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1. Divide the sequence of positive integers $1,2, \cdots$ from left to right into segments such that the first segment has $1 \times 2$ numbers, the second segment has $2 \times 3$ numbers, $\cdots$, the $n$-th segment has $n \times(n+1)$ numbers, $\cdots$. Then 2014 is in the $\qquad$ segment. | $-, 1.18$.
$$
\begin{array}{l}
\text { Let } S_{n}=1 \times 2+2 \times 3+\cdots+n(n+1) \\
=\left(1^{2}+2^{2}+\cdots+n^{2}\right)+(1+2+\cdots+n) \\
=\frac{n(n+1)(n+2)}{3} .
\end{array}
$$
If 2014 is in the $(n+1)$-th segment, since there are $S_{n}$ numbers before this segment, then $S_{n}<2014 \leqslant S_{n+1}$.
Sinc... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: x^{2}-\frac{y^{2}}{24}=1$, respectively, and let $P$ be a point on the hyperbola $C$ in the first quadrant. If $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{4}{3}$, then the radius of the incircle of $\triangle P F_{1} F_{2}$ is . $\q... | 4. 2 .
Let $\left|P F_{1}\right|=4 t$. Then $\left|P F_{2}\right|=3 t$.
Thus $4 t-3 t=\left|P F_{1}\right|-\left|P F_{2}\right|=2$
$$
\Rightarrow t=2,\left|P F_{1}\right|=8,\left|P F_{2}\right|=6 \text {. }
$$
Combining with $\left|F_{1} F_{2}\right|=10$, we know that $\triangle P F_{1} F_{2}$ is a right triangle, $P... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. If the fraction $\frac{p}{q}\left(p, q \in \mathbf{Z}_{+}\right)$ is converted to a decimal as
$$
\frac{p}{q}=0.198 \cdots,
$$
then when $q$ takes the minimum value, $p+q=$ . $\qquad$ | 6. 121 .
Given $\frac{p}{q}=0.198 \cdots5 p$.
Let $q=5 p+m\left(m \in \mathbf{Z}_{+}\right)$. Then
$$
\begin{array}{l}
\frac{p}{5 p+m}=0.198 \cdots \\
\Rightarrow 0.198(5 p+m)<p<0.199(5 p+m) \\
\Rightarrow 19.8 m<p<39.8 m .
\end{array}
$$
When $m=1$, $20 \leqslant p \leqslant 39$, taking $p=20, m=1$, $q$ is minimized... | 121 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given points $A(1,-1), B(4,0), C(2,2)$, the plane region $D$ consists of all points $P(x, y)$ that satisfy
$$
\overrightarrow{A P}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}(1<\lambda \leqslant a, 1<\mu \leqslant b)
$$
If the area of region $D$ is 8, then the minimum value of $a+b$ is . $\qquad$ | 8. 4 .
As shown in Figure 3, extend $A B$ to point $N$, and extend $A C$ to point $M$, such that
$$
|A N|=a|A B|,|A M|=b|A C| .
$$
Construct $\square A B E C$ and $\square A N G M$. Then quadrilateral $E H G F$ is a parallelogram.
From the conditions, the region $D$ composed of points $P(x, y)$ is the shaded area in... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. $A=\left[\frac{8}{9}\right]+\left[\frac{8^{2}}{9}\right]+\cdots+\left[\frac{8^{0114}}{9}\right]$ when divided by 63 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 9. 56 .
Notice that, for any positive integer $k$, $\frac{8^{2 k-1}}{9}$ and $\frac{8^{2 k}}{9}$ are not integers, and
$$
\frac{8^{2 k-1}}{9}+\frac{8^{2 k}}{9}=8^{2 k-1} \text {. }
$$
Therefore, for any positive integer $k$, we have
$$
\begin{array}{l}
{\left[\frac{8^{2 k-1}}{9}\right]+\left[\frac{8^{2 k}}{9}\right]=... | 56 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For any positive integer $n$, the function $f(n)$ is the sum of the digits (i.e., the digital sum) of $n^{2}+3 n+1$ in decimal notation. Question: Does there exist an integer $n$ such that
$$
f(n)=2013 \text { or } 2014 \text { or } 2015 \text { ? }
$$ | When $3 \mid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 1(\bmod 3) ;
$$
When $3 \nmid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 2(\bmod 3) \text {. }
$$
Thus, $3 \nmid f(n)$.
Since 312013, there does not exist an integer $n$ such that
$$
f(n)=2013 \text {. }
$$
If there exists an integer $n$ such that $f(n)=2014$, then by $2... | 2015 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. A certain meeting has 30 participants, each of whom knows at most five others; among any five people, at least two are not acquaintances. Find the largest positive integer $k$, such that in any group of 30 people satisfying the above conditions, there always exists a group of $k$ people, none of whom are acquaintanc... | 5, hence we have
$30-|X| \leqslant 5|X|$.
Therefore, $|X| \geqslant 5$.
If $|X|=5$, then by equation (1) the equality is achieved, which means the 25 edges are distributed among the five vertices in set $X$, i.e., the neighborhood of each vertex in set $X$ is a set of five points in set $V \backslash X$.
Since $|V \ba... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. A. The minimum value of the algebraic expression $\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}+9}$ is | $=、 6$. A. 13.
This problem can be transformed into finding the minimum value of the sum of distances from a point $(x, 0)$ on the $x$-axis to the points $(0,2)$ and $(12,3)$ in a Cartesian coordinate system.
Since the symmetric point of $(0,2)$ with respect to the $x$-axis is $(0,-2)$, the length of the line segment ... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Let real numbers $x_{1}, x_{2}, \cdots, x_{2014}$ satisfy
$$
\left|x_{1}\right|=99,\left|x_{n}\right|=\left|x_{n-1}+1\right| \text {, }
$$
where, $n=2,3, \cdots, 2014$. Find the minimum value of $x_{1}+x_{2}+\cdots+x_{2014}$. | 11. From the given, we have
$$
x_{n}^{2}=x_{n-1}^{2}+2 x_{n}+1(n=2,3, \cdots, 2014) \text {. }
$$
Adding the above 2013 equations, we get
$$
\begin{array}{l}
x_{2014}^{2}=x_{1}^{2}+2\left(x_{1}+x_{2}+\cdots+x_{2013}\right)+2013 \\
\Rightarrow 2\left(x_{1}+x_{2}+\cdots+x_{2014}\right) \\
\quad=x_{2014}^{2}+2 x_{2014}-2... | -5907 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Consider each permutation of $1,2, \cdots, 8$ as an eight-digit number. Then the number of eight-digit numbers that are multiples of 11 is $\qquad$ | 8. 4608 .
For each such eight-digit number $\overline{a_{1} a_{2} \cdots a_{8}}$, let
$$
A=\left\{a_{1}, a_{3}, a_{5}, a_{7}\right\}, B=\left\{a_{2}, a_{4}, a_{6}, a_{8}\right\} .
$$
Let $S(A)$ and $S(B)$ denote their digit sums, and assume $S(A) \geqslant S(B)$.
Then $S(A)+S(B)=36$.
Thus, $S(A)$ and $S(B)$ have the ... | 4608 | Combinatorics | 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) 2014
(B) 2015
(C) 2016
(D) 2017 | -1. D.
From the given, we have $a \geqslant b+1$.
Then $1014=a+b \geqslant 2 b+1$
$$
\begin{array}{l}
\Rightarrow b \leqslant 506.5 \Rightarrow b \leqslant 506 . \\
\text { Also } S=(a+b)+(c-b)+b \\
=1014+497+b=1511+b \\
\leqslant 1511+506=2017,
\end{array}
$$
Therefore, the maximum value of $S$ is 2017. | 2017 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given a set of data consisting of seven positive integers, the only mode is 6, and the median is 4. Then the minimum value of the sum of these seven positive integers is $\qquad$ | 2, 1.26.
Arrange these seven positive integers in ascending order, it is clear that the fourth number is 4. If 6 appears twice, then the other four numbers are $1, 2, 3, 5$, their sum is smaller, being 27; if 6 appears three times, the other three numbers are $1, 1, 2$, their sum is 26.
Therefore, the minimum sum of th... | 26 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given
$$
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{y^{2}+4}-2\right) \geqslant y>0 \text {. }
$$
Then the minimum value of $x+y$ is $\qquad$. | 2. 2 .
Notice,
$$
\begin{array}{l}
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}\right) \geqslant 1 \\
\Rightarrow 2 x+\sqrt{4 x^{2}+1} \\
\geqslant \frac{1}{\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}}=\sqrt{1+\frac{4}{y^{2}}}+\frac{2}{y} .
\end{array}
$$
When $x=\frac{1}{y}$,
$$
\left(2 x+\sq... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Given the sequence $\left\{a_{n}\right\}$ satisfies
$a_{1}=1, a_{2 n}=\left\{\begin{array}{ll}a_{n}, & n \text { is even; } \\ 2 a_{n}, & n \text { is odd, }\end{array}\right.$
$a_{2 n+1}=\left\{\begin{array}{ll}2 a_{n}+1, & n \text { is even; } \\ a_{n}, & n \text { is odd. }\end{array}\right.$
Find the number of posi... | Solve: From the given, we have
$$
\begin{array}{l}
a_{2014}=2 a_{1007}=2 a_{503}=2 a_{251}=2 a_{125} \\
=2\left(2 a_{62}+1\right)=2\left(4 a_{31}+1\right) \\
=2\left(4 a_{15}+1\right)=2\left(4 a_{7}+1\right) \\
=2\left(4 a_{3}+1\right)=2\left(4 a_{1}+1\right)=10 .
\end{array}
$$
In fact, $a_{n}$ can be obtained throug... | 320 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 2. If } \frac{y}{x}+\frac{x}{z}=a, \frac{z}{y}+\frac{y}{x}=b, \frac{x}{z}+\frac{z}{y}=c, \\ \text { then }(b+c-a)(c+a-b)(a+b-c)=\end{array}$ | 2.8.
Notice that,
$$
\begin{array}{l}
b+c-a=\left(\frac{z}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{y}\right)-\left(\frac{y}{x}+\frac{x}{z}\right) \\
=\frac{2 z}{y} .
\end{array}
$$
Similarly, $c+a-b=\frac{2 x}{z}, a+b-c=\frac{2 y}{x}$.
Therefore, $(b+c-a)(c+a-b)(a+b-c)$ $=\frac{2 z}{y} \cdot \frac{2 x}{z} \c... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. In the Cartesian coordinate plane $x O y$, the area of the region determined by the system of inequalities
$$
\left\{\begin{array}{l}
|x| \leqslant 2, \\
|y| \leqslant 2, \\
|| x|-| y|| \leqslant 1
\end{array}\right.
$$
is $\qquad$ | 6. 12 .
Obviously, the region is symmetric with respect to both the $x$-axis and the $y$-axis. Therefore, we can assume $x \geqslant 0, y \geqslant 0$.
From $\left\{\begin{array}{l}0 \leqslant x \leqslant 2, \\ 0 \leqslant y \leqslant 2, \\ -1 \leqslant x-y \leqslant 1\end{array}\right.$, draw the part of the region ... | 12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that Figure 1 is the graph of an even function $f(x)$, and Figure 2 is the graph of an odd function $g(x)$.
Let the number of real roots of the equations $f(f(x))=0, f(g(x))=0$,
$$
g(g(x))=0, g(f(x))=0
$$
be $a, b, c, d$ respectively. Then
$$
a+b+c+d=
$$
$\qquad$ | 3. 30 .
From the graph, we know that the range of the function $y=f(x)$ is $[-1,1]$, and the range of $y=g(x)$ is $[-2,2]$.
Notice that, the roots of the equation $f(x)=0$ are $0, x_{1}, x_{2}$, with $\left|x_{1}\right|=\left|x_{2}\right| \in(1,2)$;
The roots of the equation $g(x)=0$ are $0, x_{3}, x_{4}$, with $\le... | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the set $P=\{1,2, \cdots, 2014\}, A \cong P$. If any two numbers in set $A$ have a difference that is not a multiple of 99, and the sum of any two numbers is also not a multiple of 99, then the set $A$ can contain at most $\qquad$ elements. | 5.50.
Let the set
$$
B_{i}=\{99 \times 1+i, 99 \times 2+i, \cdots, 99 \times 20+i\} \text {, }
$$
where, $i=0,1, \cdots, 34$;
$$
B_{j}=\{99 \times 1+j, 99 \times 2+j, \cdots, 99 \times 19+j\},
$$
where, $j=35,36, \cdots, 98$.
Take any $a, b \in A$.
Since the difference between any two numbers in set $A$ is not a mul... | 50 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given
$$
f(x)=x^{2}-53 x+196+\left|x^{2}-53 x+196\right| \text {. }
$$
Then $f(20)+f(14)=$ $\qquad$ . | $$
\begin{array}{l}
\text { II.7.0. } \\
\text { Let } g(x)=x^{2}-53 x+196 \\
=(x-4)(x-49)
\end{array}
$$
Then when $x<4$ or $x>49$, $g(x)>0$;
when $4 \leqslant x \leqslant 49$, $g(x) \leqslant 0$.
Thus, when $x=4,5, \cdots, 49$,
$$
f(x)=g(x)+|g(x)|=g(x)-g(x)=0 \text {. }
$$
Therefore, $f(20)+f(14)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $0<x<\frac{\pi}{2}, \sin x-\cos x=\frac{\pi}{4}$. If $\tan x+\frac{1}{\tan x}$ can be expressed in the form $\frac{a}{b-\pi^{c}}$ ($a$, $b$, $c$ are positive integers), then $a+b+c=$ $\qquad$ . | 8. 50 .
Squaring both sides of $\sin x-\cos x=\frac{\pi}{4}$ and rearranging yields $\sin x \cdot \cos x=\frac{16-\pi^{2}}{32}$.
Therefore, $\tan x+\frac{1}{\tan x}=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$
$$
=\frac{1}{\sin x \cdot \cos x}=\frac{32}{16-\pi^{2}} \text {. }
$$
Thus, $a=32, b=16, c=2$.
Hence, $a+b+... | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { 4. If } x_{1}>x_{2}>x_{3}>x_{4}>0, \text { and the inequality } \\
\log _{\frac{x_{1}}{2}} 2014+\log _{\frac{x_{2}}{x_{3}}} 2014+\log _{x_{4} \frac{1}{4}} 2014 \\
\geqslant k \log _{\frac{x_{1}}{}} 2014
\end{array}
$$
always holds, then the maximum value of the real number $k$ is . $\qquad$ | 4.9.
From the given, we have
$$
\begin{array}{l}
\frac{\ln 2014}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln 2014}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln 2014}{\ln \frac{x_{3}}{x_{4}}} \\
\geqslant k \frac{\ln 2014}{\ln \frac{x_{1}}{x_{4}}}.
\end{array}
$$
Since $x_{1}>x_{2}>x_{3}>x_{4}>0$, we have
$$
\begin{array}{l}
\ln \frac{x... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
9. Let the positive integer $n$ satisfy $31 \mid\left(5^{n}+n\right)$. Then the minimum value of $n$ is $\qquad$ . | 9. 30 .
Given $5^{3} \equiv 1(\bmod 31)$, when $n=3 k\left(k \in \mathbf{Z}_{+}\right)$, $5^{n}+n \equiv 1+n \equiv 0(\bmod 31)$, at this time, the minimum value of $n$ is $n_{\text {min }}=30$;
When $n=3 k+1\left(k \in \mathbf{Z}_{+}\right)$,
$$
5^{n}+n \equiv 5+n \equiv 0(\bmod 31),
$$
at this time, the minimum val... | 30 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Given
$$
S_{n}=|n-1|+2|n-2|+\cdots+10|n-10| \text {, }
$$
where, $n \in \mathbf{Z}_{+}$. Then the minimum value of $S_{n}$ is $\qquad$ | 10. 112 .
From the problem, we know
$$
\begin{array}{l}
S_{n+1}-S_{n} \\
=|n|+2|n-1|+\cdots+10|n-9|- \\
{[|n-1|+2|n-2|+\cdots+10|n-10|] } \\
=|n|+|n-1|+\cdots+|n-9|-10|n-10| .
\end{array}
$$
When $n \geqslant 10$, $S_{n+1}-S_{n}>0$, thus, $S_{n}$ is monotonically increasing; $\square$
When $n=0$, $S_{1}-S_{0}>0$;
Wh... | 112 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $15 \mid \overline{\text { xaxax }}$, then the sum of all five-digit numbers $\overline{x a x a x}$ that satisfy the requirement is ( ).
(A) 50505
(B) 59595
(C) 110100
(D) 220200 | - 1. D.
Notice that, $15| \overline{\text { xaxax }} \Leftrightarrow\left\{\begin{array}{l}5 \mid \overline{\text { xaxax }}, \\ 3 \mid \overline{\text { xaxax }} .\end{array}\right.$
From $51 \overline{\text { xaxax }} \Rightarrow x=5$;
From $31 \overline{x a x a x} \Rightarrow 31(x+a+x+a+x)$
$$
\Rightarrow 3|2 a \Ri... | 220200 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
1. Let $a_{1}, a_{2}, \cdots, a_{2015}$ be a sequence of numbers taking values from $-1, 0, 1$, satisfying
$$
\sum_{i=1}^{2015} a_{i}=5 \text {, and } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040,
$$
where $\sum_{i=1}^{n} a_{i}$ denotes the sum of $a_{1}, a_{2}, \cdots, a_{n}$.
Then the number of 1's in this sequenc... | $=, 1.510$.
Let the number of -1's be $x$, and the number of 0's be $y$.
Then, from $\sum_{i=1}^{2015} a_{i}=5$, we know the number of 1's is $x+5$. Combining this with
$$
\begin{array}{c}
\text { the equation } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040, \text { we get } \\
\left\{\begin{array}{l}
x+(x+5)+y=2015, ... | 510 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If the integer $n$
satisfies
$$
(n+1)^{2} \mid\left(n^{2015}+1\right) \text {, }
$$
then the minimum value of $n$ is $\qquad$ | 3. -2016 .
From $n^{2015}+1=(n+1) \sum_{i=0}^{2014}(-1)^{i} n^{2014-i}$, we know
$$
\begin{array}{l}
(n+1)^{2} \mid\left(n^{2015}+1\right) \\
\left.\Leftrightarrow(n+1)\right|_{i=0} ^{2014}(-1)^{i} n^{2014-i} \\
\Leftrightarrow(n+1) \mid \sum_{i=0}^{2014}(-1)^{i}(-1)^{2014-i} \\
\Leftrightarrow(n+1) \mid \sum_{i=0}^{2... | -2016 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. If $x, y, z$ are real numbers, satisfying
$$
x+\frac{1}{y}=2 y+\frac{2}{z}=3 z+\frac{3}{x}=k \text{, and } x y z=3 \text{, }
$$
then $k=$ | 4. 4 .
Multiplying the three expressions yields
$$
\begin{array}{l}
k^{3}=6\left[x y z+\left(x+\frac{1}{y}\right)+\left(y+\frac{1}{z}\right)+\left(z+\frac{1}{x}\right)+\frac{1}{x y z}\right] \\
=6\left(3+k+\frac{k}{2}+\frac{k}{3}+\frac{1}{3}\right) \\
\Rightarrow k^{3}-11 k-20=0 \\
\Rightarrow(k-4)\left(k^{2}+4 k+5\ri... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) If $x, y \in [0,1]$, try to find the maximum value of
$$
x \sqrt{1-y} + y \sqrt{1-x}
$$ | $$
\begin{array}{l}
\text { I. Let } s=\sqrt{1-x}, t=\sqrt{1-y}, \\
z=x \sqrt{1-y}+y \sqrt{1-x} .
\end{array}
$$
Then $z=\left(1-s^{2}\right) t+\left(1-t^{2}\right) s=(s+t)(1-s t)$.
Also, $(1-s)(1-t)=1-(s+t)+s t \geqslant 0$, which means $s+t \leqslant 1+s t$,
thus $z \leqslant(1+s t)(1-s t)=1-s^{2} t^{2} \leqslant 1$... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. To color the eight vertices of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with four different colors, such that the two endpoints of the same edge have different colors, there are a total of coloring methods. | 8. 2652 .
First, color the four points $A, B, C, D$ above, with 84 coloring methods. Then consider the four points below, using the principle of inclusion-exclusion, the total number of methods is
$$
\begin{array}{l}
84\left\{84-C_{4}^{1}[3 \times(3+2 \times 2)+\right. \\
(3+2 \times 2)]+2\left(\frac{36}{84} \times 9+... | 2652 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? ${ }^{[5]}$
$(2012$, Girls' Mathematical Olympiad) | 【Analysis】Perform prime factorization $2012=2^{2} \times 503$.
First consider 503 I $\mathrm{C}_{2012}^{k}$.
By Corollary 3, we know that when and only when $k$ and $2012-k$ do not produce a carry when added in base 503, $\left(503, \mathrm{C}_{2012}^{k}\right)=1$.
Writing 2012 in base 503 gives $2012=(40)_{503}$.
Let ... | 1498 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a_{1}, a_{2}, \cdots, a_{2014}$ be a permutation of the positive integers $1,2, \cdots$, 2014. Denote
$$
S_{k}=a_{1}+a_{2}+\cdots+a_{k}(k=1,2, \cdots, 2014) \text {. }
$$
Then the maximum number of odd numbers in $S_{1}, S_{2}, \cdots, S_{2014}$ is $\qquad$ | 6.1511.
If $a_{i}(2 \leqslant i \leqslant 2014)$ is odd, then $S_{i}$ and $S_{i-1}$ have different parities. From $a_{2}$ to $a_{2014}$, there are at least $1007-1=1006$ odd numbers. Therefore, $S_{1}, S_{2}, \cdots, S_{2014}$ must change parity at least 1006 times.
Thus, $S_{1}, S_{2}, \cdots, S_{2014}$ must have at... | 1511 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Print 90000 five-digit numbers
$$
10000,10001, \cdots, 99999
$$
on cards, with one five-digit number on each card. Some cards (such as 19806, which reads 90861 when flipped) have numbers that can be read in two different ways, causing confusion. The number of cards that will not cause confusion is $\qquad$ cards. | 8. 88060 .
Among the ten digits $0 \sim 9$, the digits that can still represent numbers when inverted are $0, 1, 6, 8, 9$.
Since the first digit cannot be 0 and the last digit cannot be 0, the number of such five-digit numbers that can be read when inverted is $4 \times 5 \times 5 \times 5 \times 4=2000$.
Among these... | 88060 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. If a non-zero complex number $x$ satisfies $x+\frac{1}{x}=1$, then $x^{2014}+\frac{1}{x^{2014}}=$ $\qquad$ | $=, 7 .-1$. Therefore $x^{2014}+\frac{1}{x^{2014}}=2 \cos \frac{4 \pi}{3}=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $A$ be a set composed of any 100 distinct positive integers, and let
$$
B=\left\{\left.\frac{a}{b} \right\rvert\, a 、 b \in A \text { and } a \neq b\right\},
$$
$f(A)$ denotes the number of elements in set $B$. Then the sum of the maximum and minimum values of $f(A)$ is $\qquad$ . | 11. 10098.
From the problem, when the elements in set $B$ are pairwise coprime, $f(A)$ reaches its maximum value $\mathrm{A}_{100}^{2}=9900$; when the elements in set $B$ form a geometric sequence with a common ratio not equal to 1, $f(A)$ reaches its minimum value of $99 \times 2=198$.
Therefore, the sum of the maxim... | 10098 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. If $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{4028} x^{4028}$ is the expansion of $\left(x^{2}+x+2\right)^{2014}$, then
$$
2 a_{0}-a_{1}-a_{2}+2 a_{3}-a_{4}-a_{5}+\cdots+2 a_{4026}-a_{4007}-a_{4028}
$$
is $\qquad$ | 6.2 .
Let $x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$.
Then $x^{2}+x+2=1$.
Therefore, $a_{0}+a_{1} \omega+a_{2} \omega^{2}+a_{3}+a_{4} \omega+\cdots+$ $a_{4026}+a_{4027} \omega+a_{4028} \omega^{2}=1$.
Taking the conjugate of the above equation, we get
$$
\begin{array}{l}
a_{0}+a_{1} \omega... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let
$$
\begin{array}{l}
A=\{1,2, \cdots, 2014\}, \\
B_{i}=\left\{x_{i}, y_{i}\right\}(i=1,2, \cdots, t)
\end{array}
$$
be $t$ pairwise disjoint binary subsets of $A$, and satisfy the conditions
$$
\begin{array}{l}
x_{i}+y_{i} \leqslant 2014(i=1,2, \cdots, t), \\
x_{i}+y_{i} \neq x_{j}+y_{j}(1 \leqslant i<j \leqslan... | 8. 805 .
On the one hand, for any $1 \leqslant i<j \leqslant t$, we have
$$
\left\{x_{i}, y_{i}\right\} \cap\left\{x_{j}, y_{j}\right\}=\varnothing \text {. }
$$
Thus, $x_{1}, x_{2}, \cdots, x_{t}, y_{1}, y_{2}, \cdots, y_{t}$ are $2 t$ distinct integers. Therefore,
$$
\sum_{i=1}^{t}\left(x_{i}+y_{i}\right) \geqslant... | 805 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. A line $l$ is drawn through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ intersecting the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$ such that there are exactly three lines $l$, then $\lambda=$ $\qquad$. | 2.4.
Since there are an odd number of lines that satisfy the condition, by symmetry, the line perpendicular to the $x$-axis satisfies the condition, at this time,
$$
x=\sqrt{3}, y= \pm 2, \lambda=|A B|=4 \text {. }
$$ | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the set $A=\{1,2,3\}, f$ and $g$ are functions from set $A$ to $A$. Then the number of function pairs $(f, g)$ whose image sets intersect at the empty set is $\qquad$ . | 4. 42 .
When the image set of function $f$ is 1 element, if the image set of function $f$ is $\{1\}$, at this time the image set of function $g$ is a subset of $\{2,3\}$, there are $2^{3}=8$ kinds, so there are $3 \times 8=24$ pairs of functions $(f, g)$ that meet the requirements.
When the image set of function $f$ ... | 42 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $f(x)=\left(x^{2}+3 x+2\right)^{\cos \pi x}$. Then the sum of all $n$ that satisfy the equation
$$
\left|\sum_{k=1}^{n} \log _{10} f(k)\right|=1
$$
is | 5. 21 .
It is known that for integer $x$, we have
$$
f(x)=[(x+1)(x+2)]^{(-1)^{x}} \text {. }
$$
Thus, when $n$ is odd,
$$
\sum_{k=1}^{n} \log _{10} f(k)=-\log _{10} 2-\log _{10}(n+2) \text {; }
$$
When $n$ is even,
$$
\sum_{k=1}^{n} \log _{10} f(k)=-\log _{10} 2+\log _{10}(n+2) \text {. }
$$
Therefore, the $n$ that... | 21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given real numbers $a_{0}, a_{1}, \cdots, a_{2015}$, $b_{0}, b_{1}, \cdots, b_{2011}$ satisfy
$$
\begin{array}{l}
a_{n}=\frac{1}{65} \sqrt{2 n+2}+a_{n-1}, \\
b_{n}=\frac{1}{1009} \sqrt{2 n+2}-b_{n-1},
\end{array}
$$
where $n=1,2, \cdots, 2015$.
If $a_{0}=b_{2015}$, and $b_{0}=a_{2015}$, find the value ... | 10. Notice that, for any $k=1,2 \cdots, 2015$, we have
$$
\begin{array}{l}
a_{k}-a_{k-1}=\frac{1}{65} \sqrt{2 k+2}, \\
b_{k}+b_{k-1}=\frac{1}{1009} \sqrt{2 k+2} .
\end{array}
$$
Multiplying the two equations, we get
$$
\begin{array}{l}
a_{k} b_{k}-a_{k-1} b_{k-1}+a_{k} b_{k-1}-a_{k-1} b_{k} \\
=\frac{2 k+2}{65 \times ... | 62 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. In an All-Star basketball game, 27 players participate, each wearing a jersey with their favorite number, which is a non-negative integer. After the game, they line up in a 3-row, 9-column formation for fans to take photos. An eccentric fan only takes photos where the players in the frame form a rectangle of $a$ row... | Prompt: $s_{\text {min }}=2$.
Assume at most one player is photographed.
By symmetry, without loss of generality, assume no players in the first row are photographed.
For $i=1,2, \cdots, 9$, let the players in the $i$-th column of the 1st, 2nd, and 3rd rows have jersey numbers $a_{i}, b_{i}, c_{i}$, respectively, and ... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given
\[
\begin{aligned}
f(x, y)= & x^{3}+y^{3}+x^{2} y+x y^{2}- \\
& 3\left(x^{2}+y^{2}+x y\right)+3(x+y),
\end{aligned}
\]
and \( x, y \geqslant \frac{1}{2} \). Find the minimum value of \( f(x, y) \). ${ }^{[4]}$
(2011, Hebei Province High School Mathematics Competition) | 【Analysis】This problem involves finding the extremum of a bivariate function, with a very complex expression, making it difficult to find a breakthrough. Observing the symmetry in the expression, we can attempt the following transformation.
First, when $x \neq y$, multiply both sides of the function by $x-y$, yielding... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given the sequence $\left\{a_{n}\right\}$:
$$
a_{1}=2, a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}(n \geqslant 1) \text {. }
$$
Determine the periodicity of the sequence $\left\{a_{n}\right\}$. | Solving, from $a_{1}=2$, we get
$$
a_{2}=\frac{5 a_{1}-13}{3 a_{1}-7}=3, a_{3}=\frac{5 a_{2}-13}{3 a_{2}-7}=1,
$$
which means $a_{1}, a_{2}, a_{3}$ are all distinct.
From the given conditions, we have
$$
\begin{array}{l}
a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}, \\
a_{n+2}=\frac{5 a_{n+1}-13}{3 a_{n+1}-7}=\frac{7 a_{n}-13... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a b c \neq 0, a+b+c=a^{2}+b^{2}+c^{2}=2 . \\
\text { Then } \frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}=
\end{array}
$$ | 7.3.
From the given, we have $a b+b c+c a=1$.
Then $b c=1-a b-a c=1-a(b+c)$ $=1-a(2-a)=(1-a)^{2}$.
Thus, the desired value is 3. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. A positive integer that can be expressed as the difference of squares of two positive integers is called a "wise number". For example, $9=5^{2}-4^{2}, 9$ is a wise number.
(1) Try to determine which numbers among the positive integers are wise numbers, and explain the reason;
(2) In the sequence of wise numbers arr... | 14. (1) It is easy to know that the positive integer 1 cannot be expressed as the difference of squares of two positive integers, i.e., 1 is not a wise number.
For odd numbers greater than 1, we have
$$
2 k+1=(k+1)^{2}-k^{2}(k=1,2, \cdots),
$$
which means that all odd numbers greater than 1 are wise numbers.
When $k=2... | 2689 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
16. The basketball league has eight teams. Each season, each team plays two games (home and away) against each of the other teams in the league, and each team also plays four games against opponents outside the league. Therefore, in one season, the eight teams in the league play a total of ( ) games.
(A) 60
(B) 88
(C) ... | 16. B.
There are $\mathrm{C}_{8}^{2}=28$ pairs that can be formed from the eight teams in the league.
Since each pair of teams plays both a home and an away match, the eight teams in the league play a total of $28 \times 2=56$ matches.
Since each team plays four matches against opponents outside the league, the eight... | 88 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
14. Let the angle between vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ be $\frac{\pi}{3}$, the angle between vectors $\boldsymbol{c}-\boldsymbol{a}$ and $\boldsymbol{c}-\boldsymbol{b}$ be $\frac{2 \pi}{3}$, $|\boldsymbol{a}-\boldsymbol{b}|=5$, and $|\boldsymbol{c}-\boldsymbol{a}|=2 \sqrt{3}$. Then the maximum value of... | 14. 24 .
Let $\overrightarrow{O A}=a, \overrightarrow{O B}=b, \overrightarrow{O C}=c$. Then
$$
|\overrightarrow{A C}|=|c-a|=2 \sqrt{3},|\overrightarrow{A B}|=|a-b|=5 \text {. }
$$
Also, $\angle A O B=\frac{\pi}{3}, \angle A C B=\frac{2 \pi}{3}$, at this time, $O, A, C, B$ are concyclic.
By the Law of Sines, we get
$$... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Allocate 24 volunteer slots to 3 schools. Then the number of allocation methods where each school gets at least one slot and the number of slots for each school is different is $\qquad$ kinds. ${ }^{[2]}$ | Let the quotas allocated to schools A, B, and C be $x$, $y$, and $z$ respectively.
First, without considering that $x, y, z$ are pairwise distinct.
From $x+y+z=24$, we get a total number of combinations $\mathrm{C}_{23}^{2}$.
Next, we separate out the number of positive integer solutions $(x, y, z)$ that are not pairwi... | 222 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. The sum of the ages of three people, A, B, and C, represented by $x, y, z$ is 120, and $x, y, z \in (20,60)$. Then the number of ordered triples $(x, y, z)$ is $\qquad$ | 8. 1141 .
Notice a basic conclusion:
The number of positive integer solutions $(a, b, c)$ to the indeterminate equation $a+b+c=n$ is $\mathrm{C}_{n-1}^{2}$.
Thus, the number of solutions to the indeterminate equation
$$
(x-20)+(y-20)+(z-20)=60
$$
satisfying $x, y, z>20$ is $\mathrm{C}_{59}^{2}$.
Among these $\mathrm{... | 1141 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The sequence $\left\{a_{n}\right\}$ has 9 terms, where $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in\left\{2,1,-\frac{1}{2}\right\}$. Find the number of such sequences. | Prompt: Categorized Count. Answer: 491. | 491 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
435 Given a convex polygon $F$, consider all the figures that are positively homothetic to the convex polygon $F$ and smaller than $F$. Let $n(F)$ be the minimum number of such figures (allowing translation but not rotation) needed to cover the convex polygon $F$. Find the value of $n(F)$. | (1) On the one hand, let the convex polygon $F$ be the parallelogram $ABCD$. It is easy to see that the convex polygon $F$ can be covered by four smaller parallelograms that are similar to the convex polygon $F$ (as shown in Figure 3).
On the other hand, for any smaller parallelogram $F_1$ that is similar to the conve... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If numbers $1,2, \cdots, 14$ are taken in ascending order as $a_{1}, a_{2}, a_{3}$, such that $a_{2}-a_{1} \geqslant 3$, and $a_{3}-a_{2} \geqslant 3$, find the number of different ways to choose them. | From the given information, we have
$$
\begin{aligned}
a_{1} & \leqslant a_{2}-3 \leqslant a_{3}-6 \\
& \Rightarrow 1 \leqslant a_{1}<a_{2}-2<a_{3}-4 \leqslant 10 .
\end{aligned}
$$
Substitute $\left(a_{1}, a_{2}-2, a_{3}-4\right)=(x, y, z)$.
Then the number of tuples $\left(a_{1}, a_{2}, a_{3}\right)$ equals the numb... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Add a “+” or “-” in front of each number in $1,2, \cdots, 1989$. Find the minimum non-negative algebraic sum, and write down the equation. | First, prove that the algebraic sum is odd.
Consider the simplest case: all filled with " + ", at this time,
$$
1+2+\cdots+1989=995 \times 1989
$$
is odd.
For the general case, it is only necessary to adjust some " + " to " - ".
Since $a+b$ and $a-b$ have the same parity, each adjustment does not change the parity of... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\sum_{c y c} \frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}} \leqslant 8 \text {. }
$$ | To prove that since the inequality to be proved is homogeneous with respect to $a, b, c$, we may assume without loss of generality that $a+b+c=3$.
The original inequality is then $\sum_{\text {cyc }} \frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}} \leqslant 8$.
Consider the function $f(x)=\frac{(x+3)^{2}}{2 x^{2}+(3-x)^{2}}(0<x<3)... | 8 | Inequalities | proof | Yes | Yes | cn_contest | false |
8. Given real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
\left(2 x^{2}+8 x+11\right)\left(y^{2}-10 y+29\right)\left(3 z^{2}-18 z+32\right) \\
\leqslant 60 .
\end{array}
$$
Then $x+y-z=$ . $\qquad$ | 8. 0 .
Original expression
$$
\begin{array}{l}
=\left[2(x+2)^{2}+3\right]\left[(y-5)^{2}+4\right]\left[3(z-3)^{2}+5\right] \\
\leqslant 60 \\
\Rightarrow x=-2, y=5, z=3 \\
\Rightarrow x+y-z=0 .
\end{array}
$$ | 0 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $S_{n}=1+2+\cdots+n$. Then among $S_{1}, S_{2}$, $\cdots, S_{2015}$, there are. $\qquad$ that are multiples of 2015. | 10.8 .
Obviously, $S_{n}=\frac{1}{2} n(n+1)$.
For any positive divisor $d$ of 2015, it is easy to see that in the range $1 \sim 2015$, there is exactly one $n$ that satisfies $n$ being a multiple of $d$ and $n+1$ being a multiple of $\frac{2015}{d}$. Therefore, each divisor of 2015 will generate one $n$ such that $S_{... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 For any non-empty subset $X$ of the set $M=\{1,2, \cdots, 1000\}$, let $\alpha_{X}$ denote the sum of the maximum and minimum numbers in $X$. Find the arithmetic mean of all such $\alpha_{X}$. | 【Analysis】According to the problem, the required average is
$$
f=\frac{\sum_{\varnothing \neq X \subseteq M} \alpha_{X}}{2^{1000}-1} \text {. }
$$
The key to solving this is to calculate the sum in the numerator
$$
N=\sum_{\varnothing \neq X \subseteq M} \alpha_{X} \text {. }
$$
To compute such an "unordered sum", on... | 1001 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Divide a circle with a circumference of 24 into 24 equal segments, and select eight points from the 24 points, such that the arc length between any two points is not equal to 3 and 8. Find the number of different ways to select such a group of eight points. | 【Analysis】First analyze the essential structure of the data: Label 24 points in a clockwise direction as $1,2, \cdots, 24$. Then arrange them into the following $3 \times 8$ number table:
$$
\left(\begin{array}{cccccccc}
1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
17 & 20 & 23 & 2 & 5 & ... | 258 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Fill two $a$s and two $b$s into the 16 cells shown in Figure 3, with at most one letter per cell. If the same letters must not be in the same row or column, find the number of different ways to fill the cells. | Prompt: Categorized Count. Answer: 3960. | 3960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a positive number $x$ satisfies
$$
x^{10}+x^{5}+\frac{1}{x^{5}}+\frac{1}{x^{10}}=15250 \text {. }
$$
then the value of $x+\frac{1}{x}$ is | 2.3.
Let $a=x^{5}+\frac{1}{x^{5}}$. Then $x^{10}+\frac{1}{x^{10}}=a^{2}-2$.
Thus, the original equation becomes
$$
\begin{array}{l}
a^{2}+a-15252=0 \\
\Rightarrow(a-123)(a+124)=0
\end{array}
$$
$\Rightarrow a=123$ or $a=-124$ (discard).
Therefore, $x^{5}+\frac{1}{x^{5}}=123$.
Let $x+\frac{1}{x}=b>0$. Then
$$
\begin{ar... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If a positive integer has eight positive divisors, and the sum of these eight positive divisors is 3240, then this positive integer is called a "good number". For example, 2006 is a good number, because the sum of its positive divisors $1, 2, 17, 34, 59, 118, 1003, 2006$ is 3240. Find the smallest good number... | Let $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, where $\alpha_{i} \in \mathbf{Z}_{+}, p_{i} (i=1,2, \cdots, k)$ are prime numbers, and $p_{1}<p_{2}<\cdots<p_{k}$.
From $\tau(n)=\prod_{i=1}^{k}\left(1+\alpha_{i}\right)=8=2^{3}$, we know $k=1,2,3$.
(1) If $k=1$, then
$\alpha_{1}=7, n=p^{7}$ ($p$ is a prime number), $\sum_{i=... | 1614 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Question 3 Let positive real numbers $a, b, c$ satisfy $ab + bc + ca = 48$. Try to find the minimum value of $f = \left(a^{2} + 5\right)\left(b^{2} + 5\right)\left(c^{2} + 5\right)$. | Given $A=48, k=5$, hence
$$
f \geqslant 5(48-5)^{2}=9245 \text {. }
$$
When $a=5,\{b, c\}=\left\{\frac{43+\sqrt{97}}{12}, \frac{43-\sqrt{97}}{12}\right\}$,
the equality holds.
Therefore, $f_{\min }=9245$. | 9245 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 1, in the right $\triangle A B C$, it is known that $\angle A C B$ $=90^{\circ}, A C=21, B C=$ 28, and a square $A B D E$ is constructed outward with $A B$ as one side. The angle bisector of $\angle A C B$ intersects $D E$ at point $F$. Then the length of line segment $D F$ is $\qquad$ | 8. 15 .
As shown in Figure 6, let $CF$ intersect $AB$ and $AD$ at points $G$ and $O$ respectively.
Then $\angle BCO = 45^{\circ}$
$= \angle OAB$.
Thus, $O, A, C, B$
are concyclic.
Hence $\angle ABO$
$$
= \angle OCA = 45^{\circ}.
$$
Therefore, $OA = OB$.
Thus, $O$ is the center of the square $ABDE$.
By symmetry, $DF ... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $d$ be a positive divisor of 2015. Then the maximum value of the unit digit of $d^{\frac{2011}{d}}$ is $\qquad$ . | 9.7.
Notice that, $2015=5 \times 13 \times 31$.
Therefore, 2015 has eight positive divisors.
Let $G(n)$ denote the unit digit of $n$, then
$$
\begin{array}{l}
G\left(1^{2015}\right)=G\left(31^{165}\right)=1, \\
G\left(2015^{1}\right)=G\left(5^{403}\right)=G\left(65^{31}\right) \\
=G\left(155^{13}\right)=5 .
\end{array... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Given that $a$ and $b$ are real numbers, the system of inequalities about $x$
$$
\left\{\begin{array}{l}
20 x+a>0, \\
15 x-b \leqslant 0
\end{array}\right.
$$
has only the integer solutions $2, 3, 4$. Then the maximum value of $ab$ is . $\qquad$ | 10. -1200 .
From the conditions, we know that $-\frac{a}{20}<x \leqslant \frac{b}{15}$.
The integer solutions of the inequality system are only $2, 3, 4$, so $1 \leqslant-\frac{a}{20}<2,4 \leqslant \frac{b}{15}<5$, which means $-40<a \leqslant-20,60 \leqslant b<75$.
Therefore, $-3000<a b \leqslant-1200$.
Thus, when $a... | -1200 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
16. (25 points) Given $A \subseteq\{1,2, \cdots, 2014\}$, let real numbers $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3} 、 x_{1} 、 x_{2} 、 x_{3}$ satisfy
(i) $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3} \in\{-1,0,1\}$ and not all are 0;
(ii) $x_{1}, x_{2} 、 x_{3} \in A$;
(iii) If $x_{i}=x_{j}$, then $\lambda_{i} \lambda_{j} \ne... | 16. (1) Construct a good set $A$ with 503 elements.
Let $A=\{1,3,5, \cdots, 1005\}$.
If $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3}$ are all non-zero, then
$$
\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3} \equiv x_{1}+x_{2}+x_{3} \equiv 1(\bmod 2) \text {. }
$$
Thus, $\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_... | 503 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given sets
$$
A=\{1,3,5,7,9\}, B=\{2,4,6,8,10\} \text {. }
$$
If set $C=\{x \mid x=a+b, a \in A, b \in B\}$, then the number of elements in set $C$ is $\qquad$ . | ,- 1.9 .
Since $a$ is odd and $b$ is even, all elements in set $C$ are odd.
Also, the minimum value of $a+b$ is 3, and the maximum value is 19, and all odd numbers between 3 and 19 can be obtained, thus, set $C$ contains 9 elements. | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the function
$$
f(x)=\left\{\begin{array}{ll}
0, & x<0, \\
1, & x \geqslant 0 .
\end{array}\right.
$$
Then $f(f(x))=$ $\qquad$ | 2. 1 .
Since for any $x \in \mathbf{R}$, we have $f(x) \geqslant 0$, therefore, $f(f(x))=1$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given
$$
\sin \alpha+\sqrt{3} \sin \beta=1, \cos \alpha+\sqrt{3} \cos \beta=\sqrt{3} \text {. }
$$
Then the value of $\cos (\alpha-\beta)$ is $\qquad$ . | 3. 0 .
Squaring and adding the two known equations, we get
$$
\begin{array}{l}
4+2 \sqrt{3}(\sin \alpha \cdot \sin \beta+\cos \alpha \cdot \cos \beta)=4 \\
\Rightarrow \cos (\alpha-\beta)=0
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. If positive integers $m, n$ satisfy $\frac{(m+n)!}{n!}=5040$, then the value of $m!n$ is $\qquad$ . | 9. 144 .
$$
\begin{array}{l}
\text { Given } \frac{(m+n)!}{n!} \\
=(m+n)(m+n-1) \cdots(n+1), \\
5040=10 \times 9 \times 8 \times 7,
\end{array}
$$
we know $\left\{\begin{array}{l}m+n=10, \\ n+1=7\end{array} \Rightarrow\left\{\begin{array}{l}m=4, \\ n=6 .\end{array}\right.\right.$
Therefore, $m!n=144$. | 144 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. Let the monotonic increasing sequence $\left\{a_{n}\right\}$ consist of positive integers, and $a_{7}=120, a_{n+2}=a_{n}+a_{n+1}\left(n \in \mathbf{Z}_{+}\right)$. Then $a_{8}=$ . $\qquad$ | 10. 194 .
From $a_{n+2}=a_{n}+a_{n+1}$, we get $a_{7}=5 a_{1}+8 a_{2}=120, a_{8}=8 a_{1}+13 a_{2}$. Since $(5,8)=1$, and $a_{1}, a_{2}$ are both positive integers, it follows that $8\left|a_{1}, 5\right| a_{2}$.
Let $a_{1}=8 k, a_{2}=5 m\left(k, m \in \mathbf{Z}_{+}\right)$.
Then $k+m=3$.
Also, $a_{1}<a_{2}$, so, $k=1... | 194 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $n$ be the smallest positive integer satisfying the following conditions:
(1) $n$ is a multiple of 75;
(2) $n$ has exactly 75 positive divisors (including 1 and itself).
Find $\frac{n}{75}$.
(Eighth American Mathematical Invitational) | ```
Given: $n=75 k=3 \times 5^{2} k$.
\[
\begin{array}{l}
\text { By } 75=3 \times 5 \times 5 \\
=(2+1)(4+1)(4+1),
\end{array}
\]
we know the number of prime factors is at most three.
By discussing the number of prime factors, we can solve to get
\[
\left(\frac{n}{75}\right)_{\min }=432 \text {. }
\]
``` | 432 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Choose three different angles from $1^{\circ}, 2^{\circ}, \cdots, 179^{\circ}$ to form the three interior angles of a triangle. There are $\qquad$ different ways to do this. | 3.2611.
Notice that, the equation $x+y+z=180$ has $\mathrm{C}_{179}^{2}=$ 15931 sets of positive integer solutions, among which, the solution where $x=y=z$ is 1 set, and the solutions where exactly two of $x, y, z$ are equal are $88 \times 3=$ 264 sets.
Therefore, the number of selection methods is $\frac{15931-1-264}... | 2611 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $f_{0}(x)=|x|-2015$,
$$
f_{n}(x)=\left|f_{n-1}(x)\right|-1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Then the number of zeros of the function $y=f_{2015}(x)$ is
$\qquad$ | 8.4031 .
From the graph, it is easy to see that the function $y=f_{1}(x)$ has 4 zeros, the function $y=f_{2}(x)$ has 6 zeros, $\cdots \cdots$ and so on, the function $y=f_{2014}(x)$ has 4030 zeros. However, the intersection point of the function $y=f_{2014}(x)$ with the y-axis is $(0,1)$, therefore, the function $y=f_... | 4031 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. For a sequence of real numbers $x_{1}, x_{2}, \cdots, x_{n}$, define its "value" as $\max _{1 \leqslant i \leqslant n}\left\{\left|x_{1}+x_{2}+\cdots+x_{i}\right|\right\}$. Given $n$ real numbers, David and George want to arrange these $n$ numbers into a sequence with low value. On one hand, diligent David examines ... | 3. $c=2$.
If the initial numbers given are $1, -1, 2, -2$, then David arranges these four numbers as $1, -2, 2, -1$, and George can get the sequence $1, -1, 2, -2$.
Thus, $D=1, G=2$.
Therefore, $c \geqslant 2$.
We now prove: $G \leqslant 2 D$.
Let the $n$ real numbers be $x_{1}, x_{2}, \cdots, x_{n}$, and assume David... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. There are infinitely many cards, each with a real number written on it. For each real number $x$, there is exactly one card with the number $x$ written on it. Two players each select a set of 100 cards, denoted as $A$ and $B$, such that the sets are disjoint. Formulate a rule to determine which of the two players wi... | 6. There are 100 ways.
To prove the more general case, where each person selects $n$ cards, there are $n$ ways that satisfy the conditions. Let $A>B$ or $B>A$ if and only if $a_{k}>b_{k}$.
Such rules satisfy all three conditions, and different $k$ correspond to different rules, hence there are at least $n$ different r... | 100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. If the set $S=\{1,2,3, \cdots, 16\}$ is arbitrarily divided into $n$ subsets, then there must exist a subset in which there are elements $a, b, c$ (which can be the same), satisfying $a+b=c$. Find the maximum value of $n$.
【Note】If the subsets $A_{1}, A_{2}, \cdots, A_{n}$ of set $S$ satisfy the following condition... | 4. First, when $n=3$, assume there exists a partition of the set that does not satisfy the condition. Thus, there must be a subset with at least six elements, without loss of generality, let
$$
A=\left\{x_{1}, x_{2}, \cdots, x_{6}\right\}\left(x_{1}<x_{2}<\cdots<x_{6}\right) \text {. }
$$
Then $x_{6}-x_{1}, x_{6}-x_{2... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. A shape obtained by removing one unit square from a $2 \times 2$ grid is called an "L-shape". In an $8 \times 8$ grid, place $k$ L-shapes, each of which can be rotated, but each L-shape must cover exactly three unit squares of the grid, and any two L-shapes must not overlap. Additionally, no other L-shape can be pla... | 6. On one hand, using six straight lines to divide the large square grid into 16 smaller $2 \times 2$ grids, each $2 \times 2$ grid must have at least two cells covered by an L-shape. Therefore, the number of cells covered is no less than 32. Thus, the number of L-shapes is no less than $\left[\frac{32}{3}\right]+1$ $=... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Given odd prime numbers $x, y, z$ satisfying
$$
x \mid \left(y^{5}+1\right), y \mid \left(z^{5}+1\right), z \mid \left(x^{5}+1\right) \text {. }
$$
Find the minimum value of the product $x y z$. (Cheng Chuanping, problem contributor) | 7. Let's assume $x$ is the minimum of $x, y, z$.
(1) If $x=3$, then
$$
3^{5}+1=244=2^{2} \times 61 \Rightarrow z=61.
$$
Since $3 \mid (y^{5}+1)$, we have $y \equiv -1 \pmod{3}$.
Clearly, $5 \nmid (61^{5}+1)$.
After calculation, we find $11 \mid (61^{5}+1)$.
Thus, $y_{\text{min}}=11$.
Therefore, $(x y z)_{\min}=3 \time... | 2013 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. A coin collector has 100 coins that look the same. The collector knows that 30 of them are genuine, and 70 are fake, and that all genuine coins weigh the same, while all fake coins weigh different and are heavier than the genuine ones. The collector has a balance scale that can be used to compare the weight of two g... | 8. First, it is clear that 70 weighings can certainly find at least one genuine coin.
In fact, each time, one coin is placed on each side of the balance. If they weigh the same, both are genuine; if they do not, the heavier one must be a counterfeit. Thus, each weighing either finds two genuine coins or one counterfei... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
1. The function $f(x)$ defined on $\mathbf{R}$ satisfies $f\left(\frac{2 a+b}{3}\right)=\frac{2 f(a)+f(b)}{3}(a, b \in \mathbf{R})$,
and $f(1)=1, f(4)=7$.
Then $f(2015)=$ | $$
-, 1.4029 .
$$
From the function $f(x)$ being concave or convex in reverse on $\mathbf{R}$, we know its graph is a straight line, $f(x)=a x+b$.
Given $f(1)=1$ and $f(4)=7$, we have
$$
\begin{array}{l}
a=2, b=-1 \Rightarrow f(x)=2 x-1 \\
\Rightarrow f(2015)=4029 .
\end{array}
$$ | 4029 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. For any point $A(x, y)$ in the plane region $D$:
$$
\left\{\begin{array}{l}
x+y \leqslant 1, \\
2 x-y \geqslant-1, \\
x-2 y \leqslant 1
\end{array}\right.
$$
and a fixed point $B(a, b)$, both satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B} \leqslant 1$. Then the maximum value of $a+b$ is $\qquad$ | 2. 2 .
From the problem, we know that for any $A(x, y) \in D$, we have $a x+b y \leqslant 1$.
By taking
$$
(x, y)=(1,0),(0,1),(-1,-1) \text {, }
$$
we get that the fixed point $B(a, b)$ satisfies the necessary conditions
$$
\left\{\begin{array}{l}
a \leqslant 1, \\
b \leqslant 1, \\
a+b \geqslant-1
\end{array} \quad ... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Define the function on $\mathbf{R}$
$$
f(x)=\left\{\begin{array}{ll}
\log _{2}(1-x), & x \leqslant 0 ; \\
f(x-1)-f(x-2), & x>0 .
\end{array}\right.
$$
Then $f(2014)=$ | 3. 1.
First, consider the method for finding the function values on the interval $(0,+\infty)$.
If $x>3$, then
$$
\begin{array}{l}
f(x)=f(x-1)-f(x-2) \\
=f(x-1)-(f(x-1)+f(x-3)) \\
=-f(x-3) .
\end{array}
$$
Thus, for $x>6$, we have
$$
f(x)=-f(x-3)=f(x-6) \text {. }
$$
Therefore, $f(x)$ is a periodic function with a p... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. The set $X \backslash Y=\{a \mid a \in X, a \notin Y\}$ is called the difference set of set $X$ and set $Y$. Define the symmetric difference of sets $A$ and $B$ as
$$
A \Delta B=(A \backslash B) \cup(B \backslash A) \text {. }
$$
If two non-empty finite sets $S$ and $T$ satisfy $|S \Delta T|=1$, then the minimum va... | 6. 3 .
If $|S \Delta T|=1$, without loss of generality, assume
$$
\left\{\begin{array}{l}
|S \backslash T|=1, \\
|T \backslash S|=0,
\end{array}\right.
$$
i.e., $T$ is a proper subset of $S$, thus,
$$
\begin{array}{l}
|S|>|T| \geqslant 1 . \\
\text { Hence } k \geqslant|T|+1+|T| \geqslant 3 .
\end{array}
$$
$$
\text ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. In an $m$ row by 10 column grid, fill each cell with either 0 or 1, such that each column contains exactly three 1s. Let the sum of the numbers in the $i$-th row ($i=1,2, \cdots, m$) be denoted as $x_{i}$, and for any two columns, there always exists a row where the cells intersecting with these two columns are both... | 8. 5 .
Construct an application model: "10 people go to a bookstore to buy $m$ kinds of books, each person buys 3 books, and any two of them buy at least one book in common. Find the sales volume of the book that is purchased by the most people."
Let the $i(i=1,2, \cdots, m)$-th kind of book be purchased by $x_{i}$ p... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a, b$ be the roots of the quadratic equation $x^{2}-x-1=0$. Then the value of $3 a^{2}+4 b+\frac{2}{a^{2}}$ is | From the given, we have $a b=-1, a+b=1$,
$$
\begin{array}{l}
a^{2}=a+1, b^{2}=b+1 . \\
\text { Therefore, } 3 a^{3}+4 b+\frac{2}{a^{2}}=3 a^{3}+4 b+2 b^{2} \\
=3 a^{2}+3 a+6 b+2=6(a+b)+5=11 .
\end{array}
$$ | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Fill the numbers $1,2, \cdots, 36$ in a $6 \times 6$ grid, with each cell containing one number, such that the numbers in each row are in increasing order from left to right. Then the minimum value of the sum of the six numbers in the third column is $\qquad$ | 4. 63.
Let the six numbers filled in the third column, arranged in ascending order, be $A, B, C, D, E, F$.
Since the row where $A$ is located needs to fill in two numbers smaller than $A$ before it, then $A \geqslant 3$; since the row where $B$ is located needs to fill in two numbers smaller than $B$, and $A$ and the... | 63 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a$ and $b$ be distinct real numbers. If the quadratic function $f(x)=x^{2}+a x+b$ satisfies $f(a)=f(b)$, then the value of $f(2)$ is $\qquad$ . | $$
-, 1.4
$$
From the given conditions and the symmetry of the quadratic function graph, we get
$$
\begin{array}{l}
\frac{a+b}{2}=-\frac{a}{2} \Rightarrow 2 a+b=0 \\
\Rightarrow f(2)=4+2 a+b=4 .
\end{array}
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If the real number $\alpha$ satisfies $\cos \alpha=\tan \alpha$, then $\frac{1}{\sin \alpha}+\cos ^{4} \alpha=$ $\qquad$ | 2. 2 .
From the condition, we know that $\cos ^{2} \alpha=\sin \alpha$.
$$
\begin{array}{l}
\text { Then } \frac{1}{\sin \alpha}+\cos ^{4} \alpha=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\sin \alpha}+\sin ^{2} \alpha \\
=(1+\sin \alpha)+\left(1-\cos ^{2} \alpha\right) \\
=2+\sin \alpha-\cos ^{2} \alpha=2 .
\end{array}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
In the plane rectangular coordinate system $x O y$, the set of points
$$
\begin{aligned}
K= & \{(x, y) \mid(|x|+|3 y|-6) . \\
& (|3 x|+|y|-6) \leqslant 0\}
\end{aligned}
$$
corresponds to a plane region whose area is $\qquad$ | 6. 24 .
Let $K_{1}=\{(x, y)|| x|+| 3 y |-6 \leqslant 0\}$.
First, consider the part of the point set $K_{1}$ in the first quadrant, at this time, $x+3 y \leqslant 6$. Therefore, these points correspond to $\triangle O C D$ and its interior in Figure 2.
By symmetry, the region corresponding to the point set $K_{1}$ is... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. For a four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b 、 c 、 d \leqslant$ $9)$, if $a>b, b<c, c>d$, then $\overline{a b c d}$ is called a $P$ class number; if $a<c, c<d$, then $\overline{a b c d}$ is called a $Q$ class number. Let $N(P)$ and $N(Q)$ represent the number of $P$ class numbe... | 8. 285.
Let the sets of all P-type numbers and Q-type numbers be denoted as $A$ and $B$, respectively. Let the set of all P-type numbers whose unit digit is zero be denoted as $A_{0}$, and the set of all P-type numbers whose unit digit is not zero be denoted as $A_{1}$.
For any four-digit number $\overline{a b c d} \... | 285 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ be an arithmetic sequence, and
$$
\sum_{i=1}^{n}\left|a_{i}+j\right|=2028(j=0,1,2,3) \text {. }
$$
Then the maximum value of the number of terms $n$ is | 6. 52.
Since the equation $|x|=|x+1|=|x+2|$ has no solution, we have $n \geqslant 2$ and the common difference is not 0.
Assume the general term of the sequence is $a-k d(1 \leqslant k \leqslant n, d>0)$.
Construct the function $f(x)=\sum_{k=1}^{n}|x-k d|$.
The given condition is equivalent to $f(x)=2028$ having at le... | 52 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $\{x\}=x-[x]$. Then the sum of the squares of all natural numbers $m$ that satisfy
$$
[(2 m+1)\{\sqrt{2 m+1}\}]=m
$$
is $\qquad$ . | 8.0.
Let $2 m+1=n$. Then $2[n\{\sqrt{n}\}]=n-1$. Note that,
$$
\begin{array}{l}
2 n\{\sqrt{n}\}-21$ when, by $n=2[n\{\sqrt{n}\}]+1$ being odd, we know $n \geqslant 3$.
Thus, $\left|4 n-(2[\sqrt{n}]+1)^{2}\right| \leqslant \frac{4}{\sqrt{n}}+\frac{1}{n}<3$.
Therefore, $4 n-(2[\sqrt{n}]+1)^{2}= \pm 1$.
This leads to
$$
... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example: $10 f(x)$ is a continuous function defined on the interval $[0,2015]$, and $f(0)=f(2015)$. Find the minimum number of real number pairs $(x, y)$ that satisfy the following conditions:
(1) $f(x)=f(y)$;
(2) $x-y \in \mathbf{Z}_{+}$.
$(2015$, Peking University Mathematics Summer Camp) | 【Analysis】Consider a continuous function $f(x)$ on the interval $[0, n]$ that satisfies $f(0)=f(n)$. Let the minimum number of real number pairs $(x, y)$ that satisfy (1) and (2) be $F(n)$.
Obviously, $F(1)=1, F(2)=2$.
We will prove: $F(n)=n\left(n \in \mathbf{Z}_{+}\right)$.
Assume that for $n \leqslant k-1$, equation... | 2015 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 In $\triangle ABC$ with a fixed perimeter, it is known that $AB=6$, and when vertex $C$ is at a fixed point $P$, $\cos C$ has a minimum value of $\frac{7}{25}$.
(1) Establish an appropriate coordinate system and find the equation of the locus of vertex $C$;
(2) Draw a line through point $A$ intersecting the c... | (1) Establish a Cartesian coordinate system with the line $AB$ as the $x$-axis and the perpendicular bisector of segment $AB$ as the $y$-axis.
Let $|CA| + |CB| = 2a (a > 3)$ be a constant. Then the locus of point $C$ is an ellipse with foci at $A$ and $B$.
Thus, the focal distance $2c = |AB| = 6$.
Notice,
$$
\begin{ar... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Color each cell of a $12 \times 12$ grid either black or white, such that any $3 \times 4$ or $4 \times 3$ subgrid contains at least one black cell. Find the minimum number of black cells.
（Liang Yingde provided the problem） | 3. The minimum number of black cells required is $n=12$.
First, prove that $n \geqslant 12$.
Since a $12 \times 12$ grid can be divided into $\frac{12 \times 12}{3 \times 4}=12$ non-overlapping $3 \times 4$ sub-grids (excluding the boundaries), according to the problem statement, there must be at least 12 black cells.... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. How many different right-angled triangles with integer side lengths have an area that is 999 times their perimeter (considering congruent triangles as the same)? (Provided by Lin Chang)
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result d... | 5. Let the three sides of a right-angled triangle be \(a, b, c\) (with \(c\) being the hypotenuse).
From the Pythagorean triple formula, we have
\[
a = k \cdot 2uv, \quad b = k(u^2 - v^2), \quad c = k(u^2 + v^2),
\]
where the greatest common divisor of the three sides \(k\) is a positive integer, \(u\) and \(v\) are co... | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the 20 vertices of a regular 20-sided polygon inscribed in the unit circle in the complex plane correspond to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$. Then the number of distinct points corresponding to $z_{1}^{2015}, z_{2}^{2015}, \cdots, z_{20}^{2015}$ is $\qquad$ | 2. 4 .
Assume the vertices corresponding to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$ are arranged in a counterclockwise direction.
Let $z_{1}=\mathrm{e}^{\mathrm{i} \theta}$. Then
$$
\begin{array}{l}
z_{k+1}=\mathrm{e}^{\mathrm{i}\left(\theta+\frac{k}{10}\right)}(k=0,1, \cdots, 19), \\
z_{k+1}^{2015}=\mathrm... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. The function $f(x)$ defined on $\mathbf{R}$, for any real number $x$, satisfies
$$
\begin{array}{l}
f(x+3) \leqslant f(x)+3, \\
f(x+2) \geqslant f(x)+2,
\end{array}
$$
and $f(1)=2$. Let $a_{n}=f(n)\left(n \in \mathbf{Z}_{+}\right)$, then
$$
f(2015)=
$$
$\qquad$ | 5.2016.
Notice,
$$
\begin{array}{l}
f(x)+3 \geqslant f(x+3) \\
=f(x+1+2) \geqslant f(x+1)+2 \\
\Rightarrow f(x)+1 \geqslant f(x+1) . \\
\text { Also } f(x)+4 \leqslant f(x+2)+2 \leqslant f(x+4) \\
=f(x+1+3) \leqslant f(x+1)+3 \\
\Rightarrow f(x+1) \geqslant f(x)+1 .
\end{array}
$$
Therefore, $f(x+1)=f(x)+1$.
Thus, $f... | 2016 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=n^{4}+6 n^{3}+11 n^{2}+6 n \text {. }
$$
Then the sum of the first 12 terms $S_{12}=$ $\qquad$ | 7. 104832 .
Notice that,
$$
\begin{array}{l}
a_{n}=n^{4}+6 n^{3}+11 n^{2}+6 n \\
=n(n+1)(n+2)(n+3) . \\
\text { Let } f(n)=\frac{1}{5} n(n+1)(n+2)(n+3)(n+4) .
\end{array}
$$
Then $a_{n}=f(n)-f(n-1), S_{n}=f(n)$.
Therefore, $S_{12}=f(12)=104832$. | 104832 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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