cdots+\left(19^{3}-17^{3}\right) \\
=19^{3}+1=6860 .
\end{array}
$$
|
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 of $\frac{1}{x y} + \frac{1}{y z} + \frac{1}{z z}$;
(2) Prove:
$$
\begin{array}{l}
9(x+y)(y+z)(z+x) \\
\geqslant 8 x y z(x y + y z + z x) .
\end{array}
$$
|
(1) From the given equation, we have
$$
\begin{array}{l}
z\left(x^{2}-1\right)\left(y^{2}-1\right)+x\left(y^{2}-1\right)\left(z^{2}-1\right)+ \\
y\left(z^{2}-1\right)\left(x^{2}-1\right)=4 x y z \\
\Rightarrow \quad x y z(x y+y z+z x)-(x+y+z)(x y+y z+z x)+ \\
\quad(x+y+z)-x y z=0 \\
\Rightarrow \quad[x y z-(x+y+z)](x y+y z+z x-1)=0 .
\end{array}
$$
Since \( x y + y z + z x \neq 1 \), then
\( x y z - (x + y + z) = 0 \)
$$
\begin{array}{r}
\Rightarrow x y z = x + y + z \\
\Rightarrow \frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x} = 1 .
\end{array}
$$
(2) Since \( x, y, z \) are positive numbers, we have
$$
\begin{array}{l}
9(x+y)(y+z)(z+x)-8 x y z(x y+y z+z x) \\
= 9(x+y)(y+z)(z+x)- \\
8(x+y+z)(x y+y z+z x) \\
= x^{2} y + x y^{2} + z^{2} x + z x^{2} + y^{2} z + y z^{2} - 6 x y z \\
= x(y-z)^{2} + y(z-x)^{2} + z(x-y)^{2} \geqslant 0 \\
\Rightarrow 9(x+y)(y+z)(z+x) \\
\quad \geqslant 8 x y z(x y+y z+z x)
\end{array}
$$
|
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 $47-3 a b c=5(15-5)=50$
$$
\begin{aligned}
& \Rightarrow a b c=-1 . \\
& \text { And } a^{2}+a b+b^{2} \\
& =(a+b)(a+b+c)-(a b+b c+c a) \\
= & (5-c)-5=5(4-c) .
\end{aligned}
$$
Similarly,
$$
\begin{array}{l}
b^{2}+b c+c^{2}=5(4-a), \\
c^{2}+c a+a^{2}=5(4-b) . \\
\text { Therefore }\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \\
=125(4-a)(4-b)(4-c) \\
=125[64-16 \times 5+4 \times 5-(-1)] \\
=625 .
\end{array}
$$
|
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 $A D$
$$
\begin{array}{l}
\Rightarrow \angle B E D=\angle B C F \\
\Rightarrow \angle B A D=\angle B C F \\
\Rightarrow A, B, F, C \text { are concyclic. } \\
\text { Also } A B=A C \\
\Rightarrow \angle A B D=\angle A C B=\angle A F B \\
\Rightarrow \triangle A B D \backsim \triangle A F B \Rightarrow \frac{A B}{A F}=\frac{A D}{A B} \\
\Rightarrow A D \cdot A F=A B^{2}=5 .
\end{array}
$$
|
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, $a_{n}=\left\{\begin{array}{l}\sin n \theta, n=0,1 ; \\ \sin 2 \theta, n=2 k\left(k \in \mathbf{Z}_{+}\right) ; \\ \sin 3 \theta, n=2 k+1\left(k \in \mathbf{Z}_{+}\right) .\end{array}\right.$
Therefore, when $n=0$, $S_{n}=0$;
when $n=1$, $S_{n}=\frac{3}{5}$.
Hence, $S_{51}=\frac{3}{5}+25 \sin 2 \theta+25 \sin 3 \theta=48$.
|
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 $t>0$,
$$
\begin{array}{l}
R^{2}=(t+1)^{2}+\left(\frac{1}{t}-1\right)^{2} \\
=t^{2}+\frac{1}{t^{2}}+2\left(t-\frac{1}{t}\right)+2 .
\end{array}
$$
Let $u=t-\frac{1}{t}$. Then $R=\sqrt{u^{2}+2 u+4}$.
Thus, when $u=-1$, i.e., $t=\frac{-1 \pm \sqrt{5}}{2}$, or $x=\frac{1 \pm \sqrt{5}}{2}$, the minimum value of $R$ is $\sqrt{3}$.
$$
\begin{aligned}
\text { Hence } \overrightarrow{O A} \cdot \overrightarrow{O B}=(\overrightarrow{O C}+\overrightarrow{C A}) \cdot(\overrightarrow{O C}+\overrightarrow{C B}) \\
=(\overrightarrow{O C}+\overrightarrow{C A}) \cdot(\overrightarrow{O C}-\overrightarrow{C A}) \\
=\overrightarrow{O C}^{2}-\overrightarrow{C A}^{2}=1-R^{2}=-2 .
\end{aligned}
$$
|
-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 $x+y \leqslant 2016$, we have $n \leqslant 44, x \leqslant 990$.
Hence, the maximum number of red socks in the drawer is 990.
|
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
$$
\begin{array}{l}
f\left(\log _{\frac{1}{3}} 54\right)=f\left(-\log _{3} 2\right)=-f\left(\log _{3} 2\right) \\
=-\left(3^{\log _{3} 2}-1\right)=-1 .
\end{array}
$$
|
-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, then $n$ is at least
|
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. Assume that the sum of any two numbers in $A$ and $B$ is not a perfect square.
Without loss of generality, let $1 \in A$.
$$
\begin{array}{l}
\text { By } 1+3=2^{2}, 1+8=3^{2}, 1+15=4^{2} \\
\Rightarrow\{3,8,15\} \subseteq B ; \\
\text { By } 3+6=3^{2}, 3+13=4^{2}, 15+10=5^{2} \\
\Rightarrow\{6,13,10\} \subseteq A .
\end{array}
$$
But $6+10=4^{2}$, which contradicts the assumption that the sum of any two numbers in $A$ is not a perfect square. Therefore, $n=15$ meets the requirement. Thus, the minimum value of $n$ is 15.
|
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 .
$$
\begin{array}{l}
\text { Let } \boldsymbol{a}=\left(\sqrt{\frac{x^{2}}{2 y}}, \sqrt{\frac{4 y^{2}}{x}}\right), \boldsymbol{b}=(\sqrt{2 y}, \sqrt{x}) . \\
\text { Then }|\boldsymbol{a}|=\sqrt{\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}},|\boldsymbol{b}|=\sqrt{2 y+x}=\sqrt{2} \\
\boldsymbol{a} \cdot \boldsymbol{b}=x+2 y=2 .
\end{array}
$$
From $|\boldsymbol{a}||\boldsymbol{b}| \geqslant \boldsymbol{a} \cdot \boldsymbol{b} \Rightarrow \frac{x^{2}}{2 y}+\frac{4 y^{2}}{x} \geqslant 2$, equality holds if and only if $\boldsymbol{a}$ is in the same direction as $\boldsymbol{b}$.
Therefore, when $x=2 y=1$, $\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}$ achieves its minimum value of 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
$$
\begin{array}{l}
\left\{\begin{array}{l}
y^{2}=\frac{8}{3} x-\frac{8}{9}, \\
x^{2}+y^{2}-10 x+9=0
\end{array}\right. \\
\Rightarrow x^{2}-\frac{22}{3} x+\frac{73}{9}=0 \\
\Rightarrow|A B|+|A C|=x_{B}+x_{C}+\frac{2}{3}=8 \text {. } \\
\end{array}
$$
|
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 Vieta's formulas, $\frac{4 a}{3}=4 \Leftrightarrow a=3$.
|
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 calls to each other, find the maximum value of $n$.
【Analysis】Construct a graph theory model to characterize the relationships between the elements in the problem. From the conditions given in the problem, we know that there exists an odd cycle in the graph. Since the number of points is finite, by the extreme principle, we know that there must exist the shortest odd cycle $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$. Furthermore, in the shortest odd cycle, any two points $A_{i}$ and $A_{j}(j \neq i \pm 1)$ must not be connected, thus the internal structure of the odd cycle is fully understood. At this point, starting from the shortest odd cycle $C$, the entire graph can be divided into two parts: one part is the odd cycle $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$, and the other part is all the points except $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$. We only need to understand the structure of each part and the connection situation between the two parts to grasp the overall structure.
|
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 of edges $n$ in graph $G$.
Let the length of the shortest odd cycle $C$ in the graph be $2k+1$.
Since there are no triangles in graph $G$, we have $k \geqslant 2$.
Let $C$ be $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$. Then, for any $i, j (1 \leqslant i, j \leqslant 2 k+1)$, $A_{i}$ and $A_{j} (j \neq i \pm 1)$ are not connected $\left(A_{0}=A_{2 k+2}=A_{1}\right)$ (otherwise, there would be a smaller odd cycle, contradicting the assumption). Apart from $A_{1}, A_{2}, \cdots, A_{2 k+1}$,
the remaining
$$
21-(2 k+1)=2(10-k)
$$
points have no triangles.
By Turán's theorem, the maximum number of edges among these $2(10-k)$ points is $\frac{1}{4}(20-2 k)^{2}=(10-k)^{2}$. Additionally, each of these $2(10-k)$ points cannot be connected to two adjacent vertices in $C$ (otherwise, there would be a triangle, contradicting the assumption). Therefore, each of these $2(10-k)$ points can be connected to at most $k$ vertices in $C$. Hence, the number of edges $n$ in the graph satisfies
$$
\begin{array}{l}
n \leqslant 2 k+1+(10-k)^{2}+2(10-k) k \\
=102-(k-1)^{2} \leqslant 102-(2-1)^{2}=101 .
\end{array}
$$
Suppose the 21 vertices of $G$ are $A_{1}, A_{2}, \cdots, A_{21}$,
$$
\begin{array}{l}
X=\left\{A_{1}, A_{2}, \cdots, A_{5}\right\}, \\
Y=\left\{A_{6}, A_{8}, A_{10}, \cdots, A_{20}\right\}, \\
Z=\left\{A_{7}, A_{9}, A_{11}, \cdots, A_{21}\right\} .
\end{array}
$$
In $X$, $A_{i}$ is connected to $A_{i+1} (i=1,2,3,4)$ and $A_{5}$ is connected to $A_{1}$. In $Y$, each point is connected to each point in $Z$, and each point in $Y$ is connected to two points in $X$, $A_{1}$ and $A_{3}$. Each point in $Z$ is connected to two points in $X$, $A_{2}$ and $A_{4}$. No other pairs of points are connected. Then the number of edges in graph $G$ is
$$
n=5+8 \times 8+2 \times 8 \times 2=101 \text {, }
$$
and graph $G$ contains an odd cycle $A_{1} A_{2} \cdots A_{5} A_{1}$, but no triangles.
In conclusion, the maximum value of $n$ is 101.
|
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
$$
B\left(y_{0}, x_{0}\right), C\left(-x_{0},-y_{0}\right), D\left(-y_{0},-x_{0}\right) \text {. }
$$
Thus, quadrilateral $A B C D$ is a rectangle, and its area is
$$
\begin{array}{l}
S=|A B||A D| \\
= \sqrt{2\left(x_{0}-y_{0}\right)^{2}} \sqrt{2\left(x_{0}+y_{0}\right)^{2}} \\
= 2 \sqrt{x_{0}^{2}+y_{0}^{2}-2 x_{0} y_{0}} \sqrt{x_{0}^{2}+y_{0}^{2}+2 x_{0} y_{0}} \\
=2 \sqrt{10-8} \sqrt{10+8}=12 .
\end{array}
$$
|
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 directly.
|
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 $f^{\prime}(x)=-x^{2}+\frac{38}{3} x$.
By $f^{\prime}(x)=0 \Rightarrow x=0$ or $\frac{38}{3}$.
Thus, $f(n)\left(n \in \mathbf{Z}_{+}\right)$ is increasing in the interval $[1,12]$ and decreasing in the interval $[13,+\infty)$.
Notice that, $f(12)=336, f(13)=338$. Therefore, the maximum value is 338.
|
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.
$$
\begin{array}{l}
\text { Then }(1+2 \mathrm{i}) \alpha^{2}+(p+q \mathrm{i}) \alpha+1-2 \mathrm{i}=0 \\
\Rightarrow\left\{\begin{array}{l}
\alpha^{2}+p \alpha+1=0, \\
2 \alpha^{2}+q \alpha-2=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
p=-\left(\alpha+\frac{1}{\alpha}\right), \\
q=-2\left(\alpha-\frac{1}{\alpha}\right)
\end{array}\right. \\
\Rightarrow|m|^{2}=p^{2}+q^{2} \\
=\left(\alpha+\frac{1}{\alpha}\right)^{2}+4\left(\alpha-\frac{1}{\alpha}\right)^{2} \\
=5\left(\alpha^{2}+\frac{1}{\alpha^{2}}\right)-6 \geqslant 4 \\
\Rightarrow|m| \geqslant 2 .
\end{array}
$$
When and only when $\alpha^{4}=1$, i.e., $\alpha= \pm 1$, the equality holds, at this time, $m= \pm 2$.
Therefore, the minimum value of $|m|$ is 2.
|
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 attained at the endpoints of the interval $[k, k+1]$.
Thus, the minimum term of the sequence $\{f(k)\}(1 \leqslant k \leqslant 2016)$ is the desired value.
Notice that,
$$
\begin{array}{l}
f(k+1)>f(k) \\
\Leftrightarrow k^{2}+k-2017 \times 1009>0 \\
\Leftrightarrow k \geqslant 1427 . \\
\text { Hence } f(x)_{\min }=f(1427)=801730806 .
\end{array}
$$
|
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 edges of 1008 $2016 \times 1$ rectangles and the aforementioned $2016 \times 2016$ rectangle.
Thus, $M \leqslant 2017$.
Next, assume that the grid lines of a $2016 \times 2016$ square grid $A$ are covered by the edges of $M$ rectangles as required by the problem.
We will prove: $M \geqslant 2017$.
We call the top (bottom) edge of a rectangle the "head" ("foot").
For each rectangle in the grid $A$, we label it with a pair $(a, b)$: if the head of the rectangle is at the top of the grid $A$, we set $a=1$, otherwise $a=0$; if the foot of the rectangle is at the bottom of the grid $A$, we set $b=1$, otherwise $b=0$.
Among the $M$ rectangles, let the number of rectangles labeled $(1,1)$, $(1,0)$, $(0,1)$, and $(0,0)$ be $x$, $y$, $z$, and $w$, respectively.
There are 2015 horizontal grid lines inside the grid $A$, so $y+z+2w \geqslant 2015$;
There are 2017 vertical grid lines intersecting with the top of the grid $A$, so $2x+2y \geqslant 2017 \Rightarrow x+y \geqslant 1009$.
Similarly, $x+z \geqslant 1009$.
Thus, $2x+y+z \geqslant 2018$
$$
\begin{array}{l}
\Rightarrow 2(x+y+z+w) \geqslant 4033 \\
\Rightarrow M=x+y+z+w \geqslant 2017 .
\end{array}
$$
|
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 $p + 2 \leq \frac{a + b + c}{3}$;
(2) If the Leyden quadruple $(p, a, b, c)$ satisfies
$$
p + 2 = \frac{a + b + c}{3},
$$
find all possible values of $p$. ${ }^{[1]}$
|
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 properties:
(i) $p$ is an odd prime;
(ii) $a_{1}, a_{2}, \cdots, a_{k}$ are distinct;
(iii) $p \mid\left(a_{1} a_{2}+1\right), p \mid\left(a_{2} a_{3}+1\right), \cdots, p \mid\left(a_{k-1} a_{k}+1\right), p \mid\left(a_{k} a_{1}+1\right)$.
Then the $(k+1)$-tuple of positive integers $\left(p, a_{1}, a_{2}, \cdots, a_{k}\right)$ is called a "Leiden $(k+1)$-tuple".
Thus, after investigating the original problem, Proposition 1 is obtained.
Proposition 1 For each Leiden quadruple $\left(p, a_{1}, a_{2}, a_{3}\right)$,
we have $\frac{a_{1}+a_{2}+a_{3}}{3} \geqslant p+2$; equality holds if and only if $p=5$.
Proof of Proposition 1 Without loss of generality, assume $a_{1}p+k-1$.
Proof of Proposition 2 Without loss of generality, assume
$a_{1}2+k-3+p=p+k-1$.
Through the above calculations, it is found that the inequality in Proposition 2
is not "tight" enough. Therefore, the inequality on the right is strengthened and generalized, leading to the following series of corollaries.
Corollary 1 For each Leiden $(k+1)$-tuple $\left(p, a_{1}, a_{2}, \cdots, a_{k}\right)$, we have
$\frac{a_{1}+a_{2}+\cdots+a_{k}}{k} \geqslant 2+\frac{k-1}{2} p$,
with equality holding if and only if $p=5$.
It is easy to see that the proof of Corollary 1 is the same as that of Proposition 2. The value of $p$ for which the inequality holds with equality can be derived similarly to the proof of Proposition 1, and readers can complete it themselves.
Corollary 2 For each Leiden $(k+1)$-tuple $(p, a_{1}, a_{2}, \cdots, a_{k})$, we have
$\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}}{k}>\frac{(4+(k-1) p)^{2}}{4}$. (1)
Proof of Corollary 2 By the Cauchy-Schwarz inequality,
$\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}}{k}$
$\geqslant\left(\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}\right)^{2}$
$\geqslant\left(2+\frac{k-1}{2} p\right)^{2}$
Since $a_{1}, a_{2}, \cdots, a_{k}$ are distinct, the equality in the inequality cannot be achieved.
Therefore, equation (1) is proved.
Corollary 3 For each Leiden $(k+1)$-tuple $(p, a_{1}, a_{2}, \cdots, a_{k})$ and $n \in \mathbf{Z}_{+}$, we have
$\frac{a_{1}^{n}+a_{2}^{n}+\cdots+a_{k}^{n}}{k}>\left(\frac{4+(k-1) p}{2 k}\right)^{n}$.
Proof of Corollary 3 By the Hölder inequality,
$\left(\sum_{i=1}^{k} a_{i}^{n}\right)^{\frac{1}{n}}\left(\sum_{i=1}^{k} 1^{\frac{n}{n-1}}\right)^{\frac{n-1}{n}} \geqslant \sum_{i=1}^{k} a_{i}$
$\Rightarrow a_{1}^{n}+a_{2}^{n}+\cdots+a_{k}^{n}$
Since $a_{1}, a_{2}, \cdots, a_{k}$ are distinct, the equality in the inequality cannot be achieved.
Therefore, equation (2) is proved.
|
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 National High School Mathematics Olympiad)
Reference [1] provides a detailed solution to this problem. Through further research, it is found that a simpler solution can also be given to find the values of
$$
a^{2} + b^{2} + c^{2}, \quad a^{2} b + b^{2} c + c^{2} a, \quad a b^{2} + b c^{2} + c a^{2}
$$
|
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),
$\Rightarrow x_{1} x_{2}=b=-\frac{b-c}{a-c} \Rightarrow \frac{c-a}{c-b}=-\frac{1}{b}=x_{3}$.
Similarly, $x_{1}=-\frac{1}{c}, x_{2}=-\frac{1}{a}$, and
$x_{1} x_{2}=\frac{1}{a c}=b \Rightarrow a b c=1$.
Since $x_{2}=-\frac{1}{a}$ is a root of equation (1),
$\Rightarrow \frac{1}{a^{2}}-1+b=0 \Rightarrow \frac{1}{a^{2}}=1-b$.
Similarly, $\frac{1}{b^{2}}=1-c, \frac{1}{c^{2}}=1-a$.
Thus, $\frac{1}{a^{2}} \cdot \frac{1}{b^{2}} \cdot \frac{1}{c^{2}}=(1-b)(1-c)(1-a)$.
Combining with equation (4) we get
$1=1-a-b-c+a b+b c+c a-a b c$
$\Rightarrow a b+b c+c a=1+a+b+c$.
Since $x_{2}=-\frac{1}{a}$ is a root of equation (2), we have
$\frac{1}{a^{2}}-\frac{b}{a}+c=0 \Rightarrow 1-a b+a^{2} c=0$.
Combining with equation (4) we get
$a b c-a b+a^{2} c=0(a \neq 0)$
$\Rightarrow b c-b+c a=0$.
Similarly, $a b-a+b c=0, c a-c+a b=0$.
Adding these three equations, we get
$2(a b+b c+c a)=a+b+c$.
Combining equations (5) and (6) we get
$a+b+c=-2, a b+b c+c a=-1$.
Then $a^{2}+b^{2}+c^{2}$
$=(a+b+c)^{2}-2(a b+b c+c a)=6$.
Since $x_{1}=-\frac{1}{c}$ is a root of equation (3),
$\Rightarrow \frac{1}{c^{2}}-1+a=0 \Rightarrow 1-c^{2}+c^{2} a=0$.
Similarly, $1-b^{2}+b^{2} c=0,1-a^{2}+a^{2} b=0$.
Adding these three equations, we get
$a^{2} b+b^{2} c+c^{2} a=a^{2}+b^{2}+c^{2}-3=3$.
Since $x_{3}=-\frac{1}{b}$ is a root of equation (3), we have
$$
\frac{1}{b^{2}}-\frac{c}{b}+a=0 \Rightarrow 1-b c+b^{2} a=0 .
$$
Similarly, $1-a b+a^{2} c=0,1-c a+c^{2} b=0$.
Adding these three equations, we get
$b^{2} a+a^{2} c+c^{2} b=a b+b c+c a-3=-4$.
|
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 total number of people who won awards in the unit is $\qquad$ .
|
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 $1.5 x + y + 0.5 z = 100$
$$
\begin{array}{l}
\Rightarrow (x + y + z) + 0.5 x - 0.5 z = 100 \\
\Rightarrow x + y + z = 100 + 0.5(z - x). \\
\text{Given } 93 \leqslant z - x < 96 \\
\Rightarrow 46.5 \leqslant 0.5(z - x) < 48 \\
\Rightarrow 146.5 \leqslant x + y + z < 148.
\end{array}
$$
Since the number of people is a positive integer, hence $x + y + z = 147$.
|
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
$$
\begin{array}{l}
|\overrightarrow{P A}+\overrightarrow{P B}|^{2}=\lambda^{2}|\overrightarrow{P C}|^{2} \\
=|\overrightarrow{P A}|^{2}+|\overrightarrow{P B}|^{2}-2|\overrightarrow{P A}||\overrightarrow{P B}| \cos \frac{C}{2} \\
=R^{2}+R^{2}-2 R^{2} \cdot \frac{1}{2}=R^{2}=\lambda^{2} R^{2} \\
\Rightarrow \lambda= \pm 1 .
\end{array}
$$
From the problem, we get $\lambda=-1$.
|
-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: $\left|C_{m} \cap C_{n}\right| \leqslant 1(m \neq n)$.
In fact, suppose there exist $C_{m}, C_{n}(m \neq n), \left|C_{m} \cap C_{n}\right| \geqslant 2$. Then there exist $k_{1}, k_{2}, l_{1}, l_{2}$ such that
$$
\text { and } \begin{array}{l}
a_{k_{1}}+b_{m}, a_{k_{2}}+b_{m}, a_{l_{1}}+b_{n}, a_{l_{2}}+b_{n} \in b_{m}=a_{k_{2}}+b_{n}, a_{l_{1}}+b_{m}=a_{l_{2}}+b_{n}, \\
\\
a_{k_{1}} \neq a_{k_{2}}, a_{l_{1}} \neq a_{l_{2}} .
\end{array}
$$
Then $a_{k_{1}}-a_{k_{2}}=a_{l_{1}}-a_{l_{2}}$
$$
\Rightarrow a_{k_{1}}+a_{l_{2}}=a_{l_{1}}+a_{k_{2}},
$$
which implies $\left\{a_{k_{1}}, a_{l_{1}}\right\}=\left\{a_{k_{2}}, a_{l_{2}}\right\}$.
By conclusion (1), we know $a_{k_{1}}=a_{l_{1}}, a_{k_{2}}=a_{l_{2}}$, which is impossible, otherwise,
$$
\begin{array}{l}
a_{k_{1}}+b_{m} \in C_{n}, a_{l_{2}}+b_{n} \in C_{m} \\
\Rightarrow a_{k_{1}}=b_{n}, a_{b_{2}}=b_{m},
\end{array}
$$
which means $C_{m} \cap C_{n}$ can have at most one element, a contradiction.
Thus, $\left|C_{m} \cap C_{n}\right| \leqslant 1(m \neq n)$.
By the principle of inclusion-exclusion,
$$
\begin{array}{l}
\quad|A+B|=\left|\bigcup_{k=1}^{16} C_{k}\right| \\
\geqslant \sum_{k=1}^{16}\left|C_{k}\right|-\sum_{1 \leqslant m<n \leqslant 16}\left|C_{m} \cap C_{n}\right| \\
=320-120=200 . \\
\text { Let } A=\left\{2,2^{2}, \cdots, 2^{20}\right\}, \\
B=\left\{2,2^{2}, \cdots, 2^{16}\right\} .
\end{array}
$$
It is easy to verify that if $m \neq n \neq k$, then
$$
C_{m} \cap C_{n}=\left\{2^{m}+2^{n}\right\}, C_{m} \cap C_{n} \cap C_{k}=\varnothing .
$$
Therefore, $|A+B|=200$.
In conclusion, the minimum value of $|A+B|$ is 200.
|
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, $n=\left[\frac{3}{\sqrt{3}-1}\right]=4$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
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
$$
S \geqslant 1+2+\cdots+12=78 \text {. }
$$
Note that, $S \mid 2016$, and 78 is not a factor of 2016.
Also, $2016=2^{5} \times 3^{2} \times 7$, its smallest positive factor greater than 78 is
$$
2^{2} \times 3 \times 7=84,2016=84 \times 24 .
$$
Thus, $d \leqslant 24$, and $d=24$ can be achieved, by setting
$$
\left(a_{1}, a_{2}, \cdots, a_{11}, a_{12}\right)=(1,2, \cdots, 11,18) .
$$
|
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,
$$
\begin{array}{l}
f(x)=2\left|x-\frac{1}{2}\right|+3\left|x-\frac{2}{3}\right|+ \\
4\left|x-\frac{3}{4}\right|+5\left|x-\frac{4}{5}\right| \\
=2\left(\left|x-\frac{1}{2}\right|+\left|x-\frac{4}{5}\right|\right)+ \\
\quad 3\left(\left|x-\frac{2}{3}\right|+\left|x-\frac{4}{5}\right|\right)+4\left|x-\frac{3}{4}\right| \\
\geqslant 2\left|\left(x-\frac{1}{2}\right)-\left(x-\frac{4}{5}\right)\right|+3\left|\left(x-\frac{2}{3}\right)-\left(x-\frac{4}{5}\right)\right| \\
=2\left|\frac{4}{5}-\frac{1}{2}\right|+3\left|\frac{4}{5}-\frac{2}{3}\right|=1 .
\end{array}
$$
When and only when
$$
\begin{array}{l}
\left(x-\frac{1}{2}\right)\left(x-\frac{4}{5}\right) \leqslant 0, \\
\left(x-\frac{2}{3}\right)\left(x-\frac{4}{5}\right) \leqslant 0, x-\frac{3}{4}=0,
\end{array}
$$
i.e., $x=\frac{3}{4}$, the equality holds.
Thus, $f(x)_{\text {min }}=f\left(\frac{3}{4}\right)=1$.
|
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,
$$
\begin{array}{l}
|z|=\left|\left(z_{1}-z_{3}\right)\left(z_{2}-z_{4}\right)\right| \\
=\left|z_{1} z_{2}-z_{1} z_{4}-z_{3} z_{2}+z_{3} z_{4}\right| \\
=\left|\left(z_{1}-z_{2}\right)\left(z_{3}-z_{4}\right)+\left(z_{1}-z_{4}\right)\left(z_{2}-z_{3}\right)\right| \\
\leqslant\left|\left(z_{1}-z_{2}\right)\left(z_{3}-z_{4}\right)\right|+\left|\left(z_{1}-z_{4}\right)\left(z_{2}-z_{3}\right)\right| \\
=1 \times 2+3 \times 4=14 .
\end{array}
$$
Therefore, $|z|_{\max }=14$.
|
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
$1-\frac{2}{x+1} \geqslant 1-\frac{2}{5+1}=\frac{2}{3} \Rightarrow \frac{y+z}{y z-1} \geqslant \frac{2}{3}$.
However, when $z \geqslant y \geqslant x \geqslant 5$,
$$
\frac{y+z}{y z-1} \leqslant \frac{5+5}{25-1}=\frac{5}{12}<\frac{2}{3},
$$
which is a contradiction. Therefore, $x=2,3,4$.
When $x=2$,
$$
\begin{array}{l}
y z-1=3(y+z) \\
\Rightarrow(y-3)(z-3)=10 \\
\Rightarrow(y, z)=(4,13),(5,8) .
\end{array}
$$
When $x=3$,
$$
\begin{array}{l}
y z-1=2(y+z) \\
\Rightarrow(y-2)(z-2)=5 \\
\Rightarrow(y, z)=(3,7) .
\end{array}
$$
When $x=4$,
$$
\begin{array}{l}
3 y z-3=5(y+z) \\
\Rightarrow 3=5\left(\frac{1}{y}+\frac{1}{z}\right)+\frac{3}{y z} \leqslant 5\left(\frac{1}{4}+\frac{1}{4}\right)+\frac{3}{16},
\end{array}
$$
which is a contradiction.
Thus, the ordered triples $(x, y, z)$ that satisfy $z \geqslant y \geqslant x$ are
$$
(x, y, z)=(2,4,13),(2,5,8),(3,3,7) \text {. }
$$
By permuting the order, we can obtain $6+6+3=15$ solutions.
In conclusion, the number of ordered positive integer solutions to the equation is 15.
|
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 $\left(\mathrm{e}^{-\frac{3}{2}},+\infty\right)$.
Also, $00 .
\end{array}
$$
Therefore, the number of zeros of the function $f(x)$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $z \in \mathbf{C}$. If the equation $x^{2}-2 z x+\frac{3}{4}+\mathrm{i}=0$ (where $\mathrm{i}$ is the imaginary unit) has real roots, then the minimum value of $|z|$ is $\qquad$ .
|
7.1 .
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R}), x=x_{0}$ be a real root of the original equation.
$$
\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 x_{0}=\frac{1}{2 b}, a=\frac{3 b^{2}+1}{4 b} \\
\Rightarrow|z|^{2}=a^{2}+b^{2}=\left(\frac{3 b^{2}+1}{4 b}\right)^{2}+b^{2} \\
\quad=\frac{25}{16} b^{2}+\frac{1}{16 b^{2}}+\frac{3}{8} \\
\quad \geqslant \frac{5}{8}+\frac{3}{8}=1,
\end{array}
$$
When and only when $b= \pm \frac{\sqrt{5}}{5}$, the equality holds.
Thus, the minimum value of $|z|$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $f(x)$ be a function defined on $\mathbf{R}$, if $f(0)$ $=1008$, and for any $x \in \mathbf{R}$, it satisfies
$$
\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}
$$
Then, $f(2016)$
$$
\begin{array}{l}
=\sum_{k=0}^{167}(f(12 k+12)-f(12 k))+f(0) \\
=6 \sum_{k=0}^{167}(12 k+5)+1008 \\
=6 \times \frac{(2009+5) \times 168}{2}+1008 \\
=1008 \times 1008 .
\end{array}
$$
$$
\text { Therefore, } \frac{f(2016)}{2016}=\frac{1008}{2}=504 \text {. }
$$
|
504
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $f(x)$ is a periodic function on $\mathbf{R}$ with the smallest positive period of 2, and when $0 \leqslant x<2$, $f(x)=x^{3}-x$. Then the number of intersections between the graph of the function $y=f(x)$ and the $x$-axis in the interval $[0,6]$ is $\qquad$ .
|
2.7.
When $0 \leqslant x<2$, let $f(x)=x^{3}-x=0$, we get $x=0$ or 1.
According to the properties of periodic functions, given that the smallest period of $f(x)$ is 2, we know that $y=f(x)$ has six zeros in the interval $[0,6)$.
Also, $f(6)=f(3 \times 2)=f(0)=0$, so $f(x)$ has 7 intersection points with the $x$-axis in the interval $[0,6]$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. It is known that Team A and Team B each have several people. If 90 people are transferred from Team A to Team B, then the total number of people in Team B will be twice that of Team A; if some people are transferred from Team B to Team A, then the total number of people in Team A will be 6 times that of Team B. Then, the original minimum number of people in Team A is.
|
8. 153.
Let the original number of people in team A and team B be $a$ and $b$ respectively.
Then $2(a-90)=b+90$.
Suppose $c$ people are transferred from team B to team A. Then $a+c=6(b-c)$.
From equations (1) and (2), eliminating $b$ and simplifying, we get
$$
\begin{array}{l}
11 a-7 c=1620 \\
\Rightarrow c=\frac{11 a-1620}{7}=a-232+\frac{4(a+1)}{7} \text {. }
\end{array}
$$
Since $a$ and $c$ are positive integers, thus,
$$
c=\frac{11 a-1620}{7} \geqslant 1 \Rightarrow a \geqslant 148 \text {. }
$$
Also, $7 \mid 4(a+1),(4,7)=1$, hence, $7 \mid (a+1)$.
Therefore, the minimum value of $a$ is 153.
Thus, the original number of people in team A is at least 153.
|
153
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
518 On a circle, initially write 1 and 2 at opposite positions. Each operation involves writing the sum of two adjacent numbers between them, for example, the first operation writes two 3s, the second operation writes two 4s and two 5s. After each operation, the sum of all numbers becomes three times the previous sum. After performing the operation sufficiently many times, find the sum of the number of 2015s and 2016s written.
|
Observe the pattern, and conjecture that after a sufficient number of operations, the $n$ numbers written equal $\varphi(n)$. After each operation, the property that adjacent numbers are coprime remains unchanged, and the new number written each time is the sum of its two neighbors. Therefore, the number we are looking for should be
$$
\varphi(2015)+\varphi(2016)=2016.
$$
Since the number written each time is the sum of the two numbers on its sides, observing the numbers on the circle in a clockwise direction, it is only necessary to prove that for any positive integer $n$ and any positive integer $k$ coprime with $n$, $n$ is written as the sum of its left neighbor $k$ and right neighbor $n-k$ exactly once.
We will prove this using the second principle of mathematical induction.
When $n=1,2,3$, it is obviously true.
When $n>3$, since $k$ and $n-k$ are coprime, they are not equal.
If $k>n-k$, according to the induction hypothesis, $k$ is written exactly once as the sum of its left neighbor $2k-n$ and right neighbor $n-k$;
If $k<n-k$, according to the induction hypothesis, $n-k$ is written exactly once as the sum of its left neighbor $k$ and right neighbor $n-2k$.
Therefore, by the second principle of mathematical induction, the conclusion holds.
|
2016
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If $f(x)=\sum_{k=0}^{4034} a_{k} x^{k}$ is the expansion of $\left(x^{2}+x+2\right)^{2017}$, then $\sum_{k=0}^{1344}\left(2 a_{3 k}-a_{3 k+1}-a_{3 k+2}\right)=$ $\qquad$
|
8. 2 .
Let $x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$. Then
$$
\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_{3 k+1} \omega^{2}+a_{3 k+2} \omega\right)=1 \text {. }
$$
Adding the two equations, we get
$$
\sum_{k=0}^{1344}\left(2 a_{3 k}-a_{3 k+1}-a_{3 k+2}\right)=2 \text {. }
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Define the sequence $\left\{a_{n}\right\}: a_{n}$ is the last digit of $1+2+\cdots+n$, and $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then $S_{2016}=$ $\qquad$ .
|
6.7 066 .
From the problem, we know
$$
\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_{20} . \\
\text { Also, } S_{20}=a_{1}+a_{2}+\cdots+a_{20} \\
=1+3+6+0+5+1+8+6+5+5+ \\
\\
6+8+1+5+0+6+3+1+0+0 \\
=70, \\
S_{16}=66,
\end{array}
$$
Thus, $S_{2016}=S_{16}+100 S_{20}=7066$.
|
7066
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. From five positive integers $a, b, c, d, e$, any four are taken to find their sum, resulting in the set of sums $\{44,45,46,47\}$, then $a+b+c+d+e=$ $\qquad$ .
|
2. 57 .
From five positive integers, if we take any four to find their sum, there are five possible ways, which should result in five sum values. Since the set $\{44,45, 46,47\}$ contains only four elements, there must be two sum values that are equal. Therefore,
$$
\begin{array}{l}
44+44+45+46+47 \\
\leqslant 4(a+b+c+d+e) \\
\leqslant 44+45+46+47+47 \\
\Rightarrow 226 \leqslant 4(a+b+c+d+e) \leqslant 229 .
\end{array}
$$
Since $a, b, c, d, e$ are integers, we get
$$
\begin{array}{l}
4(a+b+c+d+e)=228 \\
\Rightarrow a+b+c+d+e=57
\end{array}
$$
|
57
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given that $M N$ is a moving chord of the circumcircle of equilateral $\triangle A B C$ with side length $2 \sqrt{6}$, $M N=4$, and $P$ is a moving point on the sides of $\triangle A B C$. Then the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is
|
9.4.
Let the circumcenter of $\triangle ABC$ be $O$.
It is easy to find that the radius of $\odot O$ is $r=2 \sqrt{2}$.
Also, $MN=4$, so $\triangle OMN$ is an isosceles right triangle, and
$$
\begin{aligned}
& \overrightarrow{O M} \cdot \overrightarrow{O N}=0,|\overrightarrow{O M}+\overrightarrow{O N}|=4 . \\
& \text { Let }|\overrightarrow{O P}|=x . \\
& \text { Then } \sqrt{2} \leqslant x \leqslant 2 \sqrt{2} \text {, and } \\
& \overrightarrow{M P} \cdot \overrightarrow{P N}=(\overrightarrow{O P}-\overrightarrow{O M}) \cdot(\overrightarrow{O N}-\overrightarrow{O P}) \\
& =-\overrightarrow{O P}{ }^{2}+(\overrightarrow{O M}+\overrightarrow{O N}) \cdot \overrightarrow{O P}-\overrightarrow{O M} \cdot \overrightarrow{O N} \\
= & -x^{2}+(\overrightarrow{O M}+\overrightarrow{O N}) \cdot \overrightarrow{O P} \\
\leqslant & -x^{2}+|\overrightarrow{O M}+\overrightarrow{O N}||\overrightarrow{O P}| \\
= & -x^{2}+4 x=4-(x-2)^{2} \leqslant 4,
\end{aligned}
$$
The equality holds if and only if $x=2$ and $\overrightarrow{O P}$ is in the same direction as $\overrightarrow{O M}+\overrightarrow{O N}$.
Therefore, the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is 4.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Arrange the numbers $2, 3, 4, 6, 8, 9, 12, 15$ in a row so that the greatest common divisor of any two adjacent numbers is greater than 1. The total number of possible arrangements is ( ) .
(A) 720
(B) 1014
(C) 576
(D) 1296
|
4. D.
First, divide the eight numbers into three groups:
I $(2,4,8)$, II $(3,9,15)$, III $(6,12)$.
Since the numbers in group I and group II have no common factors, the arrangement that satisfies the condition must be:
(1) After removing the numbers 6 and 12, the remaining numbers are divided into three parts and arranged (in sequence) as I, II, I or II, I, II. In this case, there are $2 \times 3! \times 3! \times 2 = 144$ ways to arrange these six numbers. And 6 and 12 can be placed at the intersection of different parts in two different ways, totaling $2 \times 144 = 288$ ways.
(2) After removing the numbers 6 and 12, the remaining numbers are divided into two parts and arranged as I, II or II, I. In this case, there are $2 \times 3! \times 3!$ ways to arrange these six numbers. If 6 and 12 are both at the boundary between group I and group II, there are two ways to arrange them. Otherwise, only one is at the boundary, and the other has six possible positions, totaling $2 \times 3! \times 3! \times (2 + 2 \times 6) = 1008$.
Therefore, the total number of arrangements is $1008 + 288 = 1296$.
|
1296
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
9. As shown in Figure 1, in $\triangle A B C$,
$$
\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}
$$
Then the eccentricity of the hyperbola passing through point $C$ and having $A$ and $H$ as its foci is . $\qquad$
|
9. 2 .
Given $\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow(\overrightarrow{C B}-\overrightarrow{C A}) \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow A C=B C$.
From $\overrightarrow{A H} \cdot \overrightarrow{B C}=0 \Rightarrow A H \perp B C$.
Since $\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}$, therefore,
$\sin \frac{C}{2}=\frac{\sqrt{5}}{5}, \tan \frac{C}{2}=\frac{1}{2}$.
Then $\tan C=\frac{2 \tan \frac{C}{2}}{1-\tan ^{2} \frac{C}{2}}=\frac{4}{3}$.
In the right triangle $\triangle A C H$, let's assume $C H=3$.
Then $A H=4, A C=B C=\sqrt{A H^{2}+C H^{2}}=5$.
Therefore, the eccentricity of the hyperbola with foci at $A$ and $H$ is
$$
e=\frac{A H}{A C-C H}=2 .
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure $2, P$ is a point on the incircle of square $A B C D$, and let $\angle A P C=\alpha$, $\angle B P D=\beta$. Then $\tan ^{2} \alpha+\tan ^{2} \beta$ $=$
|
4. 8 .
As shown in Figure 5, establish a Cartesian coordinate system.
Let the equation of the circle be $x^{2}+y^{2}=r^{2}$.
Then the coordinates of the vertices of the square are
$$
A(-r,-r), B(r,-r), C(r, r), D(-r, r) \text {. }
$$
If $P(r \cos \theta, r \sin \theta)$, then the slopes of the lines $P A, P B, P C$, and $P D$ are respectively
$$
\begin{array}{l}
k_{P A}=\frac{1+\sin \theta}{1+\cos \theta}, k_{P B}=-\frac{1+\sin \theta}{1-\cos \theta}, \\
k_{P C}=\frac{1-\sin \theta}{1-\cos \theta}, k_{P D}=-\frac{1-\sin \theta}{1+\cos \theta} \text {. } \\
\text { Therefore, } \tan ^{2} \alpha=\left(\frac{k_{P C}-k_{P A}}{1+k_{P A} k_{P C}}\right)^{2}=4(\cos \theta-\sin \theta)^{2} \text {, } \\
\tan ^{2} \beta=\left(\frac{k_{P D}-k_{P B}}{1+k_{P B} k_{P D}}\right)^{2}=4(\cos \theta+\sin \theta)^{2} \text {. } \\
\end{array}
$$
Therefore, $\tan ^{2} \alpha+\tan ^{2} \beta=8$.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If three numbers are taken simultaneously from the 14 integers $1,2, \cdots, 14$, such that the absolute difference between any two numbers is not less than 3, then the number of different ways to choose is $\qquad$
|
3. 120 .
Let the three integers taken out be $x, y, z (x<y<z)$.
$$
\begin{array}{l}
\text { Let } a=x, b=y-x-2, \\
c=z-y-2, d=15-z .
\end{array}
$$
Thus, $a, b, c, d \geqslant 1$.
If $a, b, c, d$ are determined, then $x, y, z$ are uniquely determined.
Since $a+b+c+d=11$, it is equivalent to dividing 11 identical balls into four piles.
Using the stars and bars method, we get $C_{10}^{3}=120$.
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\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}=\alpha_{1}\right) .
\end{array}
$$
|
Let $a_{i}=\frac{\alpha_{i+1}}{\alpha_{i}}(i=1,2, \cdots, n)$.
Then $\prod_{i=1}^{n} a_{i}=1$, and $a_{i}>0$.
The inequality to be proved becomes
$\sum_{i=1}^{n} \frac{1-a_{i}}{1+a_{i}^{2}} \geqslant 0$.
Let $x_{i}=\ln a_{i}(i=1,2, \cdots, n)$.
Then $\sum_{i=1}^{n} \frac{1-\mathrm{e}^{x_{i}}}{1+\left(\mathrm{e}^{x_{i}}\right)^{2}} \geqslant 0$, and $\sum_{i=1}^{n} x_{i}=0$.
Let $f(x)=\frac{1-\mathrm{e}^{x}}{1+\left(\mathrm{e}^{x}\right)^{2}}$.
Differentiation shows that $f(x)$ is decreasing on the interval $(-\infty, \ln (1+\sqrt{2}))$ and increasing on the interval $(\ln (1+\sqrt{2}),+\infty)$.
(1) If there exists $x_{i}x>y$ such that $f^{\prime \prime}(x)>0$.
(2) If there exist $x_{i} 、 x_{j}2$ such that $a_{2}^{8}>2 a_{2}^{7}$.
Hence, inequality (2) $>0$, holds.
(ii) When $0 \leqslant a_{2} \leqslant 2$, $2 a_{2}^{5} \geqslant a_{2}^{6}$.
Hence, inequality (2) $\geqslant a_{2}^{6}\left(a_{2}-1\right)^{2} \geqslant 0$, holds.
(iii) When $a_{2}<0$,
$$
\begin{array}{l}
\text { Inequality (2) }=a_{2}^{8}-2 a_{2}^{7}+2 a_{2}^{5}+a_{2}^{4}+a_{2}^{2}+ \\
2 a_{2}^{2}\left(a_{2}+1\right)^{2}+\left(a_{2}+2\right)^{2} \\
\geqslant a_{2}^{8}-2 a_{2}^{7}+2 a_{2}^{5}+a_{2}^{4}+a_{2}^{2} \\
=a_{2}^{2}\left(a_{2}^{3}+1\right)^{2}-2 a_{2}^{7}+a_{2}^{4} \geqslant 0, \\
\end{array}
$$
also holds.
Therefore, when $n=5$, the original inequality holds.
Thus, the maximum value of $n$ is 5.
|
5
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given ten points in space, where no four points lie on the same plane. Some points are connected by line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. ${ }^{[1]}$
(2016, National High School Mathematics Joint Competition)
|
Proof Let $v$ be a vertex in graph $G$, and $N_{i}(v)$ denote the set of points at a distance $i$ from $v$. For example,
$$
N_{0}(v)=\{v\},
$$
$N_{1}(v)=\{u \mid u$ is adjacent to $v\}$,
$N_{2}(v)=\{w \mid w$ is adjacent to $u$, not adjacent to $v$, and $u$ is adjacent to $v\}$.
First, we prove a lemma.
Lemma Let $G=(V, E)$ be a graph with girth $g(G) \geqslant 5$, average degree, number of vertices, and number of edges be $d$, $n$, and $e$ respectively. Then there exists a vertex $v$ such that
$$
\left|N_{2}(v)\right| \geqslant d^{2}-\operatorname{deg} v.
$$
Proof Assume the conclusion does not hold. Then for any vertex $v$, we have
$$
\left|N_{2}(v)\right|<d^{2}-\operatorname{deg} v.
$$
Since $g(G) \geqslant 5$, for any vertex $v$, if $w \in N_{2}(v)$, then $w$ is adjacent to exactly one vertex in $N_{1}(v)$. Because $w \in N_{2}(v)$, there is at least one vertex in $N_{1}(v)$ that is adjacent to it. If there were two vertices adjacent to $w$, say $u_{1}$ and $u_{2}$, then $v, u_{1}, w, u_{2}$ would form a cycle of length 4, contradicting $g(G) \geqslant 5$.
Therefore, for any vertex $v$, we have
$\left|N_{2}(v)\right|=\sum_{u \in N_{1}(v)}(\operatorname{deg} u-1)$.
Thus, $\sum_{u \in N_{1}(v)}(\operatorname{deg} u-1)<d^{2}-\operatorname{deg} v$.
Summing over all $v$ gives
$\sum_{v \in V} \sum_{u \in N_{1}(v)}(\operatorname{deg} u-1)<\sum_{v \in V}\left(d^{2}-\operatorname{deg} v\right) = n d^{2}-2 e$.
The left-hand side is $\sum_{v \in V}(\operatorname{deg} v)(\operatorname{deg} v-1)$, so
$$
\sum_{v \in V} \operatorname{deg}^{2} v < n d^{2} = n\left(\frac{\sum_{v \in V} \operatorname{deg} v}{n}\right)^{2}.
$$
Let $\boldsymbol{a}=(1,1, \cdots, 1)$,
$\boldsymbol{b}=\left(\operatorname{deg} v_{1}, \operatorname{deg} v_{2}, \cdots, \operatorname{deg} v_{n}\right)$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
|\boldsymbol{a} \cdot \boldsymbol{b}|^{2}=\left(\sum_{v \in V} \operatorname{deg} v\right)^{2} \\
\leqslant|\boldsymbol{a}|^{2}|\boldsymbol{b}|^{2}=n \sum_{v \in V} \operatorname{deg}^{2} v \\
\Rightarrow \sum_{v \in V} \operatorname{deg}^{2} v \geqslant n\left(\frac{\sum_{v \in V} \operatorname{deg} v}{n}\right)^{2},
\end{array}
$$
which is a contradiction.
The lemma is proved.
Since for any vertex $v$, we have
$$
n \geqslant\left|N_{0}(v)\right|+\left|N_{1}(v)\right|+\left|N_{2}(v)\right|,
$$
by the lemma, we get
$n \geqslant 1+\operatorname{deg} v+d^{2}-\operatorname{deg} v=1+d^{2}$
$\Rightarrow d \leqslant \sqrt{n-1}$.
Since $d=\frac{1}{|V|} \sum_{v \in V} \operatorname{deg} v=\frac{2 e}{n}$, we have
$e \leqslant \frac{n \sqrt{n-1}}{2}$.
Another proof: Construct a simple 10-vertex graph $G$ with the ten points as vertices and the connecting line segments as edges.
Since graph $G$ contains no triangles or quadrilaterals, $g(G) \geqslant 5$.
By the theorem, the number of edges is at most
$\frac{10 \times \sqrt{10-1}}{2}=15$.
Additionally, the Petersen graph satisfies the conditions, so the maximum number of edges is 15.
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Arrange the numbers in the set $\left\{2^{x}+2^{y} \mid x 、 y \in \mathbf{N}, x<y\right\}$ in ascending order. Then the 60th number is $\qquad$ (answer with a number).
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
3.2064.
It is known that the number of combinations $(x, y)$ satisfying $0 \leqslant x<y \leqslant n$ is $\mathrm{C}_{n+1}^{2}$.
Notice that, $\mathrm{C}_{11}^{2}=55<60<66=\mathrm{C}_{12}^{2}$.
Therefore, the 60th number satisfies $y=11, x=4$, which means the 60th number is $2^{11}+2^{4}=2064$.
|
2064
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Find the maximum value of $n$ such that there exists an arithmetic sequence $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfying
$$
\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.
$$
Thus, the absolute value sum function $f(x)=\sum_{i=1}^{n}|x-i d|$ has three "equal value points":
$$
f(a-2)=f(a)=f(a+1)=507,
$$
It must be that $n=2 k\left(k \in \mathbf{Z}_{+}\right)$, and
$$
\begin{array}{l}
\left\{\begin{array}{l}
a-2, a, a+1 \in[k d,(k+1) d], \\
f(k d)=507
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
d \geqslant 3, \\
k^{2} d=507
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
d \geqslant 3, \\
k^{2}=\frac{507}{d} \leqslant \frac{507}{3}=169
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
d \geqslant 3, \\
k \leqslant 13 .
\end{array}\right.
\end{array}
$$
Therefore, $n \leqslant 26$, with equality holding when $d=3, k=13$.
Thus, when $a-2=k d=39$, i.e., $a=41$, the corresponding arithmetic sequence is $\{41-3 k\}(k=1,2, \cdots, 26)$. Hence, $n_{\max }=26$.
|
26
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $f(x)=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$, then the maximum value of $f(x)$ is $\qquad$
|
$-、 1.11$.
From the fact that $f(x)$ is defined, we know
$$
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 \times \frac{13-x}{4}+3 \times \frac{x}{9}\right)} \\
=11 .
\end{array}
$$
When $x=9$, the equality holds.
Therefore, the maximum value of $f(x)$ is 11.
|
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 pair among the 68 pairs is erased, the student scores one point. Find the maximum score the student can achieve.
Note: These 68 pairs can include some identical pairs. ${ }^{[3]}$
(2010, USA Mathematical Olympiad)
|
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 probability $1-p$.
Thus, the probability of scoring for each pair is at least $\min \left\{p, 1-p^{2}\right\}$. Therefore, the expected score for 68 pairs is $68 \min \left\{p, 1-p^{2}\right\}$.
Solving the equation $p = 1 - p^{2}$, we get $p = \frac{\sqrt{5} - 1}{2}$.
Therefore, the expected total score is at least
\[ 68 p = 68 \times \frac{\sqrt{5} - 1}{2} > 42. \]
Thus, there must exist a way to achieve a score of 43.
Finally, we provide an example to show that 44 points may not be feasible.
For $1 \leqslant i \leqslant 8$, each pair $(i, i)$ appears 5 times; for $1 \leqslant i \neq j \leqslant 8$, each pair $(-i, -j)$ appears 1 time: a total of $40 + 28 = 68$ pairs.
Suppose $k$ pairs of numbers from 1 to 8 are deleted. Then the first type of pairs can score $5k$ points; and among the second type of pairs, at least $\mathrm{C}_{k}^{2}$ pairs cannot be taken, so the score is $28 - \mathrm{C}_{k}^{2}$. The total score is $5k + 28 - \mathrm{C}_{k}^{2}$, and by completing the square, the maximum value is 43 points.
|
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 pressed once, the number on the display screen will change to $4x+1$. Pressing either operation key once is called one operation. The number on the display screen is currently 1. Answer the following questions:
(1) Can the number 2000 appear on the display screen after a finite number of operations? Explain your reasoning.
(2) How many integers less than 2000 can appear on the display screen after a finite number of operations?
|
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 resulting number does not have two 1s adjacent. However, 2000 in binary is $(11111010000)_2$, which has two 1s adjacent. Therefore, the number 2000 cannot appear on the display.
(2) If the second operation is performed first, followed by the first operation, it is equivalent to appending a 0 to the original number. Hence, any binary number without two 1s adjacent can be obtained after a finite number of operations.
Thus, the number of integers less than 2000 that can appear on the display is equivalent to the number of natural numbers not greater than $(10101010101)_2$ and without two 1s adjacent.
Moreover, the number of natural numbers with $k(0 \leqslant k \leqslant 6)$ 1s and not greater than $(10101010101)_2$ is $\mathrm{C}_{12-k}^{k}$. Therefore, the number of natural numbers that meet the condition is
$$
1+11+45+84+70+21+1=233 .
$$
|
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 $x \in \mathbf{Z}_{+}, x \geqslant 5$,
$$
\begin{array}{l}
(x+3)^{x}=\left(\frac{x+3}{2}\right)^{4}(x+3)^{x-4} 4^{2} \\
& (x+3)^{x}-(x-2)^{x}-2 x(x+1)(x+2)+2 \\
> & (x+3)(x+2)^{x-1}-(x+2)^{x}- \\
& 2 x(x+1)(x+2)+2 \\
= & (x+2)^{x-1}-2 x(x+1)(x+2)+2 \\
\geqslant & (x+2)^{4}-2 x(x+1)(x+2)+2 \\
= & x^{4}+6 x^{3}+18 x^{2}+28 x+18>0 .
\end{array}
$$
By taking $x=1,2,3,4$ in sequence, it is found that $f(x)=0$ if and only if $x=3$.
Therefore, this function has only one positive integer zero.
|
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 pairs of people in the graph is $\mathrm{C}_{12k}^{2}$, so the number of common handshakes between any two people is
$$
d=\frac{12k \mathrm{C}_{3k+6}^{2}}{\mathrm{C}_{12k}^{2}}=\frac{3(k+2)(3k+5)}{12k-1} \in \mathbf{Z}_{+} \text{. }
$$
Noting that $(3,12k-1)=1$,
$$
\begin{aligned}
p & =\frac{(k+2)(3k+5)}{12k-1}=k+1+\frac{11-9k^{2}}{12k-1} \\
& =k+1-11+\frac{-3k(3k-44)}{12k-1} \in \mathbf{Z}_{+},
\end{aligned}
$$
and $(3k, 12k-1)=1$.
Thus, $(12k-1) \mid (3k-44)$.
Then $3k-44 \geqslant 12k-1$ or $44-3k \geqslant 12k-1$.
Solving, we get $k \leqslant -\frac{43}{9}$ (discard) or $k \leqslant 3$.
Upon inspection, only when $k=3$, $d \in \mathbf{Z}_{+}$. At this time, $d=6$.
In summary, there are 36 people attending the meeting, and the number of common handshakes between any two people is 6.
|
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 $x$,
$$
\begin{array}{c}
S=\left[\frac{1}{1}\right]+\left[\frac{2}{1}\right]+\left[\frac{1}{2}\right]+\left[\frac{2}{2}\right]+\left[\frac{3}{2}\right]+ \\
{\left[\frac{4}{2}\right]+\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{3}{3}\right]+\left[\frac{4}{3}\right]+} \\
{\left[\frac{5}{3}\right]+\left[\frac{6}{3}\right]+\cdots}
\end{array}
$$
up to 2016 terms, where, for a segment with denominator $k$, there are $2 k$ terms $\left[\frac{1}{k}\right],\left[\frac{2}{k}\right], \cdots,\left[\frac{2 k}{k}\right]$, and only the last segment may have fewer than $2 k$ terms. Then the value of $S$ is
|
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, \( 2016-1980=36 \), so the sum includes 36 terms with the denominator 45: \(\left[\frac{1}{45}\right],\left[\frac{2}{45}\right], \cdots,\left[\frac{36}{45}\right]\), whose sum is zero.
$$
\text { Therefore, } S=\sum_{k=1}^{44}(k+2)=1078 \text {. }
$$
|
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 $a=-m+2 n+2 c, b=m-n-3 c$
$$
\begin{aligned}
\Rightarrow & |a| \leqslant|m|+2|n|+2|c| \leqslant 8, \\
& |b| \leqslant|m|+|n|+3|c| \leqslant 8 \\
\Rightarrow & |a|+|b|+|c| \leqslant 17 .
\end{aligned}
$$
It is easy to verify,
$$
f(x)=8\left(x-\frac{1}{2}\right)^{2}-1=8 x^{2}-8 x+1 \text {, }
$$
When $0 \leqslant x \leqslant 1$, it always holds that $|f(x)| \leqslant 1$.
Therefore, the maximum value sought is 17.
|
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 $\frac{1}{2} \sum_{i=1}^{10} d_{i}$.
In graph $G$, we call the figure formed by two line segments sharing a common endpoint an angle. Thus, there are $\sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}$ different angles.
By the given condition, for any angle $\angle A_{i} A_{j} A_{k}$ in graph $G$, $A_{i}$ and $A_{k}$ are not adjacent. For any two angles $\angle A_{i_{1}} A_{j_{1}} A_{k_{1}}$ and $\angle A_{i_{2}} A_{j_{2}} A_{k_{2}}$, we have
$\left\{A_{i_{1}}, A_{k_{1}}\right\} \neq\left\{A_{i_{2}}, A_{k_{2}}\right\}$.
Thus, the complement graph of $G$ has at least $\sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}$ edges.
Therefore, the total number of edges $S$ satisfies
$$
\sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}=S \leqslant \mathrm{C}_{10}^{2}-\frac{1}{2} \sum_{i=1}^{10} d_{i} \text {, }
$$
which implies $\mathrm{C}_{10}^{2} \geqslant \sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}+\frac{1}{2} \sum_{i=1}^{10} d_{i}=\frac{1}{2} \sum_{i=1}^{10} d_{i}^{2}$.
Also, $\frac{1}{2} \sum_{i=1}^{10} d_{i}^{2} \geqslant \frac{1}{2} \times \frac{1}{10}\left(\sum_{i=1}^{10} d_{i}\right)^{2}$, so
$$
\mathrm{C}_{10}^{2} \geqslant \frac{1}{2} \times \frac{1}{10}\left(\sum_{i=1}^{10} d_{i}\right)^{2} \Rightarrow \frac{1}{2} \sum_{i=1}^{10} d_{i} \leqslant 15 \text {. }
$$
A specific construction is shown in Figure 1.
|
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
$$
\begin{array}{l}
x=2016-S(x)-S(S(x))-S(S(S(x))) \\
\geqslant 2016-28-10-9=1969 .
\end{array}
$$
Thus, $x \in\{1969,1970, \cdots, 2015\}$.
Also, $x, S(x), S(S(x)), S(S(S(x)))$ have the same remainder when divided by 9, and 2016 has a remainder of 0 when divided by 9, so each number has a remainder of 0 when divided by 9.
Therefore, $x$ can only be $1971, 1980, 1989, 1998, 2007$.
Upon inspection, only $1980+18+9+9=2016$. Therefore, the solution set of the equation is $\{1980\}$.
|
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
$$
\begin{array}{l}
{\left[\frac{7^{2 k-1}}{8}\right]+\left[\frac{7^{2 k}}{8}\right]=7^{2 k-1}-1} \\
\equiv 7(-1)^{k-1}-1(\bmod 50) .
\end{array}
$$
Thus, $A \equiv 7(1-1+1-1+\cdots+1-1)-1008$ $\equiv 42(\bmod 50)$.
|
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 nodes, then this line segment is called a "basic segment". Considering all possible divisions, find the maximum and minimum number of basic segments. ${ }^{[4]}$
|
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 $v+f-e=2$.
Thus, $e=v+2015$.
Note that, the points with degree 2 in graph $G$ are exactly 4, the rest of the points have a degree of 3 or 4. Points with degree 2 are the vertices of one rectangle, points with degree 3 are the vertices of two rectangles, and points with degree 4 are the vertices of four rectangles.
On one hand, the total number of vertices of all small rectangles is $4 \times 2016$.
On the other hand, the total number of vertices of all small rectangles is no less than $2(v-4)+4=2 v-4$.
Thus, $2 v-4 \leqslant 4 \times 2016 \Rightarrow v \leqslant 4034$.
Therefore, from equation (1), we get $e \leqslant 6049$.
When the rectangle $R$ is divided into $1 \times 2016$ small rectangles, the above equality holds.
Thus, the maximum value of $e$ is $e_{\text {max }}=6049$.
Next, consider the minimum value of $e$.
Suppose the rectangle $R$ is divided using $a$ horizontal lines and $b$ vertical lines, excluding the boundaries.
Since $a$ horizontal lines and $b$ vertical lines can divide the area into at most $(a+1)(b+1)$ regions, we have
$$
(a+1)(b+1) \geqslant 2016 \text {. }
$$
Let the number of points with degree 3 be $x$, and the number of points with degree 4 be $y$. Then,
$$
v=4+x+y, 4+2 x+4 y=4 \times 2016 .
$$
Each horizontal and vertical line has its endpoints as the vertices of two rectangles, thus they are all points with degree 3, and these points are distinct.
Hence, $x \geqslant 2 a+2 b$.
Combining with equation (2), we get
$$
\begin{array}{l}
x \geqslant 2(a+1)+2(b+1)-4 \\
\geqslant 4 \sqrt{(a+1)(b+1)}-4 \\
\geqslant 4 \sqrt{2016}-4>175.59 \\
\Rightarrow x \geqslant 176 .
\end{array}
$$
Thus, from equations (3) and (4), we get
$$
v=2019+\frac{1}{2} x \geqslant 2017 \text {. }
$$
From equation (1), we immediately get $e \geqslant 4122$, where the equality holds in the $42 \times 48$ division.
Thus, $e_{\min }=4122$.
|
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
$$
\begin{array}{l}
f(x)=\boldsymbol{m} \cdot \boldsymbol{n}, \boldsymbol{m}=(2 \cos x, 1), \\
\boldsymbol{n}=(\cos x, \sqrt{3} \sin 2 x), \\
f(A)=2, b=1, S_{\triangle A B C}=\frac{\sqrt{3}}{2} . \\
\text { Then } \frac{b+c}{\sin B+\sin C}=
\end{array}
$$
|
$=13.2$.
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.
$$
\begin{array}{l}
\text { Given } S_{2015}=a_{1}+\sum_{k=1}^{1007}\left(a_{2 k}+a_{2 k+1}\right) \\
=a_{1}-1008 \\
\Rightarrow m+a_{1}-1008=-1007 \\
\Rightarrow m+a_{1}=1 . \\
\text { Also, } m a_{1}>0 \text {, so } m>0, a_{1}>0 . \\
\text { Therefore, } \frac{1}{a_{1}}+\frac{4}{m}=\left(m+a_{1}\right)\left(\frac{1}{a_{1}}+\frac{4}{m}\right) \\
=5+\frac{m}{a_{1}}+\frac{4 a_{1}}{m} \geqslant 9 .
\end{array}
$$
|
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 $f(x) \leqslant 0$ is $\left(a, x_{2}\right]$, with the length of the interval being $x_{2}-a$.
Similarly, when $b<x<a$, the interval of $x$ that satisfies $f(x) \geqslant 0$ is $\left(b, x_{1}\right]$, with the length of the interval being $x_{1}-b$.
By Vieta's formulas, we have
$$
x_{1}+x_{2}=\frac{a c+b c+2}{c}=a+b+\frac{2}{c} \text {. }
$$
Then the sum of the lengths of the intervals of $x$ that satisfy the conditions is
$$
\begin{array}{l}
x_{2}-a+x_{1}-b=a+b+\frac{2}{c}-a-b \\
=\frac{2}{c}=1024 .
\end{array}
$$
|
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
$$
x^{2}+15 x+50+\left(\frac{3-\sqrt{6}}{3}\right)^{3}=0 \text {. }
$$
By $\Delta_{1}=15^{2}-4\left(50+\left(\frac{3-\sqrt{6}}{3}\right)^{3}\right)>0$, we know the original
equation has two distinct real roots $x_{1} 、 x_{2}$.
If $b_{2}=\frac{-3-\sqrt{6}}{3}$, then
$$
x^{2}+15 x+50+\left(\frac{3+\sqrt{6}}{3}\right)^{3}=0 \text {. }
$$
By $\Delta_{2}=15^{2}-4\left(50+\left(\frac{3+\sqrt{6}}{3}\right)^{3}\right)>0$, we know the original equation has two distinct real roots $x_{3} 、 x_{4}$.
Since $x_{1} 、 x_{2} 、 x_{3} 、 x_{4}$ are all distinct, the number of distinct real solutions $x$ of the original equation is 4.
|
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, $f\left(\ln \log _{21} 5\right)=f\left(-\ln \log _{5} 21\right)$
$$
=40-f\left(\ln \log _{5} 21\right)=23 \text {. }
$$
|
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 $z=x+y \mathrm{i}(x, y \in \mathbf{R})$.
$$
\begin{array}{l}
\text { Then }|(a-2) z+a \mathrm{i}|^{2}-|a-2-a z \mathrm{i}|^{2} \\
=|(a-2) x+((a-2) y+a) \mathrm{i}|^{2}- \\
|a-2+y a-a x \mathrm{i}|^{2} \\
=((a-2) x)^{2}+((a-2) y+a)^{2}- \\
(a-2+y a)^{2}-(-a x)^{2} \\
= 4(1-a)\left(x^{2}+y^{2}-1\right) \\
= 4(1-a)\left(|z|^{2}-1\right) .
\end{array}
$$
If $|z|>1$, then
$$
\begin{array}{l}
|a-2-a z \mathrm{i}||a-2-a z \mathrm{i}|,
\end{array}
$$
which contradicts equation (1).
If $|z|<1$, it would also lead to a contradiction.
Therefore, $|z|=1$.
|
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 number, then all such $n$ satisfy the condition, and there are 9 such $n$;
(2) If $n$ is a two-digit number, let $n=\overline{x y}$, then
$$
S(n)=\overline{y x}, S(S(n))=\overline{x y} \text {, }
$$
There are 81 such $n$, where $x, y \neq 0$;
(3) If $n$ is a three-digit number, let $n=\overline{x y z}$, then
$$
S(n)=\overline{y z x}, S(S(n))=\overline{z x y} \text {, }
$$
Thus, $x=y=z \neq 0$, and there are 9 such $n$;
(4) If $n$ is a four-digit number, let $n=\overline{x y z w}$, then
$$
S(S(n))=\overline{z w x y} \text {, }
$$
Thus, $w=y, z=x$, and there are 36 such $n$, where $x=1,2,3,4, y \neq 0$.
In summary, the number of positive integers $n$ that satisfy the condition is 135.
|
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 the ellipse $\Gamma$ at points $C$ and $D$ respectively. The slope of line $C D$ is $k_{2}$. Write $\frac{k_{1}^{2}}{k_{2}^{2}}$ as a reduced fraction $\frac{a}{b}$ (where $a$ and $b$ are coprime positive integers). Then $a^{2}+b=$
|
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
$$
\frac{5-x_{1}}{y_{1}^{2}} y^{2}+\frac{x_{1}-1}{y_{1}} y-4=0 \text {. }
$$
By Vieta's formulas, we have
$$
y_{1} y_{3}=-\frac{4 y_{1}^{2}}{5-x_{1}}\left(y_{1} \neq 0\right) \Rightarrow y_{3}=\frac{4 y_{1}}{x_{1}-5} \text {. }
$$
Substitute into the equation of the line $A R$ to get $x_{3}=\frac{5 x_{1}-9}{x_{1}-5}$.
Similarly, $x_{4}=\frac{5 x_{2}-9}{x_{2}-5}, y_{4}=\frac{4 y_{2}}{x_{2}-5}$.
Thus, $k_{2}=\frac{y_{3}-y_{4}}{x_{3}-x_{4}}$
$$
=\frac{4\left(y_{1} x_{2}-5 y_{1}-y_{2} x_{1}+5 y_{2}\right)}{16\left(x_{2}-x_{1}\right)} \text {. }
$$
Since points $A, F, B$ are collinear, we have
$$
\begin{array}{l}
\frac{y_{1}}{x_{1}+2}=\frac{y_{2}}{x_{2}+2} \\
\Rightarrow y_{1} x_{2}-y_{2} x_{1}=2\left(y_{2}-y_{1}\right) \\
\Rightarrow k_{2}=\frac{7}{4} \cdot \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7}{4} k_{1} \\
\Rightarrow \frac{a}{b}=\frac{k_{1}^{2}}{k_{2}^{2}}=\frac{16}{49} \Rightarrow a=16, b=49 \\
\Rightarrow a^{2}+b=305 .
\end{array}
$$
|
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
$$
\begin{array}{l}
A_{1} \cup A_{5} \cup A_{6} \cup A_{7} \cup A_{9}, A_{1} \cup A_{2} \cup A_{7} \cup A_{8} \cup A_{10}, \\
A_{2} \cup A_{3} \cup A_{6} \cup A_{8} \cup A_{9}, A_{3} \cup A_{4} \cup A_{7} \cup A_{9} \cup A_{10}, \\
A_{4} \cup A_{5} \cup A_{6} \cup A_{8} \cup A_{10},
\end{array}
$$
It can be seen that these satisfy the problem's conditions and each set has 200 elements. Thus, $r \leqslant 200$.
Define $a_{i j}=\left\{\begin{array}{l}1, i \in A_{j} ; \\ 0, \text { otherwise, }\end{array} m_{i}=\sum_{j=1}^{5} a_{i j}\right.$, where $i=1,2, \cdots, 1000 ; j=1,2, \cdots, 5$.
$$
\begin{array}{l}
\text { Then } \sum_{i=1}^{1000} m_{i}=2500 . \\
\text { By } \sum_{1 \leqslant i<j \leqslant 5}\left|A_{i} \cap A_{j}\right|=\sum_{i=1}^{1000} \mathrm{C}_{m_{i}}^{2} \\
=\frac{1}{2}\left(\sum_{i=1}^{1000} m_{i}^{2}-\sum_{i=1}^{1000} m_{i}\right), \\
\sum_{i=1}^{1000} m_{i}^{2} \geqslant \frac{1}{1000}\left(\sum_{i=1}^{1000} m_{i}\right)^{2},
\end{array}
$$
we know that $\sum_{i=1}^{1000} m_{i}^{2}$ takes its minimum value when $m_{i}$ are as close to each other as possible.
Suppose there are $x$ 2's and $y$ 3's.
Then $\left\{\begin{array}{l}x+y=1000, \\ 2 x+3 y=2500\end{array} \Rightarrow x=y=500\right.$.
Thus, $\sum_{i=1}^{1000} m_{i}^{2} \geqslant 500 \times 2^{2}+500 \times 3^{2}=6500$
$\Rightarrow \sum_{1 \leqslant i<j \leqslant 5}\left|A_{i} \cap A_{j}\right| \geqslant 2000$.
Therefore, there must exist $1 \leqslant i<j \leqslant 5$, such that
$\left|A_{i} \cap A_{j}\right| \geqslant \frac{2000}{C_{5}^{2}}=200$.
Hence, $r \geqslant 200$.
|
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 odd, let the number of positive integers less than $k$ and coprime to $k$ be $\varphi(k)$. Suppose these numbers are
$$
1=c_{1}<c_{2}<\cdots<c_{\varphi(k)} \leq k-1
$$
and
$$
c_{1}+c_{2}+\cdots+c_{\varphi(k)}=\frac{\varphi(k)(k-1)}{2} \quad \text{and} \quad \varphi(1)} \varphi(k)=n-1$.
To prove that in the set $\left\{\frac{1}{n}, \frac{2}{n}, \cdots, \frac{n-1}{n}\right\}$, when the fractions are simplified to their lowest terms, the denominators can only be the divisors $k$ of $n$ that are greater than or equal to 2, and the number of fractions with denominator $k$ is $\varphi(k)$, the total number of fractions in lowest terms is
$$
\sum_{\substack{k \mid n \\ k>1}} \varphi(k)=n-1 .
$$
The lemma is proved.
By the lemma, the number of integers in the original set is
$$
\sum_{\substack{k \mid 2015 \\ k>1}} \frac{\varphi(k)}{2}=\frac{2015-1}{2}=1007 \text {. }
$$
[Note] For problems involving the greatest common divisor or the least common multiple, setting $d=(a, b)$, and $a=A d, b=B d$ (where $A$ and $B$ are coprime integers) is a common transformation. Using this transformation can simplify the problem and pave the way for solving it.
|
1007
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Red, blue, green, and white four dice, each die's six faces have numbers $1, 2, 3, 4, 5, 6$. Simultaneously roll these four dice so that the product of the numbers facing up on the four dice equals 36, there are $\qquad$ possible ways.
|
4. 48 .
$$
\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{array}
$$
possibilities.
In total, there are 48 possibilities.
|
48
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 On the plane, there are $n(n \geqslant 5)$ distinct points, each point is exactly at a distance of 1 from four other points. Find the minimum value of such $n$. [2]
|
Keep points $A, B, D, E, G$ in Figure 1, construct $\square E A B C, \square D A G F, \square B A G I$ to get points $C, F, I$, and construct $\square C I F H$ to get point $H$.
$$
\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(180^{\circ}-\angle B A G\right) \\
=\angle E A G=60^{\circ},
\end{array}
$$
we know that $\triangle B C I$ is an equilateral triangle.
Similarly, $\triangle F G I$ is an equilateral triangle.
Thus, $C E=A B=G I=F I=C H$.
And $\angle E C H=\angle G I F=60^{\circ}$, so $\triangle C E H$ is an equilateral triangle.
Similarly, $\triangle D F H$ is an equilateral triangle.
Therefore, the points that are 1 unit away from point $A$ are $B, E, D, G$; the points that are 1 unit away from point $B$ are $A, D, I, C$; the points that are 1 unit away from point $C$ are $B, E, I, H$; the points that are 1 unit away from point $D$ are $B, A, F, H$; the points that are 1 unit away from point $E$ are $A, G, H, C$; the points that are 1 unit away from point $F$ are $G, D, I, H$; the points that are 1 unit away from point $G$ are $A, E, I, F$; the points that are 1 unit away from point $H$ are $C, E, D, F$; the points that are 1 unit away from point $I$ are $B, C, F, G$. This is an example of $n=9$, as shown in Figure 2.
|
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 equal.
|
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 only if $n=1$.
For $j=1,2, \cdots, l$, let $d_{j}$ be the largest number in the $j$-th column of the grid. Without loss of generality, assume
$d_{1}>d_{2}>\cdots>d_{l}$.
Since $d_{1}, d_{2}, \cdots, d_{l}$ are all positive divisors of $n$, then
$d_{l} \leqslant \frac{n}{l}$.
And $d_{l}$ is the largest number in the $l$-th column of the grid, so
$d_{l} \geqslant \frac{s}{k} \geqslant \frac{n}{k}$.
From equations (1) and (2), we get $\frac{n}{l} \geqslant \frac{n}{k} \Rightarrow k \geqslant l$.
Thus, $k=l$.
Therefore, the equalities in equations (1) and (2) both hold.
In particular, we have $s=n$.
Hence, $n=1$, and $n=1$ satisfies the conditions.
|
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 card from the second pile and place it in the first pile (it can be the same card), such that the sum of the numbers on the cards in each pile is equal to 0. Find the maximum possible value of $n$.
|
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 operation, the first pile will have only one card. Therefore, this card must be written with 0.
For a card written with a non-zero real number $a$, place it and the card written with 0 in the first pile, and the remaining cards in the second pile. After the operation, the first pile will have exactly two cards. Therefore, it is only possible to replace the card written with 0 with a card written with $-a$.
Thus, the real numbers on all cards must be 0 and several pairs of opposite numbers.
Assume the positive real numbers written on the cards are
$$
a_{1}<a_{2}<\cdots<a_{n} \text {. }
$$
If $n \geqslant 4$, first divide the cards written with $a_{1}, a_{2}, \cdots, a_{n}$ into the first pile, and the remaining cards into the second pile. Note that, after the operation, the first pile can only exchange one card. Therefore, it is only possible to replace the card written with $a_{n}$ with a card written with $-a_{n}$ (any other exchange will result in a final sum greater than 0). Thus,
$$
a_{n}=a_{1}+a_{2}+\cdots+a_{n-1} \text {. }
$$
Next, divide the cards written with $a_{2}, a_{3}, \cdots, a_{n}$ into the first pile, and the remaining cards into the second pile. Note that, if the first pile does not replace the card written with $a_{n}$ with a card written with $-a_{n}$, the sum will definitely be greater than 0. If the first pile replaces the card written with $a_{n}$ with a card written with $-a_{n}$, the sum will definitely be less than 0. Neither satisfies the requirement.
Therefore, $n \leqslant 3$, meaning the total number of cards does not exceed 7.
|
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 integers $m, n$ (allowing $m = n$) such that $f(m+n) \neq f(m)+f(n)$.
Find the minimum value of $k$.
|
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}$ with three different colors, respectively. Thus,
(i) For any $x \equiv y \equiv 0(\bmod 3)$, we have
$x+y \equiv 0(\bmod 3)$
$\Rightarrow f(x+y)=\frac{x+y}{3}=f(x)+f(y)$;
(ii) For any $x \equiv y \equiv 1(\bmod 3)$, we have
$x+y \equiv 2(\bmod 3)$
$\Rightarrow f(x+y)=x+y=f(x)+f(y)$;
(iii) For any $x \equiv y \equiv 2(\bmod 3)$, we have
$x+y \equiv 1(\bmod 3)$
$\Rightarrow f(x+y)=x+y=f(x)+f(y)$.
Thus, condition (1) is also satisfied.
Therefore, $k=3$ meets the requirements.
Next, prove that $k=2$ does not hold.
It suffices to prove that for any function $f$ and coloring scheme satisfying condition (1) when $k=2$, we have
$f(n)=n f(1)$ (for any $n \in \mathbf{Z}_{+}$),
which contradicts condition (2).
In condition (1), take $m=n$, then
$f(2 n)=2 f(n)$ (for any $n \in \mathbf{Z}_{+}$).
Next, prove:
$f(3 n)=3 f(n)$ (for any $n \in \mathbf{Z}_{+}$).
For any positive integer $n$, by equation (2) we know
$f(2 n)=2 f(n), f(4 n)=4 f(n)$,
$f(6 n)=2 f(3 n)$.
If $n$ and $2 n$ are the same color, then
$f(3 n)=f(2 n)+f(n)=3 f(n)$,
equation (3) holds;
If $2 n$ and $4 n$ are the same color, then
$$
\begin{array}{l}
f(3 n)=\frac{1}{2} f(6 n)=\frac{1}{2}(f(4 n)+f(2 n)) \\
=3 f(n),
\end{array}
$$
equation (3) also holds.
Otherwise, $2 n$ is a different color from both $n$ and $4 n$, so $n$ and $4 n$ are the same color. In this case, if $n$ and $3 n$ are the same color, then
$$
f(3 n)=f(4 n)-f(n)=3 f(n),
$$
equation (3) holds;
If $n$ and $3 n$ are different colors, then $2 n$ and $3 n$ are the same color,
$$
f(3 n)=f(4 n)+f(n)-f(2 n)=3 f(n) \text {, }
$$
equation (3) also holds.
Thus, equation (3) is proven.
Assume the proposition (1) does not hold. Then there exists a positive integer $m$ such that $f(m) \neq m f(1)$.
Without loss of generality, take the smallest $m$, then by equations (2) and (3), $m \geqslant 5$, and $m$ is odd. Otherwise, by the minimality of $m$,
$f\left(\frac{m}{2}\right)=\frac{m}{2} f(1)$.
Thus, $f(m)=2 f\left(\frac{m}{2}\right)=m f(1)$, a contradiction.
Consider $\frac{m-3}{2}<\frac{m+3}{2}<m$ these three numbers.
Similarly, by the minimality of $m$,
$f\left(\frac{m-3}{2}\right)=\frac{m-3}{2} f(1)$,
$f\left(\frac{m+3}{2}\right)=\frac{m+3}{2} f(1)$.
Thus, $\frac{m-3}{2}$ and $\frac{m+3}{2}$ are different colors. Otherwise,
$f(m)=f\left(\frac{m-3}{2}\right)+f\left(\frac{m+3}{2}\right)=m f(1)$,
a contradiction.
Therefore, $m$ is exactly the same color as one of $\frac{m-3}{2}, \frac{m+3}{2}$.
Let $m$ be the same color as $\frac{m+3 p}{2}(p \in\{-1,1\})$.
Notice that, $\frac{m+p}{2}<m$.
Then $f(m)+f\left(\frac{m+3 p}{2}\right)=f\left(3 \times \frac{m+p}{2}\right)$
$=3 f\left(\frac{m+p}{2}\right)=\frac{3(m+p)}{2} f(1)$
$\Rightarrow f(m)=m f(1)$,
a contradiction.
Thus, proposition (1) is proven, i.e., the minimum value of $k$ is 3.
|
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_{2}, \cdots, a_{n}$.
Then $2 \leqslant a_{i} \leqslant 3(i=1,2, \cdots, n)$.
Let the total weight of all transported goods be $S$.
Thus, $2 n \leqslant S=a_{1}+a_{2}+\cdots+a_{n} \leqslant 3 n$
$\Rightarrow 2 n \leqslant 10 \leqslant 3 n \Rightarrow \frac{10}{3} \leqslant n \leqslant 5$
$\Rightarrow n=4$ or 5.
Furthermore, it is explained that 4 vehicles are not enough.
Suppose there are 13 boxes, each weighing 13 tons.
Since $\frac{10}{13} \times 4>3$, each vehicle can transport at most 3 boxes. Therefore, 4 vehicles cannot transport all the boxes. Thus, at least 5 vehicles are needed to transport all the boxes at once.
|
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{N}_{+}\right) .
\end{array}
$$
Suppose the "harmonious function" $f$ satisfies the condition: there exists a positive integer $M$, such that
(1) When $m<M$, if $i, j \in \mathbf{N}$, $i \equiv j+1(\bmod 2017)$,
then $f^{(m)}(i)-f^{(m)}(j) \not \equiv \pm 1(\bmod 2017)$;
(2) If $i, j \in \mathbf{N}, i \equiv j+1(\bmod 2017)$, then $f^{(M)}(i)-f^{(M)}(j) \equiv \pm 1(\bmod 2017)$.
Find the maximum possible value of $M$.
|
On the one hand, note that 2017 is a prime number.
Let $g$ be a primitive root modulo 2017, then the half-order of $g$ modulo 2017 is 1008.
$$
\text{Let } f(i) \equiv g(i-1)+1(\bmod 2017) \text{.}
$$
Since $(g, 2017)=1$, $g(i-1)+1$ runs through a complete residue system modulo 2017.
Thus, the mapping $f: \mathbf{N} \rightarrow A$ is surjective.
Also, $f(i+2017) \equiv f(i)(\bmod 2017)$, i.e., $f(i+2017)=f(i)$,
Hence, the defined $f$ is a "harmonious mapping".
According to the definition of $f$,
$$
f^{(k)}(i) \equiv g^{k}(i-1)+1(\bmod 2017) \text{.}
$$
At this point, from $i \equiv j+1(\bmod 2017)$, we get
$$
\begin{array}{l}
\left(g^{M}(i-1)+1\right)-\left(g^{M}(j-1)+1\right) \\
\equiv \pm 1(\bmod 2017) \\
\Leftrightarrow g^{M}(i-j) \equiv \pm 1(\bmod 2017) \\
\Leftrightarrow g^{M} \equiv \pm 1(\bmod 2017) .
\end{array}
$$
Noting that the half-order of $g$ modulo 2017 is 1008.
Thus, 1008 divides $M$.
Therefore, the required $M_{\max } \geqslant 1008$.
On the other hand, construct a convex 2017-gon $A_{0} A_{1} \cdots A_{2016}$, denoted as graph $G$. Connect lines according to the following rule: if $1 \leqslant m<M$, $i \in A$, $f^{m}(i)=a$, $f^{m}(i+1)=b$, then connect the segment $A_{a} A_{b}$.
Clearly, the connected segments are diagonals of graph $G$, and the connected segments are not repeated. Otherwise, if there exist two identical lines, i.e., there exist $i, j \in A$, and $1 \leqslant q<p<M$, such that $f^{(p)}(i)=f^{(q)}(j)$, and
$$
f^{(p)}(i+1)=f^{(q)}(j+1)
$$
or $f^{(p)}(i+1)=f^{(q)}(j-1)$.
$$
\text{Then }\left\{\begin{array}{l}
f^{(p-q)}(i)=j, \\
f^{(p-q)}(i+1) \equiv j+1 \text{ or } j-1(\bmod 2017) .
\end{array}\right.
$$
Noting that $0<p-q<M$, this contradicts the given conditions. Hence, the connected diagonals are not repeated.
Since a total of $2017(M-1)$ segments are connected, and the convex 2017-gon $G$ has $2017 \times 2017$ segments, thus,
$$
\begin{array}{l}
2017(M-1) \leqslant 2017 \times 1007 \\
\Rightarrow M \leqslant 1008 .
\end{array}
$$
In summary, the maximum value of $M$ is 1008.
|
1008
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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