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6. (2004 National College Entrance Examination - Guangdong Paper) When $0<x<\frac{\pi}{4}$, the minimum value of the function $f(x)=\frac{\cos ^{2} x}{\cos x \sin x-\sin ^{2} x}$ is
A. $\frac{1}{4}$
B. $\frac{1}{2}$
C. 2
D. 4 | 6. D $f(x)=\frac{\cos ^{2} x}{\cos x \sin x-\sin ^{2} x}=\frac{1}{\tan x-\tan ^{2} x}=\frac{1}{-\left(\tan x-\frac{1}{2}\right)^{2}+\frac{1}{4}} \geqslant 4$
6. D $f(x)=\frac{\cos^2 x}{\cos x \sin x - \sin^2 x}=\frac{1}{\tan x - \tan^2 x}=\frac{1}{-\left(\tan x - \frac{1}{2}\right)^2 + \frac{1}{4}} \geqslant 4$ | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
The fifth question: If an infinite sequence of positive real numbers $\left\{x_{n}\right\}$ satisfies: $x_{0}=1, x_{i} \geq x_{i+1}(i \in N)$, then the sequence is called a “good sequence”. Find the smallest constant $c$ such that there exists a “good sequence” $\left\{x_{n}\right\}$ satisfying: $\sum_{i=0}^{n} \frac{x... | Fifth question,
Solution method one: On the one hand, for any "good sequence" $\left\{x_{n}\right\}$, according to the mean inequality, we have:
$$
\begin{array}{l}
=2^{n} \cdot\left(\frac{1}{2^{\sum_{i=0}^{n-1}(n-1-i) 2^{n-1-1}}} \cdot x_{0}^{2^{n}}\right)^{\frac{1}{2^{n}}} \\
=2^{n} \cdot\left(\frac{1}{2^{2^{n}(n-2)+... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. Write down natural numbers starting from 1, all the way to the 198th digit:
$$
\underbrace{123456789101112 \cdots \cdots}_{198 \text { digits. }}
$$
Then the remainder of this number when divided by 9 is
(A) 4 ;
(B) 6 ;
(C.) 7 ;
(D) None of the above.
Answer $(\quad)$ | 6. (B)
Solution This number is
$1234 \cdots 9101112 \cdots 99100101102$.
Since $1+2+3+\cdots+9$ is divisible by 9, and $1+0+1+1+1+2+$ $\cdots+9+9$ (the sum of the digits of all integers from 10 to 99) is also divisible by 9, the remainder when this number is divided by 9 is
$$
1+0+0+1+0+1+1+0+2=6 .
$$ | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Example 15 Find the sum: $\sum_{k=0}^{2 n} \cos \left(\alpha+\frac{2 k \pi}{2 n+1}\right)$. | Let $z_{k}=\cos \left(\alpha+\frac{2 k \pi}{2 n+1}\right)+i \sin \left(\alpha+\frac{2 k \pi}{2 n+1}\right) (k=0,1,2, \cdots, 2 n)$.
Take the $(2 n+1)$-th roots of unity $\varepsilon_{k}=\cos \frac{2 k \pi}{2 n+1}+i \sin \frac{2 k \pi}{2 n+1} (k=0,1,2, \cdots, 2 n)$,
then
$$
\begin{aligned}
W & =z_{0}+z_{1}+z_{2}+\cdots... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. For the geometric sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1536$, common ratio $q=-\frac{1}{2}$, $\Pi_{n}$ represents the product of its first $n$ terms $\left(n \in \mathbf{N}^{*}\right)$, the value of $n$ that maximizes $\mathrm{II}_{n}$ is . $\qquad$ | 10. 12 The general term formula of the geometric sequence $\left\{a_{n}\right\}$ is $a_{n}=1536 \cdot\left(-\frac{1}{2}\right)^{n-1}$, to make the product of the first $n$ terms $\Pi_{n}$ maximum, $n$ must be an odd number or a multiple of 4. Let $\left|a_{n}\right| \geqslant 1$ get $1536 \cdot\left(\frac{1}{2}\right)^... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Positive integers $a, b$ and $c$ satisfy the equation $\frac{31}{72}=\frac{a}{8}+\frac{b}{9}-c$. What is the smallest possible value of $b$ ? | 5. Since $\frac{31}{72}=\frac{a}{8}+\frac{b}{9}-c$, then multiplying both sides by 72 gives $31=9 a+8 b-72 c$.
Rearranging, we obtain $72 c-9 a=8 b-31$.
Since $a$ and $c$ are positive integers, then $72 c-9 a$ is an integer.
Since $72 c-9 a=9(8 c-a)$, then $72 c-9 a$ must be a multiple of 9 .
Since $72 c-9 a$ is a mult... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.66 The figure below represents the solution set of the inequality $a x > 2 x$ on the number line. Then the value of the expression
$$
\sqrt{a+2 \sqrt{a-1}}+\sqrt{a-2 \sqrt{a-1}}
$$
is
(A) $2 \sqrt{a-1}$.
(B) 2 .
(C) $a$.
(D) 1
(5th "Jinyun Cup" Junior High School Mathematics Invitational, 1988) | [Solution] From $a x>2 x$ and $x<0$, we get $a<2$.
Also, for the square root to be defined, $a-1 \geqslant 0$, i.e., $a \geqslant 1$, so $1 \leqslant a<2$.
$$
\begin{aligned}
& \sqrt{a+2 \sqrt{a-1}}+\sqrt{a-2 \sqrt{a-1}} \\
= & \sqrt{(a-1)+2 \sqrt{a-1}+1}+\sqrt{(a-1)-2 \sqrt{a-1}+1} \\
= & |\sqrt{a-1}+1|+|\sqrt{a-1}-1|... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 2 (2003 Thailand Mathematical Olympiad) Find all prime numbers $p$ such that $p^{2}+2543$ has fewer than 16 distinct positive divisors. | Let $p(p>3)$ be a prime number. It is easy to prove that $p^{2}-1$ is divisible by 24.
It is easy to see that $p^{2}+2543=p^{2}-1+106 \times 24$ is a multiple of 24.
Let $p^{2}+2543=24 k(k \in \mathbf{N}, k \geqslant 107)$.
Suppose $k=2^{r} \times 3^{s} \times k^{\prime}$, where $r, s$ are non-negative integers, and $k... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. Big Ears Tutu and his good friend ZHUANG ZHUANG come to the convenience store together, where the price of lollipops is: 1 yuan for one, 4 yuan for five, and 7 yuan for nine. Tutu has enough money to buy a maximum of 11 lollipops, and ZHUANG ZHUANG has enough money to buy a maximum of 23 lollipops. So, ZHUANG ZHUAN... | $9$ | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (20 points) Natural numbers $a, b$ satisfy $23a - 13b = 1$, find the minimum value of $a + b$.
untranslated text is retained in its original form. | 【Analysis】From $23 a-13 b=1$, we get $13 b=23 a-1=26 a-(3 a+1)$, leading to the form of the solution for $b$ in terms of $a$: $b=26 a \div 13-(3 a+1) \div 13=2 a-(3 a+1) \div 13$. Since $a$ and $b$ are both natural numbers, $3 a+1$ must be divisible by 13. Clearly, the smallest value for $a$ is $4$, and $b$ simultaneou... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) A school has two classes each in the third and fourth grades. Class 3-1 has 4 more students than Class 3-2, Class 4-1 has 5 fewer students than Class 4-2, and the third grade has 17 fewer students than the fourth grade. Therefore, Class 3-1 has fewer students than Class 4-2 by ____. | 【Answer】Solution: $4+17=21$ (people)
$$
\begin{array}{l}
(21+5) \div 2 \\
= 26 \div 2 \\
= 13 \text { (people) } \\
13-4=9 \text { (people) }
\end{array}
$$
Answer: Class 1 of Grade 3 has 9 fewer people than Class 2 of Grade 4.
Therefore, the answer is: 9. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (6 points) The calculation result of the expression $2016 \times\left(\frac{1}{4}-\frac{1}{8}\right) \times\left(\frac{1}{7}-\frac{1}{9}\right)$ is | 【Solution】Solve: $2016 \times\left(\frac{1}{4}-\frac{1}{8}\right) \times\left(\frac{1}{7}-\frac{1}{9}\right)$
$$
\begin{array}{l}
=63 \times 8 \times 4 \times\left(\frac{1}{4}-\frac{1}{8}\right) \times\left(\frac{1}{7}-\frac{1}{9}\right) \\
=4 \times\left[\left(\frac{1}{4}-\frac{1}{8}\right) \times 8\right] \times\left... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. For the point sets $A=\left\{(x, y) \mid y=-3 x+2, x \in \mathbf{N}_{+}\right\}, B=\left\{(x, y) \mid y=m\left(x^{2}-x+1\right), x \in \mathbf{N}_{+}\right\}$, prove: there exists a unique non-zero integer, such that $A \cap B \neq \varnothing$. | 10. If $A \cap B \neq \phi$, then there must exist $\left(x_{0}, y_{0}\right)$ such that
$$
\left\{\begin{array}{l}
y_{0}=-3 x_{0}+2 \\
y_{0}=m\left(x_{0}^{2}-x_{0}+1\right)
\end{array}\right.
$$
Thus, we only need to discuss the conditions under which the above system of equations has a solution that meets the condit... | -1 | Algebra | proof | Yes | Yes | olympiads | false |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$ and $a_{1}=9$, with the sum of its first $n$ terms being $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is
A. 5
B. 6
C. 7
D. 8 | 2. C From $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right), a_{1}-1=8$ we know that the sequence $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$, so $S_{n}-n=\frac{8 \times\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{1+\frac{1}{3}}=6-6 \times\left(-\frac{1}{3}\right... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
13.6.3 * The equation $y^{2}-4 x-2 y+1=0$, represents what kind of curve? If the line $y=k x+2$ is tangent to the curve represented by the equation, what should $k$ be? | Analysis one: Transform the equation $y^{2}-4 x-2 y+1=0$ into $x=\frac{1}{4}(y-1)^{2}$. Therefore, the original equation represents $(2 k-4) x+1=0$. (3) For the line to be a tangent to the parabola, the system of equations must have the same real solution, meaning equation (3) must have two identical real roots (i.e., ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For example, how many pairs of integer solutions $(x, y)$ does the equation $\sqrt{x}+\sqrt{y}=\sqrt{200300}$ have? (2003 Singapore Mathematical Olympiad) | Analysis We use squaring to remove the square root.
Solution From the given, we have: $\sqrt{x}=\sqrt{200300}-\sqrt{y}$, hence $x=200300+y-20 \sqrt{2003 y}$.
Since 2003 is a prime number, we can set $y=2003 a^{2}(0 \leqslant a \leqslant 10)$, then $x=2003(10-a)^{2}$. Therefore, there are 11 pairs of integer solutions.
... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Let $x, y \in \mathbf{R}$, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{2003}+2002(x-1)=-1, \\
(y-2)^{2003}+2002(y-2)=1,
\end{array}\right.
$$
then $x+y=$ $\qquad$ | 7. Construct the function $f(t)=t^{2003}+2002 t$, it is easy to see that $f(t)$ is an odd function on $\mathbf{R}$, and also a monotonically increasing function. From this, we can get $f(x-1)=-f(y-2)$, which means $f(x-1)=f(2-y)$, hence $x-1=2-y, x+y=3$. Therefore, the answer is 3.
8. Since $x$ is an acute angle, then ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. A three-digit number $a b c$ satisfies $a b c=a+b^{2}+c^{3}$. Then the number of three-digit numbers $\overline{a b c}$ that satisfy the condition is $\qquad$ . | 8. 4 .
$$
\begin{array}{l}
\text { Given } \\
\overline{a b c}=a+b^{2}+c^{3} \\
\Leftrightarrow 100 a+10 b+c=a+b^{2}+c^{3} \\
\Leftrightarrow(c-1) c(c+1)=99 a+b(10-b) .
\end{array}
$$
Therefore, $99 a \leqslant(c-1) c(c+1) \leqslant 99 a+25$.
When $a=1$,
$$
99 \leqslant(c-1) c(c+1) \leqslant 124 \Rightarrow c=5 \text... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. (8 points) If the difference between two prime numbers is exactly 2, these two prime numbers are called a pair of twin primes.
For example, 3 and 5 are a pair of twin primes, and 29 and 31 are also a pair of twin primes. In number theory research, twin primes are one of the hottest research topics. Chinese-American ... | 【Answer】Solution: Among the integers not exceeding 100, the following 8 pairs: 3, 5; 5, 7; 11, 13; 17, 19; 29, 31; 41, 43; 59, 61; 71, 73 are twin primes. Therefore, the answer is 8. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Given a positive real number $x$ satisfies
$$
x^{3}+x^{-3}+x^{6}+x^{-6}=2754 \text {. }
$$
then $x+\frac{1}{x}=$ $\qquad$ | 3. 4 .
Transform the given equation into
$$
\begin{array}{l}
\left(x^{3}+x^{-3}\right)^{2}+\left(x^{3}+x^{-3}\right)-2756=0 \\
\Rightarrow\left(x^{3}+x^{-3}+53\right)\left(x^{3}+x^{-3}-52\right)=0 .
\end{array}
$$
Notice that, $x^{3}+x^{-3}+53>0$.
Therefore, $x^{3}+x^{-3}=52$.
Let $b=x+x^{-1}$.
Then $x^{3}+x^{-3}=\le... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A rope, the first time cut off $\frac{1}{3}$ of its total length, the second time cut off $30\%$ of the remaining part. If the parts cut off twice are 0.4 meters more than the remaining part, then the original length of the rope is $\qquad$ meters. | 【Solution】Solution: The second cut represents the following fraction of the total length:
$$
\begin{array}{l}
\left(1-\frac{1}{3}\right) \times 30 \% \\
= \frac{2}{3} \times \frac{3}{10} \\
= \frac{1}{5}, \\
0.4 \div\left[\frac{1}{3}+\frac{1}{5}-\left(1-\frac{1}{3}-\frac{1}{5}\right)\right] \\
= 0.4 \div\left[\frac{8}{... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let $p, \alpha, m \in \mathbf{Z}_{+}$, denote $p^{\alpha} \| m$ to mean $p^{\alpha} \mid m$, but $p^{\alpha+1} \nmid m$. For any positive integer $n$, the maximum value of the positive integer $\alpha$ satisfying $2^{\alpha} \|\left(3^{n}+1\right)$ is $\qquad$ | 4. 1 .
Notice that the square of an odd number is congruent to 1 modulo 4.
If $n=2 k\left(k \in \mathbf{Z}_{+}\right)$, then
$$
3^{n}+1=3^{2 k}+1=\left(3^{k}\right)^{2}+1 \equiv 2(\bmod 4) \text {. }
$$
At this point, $2 \|\left(3^{n}+1\right)$.
If $n=2 k+1\left(k \in \mathbf{Z}_{+}\right)$, then
$$
\begin{array}{l}
... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. There are 11 ones, 22 twos, 33 threes, and 44 fours on the blackboard. Perform the following operation: each time, erase 3 different numbers, and write 2 more of the fourth number that was not erased. For example: if you erase 1 one, 1 two, and 1 three, then write 2 more fours. After several operations, only 3 numbe... | Analysis: Operation problem.
By backtracking the simplest case, there are only 3 numbers left on the blackboard, and it is impossible to continue the operation according to the rules. There are only two situations: 2 $\mathrm{a}$s and one b, or 3 $\mathrm{a}$s. We perform the operation according to the rules. When only... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
19. In the park, there are several dogs and some ducks, and they have a total of 22 legs. If each dog hides its two hind legs, and all the ducks jump into the water, the number of visible legs decreases by 14. There are $\qquad$ ducks in the park. | $3$ | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
14. From the 20 numbers 11, 12, 13, 14, ... 30, at least ( ) numbers must be taken to ensure that there are two numbers whose sum is a multiple of ten. | 【Analysis】(11, 29) (12, 28) (13, 27) (14, 26) (15, 25) (16, 24) (17, 23) (18, 22) (19, 21) (20, 30), there are 10 drawers. By the drawer principle, selecting 11 numbers, there must be 2 numbers falling into the same drawer, and the sum of the 2 numbers in the same drawer is a multiple of ten. If taking 10 numbers, one ... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. As shown in the figure, it is a chessboard. The chess piece "Knight" can move from one square, jumping 2 squares horizontally or vertically, and at the same time jumping 1 square in the other direction, forming an "L" shape. For example, from point A in the figure, it can reach any of the B positions in one step. T... | $6$ | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Ahua and Ahua did the same number of questions. For each correct answer, 10 points are added, and for each wrong answer, 5 points are deducted. In the end, Ahua's score was 30 points higher than Aflower's. It is known that Ahua got 5 questions right, so Aflower got $\qquad$ questions right. | 【Analysis】 $30 \div(10+5)=2$ (questions), Ahua got $5-2=3$ questions correct. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Example 2.4.1] There is a rectangular iron sheet with dimensions $80 \times 50$. We need to cut out a square of the same size from each of the four corners and then fold it to form an open-top box. What should be the side length of the square cut out from each corner to maximize the volume of the open-top box? | We can set the length of the cut edges to be $x$, and the volume of the lidless box to be $V$, then
$$
V=x(80-2 x)(50-2 x) \text {. }
$$
How can we find the maximum value of $V$? To find the maximum value of the product, using the arithmetic mean-geometric mean inequality, we should make the "sum" a constant. Generall... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the tetrahedron $P-ABC$, it is known that $BC=3$, $CA=4$, $AB=5$. If the dihedral angles between the three lateral faces and the base are all $45^{\circ}$, then the volume of the tetrahedron $P-ABC$ is $\qquad$ | 4.2.
Let $P O \perp$ plane $A B C$ at point $O, O D \perp B C$ at point $D, O E \perp C A$ at point $E, O F \perp A B$ at point $F$.
Let $O P=h$.
Then $\angle P D O=\angle P E O=\angle P F O=45^{\circ}$.
Thus, $O D=O E=O F=h \cot 45^{\circ}=h$.
In $\triangle A B C$,
$$
\begin{array}{l}
3 O D+4 O E+5 O F=2 S_{\triangle... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Given that the opposite number of a root of $x^{2}-4 x+b=0$ is a root of $x^{2}+4 x-b=0$, the positive root of $x^{2}+$ $b x-4=0$ is $\qquad$ . | 8. 2 .
Let $x_{0}$ be a root of $x^{2}-4 x+b=0$, then $x_{0}^{2}-4 x_{0}+b=0$. Also, $-x_{0}$ is a root of $x^{2}+4 x-b=0$, then $x_{0}^{2}-4 x_{0}-b=0$. Therefore, $b=0$, the equation $x^{2}+b x-4=0$ becomes $x^{2}-4=0$, the positive root is 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(7) Let the function $f(x)=\left\{\begin{array}{l}\frac{1}{p}\left(x=\frac{q}{p}\right), \\ 0\left(x \neq \frac{q}{p}\right),\end{array}\right.$ where $p, q$ are coprime (prime), and $p \geqslant 2$. Then the number of $x$ values that satisfy $x \in[0,1]$, and $f(x)>\frac{1}{5}$ is $\qquad$ | (7) 5 Hint: Clearly, $x=\frac{q}{p}$ (otherwise $f(x)=0$). At this point, from $f(x)=\frac{1}{p}=\frac{1}{5}$, we get $p<5$, i.e., $p=2,3,4$.
When $p=2$, $x=\frac{1}{2}$; when $p=3$, $x=\frac{1}{3}, \frac{2}{3}$; when $p=4$, $x=\frac{1}{4}, \frac{3}{4}$. Therefore, there are 5 values of $x$ that satisfy the condition. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest positive integer $n$ such that $x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2002}$ has integer solutions.
(IMO Shortlist, 43rd) | Analysis We first estimate the value of $n$, then construct a solution.
Solution $2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9), 2002=667 \times 3+1$,
so $2002^{2002} \equiv 4^{2002} \equiv 4(\bmod 9)$.
Also, $x^{3} \equiv 0, \pm 1(\bmod 9)$, where $\quad x \in \mathbf{Z}$.
Thus, $x_{1}{ }^{3}, x_{1}{ }^{3}+x_{2}{ }^... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 1 Given $a b c=1$, find the value of $\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1}$. | Replace 1 with $a b c$ in the problem, we get
$$
\text { original expression }=\frac{a}{a b+a+a b c}+\frac{b}{b c+b+1}+\frac{b c}{b c+b+1}=\frac{b c+b+1}{b c+b+1}=1 \text {. }
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. For any real numbers $x, y$, define the operation $x * y$ as $x * y=a x+b y+c x y$, where $a, b, c$ are constants, and the operations on the right side of the equation are the usual real number addition and multiplication. It is known that $1 * 2=3, 2 * 3=4$, and there is a non-zero real number $d$, such that for an... | 4. For any real number $x$, we have
$$
\begin{array}{cc}
& \left.x * d=a x+b d+c d x=x_{1} \quad \text { (where } d \neq 0\right), \\
\therefore & 0 * d=b d=0 . \\
\because & d \neq 0, \quad \therefore \quad b=0 .
\end{array}
$$
Thus, from
we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
1 * 2=a+2 b+2 c=3, \\
2 * ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(5) Let $f(x)$ be an odd function defined on $\mathbf{R}$, $f(1)=2$, when $x>0$, $f(x)$ is an increasing function, and for any $x, y \in \mathbf{R}$, we have $f(x+y)=f(x)+$ $f(y)$, then the maximum value of the function $f(x)$ on the interval $[-3,-2]$ is $\qquad$. | $5-4$ Hint: Since $f(x)$ is an odd function and is increasing on $(0,+\infty)$, $f(x)$ is also increasing on $(-\infty, 0)$, thus $f(-3) \leqslant$ $f(x) \leqslant f(-2)$.
Also, $f(2)=f(1)+f(1)=4$, so $f(-2)=-f(2)=-4$.
Therefore, the maximum value of the function $f(x)$ on $[-3,-2]$ is -4. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) As shown in the figure, it is a rectangular paper sheet with a grid of $3 \times 4$. The front side of the rectangular paper sheet is gray, and the back side is red. The grid consists of identical small squares. Cut the rectangle along the grid lines into two cards of the same shape. If the shape and the... | 【Analysis】First analyze the subtraction to the left, then derive the cutting methods to the right based on the left-right symmetry, subtract the overlapping cutting methods, and thus obtain the total number of different cutting methods.
【Solution】Solution: First consider cutting from the front, the middle thick line m... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. The function $y=2 \sin (\omega x+\varphi)(\omega>0,0 \leqslant \varphi<\pi)$ is an even function, and part of its graph is shown in Figure 1-2-5. The distance between points $A$ and $B$ is $4 \sqrt{2}$. Then, one of the symmetry axes of the function is
A. $x=\frac{\pi}{2}$
B. $x=2$
C. $x=4$
D. $x=\pi$ | 4. C Given the function $y=2 \sin (\omega x+\varphi)(\omega>0, 0 \leqslant \varphi<\pi)$ is an even function, so $2 \sin (\omega x+\varphi)=2 \sin (-\omega x+\varphi)$.
Expanding gives $\sin \omega x \cos \varphi+\cos \omega \pi \sin \varphi=\sin \omega x \cos \varphi-\cos \omega x \sin \varphi$.
$\sin \omega x \cos \v... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
Numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=4$, then the maximum value of $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$ is | 12. 12.
Garden $(a \quad b)^{\prime}+(b-c)^{2}+(c-a)^{2}=2\left(a^{2}+b^{2}+c^{2}\right)-(2 a b+2 b c+2 c a)=3\left(a^{2}+b^{2}+c^{2}\right)$ $-(a-b+c)^{2} \leqslant 3 \times 4=12$. When $a+b+c=0$ and $a^{2}+b^{2}-c^{2}=4$, equality is achieved (for example, take $a=0$, $b=-(-\sqrt{2})$. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Given the sets $M=\{x, x y, \lg x y\}, N=\{0,|x|, y\}$, and $M=N$, then the value of $\left(x+\frac{1}{y}\right)+\left(x^{2}+\frac{1}{y^{2}}\right)+\left(x^{3}+\right.$ $\left.\frac{1}{y^{3}}\right)+\cdots+\left(x^{2 n+1}+\frac{1}{y^{2 n+1}}\right)$ is $\qquad$. | 9. From the equality of sets, we know that the elements of the two sets are the same. Thus, there must be an element 0 in $M$. Also, by the definition of logarithms, $x y \neq 0$, so $x, y$ are not zero. Therefore, only $\lg x y=0$, which means $x y=1$. Then, $M=\{x, 1,0\}, N=\left\{0,|x|, \frac{1}{x}\right\}$. By $M=N... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. The figure on the right can be drawn in one stroke, there are $\qquad$ different ways to draw it in one stroke (if the starting point, ending point, or order is different, it counts as a different way).
The translation preserves the original text's line breaks and format. | 【Analysis】12 kinds
Exam point: One-stroke drawing;
First, name each point in the figure as follows:
Since points $A, B$ are both odd points, the drawing method must start from $A$ and end at $B$, or start from $B$ and end at $A$, and it is not difficult to think that the number of drawing methods for these two cases is... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
69. Let $f(x)=(x-1)(x-3)(x-5)(x-7)(x-9)$, then $f(0)+f(1)+f(2)+\ldots+f(10)=$ | Answer: 0.
Solution: From the given conditions, we get $f(1)=f(3)=f(5)=f(7)=f(9)=0$,
and since
$$
f(0)+f(10)=0, \quad f(2)+f(8)=0, \quad f(4)+f(6)=0,
$$
thus
$$
f(0)+f(1)+f(2)+\ldots+f(10)=0 .
$$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Given the sequence $\left\{a_{n}\right\}$, the odd terms form an arithmetic sequence with a common difference of $d_{1}$, and the even terms form an arithmetic sequence with a common difference of $d_{2}$. For any $n \in \mathbf{N}^{*}$, it is true that $a_{n}<a_{n+1}$. If $a_{1}=1, a_{2}=2$, and the sum of the firs... | $$
\begin{array}{l}
a_{1}+a_{3}+a_{5}+a_{7}+a_{9}=5 a_{1}+10 d_{1}=5+10 d_{1}, a_{2}+a_{4}+a_{6}+a_{8}+a_{10}=5 a_{2}+10 d_{2} \\
=10+10 d_{2} \Rightarrow 15+10\left(d_{1}+d_{2}\right)=75 \Rightarrow d_{1}+d_{2}=6 .
\end{array}
$$
For $n=2 k-1$, $a_{1}+(k-1) d_{1}1-d_{1}\end{array}\right.$ holds for any $k \in \mathbf... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.15 $\star \star$ For what minimum value of the natural number $n$ does the equation
$$
\left[\frac{10^{n}}{x}\right]=1989
$$
have an integer solution? | The smallest $n=7$. Clearly, $n \geqslant 4$, and $1989 \leqslant \frac{10^{n}}{x}<1990$, which means $\frac{10^{n}}{1990}<x \leqslant \frac{10^{n}}{1989}$. It can be verified that for $n=4,5,6$, the interval $\left(\frac{10^{n}}{1990}, \frac{10^{n}}{1989}\right]$ does not contain any integers. When $n=7$, $x=5026$ and... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given real numbers $x, y$ satisfy
$$
\frac{4}{x^{4}}-\frac{2}{x^{2}}=3, y^{4}+y^{2}=3 \text {. }
$$
Then the value of $\frac{4}{x^{4}}+y^{4}$ is $\qquad$ | By observation, we know that $-\frac{2}{x^{2}}$ and $y^{2}$ are two different solutions of the equation $m^{2}+m=3$.
By Vieta's formulas, we have
$$
\begin{array}{l}
\frac{4}{x^{4}}+y^{4}=m_{1}^{2}+m_{2}^{2}=\left(m_{1}+m_{2}\right)^{2}-2 m_{1} m_{2} \\
=(-1)^{2}-2(-3)=7 .
\end{array}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. 7511 sets $M_{1}, M_{2}, \cdots, M_{11}$. Each set has 5 elements, and the intersection of any two sets is non-empty. Find the minimum possible value of the maximum number of sets that have a common element. | [Solution] Let $n(x)$ denote the number of sets containing the element $x$. Let $T$ be the set of all elements in the 11 sets $M_{1}, M_{2}, \cdots, M_{11}$, i.e.,
$$
T=\bigcup_{i=1}^{11} M_{i} \text {. }
$$
From the problem, we know that,
$$
\sum_{x \in T} n(x)=5 \cdot 11=55.
$$
$C_{n(x)}^{2}$ is the number of pairs ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. In a certain archery competition, the full score is 10 points, and in the preliminary stage, 50% of all participants are eliminated. It is known that the average score of the contestants who advance to the finals is 2 points higher than the average score of all contestants, and the average score of the contestants w... | 【Analysis】Suppose there are $2 n$ people in total, then the number of participants who enter the semi-final is $n$ people, and the number of participants who are eliminated is also $n$ people. The average score of all participants is 6 points, so the total score is $6 \times 2 n=12 n$ points. The total score of the par... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16.21 In $\triangle A B C$, $\angle A>\angle B>\angle C, \angle A \neq 90^{\circ}$, draw a line to divide $\triangle A B C$ into two parts, such that one part is similar to $\triangle A B C$. The number of such non-parallel lines is
(A) 3.
(B) 4.
(C) 5.
(D) 6.
(China Guangzhou, Wuhan, Fuzhou, and other five cities juni... | ［Solution］Such straight lines can be divided into three categories:
(1) Those parallel to one side of the triangle, totaling three;
(2) Two lines starting from the largest angle $\angle A$ meet the requirement, see the following figure (1) (note $\angle A$ $\left.\neq 90^{\circ}\right):$
(3) One line starting from the ... | 6 | Geometry | MCQ | Yes | Yes | olympiads | false |
$28 \cdot 2$ In the following expression, different letters represent different digits, then $C$ is
\begin{tabular}{rrr}
$A$ & $B$ & $C$ \\
$A$ & $B$ \\
+ & & $A$ \\
\hline \multirow{3}{3}{$\quad 0$} & 0
\end{tabular}
(A) 1 .
(B) 3 .
(C) 5 .
(D) 7 .
(E) 9 .
(7th American Junior High School Mathematics Examination, 1991... | [Solution 1] From the equation, we can see that $A=3$ or 2 (if there is a carry, it can be at most 1).
$\because B+A$ plus the carry digit from $A+B+C$ (which can be at most 2) and the last digit is 0, it will carry over, so $A=2$.
Given $A=2$, we know $A+B+C \leqslant 20$, and the carry does not exceed 2. Therefore, ... | 1 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
(8) As shown in the figure, in $\triangle A B C$, $A B=3, A C=$ 5. If $O$ is the circumcenter of $\triangle A B C$, then the value of $\overrightarrow{A O} \cdot \overrightarrow{B C}$ is $\qquad$ | 88 Prompt: As shown in the figure, let $D$ be the midpoint of $B C$, and connect $O D, A D$, then $O D \perp B C$. Therefore,
$$
\begin{aligned}
\overrightarrow{A O} \cdot \overrightarrow{B C} & =(\overrightarrow{A D}+\overrightarrow{D O}) \cdot \overrightarrow{B C} \\
& =\overrightarrow{A D} \cdot \overrightarrow{B C}... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
18.6.5 ** Let $a, b, c, d$ be odd numbers, $0<a<b<c<d$, and $ad=bc$. Prove: If for some integers $k$ and $m$ there are $a+d=2^k, b+c=2^m$, then $a=1$.
---
The translation is provided as requested, maintaining the original formatting and structure. | Since $a((a+d)-(b+c))=a^{2}+da-ab-ac=a^{2}+bc-ab-ac=(a-b)(a-c)>0$, it follows that $a+d>b+c$, which means $2^{k}>2^{m}, k>m$. Also, from $ad=bc$ we have $a(2^{k}-a)-b(2^{m}-b)$, thus $2^{m}(b-2^{k-m}a)=b^{2}-a^{2}=(b+a)(b-a)$, hence $2^{m}$ divides $(b+a)(b-a)$. However, $b+a$ and $b-a$ cannot both be divisible by 4 (s... | 1 | Number Theory | proof | Yes | Yes | olympiads | false |
9. (Adapted from the 1st "Hope Cup" Senior High School Competition) Let the function $f(n)=k$, where $n$ is a natural number, and $k$ is the digit at the $n$-th position after the decimal point of the irrational number $\pi=3.1415926535 \cdots$, with the rule that $f(0)=3$. Let $F_{n}=$ $\underbrace{f\{f\{f\{f\{f}(n)\}... | 9. From the problem, for any $n, f(n)=k$ is a function defined on the set of non-negative integers, taking values in $\{0,1,2,3,4,5,6,7,8,9\}$.
Let $f_{(2)}(n)=f(f(n)), f_{(3)}(n)=f\{f(f(n))\}, f_{(k)}(n)$ denote the $k$-fold composition of $f$.
Then $f_{(2)}(n)$ is a function defined on $\{0,1,2,3,4,5,6,7,8,9\}$, taki... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. The number of real roots of the equation $8 x\left(2 x^{2}-1\right)\left(8 x^{4}-8 x^{2}+1\right)=1$ is | 7. 7 detailed solutions: Let $x=\cos \theta$, then we get $8 \cos \theta \cos 2 \theta \cos 4 \theta=1$, so $\sin \theta=\sin 8 \theta \neq 0$, thus $\theta=\frac{k \pi}{8-(-1)^{k}}, k \in \mathbf{Z}$, hence $\theta=\frac{\pi}{9}, \frac{2 \pi}{7}, \frac{\pi}{3}, \frac{4 \pi}{7}, \frac{5 \pi}{9}, \frac{6 \pi}{7}, \frac{... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest positive integer $n$, such that $x_{1}{ }^{3}+x_{2}{ }^{3}+\cdots+x_{n}{ }^{3}=2002^{2002}$ has integer solutions. | Analysis We first estimate the value of $n$, then construct a solution.
Solution $2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9), 2002=667 \times 3+1$,
so $2002^{2002} \equiv 4^{2002} \equiv 4(\bmod 9)$.
Also, $x^{3} \equiv 0, \pm 1(\bmod 9)$, where $\quad x \in \mathbf{Z}$.
Thus, $x_{1}{ }^{3}, x_{1}{ }^{3}+x_{2}{ }^... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. The equation $(a-1)(\sin 2 x+\cos x)+(a+1) \cdot(\sin x-\cos 2 x)=0$ (parameter $a<0$) has $\qquad$ solutions in the interval $(-\pi, \pi)$. | 8,4
The original equation can be transformed into $\sin \left(\frac{x}{2}+\frac{\pi}{2}\right)\left[\cos \left(\frac{3 x}{2}-\frac{\pi}{4}\right)+\frac{a+1}{a-1} \sin \left(\frac{3 x}{2}-\frac{\pi}{4}\right)\right]=0$.
(1) The equation $\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)=0$ has a solution $x=-\frac{\pi}{2}$ in... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 8 Let $R_{n}=\frac{1}{2}\left(a^{n}+b^{n}\right), a=3+2 \sqrt{2}, b=3-2 \sqrt{2}, n=1,2, \cdots$ What is the units digit of $R_{12345}$?
(41st AHSME Problem) | Given that $a$ and $b$ are the roots of the equation $x^{2}=6 x-1$, according to Newton's formula, we have
$$
R_{n+2}=6 R_{n+1}-R_{n} \text {. }
$$
It is easy to see that $R_{1}=\frac{1}{2}(a+b)=3, R_{2}=\frac{1}{2}\left(a^{2}+b^{2}\right)=17$. For convenience in writing, let $D_{n}$ represent the unit digit of $R_{n}... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Given $a, b \in \mathbf{R}$, the circle $C_{1}: x^{2}+y^{2}-2 x+4 y-b^{2}+5=0$ intersects with $C_{2}: x^{2}+y^{2}-2(a-6) x-2 a y$ $+2 a^{2}-12 a+27=0$ at two distinct points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and $\frac{y_{1}+y_{2}}{x_{1}+x_{2}}+\frac{x_{1}-x_{2}}{y_{1}-y_{2}}$ $=0$, then $a=... | 2. $a=4$ Detailed explanation: Let the midpoint of $A B$ be $M$, and the origin of coordinates be $O$. Then, from $\frac{y_{1}+y_{2}}{x_{1}+x_{2}}+\frac{x_{1}-x_{2}}{y_{1}-y_{2}}=0$ we get $M O \perp A B$, so $O, M, C_{1}, C_{2}$ are collinear. From $k_{\alpha_{1}}=k_{\alpha_{2}}$ we get $a=4$. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
79. As shown in the figure, in rectangle $A B C D$, $A B=4, B C=6, E, F, G, H$ are on $A B, B C, C D, D A$ respectively, and $A E: E B=3: 1, B F: F C=2: 1, D G: G C=1: 3, A H: H D=1: 2, P$ is on $H F$, and the area of quadrilateral $A E P H$ is 5, find the area of quadrilateral $P F C G$. | Answer: 8.
Solution: As shown in the figure, connect $E H, E F, G H, G F$ and use $S_{1}, S_{2} \cdots S_{8}$ to represent the areas of these 8 small triangles.
It is easy to know that $A E=3, E B=1, B F=4, F C=2, C G=3, D G=1, H D=4, A H=2$,
$$
\begin{array}{c}
S_{1}+S_{2}=5, \quad S_{1}=\frac{1}{2} \times 2 \times 3... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 5 Given the set $M=\left\{x \mid x^{2}-2 x+m=0, x \in \mathbf{R}\right\}$ is non-empty, then the sum of all elements in set $M$ is $\qquad$ . | When $m=1$, the set $M$ contains only the unique element 1;
When $m \neq 1$, the set $M$ contains two different elements $x_{1}$ and $x_{2}$, and $x_{1}+x_{2}=2$. Therefore, the sum of all elements in the set is 1 or 2.
Solution Guide: Generally, for a non-empty set $\left\{x \mid a x^{2}+b x+c=0, x \in \mathbf{R}\rig... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The number of integer solutions $(x, y)$ for the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{7}$ is ( ).
(A) 5
(B) 6
(C) 7
(D) 8 | 4. A.
When $x=7$, there is no solution.
When $x \neq 7$, $y=\frac{7 x}{x-7}$.
Thus, $(x-7)|7 x \Rightarrow(x-7)| 49$
$$
\begin{array}{l}
\Rightarrow x=-42,6,8,14,56 \\
\Rightarrow(x, y)=(-42,6),(6,-42),(8,56), \\
(14,14),(56,8) .
\end{array}
$$
In summary, there are 5 solutions. | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 4 Try to find the minimum value of the function $f(x, y)=6\left(x^{2}+y^{2}\right)(x+y)-4\left(x^{2}+x y+y^{2}\right)-3(x+y)+5$ in the region $D=\{(x, y) \mid x>0, y>0\}$. | Notice that $x>0, y>0$, without loss of generality, let $x+y=k(k>0)$, it is obvious that
$$
x \cdot y \leqslant \frac{1}{4}(x+y)^{2}=\frac{1}{4} k^{2}, x^{2}+y^{2} \geqslant \frac{1}{2}(x+y)^{2}=\frac{1}{2} k^{2} \text {. }
$$
Consider taking $k=1$, that is, when $x+y \leqslant 1$, we have $x y \leqslant \frac{1}{4}(x... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (8 points) The 6 vertices of the hexagram coincide exactly with the 6 vertices of a regular hexagon. Then the area of the shaded part is $\qquad$ times the area of the blank part.
The translation is provided as requested, maintaining the original formatting and structure. | 【Answer】Solution: According to the analysis, as shown in the figure, divide the figure into triangles of equal area. The shaded part consists of 18 small triangles, while the blank part has 6 small triangles. Therefore, the area of the shaded part is $18 \div 6=3$ times the area of the blank part. The answer is: 3. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. For all $m, n$ satisfying $1 \leqslant n \leqslant m \leqslant 5$, the polar equation
$$
\rho=\frac{1}{1-C_{m}^{n} \cos \theta}
$$
represents the number of different hyperbolas is
(A) 15 ;
(B) 10 ;
(C) 7 ;
(D) 6 . | 3. (D)
The polar equation of a conic section $\rho=\frac{p}{1-e \cos \theta}$,
represents a hyperbola if and only if $e>1$. Therefore, the equation in this problem must satisfy
$$
C_{m}^{n}>1 \text {, }
$$
Given that
$$
1 \leqslant n \leqslant m \leqslant 5,
$$
we can obtain 6 different values:
$$
C_{3}^{1}, C_{3}^{... | 6 | Geometry | MCQ | Yes | Yes | olympiads | false |
3. (10 points) There is a pasture, 10 cows can finish the grass in 8 days; 15 cows, if one less cow starts from the second day, can finish in 5 days. Then the grass that grows on the pasture every day is enough for $\qquad$ cows to eat for one day.
| 【Analysis】By changing the perspective, 15 cows, if one less cow starts eating from the second day, can finish eating in 5 days, can be converted into 13 cows eating for 5 days to solve the problem.
【Solution】Solution: According to the problem, we have:
$$
10 \times 8-(15+14+13+12+11)=15 \text { (units). }
$$
15 cows, i... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (8 points) In a cage, there are 30 crickets and 30 grasshoppers. The Red-Haired Magician, with each transformation, turns 4 grasshoppers into 1 cricket; the Green-Haired Magician, with each transformation, turns 5 crickets into 2 grasshoppers. After the two magicians have transformed a total of 18 times, there are o... | 3. (8 points) In a cage, there are 30 crickets and 30 grasshoppers. The Red-Haired Magician, with each transformation, turns 4 grasshoppers into 1 cricket; the Green-Haired Magician, with each transformation, turns 5 crickets into 2 grasshoppers. After the two magicians have performed a total of 18 transformations, the... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Let the equation
$$
1+x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots-\frac{x^{2018}}{2018}=0
$$
have all its real roots within the interval $[a, b](a, b \in \mathbf{Z}$, $a<b)$. Then the minimum value of $b-a$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 5. C.
Let $f(x)=1+\sum_{i=1}^{2018}(-1)^{i-1} \frac{x^{i}}{i}$.
Then $f^{\prime}(x)=\sum_{i=0}^{2017}(-1)^{i} x^{i}$.
When $x<1$,
$$
f^{\prime}(x)=\sum_{i=0}^{1008} x^{2 i}(1-x)>0,
$$
hence $f(x)$ is an increasing function;
When $x=1$, $f^{\prime}(x)=0$;
When $x>1$,
$$
f^{\prime}(x)=\sum_{i=0}^{1008} x^{2 i}(1-x)<0, \... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
A1 Determine the positive integer $n$ such that $8^{4}=4^{n}$. | Solution: The answer is $n=6$.
Solution 1: Note that $8^{4}=\left(2^{3}\right)^{4}=2^{12}=4^{6}$. Therefore, $n=6$.
Solution 2: We write $8^{4}$ and $4^{n}$ as an exponent with base 2 .
$$
\begin{array}{ll}
8^{4} & =4^{n} \\
\left(2^{3}\right)^{4} & =\left(2^{2}\right)^{n} \\
2^{12} & =2^{2 n}
\end{array}
$$
Therefo... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\sin y \cos y+a=0$.
Then $\cos (x+2 y)=$ $\qquad$ . | Let $f(x)=x^{3}+\sin x-2 a$, then $f^{\prime}(x)=3 x^{2}+\cos x \geqslant 0, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ $\Rightarrow f(x)$ is an odd function and monotonically increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Also, $4 y^{3}+\sin y \cos y+a=0 \Rightarrow(-2 y)^{3}+\sin (-2 y)-2 a=0$,
th... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. (16 points) Let $n$ be a positive odd number, and the complex number $z$ satisfies $z^{2^{n}-1}=1$. Find $Z_{n}=\prod_{k=0}^{n-1}\left(z^{2^{k}}+\frac{1}{z^{2^{k}}}-1\right)$.
| Sure, here is the translated text:
```
9. Since $n$ is a positive odd number, we have
$$
2^{n}-1 \equiv(-1)^{n}-1 \equiv-2(\bmod 3) \text {. }
$$
Thus, $z^{3} \neq 1$.
Therefore, $z+\frac{1}{z}+1 \neq 0$.
By the difference of squares formula, we get
$$
\begin{array}{l}
\left(z+\frac{1}{z}+1\right) Z_{n} \\
=\left(z+\... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. (6 points) The product of 2013 eights is a very large number. The unit digit of this number is $\qquad$ | 【Solution】Solution: Starting from 2 eights, the unit digits of the product of multiple eights are $4,2,6,8$, these 4 numbers cycle;
$$
\begin{aligned}
& (2013-1) \div 4, \\
= & 2012 \div 4, \\
= & 503 ;
\end{aligned}
$$
There is no remainder, so the unit digit of the product of 2013 eights is the same as that of the p... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
18. Given the function $f(x)=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x$ is an even function, $g(x)=2^{x}+\frac{a+b}{2^{x}}$ is an odd function, where $a, b$ are constants, then $\sum_{k=1}^{2008}\left(a^{k}+b^{k}\right)$ is $\qquad$. | 18. Fill in 1. Reason: From $f(-x)=f(x)$ and $g(-x)=-g(x)$, we have $\log _{\frac{1}{3}}\left(3^{-x}+1\right)-\frac{1}{2} a b=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x$ and $2^{-x}+\frac{a+b}{2^{-x}}=-2\left(2^{x}+\frac{a+b}{2^{x}}\right)$, which implies $a b=1$ and $a+b=-1$.
Thus, $a$ and $b$ are the ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. If the inequality
$$
\left|a x^{2}+b x+a\right| \leqslant x
$$
holds for all $x \in[1,2]$, then the maximum value of $3 a+b$ is
$\qquad$ . | 4.3.
$$
\begin{array}{l}
\text { Given }\left|a x^{2}+b x+a\right| \leqslant x \\
\Rightarrow\left|a\left(x+\frac{1}{x}\right)+b\right| \leqslant 1,
\end{array}
$$
Given $x \in[1,2] \Rightarrow t=x+\frac{1}{x} \in\left[2, \frac{5}{2}\right]$.
Then $|2 a+b| \leqslant 1$, and $\left|\frac{5}{2} a+b\right| \leqslant 1$.
... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) A deck of playing cards, excluding the joker, has 4 suits totaling 52 cards, with each suit having 13 cards, numbered from 1 to 13. Feifei draws 2 hearts, 3 spades, 4 diamonds, and 5 clubs. If the sum of the face values of these 14 cards Feifei drew is exactly 35, then how many of them are 1?
$\qquad$ ... | 【Solution】Solution: According to the analysis, the minimum sum of the two hearts is $1+2=3$, the minimum sum of the three spades is $1+2+3=6$,
the minimum sum of the four diamonds is $1+2+3+4=10$, and the sum of the five clubs is: $1+2+3+4+5=15$, so the minimum sum of the 14 cards is: $3+6+10+15=34$,
(1) If only $1 \s... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14. Given that $a, b, c, d$ are rational numbers, $|a-b| \leq 9,|c-d| \leq 16$, and $|a-b-c+d|=25$, then $|d-c|-|b-a|=$ $\qquad$ . | answer: 7 | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
85. Mr. Zhou and Mr. Wang are walking along the school's 400-meter circular track. They start from the same point and walk in opposite directions. Mr. Zhou walks 120 meters per minute, and Mr. Wang walks 80 meters per minute. After 10 minutes, they have met $\qquad$ times. | Reference answer: 5 | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$4 \cdot 61$ Given that in the subway network, each line has at least 4 stations, of which no more than 3 are transfer stations, and at each transfer station, no more than two lines intersect. If from any station, one can reach any other station with at most two transfers, how many lines can this network have at most? | ［Solution］Fix one line, on which there are at most 3 stations available for transfer, so from this line, one can transfer to at most 3 other lines. And from each of these 3 lines, one can transfer to at most 2 new lines. Therefore, the total number of lines is no more than $1+3+3 \times 2=10$.
On the other hand, the n... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 6 Arrange the terms of the arithmetic sequence $2,6,10,14, \cdots, 2006$ tightly together to form a “large number”: $A=261014 \cdots 2006$. Find the remainder when $A$ is divided by 9. | First, we prove a lemma.
Lemma When a positive integer $n$ is arbitrarily divided into several groups, each group being a number formed by several consecutive digits of $n$, let the numbers corresponding to each group (from right to left) be $A_{1}, A_{2}, \cdots, A_{t}$, then
$$
n \equiv A_{1}+A_{2}+\cdots+A_{t}(\bmod... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10 As shown in the figure, in the triangular prism $A B C-A_{1} B_{1} C_{1}$, the lateral edge $A A_{1} \perp$ plane $A B C$, triangle $A B C$ is an equilateral triangle with a side length of 2, $M$ is a point on $A A_{1}$, $A A_{1}=4, A_{1} M=$ 1. $P$ is a point on the edge $B C$ such that the shortest path from $P$ a... | 10 (1) Since the lateral edge $A A_{1} \perp$ plane $A B C$, and triangle $A B C$ is an equilateral triangle, therefore, all the lateral faces are congruent rectangles. As shown in the figure, rotate the lateral face $B C_{1}$ by $120^{\circ}$ so that it lies in the same plane as the lateral face $A C_{1}$, and point $... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
30-39 In decimal, a six-digit number is called a "special square number" if it satisfies the following conditions:
(i) None of its digits are zero;
(ii) It is a perfect square;
(iii) The first two digits, the middle two digits, and the last two digits are all perfect squares when considered as two-digit numbers.
How ma... | [Solution] If $N$ is a special square number, by the problem's condition, there exist six positive one-digit integers $A$, $B$, $C$, $a$, $b$, $c$ satisfying
$$
N=10^{4} A^{2}+10^{2} B^{2}+C^{2}=\left(10^{2} a+10 b+c\right)^{2} .
$$
Let $d=10 b+c$, then $d \leqslant 99$. This necessarily implies $A=a$,
otherwise, if $... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
As shown in the figure, the vertices of the shaded square are the midpoints of the sides of the larger square $\mathrm{EFGH}$. Semi-circles are drawn outward with half the sides of the larger square as diameters, and then semi-circles are drawn outward with the sides of the shaded square as diameters, forming 8 "cresce... | 3. Answer: 10
Analysis: As shown in the figure, connect $\mathrm{AB}$ and $\mathrm{CD}$ intersecting at 0. It is easy to obtain the area of triangle $\mathrm{ACH}$ using the Pythagorean theorem and the formula for the area of a semicircle, which means the area of triangle $\mathrm{AOC}$ is equal to the sum of the area... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
(1) The sum of the x-coordinates of the points where the graph of the function $y=x^{2}-2011|x|+2012$ intersects the x-axis is $\qquad$ . | (10 Hint: The original problem can be transformed into finding the sum of all real roots of the equation
$$
x^{2}-2011|x|+2012=0
$$
If a real number $x_{0}$ is a root of equation (1), then its opposite number $-x_{0}$ is also a root of equation (1).
Therefore, the sum of all real roots of the equation is 0, that is, t... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. (10 points) 2012 students line up in a row to report numbers. If a student reports a single-digit number, the next student reports double that number; if a student reports a two-digit number, the next student reports the sum of the unit digit and 5. It is known that the first student reports 1, and by the time it r... | 【Solution】Solution: According to the rules, write down the numbers reported by the first few students: 1, $2,4,8,16,11,6,12,7$, $14,9,18,13,8,16 \cdots$ It can be observed that starting from the 5th student, every 10 students form a cycle, so the number reported by the 99th student is 7;
Since the last student reporte... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 7 Given the function $y=\sqrt{1+\sin x}+\sqrt{1-\sin x}$, find the maximum value of $y$.
(Singapore Competition Question) | Let $u=\sqrt{1+\sin x}, v=\sqrt{1-\sin x}$, then we have the line equation: $u+v-y=0$, and the circle equation: $u^{2}+v^{2}=2$.
From the given, we know that the line and the circle have a common point $(u, v)$, thus
$\frac{|y|}{\sqrt{2}} \leqslant \sqrt{2}$, i.e., $|y| \leqslant 2$, equality holds if and only if $x=0$... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10、Ants A, B, and C crawl at a speed ratio of $8: 6: 5$. They crawl along a circle from the same point and in the same direction at the same time. When they first return to the starting point together, the crawling ends. How many times does Ant A catch up with Ant B (including the end moment)? | 10 Solution: (1) The speed ratio of ants A, B, and C is $8: 6: 5$, so, when they first return to the starting point at the same time, ant A has moved 8 laps, and ant B has moved 6 laps. (2) Ant A catches up to ant B once for every extra lap it moves, so, A catches up to B 2 times in total. Answer: When the three ants f... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (2004 National High School Mathematics League) Let point $O$ be inside $\triangle A B C$, and $\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\mathbf{0}$. Find the ratio of the area of $\triangle A B C$ to the area of $\triangle A O C$. | 2. Extend $O B$ to $O B^{\prime}$ such that $\left|\overrightarrow{O B^{\prime}}\right|=2|\overrightarrow{O B}|$, and extend $C$ to $C^{\prime}$ such that $\left|\overrightarrow{O C^{\prime}}\right|=3|\overrightarrow{O C}|$. By vector knowledge, $O$ is the centroid of $\triangle A B^{\prime} C^{\prime}$, so $S_{A O B^{... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{2}=3$, and $a_{n+2}=\left|a_{n+1}-a_{n}\right|\left(n \in \mathbf{Z}_{+}\right)$.
Then $a_{2014}=$ $\qquad$ | -1.1 .
It is easy to see that $a_{1}=1, a_{2}=3, a_{3}=2, a_{4}=1, a_{5}=1$,
$$
a_{6}=0, a_{7}=1, a_{8}=1, a_{9}=0, \cdots \cdots
$$
Therefore, starting from the fourth term, every three consecutive terms periodically take the values $1, 1, 0$.
Thus, $a_{2014}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.34 A particle is placed at a point P on the parabola $y=x^{2}-x-6$ (the y-coordinate of P is 6), if the particle can freely roll along the parabola to point Q (the y-coordinate of Q is -6), then the shortest horizontal distance the particle moves is
(A) 5 .
(B) 4 .
(C) 3 .
(D) 2 .
(E) 1 .
(9th American High School Ma... | [Solution] Suppose the coordinates of point $P$ are $(4,6)$ or $(-3,6)$; and the coordinates of point $Q$ are $(0,-6)$ or $(1,-6)$.
By symmetry and according to the problem, the particle rolls from $(4,6)$ to $(1,-6)$ or from $(-3,6)$ to $(0,-6)$.
The horizontal distance the particle moves is
$4-1=3$ or $0-(-3)=3$.
Th... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
The teacher said to Xu Jun: "Two years ago, my age was three times your age." Xu Jun said to the teacher: "In eight years, your age will be twice mine." Xu Jun is $\qquad$ years old this year. | (Method One)
2 years ago, the teacher's age was 3 times Xu Jun's age, the age difference was 2 times Xu Jun's age (3-1=2); 8 years later, the teacher's age will be 2 times Xu Jun's age, the age difference will be 1 time Xu Jun's age (2-1=1); 8 years later, Xu Jun's age will be 2 times his age 2 years ago;
8 years later... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 1 Calculate: $\lg ^{2} 2 \cdot \lg 250+\lg ^{2} 5 \cdot \lg 40$. | \begin{array}{l} \text { Original expression }=\lg ^{2} 2 \cdot \lg \left(5^{2} \times 10\right)+\lg ^{2} 5 \cdot \lg \left(2^{2} \times 10\right) \\ =\lg ^{2} 2 \cdot(2 \lg 5+1)+\lg ^{2} 5 \cdot(2 \lg 2+1) \\ =\lg ^{2} 2+2 \lg 2 \cdot \lg ^{2} 5+2 \lg 5 \cdot \lg ^{2} 2+\lg ^{2} 5 \\ =\lg ^{2} 2+2 \lg 2 \cdot \lg 5 \c... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
78. A person is walking along the road, and a car comes from the opposite direction. He asks the driver: "Was there a bicycle behind you?" The driver answers: "I passed a bicycle 10 minutes ago." The person continues walking for 10 minutes and meets the bicycle. If the bicycle's speed is 3 times the walking speed, then... | answer: 7 | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
13.9 Given several sets with identical contents, each set consists of 4 objects with integer gram weights. Now, we want to use these objects as weights to measure every integer gram weight from 1 gram to 1985 grams (including 1985 grams). To minimize the total weight of the objects used as weights, how many different w... | ［Solution］Obviously, the total weight of the objects used as weights cannot be less than 1985 grams, and
$$
1985=5 \times 397 .
$$
Thus, the minimum total weight of 4 objects in each set is 5. The only selection of 4 objects with integer gram weights is $\{1,1,1,2\}$.
Therefore, there is only one way to select the wei... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The number of rational solutions $(x, y, z)$ for the system of equations $\left\{\begin{array}{l}x+y+z=0, \\ x y z+z=0, \\ x y+y z+x z+y=0\end{array} \quad\right.$ is
A. 1
B. 2
C. 3
D. 4 | Solve
$x y z+z=z(x y+1)=0 \Rightarrow z=0$ or $x y=-1$.
When $z=0$, $\left\{\begin{array}{l}x+y=0, \\ x y+y=y(x+1)=0\end{array} \Rightarrow\left\{\begin{array}{l}x=0, \\ y=0\end{array}\right.\right.$ or $\left\{\begin{array}{l}x=-1, \\ y=1 .\end{array}\right.$
When $x y=-1$, $\left\{\begin{array}{l}(x+y)^{2}=y-1, \\ x ... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
31-14 Let $a, b$ be natural numbers not exceeding 10. Then, the number of pairs $(a, b)$ that make the solution of the equation $a x=b$ greater than $\frac{1}{4}$ and less than $\frac{1}{3}$ is
(A) 2.
(B) 3.
(C) 4.
(D) 1.
（1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989) | [Solution] From the problem, we have $x=\frac{b}{a}$, then $\frac{1}{4}<\frac{b}{a}<\frac{1}{3}$, which means $3 b<a<4 b$. Also, $a, b$ are natural numbers not exceeding 10, so $3 b<10$, hence $b \leqslant 3$.
If $b=1$, then there is no solution for $a$;
If $b=2$, then $6<a<8$, we get $\quad a=7$;
If $b=3$, then $9<a<1... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
$31 \cdot 49$ satisfies the system of equations $\left\{\begin{array}{l}x y+x z=255 \\ x y+y z=31\end{array}\right.$, the number of positive integer solutions $(x, y, z)$ is
(A) 3.
(B) 2.
(C) 1.
(D) 0.
(Chinese Hubei Province Huanggang Junior High School Mathematics Competition, 1993) | ［Solution］The given system of equations can be transformed into
$$
\left\{\begin{array}{l}
x(y+z)=5 \cdot 3 \cdot 17, \\
y(x+z)=1 \cdot 31 .
\end{array}\right.
$$
Assuming the solutions to the equations are positive integers, then $x+z>1$. From (2), we know $y=1$, and $x+z=31$. The system of equations can be simplifie... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
34. As shown in Figure 11, the areas of square $A B C D$ and square $E F G H$ are 16 square centimeters and 4 square centimeters, respectively. $A H$ is 1 centimeter, and the area of triangle $D E G$ is $\qquad$ square centimeters. | Reference answer: 7 | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) If the expression $\frac{1}{1 \times 2}-\frac{1}{3 \times 4}+\frac{1}{5 \times 6}-\frac{1}{7 \times 8}+\cdots+\frac{1}{2007 \times 2008}$ is converted to a decimal, then the first digit after the decimal point is $\qquad$ . | 【Analysis】Based on the periodicity of addition and subtraction of the fractional sequence operation symbols, group the fractional sequence to find approximate values, and perform estimation.
【Solution】Solution: $\frac{1}{1 \times 2}-\frac{1}{3 \times 4} \approx 0.416$
$$
\begin{array}{l}
\frac{1}{5 \times 6}-\frac{1}{7... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. (10 points) There are 11 square formations, each consisting of the same number of soldiers. If 1 general is added, a large square formation can be formed. The minimum number of soldiers in one of the original square formations is $\qquad$.
| 【Analysis】This problem examines the issue of square arrays.
【Solution】Solution: According to the problem, let the original square array have $a$ soldiers, then $a$ and $11 a+1$ are perfect squares,
When $a=1$, $11 a+1=12$, which does not meet the requirement;
When $a=4$, $11 a+1=45$, which does not meet the requiremen... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2 Answer the following two questions and justify your answers:
(1) What is the last digit of the sum $1^{2012}+2^{2012}+3^{2012}+4^{2012}+5^{2012}$ ?
(2) What is the last digit of the sum $1^{2012}+2^{2012}+3^{2012}+4^{2012}+\cdots+2011^{2012}+2012^{2012}$ ? | Solution: The final digit of a power of $k$ depends only on the final digit of $k$, so there are 10 cases to consider. These are easy to work out. For $k$ ending in 1 , the final digits are $1,1,1,1, \ldots$. For $k$ ending in 2 they are $2,4,8,6,2,4,8,6, \ldots$, et cetera. In fact all 10 possible final digits repeat ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
84. There are two boxes of apples. After taking 5 apples from the first box and putting them into the second box, the first box still has 2 more apples than the second box. Originally, the first box had $\qquad$ more apples than the second box. | answer: 12 | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In the isosceles trapezoid $A B C D$, $A B / / C D$, and $A B > C D$. Let the eccentricity of the hyperbola with foci at $A, B$ and passing through point $D$ be $e_{1}$, and the eccentricity of the ellipse with foci at $C, D$ and passing through point $A$ be $e_{2}$, then $e_{1} e_{2}=$ | 5. 1 .
As shown in Figure 3, let $AB = x, CD = y, AD = BC = z, AC = BD = u$, then
$$
e_{1} = \frac{AB}{BD - AD} = \frac{x}{u - z}, \quad e_{2} = \frac{CD}{AD + AC} = \frac{y}{u + z}.
$$
Therefore,
$$
e_{1} e_{2} = \frac{xy}{u^2 - z^2}.
$$
Draw $DH \perp AB$, with the foot of the perpendicular at $H$. By the symmetry... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. $\frac{1}{1+\sqrt[4]{3}}+\frac{1}{1-\sqrt[4]{3}}+\frac{2}{1+\sqrt{3}}$ The value is
(A) 1 ;
(B) -1 ;
(C) 2 ;
(D) -2 . | 1. (D)
$$
\text { Sol } \begin{aligned}
\text { Original expression } & =\frac{2}{1-\sqrt{3}}+\frac{2}{1+\sqrt{3}} \\
& =\frac{2+2 \sqrt{3}}{-2}+\frac{2-2 \sqrt{3}}{-2} \\
& =-2 .
\end{aligned}
$$ | -2 | Algebra | MCQ | Yes | Yes | olympiads | false |
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