problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
3. In $\triangle A B C$, it is known that $\angle A=2 \angle B, C D$ is the angle bisector of $\angle C$, $A C=16, A D=8$. Then $B C=$ | 3. 24 .
From $\angle A=2 \angle B$, we know $B C>A C$.
As shown in Figure 5, take a point $E$ on side $B C$ such that $E C=A C$,
and connect $D E$.
Then $\triangle C E D \cong \triangle C A D$
$$
\begin{array}{l}
\Rightarrow E D=A D, \\
\angle C E D=\angle C A D . \\
\text { Hence } \angle B D E=\angle C E D-\angle D... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given prime numbers $p$ and $q$, such that $p^{3}-q^{5}=(p+q)^{2}$. Then $\frac{8\left(p^{2013}-p^{2010} q^{5}\right)}{p^{2011}-p^{2009} q^{2}}=$ $\qquad$. | 4. 140.
If $p$ and $q$ have the same remainder when divided by 3, and since $p$ and $q$ are both prime numbers:
If the remainder is 0 when divided by 3, then $p=q=3$, at this time,
$$
p^{3}-q^{5}=0 \Rightarrow 3^{3}>q^{5} \Rightarrow q^{5}<27,
$$
such a prime number $q$ does not exist.
Therefore, it can only be $q=3$... | 140 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=1$, and $a_{2 n}=a_{n}, a_{2 n+1}=a_{n}+1\left(n \in \mathbf{Z}_{+}\right)$.
Then $a_{2013}=$ . $\qquad$ | 2.9.
From the problem, we know
$$
\begin{array}{l}
a_{2013}=a_{1006}+1=a_{503}+1=a_{251}+2 \\
=a_{125}+3=a_{62}+4=a_{31}+4=a_{15}+5 \\
=a_{7}+6=a_{3}+7=a_{1}+8=9 .
\end{array}
$$ | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. From the set $A=\{1,2, \cdots, 30\}$, select five different numbers such that these five numbers form an arithmetic sequence. The number of different arithmetic sequences obtained is $\qquad$ . | 7. 196.
Let the five numbers be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$, with a common difference of $d$.
From $a_{5}-a_{1}=4d$, we know $a_{1} \equiv a_{5}(\bmod 4)$.
Divide the set $A=\{1,2, \cdots, 30\}$ into four categories based on the remainder modulo 4:
$$
\begin{array}{l}
B=\{1,5,9, \cdots, 29\}, \\
C=\{2,6,10, \c... | 196 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $M$ be a moving point on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. Given points $F(1,0)$ and $P(3,1)$. Then the maximum value of $2|M F|-|M P|$ is $\qquad$. | 2.1.
Notice that $F$ is the right focus of the ellipse, and the right directrix of the ellipse is $l: x=4$. Then $2|M F|$ is the distance from point $M$ to $l$.
Draw a perpendicular line $M A$ from point $M$ to $l$, and draw a perpendicular line $P B$ from point $P$ to $M A$, where $A$ and $B$ are the feet of the per... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $x, y \in \mathbf{R}$, and $x^{2}+y^{2} \leqslant 1$. Then the maximum value of $x+y-x y$ is $\qquad$ . | 3. 1 .
Notice that, $x+y-x y=x(1-y)+y$.
When $y$ is fixed, the expression can only achieve its maximum value when $x$ is as large as possible; similarly, when $x$ is fixed, $y$ should also be as large as possible. Therefore, we might as well assume that $x$ and $y$ are non-negative, and $x^{2}+y^{2}=1$.
Let $x+y=t(t \... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $x_{n}=\sum_{k=1}^{2013}\left(\cos \frac{k!\pi}{2013}\right)^{n}$. Then $\lim _{n \rightarrow \infty} x_{n}=$ | 4. 1953.
Notice that, $2013=3 \times 11 \times 61$.
Therefore, when $1 \leqslant k \leqslant 60$,
$$
\frac{k!}{2013} \notin \mathbf{Z}, \cos \frac{k!\pi}{2013} \in(-1,1).
$$
Hence $\lim _{n \rightarrow \infty} \cos ^{n} \frac{k!\pi}{2013}=0$.
And when $k \geqslant 61$, $\frac{k!}{2013}$ is an integer, and always an e... | 1953 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. If the sum of all elements in a non-empty subset of $\{1,2, \cdots, 9\}$ is a multiple of 3, the subset is called a "Jin state subset". Then the number of such Jin state subsets is $\qquad$ . | 5. 175.
If a proper subset is a peculiar subset, then its complement is also a peculiar subset. Therefore, we only need to consider peculiar subsets with fewer than or equal to 4 elements.
Thus, there are 3 peculiar subsets with only 1 element; there are $\mathrm{C}_{3}^{2}+3 \times 3=12$ peculiar subsets with exactl... | 175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. The function $f: \mathbf{N} \rightarrow \mathbf{N}$, such that for all $n \in \mathbf{N}$, we have
$$
\begin{array}{c}
f(f(n))+f(n)=2 n+3 \text {, and } f(0)=1 . \\
\text { Then } \frac{f(6) f(7) f(8) f(9) f(10)}{f(1)+f(2)+f(3)+f(4)+f(5)}=
\end{array}
$$ | 3. 2772 .
Substituting $n=0$ into equation (1) yields
$$
f(1)+1=3 \Rightarrow f(1)=2 \text {. }
$$
Furthermore, let $n=1$.
From equation (1), we get $f(2)=3$.
Similarly, $f(n)=n+1(n=3,4, \cdots, 10)$.
$$
\text { Hence } \frac{f(6) f(7) f(8) f(9) f(10)}{f(1)+f(2)+f(3)+f(4)+f(5)}=2772 \text {. }
$$ | 2772 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the function
$$
f(x)=A \cos \left(\omega x+\frac{\pi}{4} \omega\right)(A>0)
$$
is decreasing on $\left(0, \frac{\pi}{8}\right)$. Then the maximum value of $\omega$ is | Ni,6.8.
Assume $\omega>0$.
To make $f(x)$ a decreasing function in $\left(0, \frac{\pi}{8}\right)$, combining the image of the cosine-type function, we must have
$$
\begin{array}{l}
\frac{T}{2} \geqslant \frac{\pi}{8} \Rightarrow \frac{\pi}{\omega} \geqslant \frac{\pi}{8} \Rightarrow \omega \leqslant 8 . \\
\text { Whe... | 8 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
1. For the real number $x$, the functions are
$$
f(x)=\sqrt{3 x^{2}+7}, g(x)=x^{2}+\frac{16}{x^{2}+1}-1,
$$
then the minimum value of the function $g(f(x))$ is . $\qquad$ | $-, 1.8$.
From the problem, we have
$$
g(f(x))=3 x^{2}+7+\frac{16}{3 x^{2}+8}-1.
$$
Let $t=3 x^{2}+8(t \geqslant 8)$. Then
$$
h(t)=g(f(x))=t+\frac{16}{t}-2 \text{. }
$$
It is easy to see that $h(t)$ is a monotonically increasing function on the interval $[8,+\infty)$. Therefore, $h(t) \geqslant h(8)=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. If $x, y$ are two different real numbers, and
$$
x^{2}=2 x+1, y^{2}=2 y+1 \text {, }
$$
then $x^{6}+y^{6}=$ $\qquad$ . | 4. 198 .
$$
\text { Let } S_{n}=x^{n}+y^{n} \text {. }
$$
From $x^{2}=2 x+1, y^{2}=2 y+1$, we get
$$
x^{n+2}=2 x^{n+1}+x^{n}, y^{n+2}=2 y^{n+1}+y^{n} \text {. }
$$
Thus, $S_{n+2}=2 S_{n+1}+S_{n}$.
Also, $S_{1}=2, S_{2}=6$, so
$$
S_{3}=14, S_{4}=34, S_{5}=82, S_{6}=198
$$ | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $a_{1}, a_{2}, \cdots, a_{10}$ and $b_{1}, b_{2}, \cdots, b_{10}$ are 20 distinct real numbers. If the equation
$$
\begin{array}{l}
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{10}\right| \\
=\left|x-b_{1}\right|+\left|x-b_{2}\right|+\cdots+\left|x-b_{10}\right|
\end{array}
$$
has a finite numb... | 7.9.
$$
\text { Let } \begin{aligned}
f(x)= & \left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{10}\right|- \\
& \left|x-b_{1}\right|-\left|x-b_{2}\right|-\cdots-\left|x-b_{10}\right| .
\end{aligned}
$$
Thus, by the problem statement, $f(x)=0$.
Let $c_{1}<c_{2}<\cdots<c_{20}$ be the elements of the set
$$
\l... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. If the remainder of $\underbrace{11 \cdots 1}_{n+1 \uparrow} 1$ divided by 3102 is 1, then the smallest positive integer $n$ is $\qquad$ . | 8. 138 .
Notice that, $3102=2 \times 3 \times 11 \times 47$.
From $\underbrace{11 \cdots 1}_{n+1 \uparrow}=3102 k+1(k \in \mathbf{Z})$, we know $\underbrace{11 \cdots 10}_{n \uparrow}=3102 k$.
Thus, $\underbrace{11 \cdots 10}_{n \uparrow}$ is divisible by $2, 3, 11, 47$.
(1) For any positive integer $n$, obviously, $\... | 138 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
II. (16 points) Given the sequence $\left\{F_{n}\right\}$ satisfies
$$
\begin{array}{l}
F_{1}=F_{2}=1, \\
F_{n+2}=F_{n+1}+F_{n}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$
If $F_{a} γ F_{b} γ F_{c} γ F_{d}(a<b<c<d)$ are the side lengths of a convex quadrilateral, find the value of $d-b$. | From the given, we know that $F_{a}+F_{b}+F_{c}>F_{d}$.
If $c \leqslant d-2$, then
$$
F_{a}+\left(F_{b}+F_{c}\right) \leqslant F_{a}+F_{d-1} \leqslant F_{d},
$$
which is a contradiction.
Therefore, $c=d-1$.
Thus, the side lengths of the quadrilateral are $F_{a} γ F_{b} γ F_{d-1} γ F_{d}$.
If $b \leqslant d-3$, then
$$... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Place 11 identical balls into six distinct boxes so that at most three boxes are empty. The number of ways to do this is $\qquad$. | 10. 4212 .
If there are no empty boxes, there are $\mathrm{C}_{10}^{5}$ ways to place them; if there is one empty box, there are $\mathrm{C}_{6}^{1} \mathrm{C}_{10}^{4}$ ways to place them; if there are two empty boxes, there are $\mathrm{C}_{6}^{2} \mathrm{C}_{10}^{3}$ ways to place them; if there are three empty box... | 4212 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
13. Let the equation $x^{2}-m x-1=0$ have two real roots $\alpha, \beta (\alpha<\beta)$, and the function $f(x)=\frac{2 x-m}{x^{2}+1}$.
(1) Find the value of $\alpha f(\alpha)+\beta f(\beta)$;
(2) Determine the monotonicity of $f(x)$ in the interval $(\alpha, \beta)$, and provide a proof;
(3) If $\lambda, \mu$ are both... | Three, 13. (1) Given that $\alpha, \beta$ are the roots of the equation $x^{2}-m x-1=0$, we know
$$
\begin{array}{l}
\alpha+\beta=m, \alpha \beta=-1 . \\
\text { Then } f(\alpha)=\frac{2 \alpha-m}{\alpha^{2}+1}=\frac{2 \alpha-(\alpha+\beta)}{\alpha^{2}-\alpha \beta} \\
=\frac{\alpha-\beta}{\alpha(\alpha-\beta)}=\frac{1... | 2 | Algebra | proof | Yes | Yes | cn_contest | false |
Example 4: There are 2013 balls placed around a circle, numbered $1, 2, \cdots, 2013$ in a clockwise direction. Starting from a certain ball $a$, and then taking every other ball in a clockwise direction, this process continues until only one ball remains on the circle. For what value of the number of ball $a$ will the... | First, consider a special case.
When the ball-picking procedure starts with the ball numbered 1, then when the person picking the balls reaches the next ball to be taken (numbered 3), it is essentially as if the ball numbered 2 has been placed at the farthest point in the direction of his movement. And when he takes th... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. For a positive integer $n$, let $1 \times 2 \times \cdots \times n=n$!. If $M=1!\times 2!\times \cdots \times 10!$, then the number of perfect cubes among the positive divisors of $M$ is ( ).
(A) 468
(B) 684
(C) 846
(D) 648 | 2. A.
Notice that,
$$
\begin{array}{l}
M=1! \times 2! \times \cdots \times 10! \\
=2^{9} \times 3^{8} \times 4^{7} \times 5^{6} \times 6^{5} \times 7^{4} \times 8^{3} \times 9^{2} \times 10 \\
=2^{9+2 \times 7+5+3 \times 3+1} \times 3^{8+5+2 \times 2} \times 5^{6+1} \times 7^{4} \\
=2^{38} \times 3^{17} \times 5^{7} \... | 468 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
2. The number of all positive integer solutions $(x, y, z)$ for the equation $x y+1000 z=2012$ is. $\qquad$ | 2. 18 .
First consider the positive integer solutions for $x \leqslant y$.
When $z=1$, $x y=1012=4 \times 253=2^{2} \times 11 \times 23$. Hence, $(x, y)=(1,1012),(2,506),(4,253)$,
$$
(11,92),(22,46),(23,44) \text {. }
$$
When $z=2$, $x y=12=2^{2} \times 3$. Hence, $(x, y)=(1,12),(2,6),(3,4)$.
Next, consider the posit... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Xiao Li and Xiao Zhang are running at a constant speed on a circular track. They start from the same place at the same time. Xiao Li runs clockwise and completes a lap every 72 seconds, while Xiao Zhang runs counterclockwise and completes a lap every 80 seconds. At the start, Xiao Li has a relay baton, and each time... | 3. 720 .
The time taken from the start to the first meeting is
$$
\frac{1}{\frac{1}{72}+\frac{1}{80}}=\frac{720}{19}
$$
seconds, during which Xiao Li runs $\frac{10}{19}$ laps, and Xiao Zhang runs $\frac{9}{19}$ laps.
Divide the circular track into 19 equal parts, and number the points clockwise from the starting poin... | 720 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: At each vertex of a regular 2009-gon, a non-negative integer not exceeding 100 is placed. Adding 1 to the numbers at two adjacent vertices is called an operation on these two adjacent vertices. For any given two adjacent vertices, the operation can be performed at most $k$ times. Find the minimum value of $k... | Solve $k_{\min }=100400$.
Let the numbers at vertices $A_{1}, A_{2}, \cdots, A_{2009}$ be $a_{1}, a_{2}, \cdots, a_{2009}$ respectively.
First, take $a_{2}=a_{4}=a_{2008}=100, a_{1}=a_{3}=a_{2009}=0$.
For each operation, assign
$$
S=\left(a_{2}-a_{3}\right)+\left(a_{4}-a_{5}\right)+\cdots+\left(a_{2008}-a_{2009}\right)... | 100400 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given that $n$ is a positive integer. If in any permutation of $1,2, \cdots, n$, there always exists a sum of six consecutive numbers greater than 2013, then $n$ is called a "sufficient number". Find the smallest sufficient number.
---
Please note that the term "ζΈ©ι₯±ζ°" is translated as "sufficient nu... | Three, the smallest well-fed number is 671.
First, prove: 671 is a well-fed number.
For any permutation $b_{1}, b_{2}, \cdots, b_{671}$ of $1,2, \cdots, 671$, divide it into 112 groups, where each group contains six numbers, and the last two groups share one number, i.e.,
$$
\begin{array}{l}
A_{1}=\left(b_{1}, b_{2}, \... | 671 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. From $1,2, \cdots, 100$ choose three different numbers such that they cannot form the three sides of a triangle. The number of different ways to do this is. | 2. 82075.
Let these three numbers be $i, j, k(1 \leqslant i<j<k \leqslant 100)$. Then $k \geqslant i+j$.
Therefore, the number of different ways to choose these three numbers is
$$
\begin{array}{l}
\sum_{i=1}^{100}\left[\sum_{j=i+1}^{100}\left(\sum_{k=i+j}^{100} 1\right)\right]=\sum_{i=1}^{100}\left[\sum_{j=i+1}^{100-... | 82075 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the set
$$
A=\left\{x \mid x=a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}\right\} \text {, }
$$
where, $a_{i} \in\{1,2, \cdots, 6\}(i=0,1,2,3)$. If positive integers $m, n \in A$, and $m+n=2014(m>n)$, then the number of pairs of positive integers $(m, n)$ that satisfy the condition is $\qquad$.
| 3. 551.
Notice that, $2014=5 \times 7^{3}+6 \times 7^{2}+5$.
Given $m, n \in A$, let
$$
\begin{array}{l}
m=a \times 7^{3}+b \times 7^{2}+c \times 7+d, \\
n=a^{\prime} \times 7^{3}+b^{\prime} \times 7^{2}+d^{\prime},
\end{array}
$$
where $a, b, c, d, a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime} \in\{1,2, \cdots, 6\}... | 551 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { II. (40 points) Given } x_{i} \in\{\sqrt{2}-1, \sqrt{2}+1\} \\
(i=1,2, \cdots, 2013) \text {, let } \\
S=x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{2012} x_{2013} .
\end{array}
$$
Find the number of different integer values that $S$ can take. | Because $(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$,
$$
(\sqrt{2}+1)^{2}=3+2 \sqrt{2}, (\sqrt{2}-1)(\sqrt{2}+1)=1,
$$
Therefore, $x_{i} x_{i+1} \in\{3-2 \sqrt{2}, 3+2 \sqrt{2}, 1\}$.
Let the sum $S$ contain $a$ instances of $3+2 \sqrt{2}$, $b$ instances of $3-2 \sqrt{2}$, and $c$ instances of 1. Then $a, b, c \in \mathbf{N}$, and... | 1005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the set $X=\{1,2, \cdots, 100\}$, and the function $f: X \rightarrow X$ satisfies the following conditions:
(1) For any $x \in X$, $f(x) \neq x$;
(2) For any 40-element subset $A$ of $X$, $A \cap f(A) \neq \varnothing$.
Find the smallest positive integer $k$ such that for any function $f$ satisfying the above c... | 5. First consider the function $f: X \rightarrow X$ defined as follows:
For $i=1,2, \cdots, 30, j=91,92, \cdots, 99$, define
$f(3 i-2)=3 i-1, f(3 i-1)=3 i$,
$f(3 i)=3 i-2, f(j)=100, f(100)=99$.
Clearly, the function $f$ satisfies condition (1).
For any 40-element subset $A$ of the set $X$, either there exists an integ... | 69 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. For the convex pentagon $A B C D E$, the side lengths are sequentially $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. It is known that a quadratic trinomial in $x$ satisfies:
When $x=a_{1}$ and $x=a_{2}+a_{3}+a_{4}+a_{5}$, the value of the quadratic trinomial is 5;
When $x=a_{1}+a_{2}$, the value of the quadratic trinomial i... | 4. 0 .
Let the quadratic trinomial be
$$
f(x)=a x^{2}+b x+c(a \neq 0) \text {, }
$$
and let the axis of symmetry of its graph be $x=x_{0}$.
By the problem, we know
$$
f\left(a_{1}\right)=f\left(a_{2}+a_{3}+a_{4}+a_{5}\right)=5,
$$
and $\square$
$$
\begin{array}{l}
a_{1} \neq a_{2}+a_{3}+a_{4}+a_{5} . \\
\text { Henc... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. A three-digit number that is divisible by 35 and whose digits sum to 15 is | 5. 735.
Since the sum of the digits is 15, the three-digit number is a multiple of 3. Also, since the three-digit number is a multiple of 35, this three-digit number is a multiple of $35 \times 3=105$.
Starting from 105, list the three-digit multiples of 105:
$$
105,210,315,420,525,630,735,840, 945 \text {, }
$$
Amon... | 735 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Find the smallest integer \( n (n > 1) \), such that there exist \( n \) integers \( a_{1}, a_{2}, \cdots, a_{n} \) (allowing repetition) satisfying
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2013 .
$$
12. (20 points) Let positive integers \( a, b, c, d \) satisfy
$$
a^{2}=c(d+13), b^{2}=c(d-1... | 11. Since $a_{1} a_{2} \cdots a_{n}=2013$, it follows that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. From $a_{1}+a_{2}+\cdots+a_{n}=2013$ being odd, we know that $n$ is odd (otherwise, the sum of an even number of odd numbers should be even, which is a contradiction).
If $n=3$, then $a_{1}+a_{2}+a_{3}=a_{1} a_... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Write all positive integers in ascending order in a row. Then the 2013th digit from left to right is | $-1.7$
All single-digit numbers occupy 9 positions, all two-digit numbers occupy $2 \times 90=180$ positions, and next come the three-digit numbers in sequence. Since $2013-9-180=1824$, and $\frac{1824}{3}=608$, because $608+99=707$, the 2013th digit is the last digit of the three-digit number 707, which is 7. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $a, b, c$ are distinct positive integers. If the set
$$
\{a+b, b+c, c+a\}=\left\{n^{2},(n+1)^{2},(n+2)^{2}\right\} \text {, }
$$
where $n \in \mathbf{Z}_{+}$. Then the minimum value of $a^{2}+b^{2}+c^{2}$ is | 6. 1297 .
From $n^{2}+(n+1)^{2}+(n+2)^{2}=2(a+b+c)$ being even, we know that among $n, n+1, n+2$, there are two odd numbers and one even number, which means $n$ is odd.
Obviously, $n>1$.
Without loss of generality, let $a<b<c$.
If $n=3$, then
$$
\begin{array}{l}
a+b=9, a+c=16, b+c=25 \\
\Rightarrow a+b+c=25 .
\end{arr... | 1297 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. The minimum value of the function $f(x)=\sum_{k=1}^{2013}|x-k|$ is | 7.1013042.
Notice that, the median of $1,2, \cdots, 2013$ is 1007.
Therefore, when $x=1007$, the function reaches its minimum value
$$
\begin{array}{l}
f(1007)=2(1+2+\cdots+1006) \\
=1006 \times 1007=1013042
\end{array}
$$ | 1013042 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If three non-zero and distinct real numbers $a, b, c$ satisfy $\frac{1}{a}+\frac{1}{b}=\frac{2}{c}$, then $a, b, c$ are called "harmonic"; if they satisfy $a+c=2b$, then $a, b, c$ are called "arithmetic".
Given the set $M=\{x|| x | \leqslant 2013, x \in \mathbf{Z}\}$, the set $P$ is a three-element subset of set $M... | 3.1006.
If $a, b, c$ are both harmonic and arithmetic, then
$$
\left\{\begin{array}{l}
\frac{1}{a} + \frac{1}{b} = \frac{2}{c}, \\
a + c = 2b
\end{array} \Rightarrow \left\{\begin{array}{l}
a = -2b \\
c = 4b
\end{array}\right.\right.
$$
Thus, the good set is of the form $\{-2b, b, 4b\}(b \neq 0)$.
Since the good set ... | 1006 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $x, y$ satisfy $x y+1=4 x+y$, and $x>1$. Then the minimum value of $(x+1)(y+2)$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4.27.
From the problem, we know $y=\frac{4 x-1}{x-1}$.
$$
\begin{array}{l}
\text { Therefore, }(x+1)(y+2)=(x+1)\left(\frac{4 x-1}{x-1}+2\right) \\
\quad= \frac{3(x+1)(2 x-1)}{x-1} .
\end{array}
$$
Let $x-1=t>0$. Then
$$
\begin{array}{l}
(x+1)(y+2)=\frac{3(t+2)(2 t+1)}{t} \\
=6\left(t+\frac{1}{t}\right)+15 \geqslant ... | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
7. Equation
$$
\sin \pi x=\left[\frac{x}{2}-\left[\frac{x}{2}\right]+\frac{1}{2}\right]
$$
The sum of all real roots of the equation in the interval $[0,2 \pi]$ is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 7.12. Let $\left\{\frac{x}{2}\right\}=\frac{x}{2}-\left[\frac{x}{2}\right]$. Then for any real number $x$, we have $0 \leqslant\left\{\frac{x}{2}\right\}<1$.
Thus, the original equation becomes
$$
\sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right] \text {. }
$$
(1) If $0 \leqslant\left\{\frac{x}{2}\right\}<... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $f(x)$ is an increasing function on $\mathbf{R}$, and for any $x \in$ $\mathbf{R}$, we have
$$
f\left(f(x)-3^{x}\right)=4 .
$$
Then $f(2)=$ | 8. 10 .
From the problem, we know that $f(x)-3^{x}$ is a constant. Let's assume $f(x)-3^{x}=m$.
Then $f(m)=4, f(x)=3^{x}+m$.
Therefore, $3^{m}+m=4 \Rightarrow 3^{m}+m-4=0$.
It is easy to see that the equation $3^{m}+m-4=0$ has a unique solution $m=1$.
Thus, $f(x)=3^{x}+1$,
and hence, $f(2)=3^{2}+1=10$. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that the elements of set $A$ are all integers, the smallest is 1, and the largest is 200, and except for 1, every number in $A$ is equal to the sum of two numbers (which may be the same) in $A$. Then the minimum value of $|A|$ is $\qquad$ ( $|A|$ represents the number of elements in set $A$). | 9. 10 .
It is easy to know that the set
$$
A=\{1,2,3,5,10,20,40,80,160,200\}
$$
meets the requirements, at this time, $|A|=10$.
Next, we will show that $|A|=9$ does not meet the requirements.
Assume the set
$$
A=\left\{1, x_{1}, x_{2}, \cdots, x_{7}, 200\right\},
$$
where $x_{1}<x_{2}<\cdots<x_{7}$ meets the require... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
15. Given the set
$$
P=\left\{x \mid x=7^{3}+a \times 7^{2}+b \times 7+c, a γ b γ c\right. \text { are positive integers not }
$$
exceeding 6 $\}$.
If $x_{1}, x_{2}, \cdots, x_{n}$ are $n$ elements in set $P$ that form an arithmetic sequence, find the maximum value of $n$. | 15. (1) Clearly,
$$
\begin{array}{l}
7^{3}+7^{2}+7+1, 7^{3}+7^{2}+7+2, \\
7^{3}+7^{2}+7+3, 7^{3}+7^{2}+7+4, \\
7^{3}+7^{2}+7+5, 7^{3}+7^{2}+7+6
\end{array}
$$
These six numbers are in the set $P$ and form an arithmetic sequence.
(2) Prove by contradiction: Any seven different numbers in the set $P$ cannot form an arit... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. Given a convex quadrilateral $ABCD$ whose four interior angles form an arithmetic sequence. In $\triangle ABD$ and $\triangle BCD$, if $\angle ABD = \angle CDB$, $\angle DAB = \angle CBD$, and the three interior angles of these two triangles also form an arithmetic sequence, then the maximum possible value of the s... | 13. D.
Assume the four interior angles of the quadrilateral are
$$
\alpha, \alpha+d, \alpha+2d, \alpha+3d \text{.}
$$
Since the sum of the interior angles of a quadrilateral is $360^{\circ}$, we have
$$
\alpha+\alpha+3d=\alpha+d+\alpha+2d=180^{\circ} \text{.}
$$
Given that $\angle ABD = \angle CDB$ and $\angle DAB =... | 240 | Geometry | MCQ | Yes | Yes | cn_contest | false |
19. In $\triangle A B C$, it is known that $A B=13, B C=14, C A=$ 15, points $D, E, F$ are on sides $B C, C A, D E$ respectively, and $A D \perp$ $B C, D E \perp A C, A F \perp B F$. If the length of segment $D F$ is $\frac{m}{n}$ $\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$, then $m+n=(\quad)$.
(A) 18
(B) 21
(C) 24... | 19. B.
As shown in Figure 3, with $D$ as the origin, the line $BC$ as the $x$-axis, and the line $AD$ as the $y$-axis, we establish a Cartesian coordinate system $x D y$.
It is easy to know that $p_{\triangle ABC}=13+14+15=42$,
$S_{\triangle ABC}=\sqrt{21(21-13) \times(21-14) \times(21-15)}=84$.
Thus, $AD=\frac{2 S_{\... | 21 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $A=\{2,4, \cdots, 2014\}, B$ be any non-empty subset of $A$, and $a_{i} γ a_{j}$ be any two elements in set $B$. There is exactly one isosceles triangle with $a_{i} γ a_{j}$ as side lengths. Then the maximum number of elements in set $B$ is $\qquad$ | 7. 10 .
By symmetry, without loss of generality, assume $a_{i}<a_{j}$. Then there must exist an isosceles triangle with $a_{j}$ as the waist and $a_{i}$ as the base, and there is only one isosceles triangle with $a_{i}$ and $a_{j}$ as side lengths.
Thus, $a_{i}+a_{i} \leqslant a_{j} \Rightarrow a_{j} \geqslant 2 a_{i}... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 2, given that $D$ is the intersection of the tangents to the circumcircle $\odot O$ of $\triangle A B C$ at points $A$ and $B$, the circumcircle of $\triangle A B D$ intersects line $A C$ and segment $B C$ at another point $E$ and $F$ respectively, and $C D$ intersects $B E$ at point $G$. I... | Let $C D$ intersect $A B$ at point $M$.
If the center $O$ is not on the line segment $B C$, from $\frac{B C}{B F}=2$, we know that $F$ is the midpoint of $B C$.
Then, by the perpendicular diameter theorem,
$$
F O \perp B C \Rightarrow \angle O F B=90^{\circ} \text {. }
$$
Since $D A$ and $D B$ are tangent to $\odot O$... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. In $\triangle A B C$, it is known that $a, b, c$ are the sides opposite to $\angle A$, $\angle B$, and $\angle C$ respectively, and $a c+c^{2}=b^{2}-a^{2}$. If the longest side of $\triangle A B C$ is $\sqrt{7}$, and $\sin C=2 \sin A$, then the length of the shortest side of $\triangle A B C$ is $\qquad$. | 6. 1 .
From $a c+c^{2}=b^{2}-a^{2}$
$\Rightarrow \cos B=-\frac{1}{2} \Rightarrow \angle B=\frac{2 \pi}{3}$.
Thus, the longest side is $b$.
Also, $\sin C=2 \sin A \Rightarrow c=2 a$.
Therefore, $a$ is the shortest side.
By the cosine rule,
$$
(\sqrt{7})^{2}=a^{2}+4 a^{2}-2 a \times 2 a \times\left(-\frac{1}{2}\right) \... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then $\sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=$ $\qquad$ . | 10.2013.
Obviously, when $k \geqslant 11$, $\sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=0$.
$$
\begin{array}{l}
\text { Hence } \sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=\sum_{k=0}^{10}\left[\frac{2013+2^{k}}{2^{k+1}}\right] \\
= 1007+503+252+126+63+ \\
31+16+8+4+2+1 \\
= 2013 .
\end{array... | 2013 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
12. Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has one focus at $F_{1}(-\sqrt{3}, 0)$, and passes through the point $H\left(\sqrt{3}, \frac{1}{2}\right)$. Let the upper and lower vertices of the ellipse $E$ be $A_{1}$ and $A_{2}$, respectively, and let $P$ be any point on the ellipse differ... | 12. From the problem, we have
$$
a^{2}-b^{2}=3, \frac{3}{a^{2}}+\frac{1}{4 b^{2}}=1 \text {. }
$$
Solving, we get $a^{2}=4, b^{2}=1$.
Thus, the equation of the ellipse $E$ is $\frac{x^{2}}{4}+y^{2}=1$.
From this, we know the points $A_{1}(0,1), A_{2}(0,-1)$. Let point $P\left(x_{0}, y_{0}\right)$.
Then $l_{P A_{1}}: y... | 2 | Geometry | proof | Yes | Yes | cn_contest | false |
15. Given that $a, b, c$ are positive real numbers. Prove:
$$
\frac{\sqrt{a^{2}+3 b c}}{a}+\frac{\sqrt{b^{2}+3 a c}}{b}+\frac{\sqrt{c^{2}+3 a b}}{c} \geqslant 6 \text {. }
$$ | 15. Notice,
$$
\begin{array}{l}
\frac{\sqrt{a^{2}+3 b c}}{a}=\frac{\sqrt{a^{2}+b c+b c+b c}}{a} \\
\geqslant \frac{\sqrt{4 \sqrt[4]{a^{2} b^{3} c^{3}}}}{a}=\frac{2 \sqrt[8]{a^{2} b^{3} c^{3}}}{a} .
\end{array}
$$
Similarly, $\frac{\sqrt{b^{2}+3 a c}}{b} \geqslant \frac{2 \sqrt[8]{a^{3} b^{2} c^{3}}}{b}$,
$$
\begin{arr... | 6 | Inequalities | proof | Yes | Yes | cn_contest | false |
1. If every prime factor of 2013 is a term in a certain arithmetic sequence $\left\{a_{n}\right\}$ of positive integers, then the maximum value of $a_{2013}$ is $\qquad$ | $-, 1.4027$.
Notice that, $2013=3 \times 11 \times 61$.
If $3, 11, 61$ are all terms in a certain arithmetic sequence of positive integers, then the common difference $d$ should be a common divisor of $11-3=8$ and $61-3=58$. To maximize $a_{2013}$, the first term $a_{1}$ and the common difference $d$ should both be as ... | 4027 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. If $a, b, c > 0, \frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1$, then the minimum value of $a+2b+3c$ is . $\qquad$ | 2. 36 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
a+2 b+3 c=(a+2 b+3 c)\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\right) \\
\geqslant(1+2+3)^{2}=36 .
\end{array}
$$ | 36 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. If the distances from the center of the ellipse to the focus, the endpoint of the major axis, the endpoint of the minor axis, and the directrix are all positive integers, then the minimum value of the sum of these four distances is $\qquad$ .
| 5.61.
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the distances from the center $O$ of the ellipse to the endpoints of the major axis, the endpoints of the minor axis, the foci, and the directrices are $a$, $b$, $c$, $d$ respectively, and satisfy
$$
c^{2}=a^{2}-b^{2}, d=\frac... | 61 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a composite number $k(1<k<100)$. If the sum of the digits of $k$ is a prime number, then the composite number $k$ is called a "pseudo-prime". The number of such pseudo-primes is . $\qquad$ | 7.23.
Let $S(k)$ denote the sum of the digits of $k$, and $M(p)$ denote the set of composite numbers with a pseudo-prime $p$.
When $k \leqslant 99$, $S(k) \leqslant 18$, so there are 7 prime numbers not exceeding 18, which are $2, 3, 5, 7, 11, 13, 17$.
The composite numbers with a pseudo-prime of 2 are $M(2)=\{20\}$.... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Make a full permutation of the elements in the set $\{1,2, \cdots, 8\}$, such that except for the number at the far left, for each number $n$ on the right, there is always a number to the left of $n$ whose absolute difference with $n$ is 1. The number of permutations that satisfy this condition is $\qquad$ | 8. 128 .
Suppose for a certain permutation that satisfies the conditions, the first element on the left is $k(1 \leqslant k \leqslant 8)$. Then, among the remaining seven numbers, the $8-k$ numbers greater than $k$, $k+1, k+2, \cdots, 8$, must be arranged in ascending order; and the $k-1$ numbers less than $k$, $1,2, ... | 128 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $g(x)=\sum_{i=1}^{n} \mathrm{C}_{n}^{i} \frac{i x^{i}(1-x)^{n-i}}{n}$. Then $g(2014)=$ $\qquad$ | $$
-, 1.2014
$$
From $\frac{r}{n} \mathrm{C}_{n}^{r}=\mathrm{C}_{n-1}^{r-1}$, we get
$$
g(x)=x \sum_{k=0}^{n-1} \mathrm{C}_{n-1}^{k} x^{k}(1-x)^{n-k}=x \text {. }
$$
Thus, $g(2014)=2014$. | 2014 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $a, b, c \in \mathbf{R}_{+}$, and
$$
a+b+c=12, a b+b c+c a=45 \text{. }
$$
Then $\min \max \{a, b, c\}=$ $\qquad$ | 6. 5 .
Let $a=\max \{a, b, c\}$.
From $a+b+c=12$, we get $a \geqslant 4$.
$$
\begin{array}{l}
\text { By }(a-b)(a-c) \geqslant 0 \\
\Rightarrow a^{2}-a(12-a)+b c \geqslant 0 \\
\Rightarrow b c \geqslant 12 a-2 a^{2} . \\
\text { Also } 45=a b+b c+c a=b c+a(12-a) \\
\geqslant 12 a-2 a^{2}+a(12-a),
\end{array}
$$
Then ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. The unit digit of $\left[\frac{10^{10000}}{10^{100}+9}\right]$ is | 8. 1 .
Notice that,
$$
\frac{10^{10000}}{10^{100}+9}=\frac{\left(10^{100}\right)^{100}-3^{200}}{10^{100}+9}+\frac{3^{200}}{10^{100}+9} \text {. }
$$
And $\left(10^{100}\right)^{100}-3^{200}=\left[\left(10^{100}\right)^{2}\right]^{50}-\left(9^{2}\right)^{50}$, so $\left[\left(10^{100}\right)^{2}-9^{2}\right] \mid\left... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Given $\odot O: x^{2}+y^{2}=4$, circle $M$ :
$$
(x-5 \cos \theta)^{2}+(y-5 \sin \theta)^{2}=1(\theta \in \mathbf{R}) \text {, }
$$
Through any point $P$ on circle $M$, draw two tangents $P E$ and $P F$ to $\odot O$, with the points of tangency being $E$ and $F$. Try to find the minimum value of $\overri... | 9. The center of circle $M$ is on the circle $x^{2}+y^{2}=25$.
Let $|P E|=|P F|=d$.
In the right triangle $\triangle P E O$, it is easy to see that
$4 \leqslant|P O| \leqslant 6,|O E|=2$.
Thus, $2 \sqrt{3} \leqslant d \leqslant 4 \sqrt{2}$.
Also, $\overrightarrow{P E} \cdot \overrightarrow{P F}=|\overrightarrow{P E}||\... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let
$$
\begin{array}{l}
f(x)=a_{1} x^{2013}+a_{2} x^{2012}+\cdots+a_{2013} x+a_{2014} \\
=x^{13}\left(x^{10}+x^{2}+x\right)^{2000}, \\
b_{0}=1, b_{1}=2, b_{n+2}+b_{n}=b_{n+1}(n \in \mathbf{N}) .
\end{array}
$$
Find the value of $\sum_{i=1}^{2013} a_{i} b_{i}$. | 11. From the recurrence relation of $\left\{b_{n}\right\}$, we have
$$
\begin{array}{l}
b_{2}=1, b_{3}=-1, b_{4}=-2, b_{5}=-1, \\
b_{6}=1=b_{0}, b_{7}=2=b_{1} .
\end{array}
$$
Thus, $b_{6 k}=1, b_{6 k+1}=2, b_{6 k+2}=1, b_{6 k+3}=-1$, $b_{6 k+4}=-2, b_{6 k+5}=-1(k \in \mathbf{N})$.
Let $\lambda=\frac{1}{2}+\frac{\sqrt... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ distinct positive integers, none of whose decimal representations contain the digit 9. Prove:
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<30 .
$$ | Prove that if all positive integers in decimal representation without the digit 9 are arranged as $b_{1}, b_{2}, \cdots$, then
$$
\begin{array}{l}
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots \\
\leqslant c+\frac{9}{10} c+\left(\frac{9}{10}\right)^{2} c+\left(\frac{9}{10... | 30 | Number Theory | proof | Yes | Yes | cn_contest | false |
Let the function $f(x)$ satisfy $f(1)=1, f(4)=7$, and for any $a, b \in \mathbf{R}$, we have
$$
f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3} .
$$
Then $f(2014)=$ $\qquad$
(A) 4027
(B) 4028
(C) 4029
(D) 4030 | Solution 1 From $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$, we get
$f(2)=f\left(\frac{4+2 \times 1}{3}\right)=\frac{f(4)+2 f(1)}{3}=3$,
$f(3)=f\left(\frac{1+2 \times 4}{3}\right)=\frac{f(1)+2 f(4)}{3}=5$.
Thus, we conjecture that $f(n)=2 n-1\left(n \in \mathbf{Z}_{+}\right)$.
We prove this by mathematical in... | 4027 | Algebra | MCQ | Yes | Yes | cn_contest | false |
15. Given
$$
3^{p}+3^{4}=90,2^{r}+44=76,5^{3}+6^{3}=1421 \text {. }
$$
Then $p r s=(\quad)$.
(A) 27
(B) 40
(C) 50
(D) 70
(E) 90 | 15. B. From the known information, we get $p=2, r=5, s=4$. Therefore, prs $=40$. | 40 | Algebra | MCQ | Yes | Yes | cn_contest | false |
8. If the complex number $x$ satisfies $x+\frac{1}{x}=-1$, then $x^{2013}+\frac{1}{x^{2013}}=$ $\qquad$ . | 8. 2 .
Given that $x^{2}+x+1=0$.
Since the discriminant $\Delta=-3<0$, $x$ is a complex number.
Also, $x^{3}-1=(x-1)\left(x^{2}+x+1\right)=0$
$\Rightarrow x^{3}=1$.
Therefore, $x^{2013}+\frac{1}{x^{2013}}=\left(x^{3}\right)^{671}+\frac{1}{\left(x^{3}\right)^{671}}=1+1=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. A science and technology innovation competition sets first, second, and third prizes (all participants will receive an award), and the probabilities of winning the corresponding prizes form a geometric sequence with the first term $a$ and a common ratio of 2. The corresponding prizes form an arithmetic sequence with... | 3. 500 .
Let the prize money obtained be $\xi$ yuan. Then $\xi=700,560,420$. From the problem, we know
$$
\begin{array}{l}
P(\xi=700)=a, P(\xi=560)=2 a, \\
P(\xi=420)=4 a .
\end{array}
$$
From $7 a=1$, we get $a=\frac{1}{7}$.
Therefore, $E \xi=700 \times \frac{1}{7}+560 \times \frac{2}{7}+420 \times \frac{4}{7}$ $=50... | 500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $l_{1}, l_{2}, \cdots, l_{100}$ are 100 distinct and coplanar lines. If the lines numbered $4 k\left(k \in \mathbf{Z}_{+}\right)$ are parallel to each other, and the lines numbered $4 k-1$ all pass through point $A$, then the maximum number of intersection points of these 100 lines is $\qquad$ . | 7.4351 .
According to the problem, the number of combinations of any two lines out of 100 lines is
$$
\mathrm{C}_{100}^{2}-\mathrm{C}_{25}^{2}-\mathrm{C}_{25}^{2}+1=4351 .
$$ | 4351 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
14. (15 points) Let
$$
A=x^{4}+2 x^{3}-x^{2}-5 x+34 \text {. }
$$
Find the integer values of \( x \) for which \( A \) is a perfect square. | 14. Notice that,
$$
A=\left(x^{2}+x-1\right)^{2}-3(x-11) \text {. }
$$
So, when $x=11$, $A=131^{2}$ is a perfect square.
Next, we prove: there are no other integer $x$ that satisfy the condition.
(1) When $x>11$, we have $A>0$, thus, $A>\left(x^{2}+x-2\right)^{2}$.
Therefore, $\left(x^{2}+x-2\right)^{2}<A<\left(x^{2}+... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Seven balls of different colors are placed into three boxes numbered 1, 2, and 3. It is known that the number of balls in each box is not less than its number. The number of different ways to place the balls is $\qquad$ | 5.455.
(1) If the number of balls placed in boxes 1, 2, and 3 are 2, 2, and 3 respectively, then the number of different ways to place them is $\mathrm{C}_{7}^{2} \mathrm{C}_{5}^{2}=210$;
(2) If the number of balls placed in boxes 1, 2, and 3 are 1, 3, and 3 respectively, then the number of different ways to place them... | 455 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
381 A chess piece starts from one corner of an $8 \times 8$ chessboard, and moving one square in the row direction or column direction is called a step. After several steps, it passes through every square without repetition and returns to the starting point.
(1) Prove: the number of steps in the row direction is differ... | (1) Construct an $8 \times 8$ grid, and connect every adjacent point with a line segment to form a graph. Assume the distance between adjacent points is 1, and call a square with an area of 1 and vertices on the grid points a "cell."
The original problem is equivalent to: Prove that in every Hamiltonian cycle of the g... | 4 | Combinatorics | proof | Yes | Yes | cn_contest | false |
1. The sequence satisfies $a_{0}=\frac{1}{4}$, and for natural number $n$, $a_{n+1}=a_{n}^{2}+a_{n}$.
Then the integer part of $\sum_{n=0}^{201} \frac{1}{a_{n}+1}$ is $\qquad$.
(2011, National High School Mathematics League Gansu Province Preliminary) | Given: $\frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}$
$$
\Rightarrow \sum_{n=0}^{2011} \frac{1}{a_{n}+1}=\frac{1}{a_{0}}-\frac{1}{a_{2012}}=4-\frac{1}{a_{2012}} \text {. }
$$
Obviously, the sequence $\left\{a_{n}\right\}$ is monotonically increasing.
$$
\begin{array}{l}
\text { Also, } a_{1}=\frac{5}{16}, a_{2}... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $A_{1} A_{2} \cdots A_{101}$ be a regular 101-gon. Each vertex is colored either red or blue. Let $N$ be the number of obtuse triangles that satisfy the following conditions: the three vertices of the triangle are vertices of the 101-gon, the two acute vertices have the same color, and the color of the obtuse ve... | 2. Let $x_{i}=0$ or 1 represent $A_{i}$ being red or blue, respectively. For an obtuse triangle $\triangle A_{i-a} A_{i} A_{i+b}$, where vertex $A_{i}$ is the obtuse angle vertex, i.e., $a+b \leqslant 50$.
The coloring of the three vertices satisfies the condition if and only if
$$
\left(x_{i}-x_{i-a}\right)\left(x_{i}... | 32175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. (18 points) Given
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right) .
$$
(1) If there exist positive numbers $a, b, c$ such that inequality (1) holds, prove: $k>5$;
(2) If there exist positive numbers $a, b, c$ such that inequality (1) holds, and any set of positive numbers $a, b, c$ that satisfy inequality (1) are... | 11. (1) Since $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, therefore, from inequality (1) we know
$$
k(a b+b c+c a)>5(a b+b c+c a) .
$$
Noting that $a, b, c$ are positive numbers.
Thus, $a b+b c+c a>0$.
Hence $k>5$.
(2) From (1) we know $k>5$.
Since $k$ is an integer, then $k \geqslant 6$.
Let $a=1, b=1, c=2$. It is ea... | 6 | Inequalities | proof | Yes | Yes | cn_contest | false |
16. Given that one side of the square $A B C D$ lies on the line $y=2 x-17$, and the other two vertices are on the parabola $y=x^{2}$. Then the minimum value of the area of the square is $\qquad$ . | 16. 80 .
From the problem, we know that the distance from a point $\left(x, x^{2}\right)$ on the parabola to the line $l$ is
$$
d=\frac{\left|2 x-x^{2}-17\right|}{\sqrt{5}}>0 .
$$
When $x=1$, $d$ reaches its minimum value $\frac{16}{\sqrt{5}}$.
To minimize the area of the square, and by the symmetry of the square, th... | 80 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) In the sequence $\left\{x_{n}\right\}$,
$$
x_{n}=p^{n}+q^{n}, x_{1}=1, x_{3}=4 \text {. }
$$
(1) Prove: $x_{n+2}=x_{n+1}+x_{n}$;
(2) Determine the last digit of $x_{2011}$, but no proof is required. | 11. (1) Notice,
$$
\begin{array}{l}
p+q=1, p^{3}+q^{3}=4, \\
p^{3}+q^{3}=(p+q)\left(p^{2}-p q+q^{2}\right) .
\end{array}
$$
Thus, $p^{2}-p q+q^{2}=4, p^{2}+2 p q+q^{2}=1$.
$$
\begin{array}{l}
\text { Then } x_{2}=p^{2}+q^{2}=3 \\
\Rightarrow p^{2}+(1-p)^{2}=3 \\
\Rightarrow p^{2}=p+1, q^{2}=q+1 \text {. }
\end{array}
... | 6 | Algebra | proof | Yes | Yes | cn_contest | false |
2. Given that for any real number $x$ we have $a \cos x + b \cos 2x \geqslant -1$.
Then the maximum value of $a + b$ is $\qquad$ | 2. 2 .
Let $x=\frac{2 \pi}{3}$, then $a+b \leqslant 2$.
When $a=\frac{4}{3}, b=\frac{2}{3}$,
$$
\begin{array}{l}
a \cos x+b \cos 2 x=\frac{4}{3} \cos x+\frac{2}{3} \cos 2 x \\
=\frac{4}{3} \cos ^{2} x+\frac{4}{3} \cos x-\frac{2}{3} \\
=\frac{4}{3}\left(\cos x+\frac{1}{2}\right)^{2}-1 \\
\geqslant-1 .
\end{array}
$$ | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the sequence $\left\{a_{n}\right\}_{n \geqslant 1}$ satisfies
$$
a_{n+2}=a_{n+1}-a_{n} \text {. }
$$
If the sum of the first 1000 terms of the sequence is 1000, then the sum of the first 2014 terms is $\qquad$ . | 3. 1000 .
From $a_{n+2}=a_{n+1}-a_{n}$, we get
$$
a_{n+3}=a_{n+2}-a_{n+1}=\left(a_{n+1}-a_{n}\right)-a_{n+1}=-a_{n} \text {. }
$$
Therefore, for any positive integer $n$ we have
$$
a_{n}+a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+a_{n+5}=0 \text {. }
$$
Also, $2014=1000(\bmod 6)$, so
$$
S_{2014}=S_{1000}=1000 \text {. }
$$ | 1000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the function
$$
f(x)=x^{3}-6 x^{2}+17 x-5 \text {, }
$$
real numbers $a, b$ satisfy $f(a)=3, f(b)=23$. Then $a+b=$ $\qquad$ | 4.4 .
Notice,
$$
\begin{array}{l}
f(x)=x^{3}-6 x^{2}+17 x-5 \\
=(x-2)^{3}+5(x-2)+13 \text {. } \\
\text { Let } g(y)=y^{3}+5 y .
\end{array}
$$
Then $g(y)$ is an odd function and monotonically increasing.
And $f(a)=(a-2)^{3}+5(a-2)+13=3$, so, $g(a-2)=-10$.
Also, $f(b)=(b-2)^{3}+5(b-2)+13=23$, then $g(b-2)=10$.
Thus $... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. The positive integer solutions of the equation $x+y^{2}+(x, y)^{3}=x y(x, y)$ are $\qquad$ groups $((x, y)$ represents the greatest common divisor of integers $x, y)$.
| 7.4.
Let $(x, y)=d$. Then $d^{2} \mid x$.
Let $x=a d^{2}, y=b d$. Then $(a d, b)=1$.
Thus, the original equation becomes
$$
a+b^{2}+d=a b d^{2} \text {. }
$$
Therefore, $b \mid(a+d) \Rightarrow b \leqslant a+d$.
Then $a+b^{2}+d=a b d^{2}$
$$
\begin{array}{l}
=(a+d) b+(a+d) b+b\left(a d^{2}-2 a-2 d\right) \\
\geqslant... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. A middle school has 35 lights on each floor. To save electricity while ensuring the lighting needs of the corridors, the following requirements must be met:
(1) Two adjacent lights cannot be on at the same time;
(2) Any three consecutive lights cannot be off at the same time.
If you were to design different lighting... | 8. 31572 .
Notice that, $a_{n+3}=a_{n}+a_{n+1}, a_{3}=4, a_{4}=5$. Therefore, $a_{35}=31572$. | 31572 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $[x]$ denote the greatest integer not exceeding the real number $x$,
$$
a_{k}=\left[\frac{2014}{k}\right](k=1,2, \cdots, 100) \text {. }
$$
Then, among these 100 integers, the number of distinct integers is | 6.69.
When $01$, $[a+b]>[a]$.
$$
\begin{array}{l}
\text { Also } \frac{2014}{k}-\frac{2014}{k+1}=\frac{2014}{k(k+1)} \\
\Rightarrow \frac{2014}{k}=\frac{2014}{k+1}+\frac{2014}{k(k+1)} .
\end{array}
$$
When $k \leqslant 44$, $\frac{2014}{k(k+1)}>1$;
When $k \geqslant 45$, $\frac{2014}{k(k+1)}<1$.
Therefore, when $k \l... | 69 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $A$ and $B$ be two different subsets of the set $\{a, b, c, d, e\}$, such that set $A$ is not a subset of set $B$, and $B$ is not a subset of set $A$. Then the number of different ordered pairs $(A, B)$ is | 7.570.
Notice that, the set $\{a, b, c, d, e\}$ has $2^{5}$ subsets, and the number of different ordered pairs $(A, B)$ is $2^{5}\left(2^{5}-1\right)$.
If $A \subset B$, and suppose the set $B$ contains $k(1 \leqslant k \leqslant 5)$ elements, then the number of ordered pairs $(A, B)$ satisfying $A \subset B$ is
$$
\... | 570 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
385 A certain school's 2014 graduates numbered 2014 students. The school's six leaders must sign each student's graduation album. It is known that each leader must and can only use one of the three designated colored pens, and the color choices for the six leaders signing 2014 albums can be represented as a $6 \times 2... | Let 1, 2, 3 represent three colors, and the colors used by six school leaders for signing in each memorial album be represented by a six-element ordered array
$$
\left(x_{1}, x_{2}, \cdots, x_{6}\right)\left(x_{i} \in\{1,2,3\}\right)
$$
This array is referred to as the sequence group $\left(x_{1}, x_{2}, \cdots, x_{6}... | 64 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Real-coefficient polynomials $f_{i}(x)=a_{i} x^{2}+b_{i} x+c_{i}$ $\left(a_{i}>0, i=1,2, \cdots, 2011\right)$, and $\left\{a_{i}\right\} γ\left\{b_{i}\right\} γ\left\{c_{i}\right\}$ are all arithmetic sequences. If $F(x)=\sum_{i=1}^{204} f_{i}(x)$ has real roots, then at most how many polynomials in $\left\{f... | γAnalysisγFrom $\left\{a_{i}\right\}$ being an arithmetic sequence, we know
$$
\sum_{i=1}^{2011} a_{i}=(2 \times 1005+1) a_{1006}=2011 a_{1006} \text {. }
$$
Similarly, $\sum_{i=1}^{2011} b_{i}=2011 b_{1006}$,
$$
\sum_{i=1}^{2011} c_{i}=2011 c_{1006} \text {. }
$$
Therefore, $F(x)=\sum_{i=1}^{200} f_{i}(x)=2011 f_{10... | 1005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 1, in $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$ respectively. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array}
$$
t... | 3. 2 .
Solution 1 Notice that,
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} .
$$
Since points $M, O, N$ are collinear, we have,
$$
\frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 \text {. }
$$
Solution 2 Since points $M, ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Given
$$
\begin{array}{l}
A \cup B \cup C=\{a, b, c, d, e\}, A \cap B=\{a, b, c\}, \\
c \in A \cap B \cap C .
\end{array}
$$
Then the number of sets $\{A, B, C\}$ that satisfy the above conditions is. $\qquad$ | 7. 100 .
As shown in Figure 3, the set $A \cup B \cup C$ can be divided into seven mutually exclusive regions, denoted as $1, 2, \cdots, 7$.
It is known that element $c$ is in region 7, elements $a, b$ can appear in each of the regions 4, 7, with 4 possibilities; elements $d, e$ can appear in each of the regions $1, ... | 100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. Use five different colors to color the six vertices of the triangular prism $A B C-D E F$, requiring each point to be colored with one color, and the two endpoints of each edge to be colored with different colors. Then the number of different coloring methods is $\qquad$.
| 9. 1920.
Transform the adjacency relationship of the vertices of a triangular prism into the regional relationship as shown in Figure 4, then the coloring methods for vertices $A$, $B$, and $C$ are $\mathrm{A}_{5}^{3}$ kinds.
For each determined coloring scheme of points $A$, $B$, and $C$, if we temporarily do not co... | 1920 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. (15 points) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and an inscribed parallelogram with one pair of opposite sides passing through the foci $F_{1}$ and $F_{2}$ of the ellipse. Find the maximum area of the parallelogram. | 12. Given that $\left|F_{1} F_{2}\right|=2$.
As shown in Figure 5, the inscribed quadrilateral $\square A B C D$ of the ellipse has a pair of opposite sides $B C$ and $A D$ passing through the foci $F_{1}$ and $F_{2}$, respectively.
Obviously, the center of symmetry of $\square A B C D$ is the origin.
Figure 5
Let th... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Determine the least possible value of the largest term in an arithmetic sequence composed of seven distinct primes. ${ }^{[4]}$
(2005, British Mathematical Olympiad) | γAnalysisγLet these seven different prime numbers be $p_{i}(i=1,2$, $\cdots, 7)$, and
$$
\begin{array}{l}
p_{2}=p_{1}+d, p_{3}=p_{1}+2 d, p_{4}=p_{1}+3 d, \\
p_{5}=p_{1}+4 d, p_{6}=p_{1}+5 d, p_{7}=p_{1}+6 d,
\end{array}
$$
where the common difference $d \in \mathbf{Z}_{+}$.
Since $p_{2}=p_{1}+d$ and $p_{3}=p_{1}+2 d$... | 907 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have
$$
\begin{aligned}
f(x+2) & =f(x)+2, \\
\text { then } \sum_{k=1}^{2014} f(k) & =
\end{aligned}
$$ | 14.2029105.
Notice that, $f(0)=0$.
By $f(x+2)=f(x)+2$, let $x=-1$. Then $f(1)=f(-1)+2 \Rightarrow f(1)=1$.
$$
\begin{array}{l}
\text { Also, } f(2 n)=\sum_{k=1}^{n}(f(2 k)-f(2 k-2))+f(0) \\
=2 n, \\
f(2 n-1)=\sum_{k=2}^{n}(f(2 k-1)-f(2 k-3))+f(1) \\
=2 n-1, \\
\text { Therefore, } \sum_{k=1}^{2014} f(k)=\sum_{k=1}^{20... | 2029105 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
17. A courier company undertakes courier services between 13 cities in a certain area. If each courier can take on the courier services for at most four cities, to ensure that there is at least one courier between every two cities, the courier company needs at least $\qquad$ couriers. | 17. 13.
From the problem, we know that there are $\mathrm{C}_{13}^{2}$ types of express delivery services between 13 cities. Each courier can handle at most $\mathrm{C}_{4}^{2}$ types of express delivery services between four cities. Therefore, at least $\frac{\mathrm{C}_{13}^{2}}{\mathrm{C}_{4}^{2}}=13$ couriers are ... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
21. Among the 100 integers from $1 \sim 100$, arbitrarily select three different numbers to form an ordered triplet $(x, y, z)$. Find the number of triplets that satisfy the equation $x+y=3z+10$. | (1) When $3 z+10 \leqslant 101$, i.e., $z \leqslant 30$, the number of ternary tuples satisfying $x+y=3 z+10$ is
$$
S=\sum_{k=1}^{30}(3 k+9)=1665 \text{. }
$$
(2) When $3 z+10 \geqslant 102$, i.e., $31 \leqslant z \leqslant 63$, the number of ternary tuples satisfying $x+y=3 z+10$ is
$$
\begin{aligned}
T & =\sum_{k=31}... | 3194 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that the arithmetic mean of $\sin \theta$ and $\cos \theta$ is $\sin \alpha$, and the geometric mean is $\sin \beta$. Then $\cos 2 \alpha-\frac{1}{2} \cos 2 \beta=$ $\qquad$ . | 4. 0 .
From the given, we have
$$
\begin{array}{l}
\sin \alpha=\frac{\sin \theta+\cos \theta}{2}, \sin ^{2} \beta=\sin \theta \cdot \cos \theta. \\
\text { Therefore, } \cos 2 \alpha-\frac{1}{2} \cos 2 \beta \\
=1-2 \sin ^{2} \alpha-\frac{1}{2}\left(1-2 \sin ^{2} \beta\right) \\
=\frac{1}{2}-2 \sin ^{2} \alpha+\sin ^{... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. If the edge length of the cube $A_{1} A_{2} A_{3} A_{4}-B_{1} B_{2} B_{3} B_{4}$ is 1, then the number of elements in the set
$$
\left\{x \mid x=\overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}, i, j \in\{1,2,3,4\}\right\}
$$
is $\qquad$ | 5. 1 .
Solution 1 Note that,
$$
\begin{array}{l}
\overrightarrow{A_{1} B_{1}} \perp \overrightarrow{A_{i} A_{1}}, \overrightarrow{A_{1} B_{1}} \perp \overrightarrow{B_{1} B_{j}}(i, j \in\{2,3,4\}) \text {. } \\
\text { Hence } \overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}=\overrightarrow{A_{1} B_{1}}... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}^{2}=a_{n+1} a_{n}-1\left(n \in \mathbf{Z}_{+}\right) \text {, and } a_{1}=\sqrt{2} \text {. }
$$
Then the natural number closest to $\sqrt{a_{2014}}$ is $\qquad$ | 6.8.
From the given, we have
$$
\begin{array}{l}
a_{n+1}=a_{n}+\frac{1}{a_{n}} \Rightarrow a_{n+1}^{2}-a_{n}^{2}=2+\frac{1}{a_{n}^{2}} \\
\Rightarrow a_{n+1}^{2}=a_{1}^{2}+2 n+\sum_{i=1}^{n} \frac{1}{a_{i}^{2}} .
\end{array}
$$
Thus, $a_{2014}^{2}=2+2 \times 2013+\sum_{i=1}^{2013} \frac{1}{a_{i}^{2}}$
$$
>2+2 \times ... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=6$. Then the maximum value of $x+\sqrt{x y}+\sqrt[3]{x y z}$ is $\qquad$ . | 7.8.
By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{1}{4} x+y+4 z \geqslant 3 \sqrt[3]{x y z}, \\
\frac{3}{4} x+3 y \geqslant 3 \sqrt{x y} .
\end{array}
$$
Adding the two inequalities, we get
$$
\begin{array}{l}
x+4 y+4 z \geqslant 3 \sqrt{x y}+3 \sqrt[3]{x y z} \\
\Rightarrow x+\sqrt{x y}+\sqrt[3]{x y z}... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Fill the four characters β马β βδΈβ βζβ βεβ in a $5 \times 5$ grid, with at most one character in each small square. β马β βδΈβ must be filled in from left to right, and βζβ βεβ must also be filled in from left to right. β马β βδΈβ must be in the same row or in the same column from top to bottom, or βζβ βεβ must be in the sa... | 8.42100.
The problem is equivalent to:
Filling 2 $a$s and 2 $b$s in a $5 \times 5$ grid, with at most one letter in each small square, such that at least one pair of the same letters is in the same row or column.
First, consider the case where the same letters are neither in the same row nor in the same column.
The nu... | 42100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 The sequence $\left\{a_{n}\right\}$:
$1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots$.
Its construction method is:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, to get $a_{2}=1, a_{3}=2$;
Next, copy all the previous items $1,1,2$, and add the successor number 3 of 2, to get
$$
a_{4}=1, ... | From the construction method of the sequence $\left\{a_{n}\right\}$, it is easy to know that
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \cdots
$$
In general, we have $a_{2^{n}-1}=n$, meaning 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
$... | 3952 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 There are 800 points on a circle, numbered $1,2, \cdots, 800$ in a clockwise direction. They divide the circumference into 800 gaps. Choose one point and color it red, then proceed to color other points red according to the following rule: if the $k$-th point has been colored red, then move $k$ gaps in a cloc... | Consider a circle with $2n$ points in general.
(1) On a circle with $2n$ points, if the first red point is an even-numbered point, such as the $2k$-th point, then according to the coloring rule, every red point colored afterward will also be an even-numbered point. In this case, if we rename the 2nd, 4th, ..., $2k$-th,... | 25 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 2 Let real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy
(1) $-\frac{1}{\sqrt{3}} \leqslant x_{i} \leqslant \sqrt{3}(i=1,2, \cdots, 1997)$;
(2) $x_{1}+x_{2}+\cdots+x_{1997}=-318 \sqrt{3}$.
Try to find: $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$'s maximum value, and explain the reason. | Given that $f(x)=x^{12}$ is a convex function, then $\sum_{i=1}^{197} x_{i}^{12}$ achieves its maximum value when at least 1996 of $x_{1}, x_{2}, \cdots, x_{1997}$ are equal to $-\frac{1}{\sqrt{3}}$ or $\sqrt{3}$.
Assume that there are $t$ instances of $-\frac{1}{\sqrt{3}}$ and $1996-t$ instances of $\sqrt{3}$. Then t... | 189548 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$, $b$, and $c$ are constants, and for any real number $x$ we have
$$
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \text {. }
$$
then $a b c=$ . $\qquad$ | $$
\text { Two, 1. }-9 \text {. }
$$
From the problem, we know
$$
\begin{array}{c}
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \\
=x^{3}+(a+1) x^{2}+(a+b) x+b .
\end{array}
$$
Thus, $a+1=0, a+b=2, b=c$
$$
\begin{array}{l}
\Rightarrow a=-1, b=c=3 \\
\Rightarrow a b c=-9
\end{array}
$$ | -9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If $a, b$ are positive integers, and satisfy $5a+7b=50$, then $ab=$ . $\qquad$ | 3. 15 .
From the problem, we know
$$
a=\frac{50-7 b}{5}=10-b-\frac{2 b}{5} \text { . }
$$
Therefore, $a=3, b=5$.
Thus, $a b=15$. | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $S_{\triangle M B C}=4, 3 A B=2 B C$, draw the perpendicular from point $C$ to the angle bisector $B E$ of $\angle A B C$, and let the foot of the perpendicular be $D$. Then $S_{\triangle B D C}=$ $\qquad$ | 4.3.
Construct the reflection of point $C$ about $BD$ as point $F$, thus, $BC=BF$. Connect $FA$ and $FD$, then point $F$ lies on the extension of $BA$, and points $C$, $D$, and $F$ are collinear.
$$
\text{Therefore, } \frac{S_{\triangle ABC}}{S_{\triangle FBC}}=\frac{AB}{FB}=\frac{AB}{BC}=\frac{2}{3} \text{.}
$$
Sinc... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the function $y=f(x)$ is defined on $\mathbf{R}$, with a period of 3, and Figure 1 shows the graph of the function in the interval $[-2,1]$. Then $\frac{f(2014)}{f(5) f(15)}=$ $\qquad$ . | $$
\text { II,6. }-2 \text {. }
$$
From the problem statement and combining with the graph, we know that
$$
\frac{f(2014)}{f(5) f(15)}=\frac{f(1)}{f(-1) f(0)}=\frac{2}{(-1) \times 1}=-2 \text {. }
$$ | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.