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4. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, $\angle A B C=120^{\circ}$, the angle bisector of $\angle A B C$ intersects $A C$ at point $D$, and $B D=1$. Then the minimum value of $4 a+c$ is $\qquad$. | 4.9.
From the problem, we know that $S_{\triangle A B C}=S_{\triangle A B D}+S_{\triangle B C D}$. By the angle bisector property and the formula for the area of a triangle, we get
$$
\begin{array}{l}
\frac{1}{2} a c \sin 120^{\circ} \\
=\frac{1}{2} a \times 1 \times \sin 60^{\circ}+\frac{1}{2} c \times 1 \times \sin ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (6 points) Given that $a$ is a digit from 1 to 9, if the repeating decimal $0.1 \mathrm{a}=\frac{1}{\mathrm{a}}$, then $a=$ $\qquad$ | 【Solution】According to the problem, we have:
$$
\frac{10+a-1}{90}=\frac{1}{a}
$$
Simplifying, we get:
$$
\begin{array}{l}
a^{2}+9 a-90=0 \\
\quad(a+15)(a-6)=0
\end{array}
$$
Solving, we get: $a=-15$ (discard), or $a=6$, so the answer is: 6. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Let "学" and "习" represent two different digits. The product of the four-digit numbers "学学学学" and "习习习习" is a seven-digit number, and its unit and millionth digits are the same as the digit represented by "学". How many two-digit numbers can "学习" represent? $\qquad$ | Analysis: The unit and million digits of the product are "xue", so "xi" is 1,
"xue xue xue xue" = "xue" $\times 1111$, "xue xue xue xue" $\times$ "xi xi xi xi" = "xue" $\times 1111^{2}=$ "xue" $\times 1234321$, and since the million digit of the product is "xue", "xue" can only be $2,3,4$; therefore, the two-digit numb... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. If a number is not a multiple of 11, but by removing any one of its digits, it becomes a multiple of 11 (for example, 111 is such a number, as removing any of its units, tens, or hundreds digit results in a multiple of 11), such a number is defined as a "Zhonghuan number". The number of four-digit "Zhonghuan numbers... | 【Analysis】Let such a four-digit number be $\overline{a b c d}$, then according to the problem: $\left\{\begin{array}{l}\text { Move } a, \text { then } 11 \mid b+d-c \\ \text { Move } b, \text { then } 11 \mid a+d-c \\ \text { Move } c, \text { then } 11 \mid a+d-b \\ \text { Move } d, \text { then } 11 \mid a+c-b\end{... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Given $f(x)=|1-2 x|, x \in[0,1]$, then the number of solutions to the equation $f(f(f(x)))=\frac{1}{2} x$ is $(\quad)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 8
2. 【Analysis and Solution】Let $y=f(x), z=f(y), \omega=f(z)$, analyze the change of $y, z, \omega$ when $x$ changes from 0 to 1:
$$
\begin{array}{l}
x: 0 \xrightarrow{\frac{1}{2}} 1, \\
y: 1 \xrightarrow{\longrightarrow} 0 \longrightarrow, \\
z: 1 \xrightarrow{\frac{1}{2}} 0 \xrightarrow{\frac{1}{2}} 1 \xrightarrow{\f... | 8 | Number Theory | proof | Yes | Yes | olympiads | false |
4. The real root of the equation $\sin x=\lg x$ is
(A) 1 ;
(B) 2 ;
(C) 3 ;
(D) greater than 3 . | 4. (C)
The number of solutions to the equation $\sin x=\lg x$ is the number of intersection points between the sine curve $\sin x$ and the logarithmic curve $\lg x$.
First, determine the range of $x$. By the definition of $\lg x$,
$$
\begin{array}{l}
x>0 \text {, and } \quad \sin x \leqslant 1, \\
\therefore \quad \l... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 7 Given
\[
\begin{aligned}
f(x, y)= & x^{3}+y^{3}+x^{2} y+x y^{2}- \\
& 3\left(x^{2}+y^{2}+x y\right)+3(x+y),
\end{aligned}
\]
and \( x, y \geqslant \frac{1}{2} \). Find the minimum value of \( f(x, y) \).
(2011, Hebei Province High School Mathematics Competition) | 【Analysis】This problem involves finding the extremum of a bivariate function, with a very complex expression, making it difficult to find a breakthrough. Observing the symmetry in the expression, we can attempt the following transformation.
First, when $x \neq y$, multiply both sides of the function by $x-y$, yielding... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) Let the real number $a \geqslant 2$, and the roots of the equation $x^{2}-a x+1=0$ be $x_{1}, x_{2}, S_{n}=x_{1}^{n}+x_{2}^{n}$ $(n=1,2, \cdots)$.
(1) Determine the monotonicity of the sequence $\left\{\frac{S_{n}}{S_{n+1}}\right\}$, and provide a proof;
(2) Find all real values of $a$ such that for any... | 10. Given $a \geqslant 2$, it is easy to see that the roots $x_{1}, x_{2}$ of the equation $x^{2}-a x+1=0$ are both positive, and $x_{1} x_{2}=1$. Therefore, $S_{n}>0$ $(n=1,2, \cdots)$.
(1) The sequence $\left\{\frac{S_{n}}{S_{n+1}}\right\}$ is monotonically non-increasing.
In fact, we have
$$
\begin{aligned}
\frac{S_... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
$34 \cdot 6$ number 695 is written as a polynomial in factorials: $695=a_{1}+a_{2} \cdot 2!+$ $a_{3} \cdot 3!+\cdots+a_{n} \cdot n!$, where $a_{1}, a_{2}, \cdots, a_{k}$ are integers, and $0 \leqslant a_{k} \leqslant k$, then $a_{4}$ equals
(A) 0 .
(B) 1 .
(C) 2 .
(D) 3 .
(E) 4 .
(12th American High School Mathematics ... | [Solution] From $6!>695$, we know we can set
$$
695=a_{1}+a_{2} \cdot 2!+a_{3} \cdot 3!+a_{4} \cdot 4!+a_{5} \cdot 5!,
$$
i.e., $695=a_{1}+2 a_{2}+6 a_{3}+24 a_{4}+120 a_{5}$, where $0 \leqslant a_{k} \leqslant k$.
From the above, $a_{5}=5$ (otherwise the sum cannot reach 695).
Also, when $a_{5}=5$, $a_{4} \neq 4$ (n... | 3 | Number Theory | MCQ | Yes | Yes | olympiads | false |
1. For the quadratic function $f(x)=x^{2}-3 x+2$, the number of distinct real roots of the equation $f(f(x))=0$ is ( ). (A)1
(B) 2
(C) 3
(D) 4 | Answer: Choose $D$.
Because $f(f(x))=\left(x^{2}-3 x+2\right)^{2}-3\left(x^{2}-3 x+2\right)+2=x^{4}-6 x^{3}+10 x^{2}-3 x$, therefore we have $x(x-3)\left(x^{2}-3 x+1\right)=0, x_{1}=0, x_{2}=3, x_{3,4}=\frac{3 \pm \sqrt{5}}{2}$, so the original equation has 4 distinct real roots.
Note: It is also possible to discuss t... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. Let $S_{n}=1-2+3-4+\cdots+(-1)^{n-1} n, n=1,2, \cdots$, then $S_{17}+S_{33}+S_{50}$ equals
A. 0
B. 1
C. 6
D. -1
E. 2 | 1. $\mathrm{B}$
(1) When $n$ is even
$$
\begin{aligned}
S_{n} & =1-2+3-4+\cdots+(-1)^{n-1} n=(1-2)+(3-4)+\cdots+[(n-1)-n] \\
& =-1-1-\cdots-1=-\frac{n}{2}
\end{aligned}
$$
(2) When $n$ is odd
$$
\begin{array}{l}
\begin{aligned}
S_{n} & =1+\{(-2+3)+(-4+5)+\cdots+[-(n-1)+n]\} \\
& =1+\{1+1+\cdots+1\}=1+\frac{n-1}{2}=\fra... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. (3 points) In the third grade, there are two classes, Class A and Class B. If 4 students are transferred from Class A to Class B, the number of students in both classes will be equal. Class A has $\qquad$ more students than Class B. | 【Answer】Solution: $4 \times 2=8$ (people)
Answer: Class A has 8 more people than Class B. Therefore, the answer is: 8. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Does there exist integers $a<b<c<d$ satisfying
$$
\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{a}{d}=\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{b}{d} ?
$$ | 1. Exist, one set of solutions is
$$
a=-28, b=-14, c=-7, d=4 \text {. }
$$
Obviously, it is impossible for all numbers to be positive (or all negative).
Assume $a=-4, b=-2, c=-1$, and solve for the corresponding $d$ that satisfies the requirements.
Substituting into the equation, we get
$$
\begin{array}{l}
\frac{-4}{... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (3 points) There are 10 pencil cases, among which 5 contain pencils, 4 contain pens, and 2 contain both pencils and pens. The number of empty pencil cases is $\qquad$. | 【Answer】Solution: $10-(5+4-2)$,
$$
\begin{array}{l}
=10-7, \\
=3 \text { (boxes); }
\end{array}
$$
Answer: There are 3 empty pencil boxes;
Therefore, the answer is: 3. | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. In the plane of square $ABCD$, there is a point $P$ such that $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, and $\triangle PDA$ are all isosceles triangles. Then, the number of points $P$ with this property is
(A) 9;
(B) 17;
(C) 1;
(D) 5. | 7. (A)
Since the base of the isosceles triangle, with one side of the square as its waist, has endpoints on the circle with the square's vertex as the center and the side length as the radius, such circles can be drawn four times, intersecting in pairs to form 8 points that meet the requirements. The vertex of the iso... | 9 | Geometry | MCQ | Yes | Yes | olympiads | false |
7. Given positive rational numbers $a, b$, (both not 0, and $a+b+c=abc$ then $\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+$ $\frac{b}{c}+\frac{c}{a}+\frac{c}{b}-ab-bc-ca=$ $\qquad$ . | 7. -3
$\frac{a}{b}+\frac{c}{b}-a c=\frac{a+c-a b c}{b}=-1$ and the rest are similar. | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. The integer sequence $\left\{a_{n}\right\}$, has $a_{1}=1, a_{2}=2, a_{n+2}=5 a_{n+1}+a_{n}$, then $\left[\frac{a_{2}}{a_{1}}\right]\left\{\left[\frac{a_{3}}{a_{2}}\right\}\left\{\frac{a_{4}}{a_{3}}\right\} \cdots\left\{\left[\left\{\frac{a_{2025}}{a_{2024}}\right\}\left[\frac{a_{2024}}{a_{2}}\right]\right.\right.\r... | Answer: 1
$5 a_{n+1}<a_{n+2}=5 a_{n+1}+a_{n}<6 a_{n+1}$, so $\left\{\frac{a_{n+2}}{a_{n+1}}\right\}=\frac{a_{n+2}}{a_{n+1}}-5=\frac{a_{n}}{a_{n+1}}$
By the periodicity of $\left\{a_{n}\right\}$ modulo 2, we know that $a_{2024}$ is even.
From the above two points, the answer is 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=2, a_{n+1}=-\frac{\left(S_{n}-1\right)^{2}}{S_{n}}\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
where $S_{n}$ is the sum of the first $n$ terms of $\left\{a_{n}\right\}$.
(1) Prove: $\left\{\frac{1}{S_{n}-1}\right\}$ is an arithmetic seque... | 14. (1) When $n \geqslant 1$, from the condition we get
$$
\begin{array}{l}
S_{n+1}-S_{n}=-\frac{\left(S_{n}-1\right)^{2}}{S_{n}} \\
\Rightarrow S_{n+1}-1=\frac{S_{n}-1}{S_{n}} .
\end{array}
$$
Thus, $\frac{1}{S_{n+1}-1}-\frac{1}{S_{n}-1}=\frac{S_{n}}{S_{n}-1}-\frac{1}{S_{n}-1}=1$.
Also, $\frac{1}{S_{1}-1}=\frac{1}{2-... | 3 | Algebra | proof | Yes | Yes | olympiads | false |
2. If $x, y \in\left[-\frac{\pi}{6}, \frac{\pi}{6}\right], a \in R$, and satisfy $\left\{\begin{array}{l}x^{3}+\sin x-3 a=0 \\ 9 y^{3}+\frac{1}{3} \sin 3 y+a=0\end{array} \quad\right.$, then $\cos (x+3 y)=$ | Hui Gui 1 .
Analysis Transform the known conditions into $\left\{\begin{array}{l}x^{3}+\sin x-3 a=0 \\ (3 y)^{3}+\sin 3 y+3 a=0\end{array}\right.$, the function $f(t)=t^{3}+\sin t$ is an increasing function and an odd function in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, and $x, 3 y \in\left[-\frac{\pi}{2}, \frac{\... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Given $A=\{1,-1, \mathrm{i},-\mathrm{i}\}$ (i is the imaginary unit), $f(x)$ is a function with domain and range both being $A$, for any $x, y \in A$, there is $f(x y)=f(x) f(y)$. Then the number of functions $f(x)$ that satisfy the condition is $\qquad$.
| 6. 2 .
Notice that, the domain and range of $f(x)$ are both finite sets $A$. Therefore, $f(x)$ is a bijection.
Let $x=y=1$, then $f(1)=f(1) f(1)$.
By $f(1) \neq 0$, hence $f(1)=1$.
Let $x=y=-1$, then $f(1)=f(-1) f(-1)$.
By $f(1)=1$, hence $f(-1)= \pm 1$; and since $f(x)$ is a bijection, thus, $f(-1)=-1$.
Therefore, wh... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. If the polynomial $x^{2}+(a+7) x+6 a+2$ can be factored into the product of two linear factors $(x+b)$ and $(x+c)$ (where $b, c$ are integers), then the number of possible values for $a$ is $\qquad$. | $3$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. This year, Xiao Jun is 5 years old, and his father is 31 years old. In $\qquad$ years, his father's age will be 3 times Xiao Jun's age. | 【Analysis】According to "this year, Xiao Jun is 5 years old, and his father is 31 years old," the age difference between father and son is (31-5) years. Since this age difference does not change, the multiple difference is 3-1=2. Therefore, using the difference multiple formula, we can find Xiao Jun's age when his fathe... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. 10 consecutive natural numbers are arranged in ascending order. If the sum of the last 6 numbers is 15 more than twice the sum of the first 4 numbers, then the smallest number among these 10 numbers is . $\qquad$ | 【Analysis】This problem mainly examines arithmetic sequences.
【Solution】Solution: Let the smallest number be $x$, then the remaining natural numbers are $x+1, x+2, \cdots, x+9$. According to the problem, we have $2(4 x+1+2+3)+15=6 x+4+5+6+7+8+9$,
Simplifying gives $8 x+27=6 x+39$
$$
\therefore x=6 \text {, }
$$
【Comme... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
17. Let $0 \leqslant x \leqslant \pi, 0 \leqslant y \leqslant 1$. Find the minimum value of the function $f(x, y)=(2 y-1) \sin x+(1-y) \sin (1-y) x$. | 17. It is known that for all $0 \leqslant x \leqslant \pi$, we have $\sin x \geqslant 0, \sin (1-y) x \geqslant 0$.
Thus, when $\frac{1}{2} \leqslant y \leqslant 1$, $f(x, y) \geqslant 0$, and the equality holds when $x=0$.
When $0 \leqslant yx>\sin x$,
and. $\sin (x+\delta)=\sin x \cdot \cos \delta+\cos x \cdot \sin ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Given positive real numbers $x, y$ satisfy
$\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{y^{2}+4}-2\right) \geqslant y$, then the minimum value of $x+y$ is $\qquad$. | $$
\begin{array}{l}
2 x+\sqrt{4 x^{2}+1} \geqslant \\
\frac{y}{\sqrt{y^{2}+4}-2}=\frac{\sqrt{y^{2}+4}+2}{y}=\frac{2}{y}+\sqrt{\frac{4}{y^{2}}+1},
\end{array}
$$
Notice that the function $u=2 x+\sqrt{4 x^{2}+1}$ is monotonically increasing on $(0,+\infty)$, so $x \geqslant \frac{1}{y}$.
Therefore, $x+y \geqslant \frac{... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
19. Find the number of 17-complete partitions.
untranslated text retained for context:
19. 求 17 -完备分拆的个数. | 19. 8.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let $n$ be a natural number, for any real numbers $x, y, z$ there is always $\left(x^{2}+y^{2}+z^{2}\right) \leqslant n\left(x^{4}+y^{4}+z^{4}\right)$, then the minimum value of $n$ is $\qquad$ | 3. 【Analysis and Solution】Solution 1 Let $a=x^{2}, b=y^{2}, c=z^{2}$, the given inequality becomes $(a+b+c)^{2} \leqslant n\left(a^{2}+b^{2}+c^{2}\right)$.
On one hand, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c$
$\leqslant a^{2}+b^{2}+c^{2}+\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)+\left(a^{2}+c^{2}\righ... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9. In the sequence $\left\{a_{n}\right\}$, $a_{4}=1, a_{11}=9$, and the sum of any three consecutive terms is 15, then $a_{2016}=$ $\qquad$ | Solution: 5.
According to the problem, for any $n \in \mathbf{N}, a_{n}+a_{n+1}+a_{n+2}=a_{n+1}+a_{n+2}+a_{n+3}=15$.
Therefore, $a_{n+1}=a_{n}$.
Thus, $a_{1}=a_{4}=1, a_{2}=a_{11}=9, a_{3}=15-a_{1}-a_{2}=5$.
Hence, $a_{3016}=a_{3 \times 672}=a_{3}=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (7th "Hope Cup" Senior High School Competition) If the function $y=\left|\sin \left(\omega x+\frac{\pi}{3}\right)-1\right|$ has the smallest positive period of $\frac{\pi}{2}$, then the value of the positive number $\omega$ is
A. 8
B. 4
C. 2
D. 1 | 4. B Because $-1 \leqslant \sin \left(\omega \pi+\frac{\pi}{3}\right) \leqslant 1$.
So $\left|\sin \left(\omega x+\frac{\pi}{3}\right)-1\right|=1-\sin \left(\omega x+\frac{\pi}{3}\right)$.
Since $T=\frac{2 \pi}{\omega}$ and $T=\frac{\pi}{2}$, then $\omega=4$. Choose $\mathrm{B}$.
Another solution: From the graph, the ... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
17. This is a figure composed of 72 identical small quadrilaterals, some of which have been infected and turned black. When a healthy small quadrilateral (white) has at least two adjacent small quadrilaterals infected, it will also be infected and turn black, spreading sequentially. Therefore, at least \qquad additiona... | 【Analysis】As shown in the upper right figure: The four areas marked with red shading have common characteristics:
For example (1) the red diamond where the red shading is located, even if the outside of the red diamond is completely infected, the inside will not be infected. When (1) becomes a source of infection, the... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Given $\alpha, \beta \in \mathbf{R}$, the lines
$$
\frac{x}{\sin \alpha+\sin \beta}+
\frac{y}{\sin \alpha+\cos \beta}=1 \text { and } \frac{y}{\cos \alpha+\sin \beta}+\frac{y}{\cos \alpha+\cos \beta}=1
$$
intersect at a point on the line $y=-x$, then $\sin \alpha+\cos \alpha+\sin \beta$
$$
+\cos \beta=
$$ | 6. 0 .
Solution: From the given information, we can assume the intersection point of the two lines to be $\left(x_{0}\right.$, $\left.-x_{0}\right)$, and $\sin \alpha, \cos \alpha$ are the roots of the equation
$$
\frac{x_{0}}{t+\sin \beta}+\frac{-x_{0}}{t+\cos \beta}=1,
$$
which means they are the roots of the equat... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 6 Given the inequality $\sqrt{2}(2 a+3) \cos \left(\theta-\frac{\pi}{4}\right)+\frac{6}{\sin \theta+\cos \theta}-2 \sin 2 \theta<3 a+6$ holds for $\theta \in\left[0, \frac{\pi}{2}\right]$, find the range of values for $a$.
Translate the above text into English, please retain the original text's line breaks and... | Given that there are two variables in the inequality, and one of their ranges is provided, the method of separating variables is commonly adopted to find the range of the other. Noting the several trigonometric expressions related to angle $\theta$, $\cos \left(\theta-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}(\sin \the... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Example 6 Prove: $2004\left\{\frac{1}{2004} C_{2004}^{0}-\frac{1}{2003} C_{2003}^{1}+\frac{1}{2002} C_{2002}^{2}-\cdots+\frac{1}{1002} C_{1002}^{1002}\right\}=2$. | Let $a_{n}=\sum_{m \geqslant 0} \frac{(-1)^{m} n}{n-m} C_{n-m}^{m}$. Notice that $\frac{n}{n-m} C_{n-m}^{m}=C_{n-m}^{m}+\frac{m}{n-m} C_{n-m}^{m}=C_{n-m}^{m}+C_{n-m-1}^{m-1}(m \geqslant 1)$, so $a_{n}=1+\sum_{m \geqslant 1}(-1)^{m}\left[C_{n-m}^{m}+C_{n-m-1}^{m-1}\right]=\sum_{m \geqslant 0}(-1)^{m} C_{n-m}^{m}+ \sum_{... | 2 | Combinatorics | proof | Yes | Yes | olympiads | false |
13. Let the set $\{1,2,3, \cdots, 1998\}$ be divided into 999 mutually disjoint binary subsets $\left\{a_{i}, b_{i}\right\}$, and for $1 \leqslant i \leqslant 999$, we have $\left|a_{i}-b_{i}\right|=1$ or 6. Prove: The last digit of the sum $\sum_{i=1}^{999}\left|a_{i}-b_{i}\right|$ is 9. | 13. Proof: First, it is easy to see that $\left|a_{i}-b_{i}\right| \equiv \sum_{i=1}^{999} 1(\bmod 5) \equiv 4(\bmod 5)$.
$$
\text { Also } \sum_{i=1}^{999}\left|a_{i}-b_{i}\right| \equiv \sum_{i=1}^{999} a_{i}+b_{i}(\bmod 2) \equiv 1(\bmod 2) \text {. }
$$
Therefore, $\sum_{i=1}^{999}\left|a_{i}-b_{i}\right| \equiv 9... | 9 | Number Theory | proof | Yes | Yes | olympiads | false |
4.26 ** Let $a>b>0$. Prove: $\sqrt{2} a^{3}+\frac{3}{a b-b^{2}} \geqslant 10$. | Parse: Since $a b-b^{2}-b(a-b) \leqslant \frac{[b+(a-b)]^{2}}{4}=\frac{a^{2}}{4}$, therefore
$$
\begin{aligned}
\sqrt{2} a^{3}+\frac{3}{a b-b^{2}} & \geqslant \sqrt{2} a^{3}+\frac{3}{\frac{a^{2}}{4}}=\sqrt{2} a^{3}+\frac{12}{a^{2}}=\frac{\sqrt{2}}{2} a^{3}+\frac{\sqrt{2}}{2} a^{3}+\frac{1}{a^{2}}+\frac{1}{a^{2}}+\frac{... | 10 | Inequalities | proof | Yes | Yes | olympiads | false |
7. Let the line $l$ passing through the fixed point $M(a, 0)$ intersect the parabola $y^{2}=4 x$ at points $P, Q$. If $\frac{1}{|P M|^{2}}+\frac{1}{|Q M|^{2}}$ is a constant, then the value of $a$ is $\qquad$ . | Answer 2.
Solution Let the parametric equation of the line $l$ be $\left\{\begin{array}{l}x=a+t \cos \alpha \\ y=t \sin \alpha\end{array}\right.$ (where $t$ is the parameter, $\alpha$ is the angle of inclination and $\left.\alpha \in(0, \pi)\right)$, substituting into the parabola equation yields:
$$
t^{2} \sin ^{2} \a... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
21. The equation
$$
(z-1)\left(z^{2}+2 z+4\right)\left(z^{2}+4 z+6\right)=0
$$
has five complex roots $x_{k}+\mathrm{i} y_{k}\left(x_{k} 、 y_{k} \in \mathbf{R}, k=1\right.$, $2, \cdots, 5)$. Let $\Gamma$ be the ellipse passing through the five points $\left(x_{k}, y_{k}\right)(k=1,2$, $\cdots, 5)$, with eccentricity $... | 21. A.
Solving the equation, we get
$$
z_{1}=1, z_{2,3}=-1 \pm \sqrt{3} \mathrm{i}, z_{4,5}=-2 \pm \sqrt{2} \mathrm{i} \text {. }
$$
Thus, the ellipse $\Gamma$ passes through the points $(1,0)$, $(-1, \pm \sqrt{3})$, and $(-2, \pm \sqrt{2})$.
By symmetry, we know the center of the ellipse is on the $x$-axis.
Let the ... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. The equation $2^{|2 x-2|}-a \cos (1-x)=0$ has only one real solution with respect to $x$, then
A. $a=-1$
B. $a=1$
C. $a=2$
D. The value of $a$ is not unique | The graph of the function $f(x)=2^{|2 x-2|}-a \cos (1-x)$ is symmetric about the line $x=1$, so $f(1)=0$ $\Rightarrow 1-a=0 \Rightarrow a=1$, hence the answer is $B$.
| 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
1 Given the complex number $m$ satisfies $m+\frac{1}{m}=1$, then $m^{2008}+\frac{1}{m^{2009}}=$ $\qquad$ . | (1) 0 Hint: From $m+\frac{1}{m}=1$, we get $m^{2}-m+1=0$, thus
$$
m^{3}+1=(m+1)\left(m^{2}-m+1\right)=0,
$$
which means $m^{3}=-1$. Therefore,
$$
m^{2008}+\frac{1}{m^{2009}}=\left(m^{3}\right)^{669} \cdot m+\frac{m}{\left(m^{3}\right)^{670}}=-m+m=0 .
$$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. (6 points) The units digit of $3^{2014}+4^{2015}+5^{2016}$ is $\qquad$ . (Note: $a^{m}$ represents $m$ $a$s multiplied together) | 【Solution】Solution: According to the analysis, first find the unit digit of $3^{2014}$, $\because 3^{1}=3,3^{2}=9,3^{3}=27,3^{4}=81,3^{5}$ $=243$,
Obviously, the unit digit of $3^{n}$ is $3, 9, 7, 1$ appearing in a cycle of 4, and $3^{2014}=3^{503^{*} 4+2}, \therefore$ the unit digit of $3^{2014}$ is 9;
Then find the... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Given that the domain of the function $f(x)$ is $[-2,+\infty)$, and $f(4)=f(-2)=$ 1. $f^{\prime}(x)$ is the derivative of $f(x)$, and the graph of the function $y=f^{\prime}(x)$ is shown in the figure on the right. Then the area of the plane region $\left\{\begin{array}{l}a \geqslant 0, \\ b \geqslant 0, \\ f(2 a+b)... | 6. B From the graph, we know that when $x \geqslant 0$, $f^{\prime}(x) \geqslant 0$, so $f(x)$ is increasing on $[0,+\infty)$. Since $a \geqslant 0, b \geqslant 0$, it follows that $2 a+b \geqslant 0$. Therefore,
$$
f(2 a+b)<f(4) \Leftrightarrow 2 a+b<4 .
$$
The given planar region is $\left\{\begin{array}{l}a \geqsla... | 4 | Calculus | MCQ | Yes | Yes | olympiads | false |
9. In a math competition, there were three problems: A, B, and C. Among the 25 participating students, each student solved at least one problem; among those who did not solve problem A, the number of students who solved problem B is twice the number of students who solved problem C; the number of students who only solv... | 9. Let the sets of students who solved problems A, B, and C be $A$, $B$, and $C$ respectively. Then, the three sets $A$, $B$, and $C$ can be divided into 7 parts, with the number of elements in each part being $a$, $b$, $c$, $d$, $e$, $f$, and $g$, as shown in the figure. From the problem, we have
$$
\left\{\begin{arra... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Let $\omega$ be a complex cube root of unity, then the value of $(1-\omega)\left(1-\omega^{2}\right)\left(1-\omega^{4}\right)\left(1-\omega^{8}\right)$ is
A. $\omega$
B. $\bar{\omega}$
C. 9
D. 6 | 6. $\mathrm{C}$ Original expression $=(1-\omega)^{2}\left(1-\omega^{2}\right)^{2}=\left(1-\omega-\omega^{2}+\omega^{3}\right)^{2}=(3)^{2}=9$ | 9 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. Fill the numbers from $-10 \sim 14$ into a $5 \times 5$ grid such that the sum of the numbers in each row, each column, and each diagonal is the same. What is this sum?
(A)2
(B) 5
(C) 10
(D) 25
(E) 50 | 7. C.
In a $5 \times 5$ grid, the sum of the numbers in each row, each column, and each diagonal is $S$.
Notice that, no matter how these 25 integers are arranged, their total sum is
$$
\begin{array}{l}
-10-9-8-\cdots+13+14 \\
=11+12+13+14=50=5 S \\
\Rightarrow S=10 .
\end{array}
$$ | 10 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
Example 1 (1997 National High School Competition Question) Let the first term and common difference of an arithmetic sequence be non-negative integers, the number of terms is no less than 3, and the sum of all terms is $97^{2}$. Then the number of such sequences is ( ).
A. 2
B. 3
C. 4
D. 5 | Solution: Choose C. Reason: Let the first term of the arithmetic sequence be $a$, and the common difference be $d$. From the given, we have $n a+\frac{1}{2} n(n-1) d=97^{2}$, which is $n[2 a+(n-1) d]=2 \times 97^{2}$.
Since $n \geqslant 3$, $n$ can only be $97, 2 \times 97, 97^{2}, 2 \times 97^{2}$. Also, because $a \g... | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
18.83 As shown in the figure, a red cross of equal width (symmetric about the diagonals) is on a square flag, with a small blue square in the center, and the rest is white. If the entire cross (including red and blue) occupies 36% of the flag's area, what is the percentage of the blue square's area relative to the flag... | [Solution] Let the side length of the square be 10, then its area is 100. According to the problem, we have
$$
S_{\text {white }}=100 \cdot(1-36 \%)=64 \text {, }
$$
and the four small white triangles can be combined to form a square, whose side length is $\sqrt{64}=8$.
Thus, as shown in the figure, we have
$$
A B=\fr... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
10 . For real numbers $a, b$, we define the symbol $\max (a, b)$ as follows:
When $a^{3} b$, $\max (a, b)=a$; when $a<b$, $\max (a, b)=b$.
For example: $\max (3,-2)=3, \max (2,2)=2, \max (3,4)=4$. If the minimum value of the function $y=\max (x+4,-x+2)$ with respect to $x$ is $m$, then $\max (\sqrt{m}, m-1)=(\quad)$
(... | Answer: (D)
Solution: When $x+4^{3}-x+2$, i.e., $x^{3}-1$,
$$
y=x+4^{3} 3,
$$
When $x+43,
$$
So $m=y_{\text {min }}=3, \max (\sqrt{m}, m-1)=\max (\sqrt{3}, 3-1)=2$. Therefore, the answer is (D). | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. 15 children are playing hide-and-seek. Xiao Zhi is searching, and the other children are hiding. Now Xiao Zhi has already found 8 children, he still has $\qquad$ children left to find. | $6$ | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Player A and Player B are having a table tennis match, with a best-of-three format. In each game, the first to score 11 points and the opponent scoring less than 10 points wins; if the score is tied at 10, the first to score 2 more points wins. The total points scored by both A and B are 30 points. Witho... | 【Analysis】Through analysis, we can know: the total scores of A and B are both 30 points, $30<3 \times 11$, in the three rounds, one of them won two rounds, so at least two scores are not less than 11, the total score of A is $30: 30=11+9+10$, the corresponding score of B is: $30=7+11+12:$ the corresponding scores are $... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.49 Find all natural numbers $n$ such that the sum of the digits of the decimal number $2^n$ is equal to 5.
(10th All-Russian Mathematical Olympiad, 1984) | [Solution] It is not difficult to verify that the smallest value of $n$ that satisfies the problem's requirements is 5. This is because
$$
2^{5}=32, \quad 3+2=5 .
$$
Below, we prove that when $n \geqslant 6$, no $n$ satisfies the problem's requirements.
Since the last digit of $2^{n}$ is 2, 4, 6, or 8, and the sum of ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1.7 ** Let the geometric sequence $\left\{z_{n}\right\}$ be such that $z_{1}=1, z_{2}=a+b \mathrm{i}, z_{3}=b+a \mathrm{i}(a, b \in \mathbf{R}$, $a>0$ ), find the smallest value of $n$ that satisfies $z_{1}+z_{2}+\cdots+z_{n}=0$, and calculate the value of $z_{1} z_{2} \cdots z_{n}$ at this time. | From the problem, we have $(a+b \mathrm{i})^{2}=(b+a \mathrm{i})$, so we have $\left\{\begin{array}{l}a^{2}-b^{2}=b \\ 2 a b=a .\end{array}\right.$ Solving this, we get $\left\{\begin{array}{l}a=\frac{\sqrt{3}}{2}, \\ b=\frac{1}{2} .\end{array}\right.$ Therefore, $q=z_{2}=\frac{\sqrt{3}}{2}+\frac{1}{2} \mathrm{i}=\cos ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.40 Find the smallest positive integer $n$, such that in any two-coloring of $K_{n}$, there exist two monochromatic triangles of the same color with no common edge. | ［Solution 1］In the $K_{7}$ shown in the right figure, the line segments that are not drawn are all red, and the drawn ones are blue line segments. It is easy to see that any two blue triangles or any two red triangles have 1 common edge. This indicates that the smallest positive integer $n \geqslant 8$.
Below, we use ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
19. An ant moves along the edges of a unit cube at a speed of 1 unit per second, starting from any vertex. Assuming that each time the ant reaches a vertex, it turns to continue moving in any direction with equal probability (allowing for backtracking), the expected time for the ant to return to its starting point for ... | Consider the vertex sets $A=\{A\}, B=\left\{B_{1}, B_{2}, B_{3}\right\}$,
$$
C=\left\{C_{1}, C_{2}, C_{3}\right\}, D=\{D\}
$$
Considering the vertices and corresponding probabilities that the ant traverses starting from vertex $A$, let $E_{A}, E_{B}$, and $E_{C}$ represent the expected time for the ant to return to ve... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
23. (IMO China National Training Team Selection Exam, 2003) Find all functions \( f \) from the set of positive integers to the set of real numbers, such that
(1) For any \( n \geqslant 1, f(n+1) \geqslant f(n) \);
(2) For any \( m, n \) with \( (m, n)=1 \), \( f(m n)=f(m) f(n) \). | 23. Clearly, $f=0$ is a solution to the problem.
Assume $f \neq 0$, then $f(1) \neq 0$. Otherwise, for any positive integer $n$ we have $f(n)=f(1) f(n)=0$, which is a contradiction. Thus, we get $f(1)=1$. From (1), we know $f(2) \geqslant 1$. We will discuss two cases:
(i) $f(2)=1$, then we can prove $f(n)=1(\forall n... | 0 | Number Theory | proof | Yes | Yes | olympiads | false |
Example 8 Find the odd prime $p$ that satisfies the following condition: there exists a permutation $b_{1}, b_{2}, \cdots, b_{p-1}$ of $1,2, \cdots, p-1$, such that $1^{b_{1}}, 2^{b_{2}}, \cdots,(p-1)^{b_{p-1}}$ forms a reduced residue system modulo $p$. | Let $p$ be an odd prime satisfying the condition, and take $g$ as a primitive root of $p$. Since $g, g^{2}, \cdots, g^{p-1}$ form a reduced residue system modulo $p$, for each $i=1,2, \cdots, p-1$, $i$ can be represented as $g^{a_{i}}$ modulo $p$, where $a_{1}, a_{2}, \cdots, a_{p-1} \in \mathbf{Z}_{+}$, and by propert... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.103 There are 1990 piles of stones, with the number of stones in each pile being $1, 2, \cdots$ 1990. The allowed operation is to select any number of piles and remove the same number of stones from each selected pile. How many operations are required at minimum to remove all the stones? | [Solution] After each operation, we group the piles that have the same number of stones. At any time, even if we count the piles with no stones as one group, we always consider there to be $n$ groups. If in the next operation we remove the same number of stones from all piles belonging to $k$ different groups, then bec... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given positive real numbers $a$ and $b$ satisfy $a+b=1$, then $M=$ $\sqrt{1+a^{2}}+\sqrt{1+2 b}$ the integer part is | 5. 2
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let the function $f(x)$ have the domain $D=(-\infty, 0) \bigcup(0,+\infty)$, and for any $x \in D$, it holds that $f(x)=\frac{f(1) \cdot x^{2}+f(2) \cdot x-1}{x}$, then the sum of all zeros of $f(x)$ is $\qquad$ . | $$
\left\{\begin{array} { l }
{ f ( 1 ) = f ( 1 ) + f ( 2 ) - 1 , } \\
{ f ( 2 ) = \frac { 4 f ( 1 ) + 2 f ( 2 ) - 1 } { 2 } }
\end{array} \Rightarrow \left\{\begin{array}{l}
f(1)=\frac{1}{4}, \\
f(2)=1
\end{array} \Rightarrow f(x)=\frac{x^{2}+4 x-4}{4 x} .\right.\right.
$$
So the sum of all zeros of $f(x)$ is -4. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Four black $1 \times 1 \times 1$ cubes and four white $1 \times 1 \times 1$ cubes can form $\qquad$ different $2 \times 2 \times 2$ cubes (cubes that are the same after rotation are considered the same type). | 【Answer】7 | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given $a>1$. Then the minimum value of $\log _{a} 16+2 \log _{4} a$ is . $\qquad$ | 3.4.
From the operation of logarithms, we get
$$
\begin{array}{l}
\log _{a} 16+2 \log _{4} a=4 \log _{a} 2+\log _{2} a \\
=\frac{4}{\log _{2} a}+\log _{2} a .
\end{array}
$$
Since $a>1$, it follows that $\log _{2} a>0$.
By the AM-GM inequality, we have
$$
\frac{4}{\log _{2} a}+\log _{2} a \geqslant 2 \sqrt{\frac{4}{\... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Given positive real numbers $x, y (x>y)$ satisfy $x y=490$, and $(\lg x-\lg 7)(\lg y-\lg 7)=-\frac{143}{4}$, then in decimal notation, the integer part of $x$ has $\qquad$ digits.
(Zhang Jia provided the problem) | 8. 8 .
From $x y=490(x>y>0)$, we get $\lg x+\lg y=2 \lg 7+1$, which means
$$
(\lg x-\lg y)+(\lg y-\lg 7)=1 .
$$
Also, $(\lg x-\lg 7)(\lg y-\lg 7)=-\frac{143}{4}$, so $\lg x-7, \lg y-7$ are the two roots of the quadratic equation $t^{2}-t-\frac{143}{4}=0$ in terms of $t$. It is easy to see that $\lg x-\lg 7$ is the la... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In the exercise book, there is a calculation problem, giving 13 natural numbers, and asking to calculate their average (rounded to two decimal places). Xiao Ma's calculation result is 155.44, and the teacher said the last digit is wrong. So, the correct last digit is $\qquad$ . | $6$ | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Four numbers
$$
\begin{array}{l}
\sqrt{2-\sqrt{3}} \cdot \sqrt{2-\sqrt{2-\sqrt{3}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}} \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{3}}}}
\end{array}
$$
The product of these is ( ).
(A) $2+\sqrt{3}$
(B) 2
(C) 1
(D) $2-\sqrt{3}$ | - 1. C.
$$
\begin{array}{l}
\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}} \times \\
\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{3}}}} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \\
\sqrt{2^{2}-(\sqrt{2-\sqrt{2-\sqrt{3}}})^{2}} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2-\s... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. (3 points) As shown in the right figure, the length of rectangle $A B C D$ is 6 cm, and the width is 2 cm. A line segment $A E$ is drawn through point $A$ to divide the rectangle into two parts, one part being a right triangle, and the other part being a trapezoid. If the area of the trapezoid is 3 times the area of... | 3. (3 points) As shown in the right figure, the length of rectangle $A B C D$ is 6 cm, and the width is 2 cm. A line segment $A E$ is drawn through point $A$, dividing the rectangle into two parts, one being a right triangle and the other a trapezoid. If the area of the trapezoid is 3 times the area of the right triang... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Given the function
$$
f(x)=\left\{\begin{array}{ll}
2 x^{2}+4 x+1, & x<0 ; \\
\frac{2}{\mathrm{e}^{x}}, & x \geqslant 0 .
\end{array}\right.
$$
Then the number of pairs of points on the graph of $y=f(x)(x \in \mathbf{R})$ that are symmetric with respect to the origin $O$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 5. C.
The graph of the function $y=f(x)$ is shown as the solid line in Figure 4.
Construct the graph of $y=\frac{2}{\mathrm{e}^{x}}(x \geqslant 0)$ symmetric to the origin $O$ (dashed line), its analytical expression is
$$
g(x)=-\frac{2}{\mathrm{e}^{-x}}(x \leqslant 0) .
$$
Since $f(-1)=-1<-\frac{2}{\mathrm{e}}=g(-1... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
15. In a geometric sequence with a common ratio greater than 1, what is the maximum number of terms that are integers between 100 and 1000? | 15. $128,192,288,432,648,972$ are integers between 100 and 1000, forming a geometric sequence with a common ratio of $\frac{3}{2}$. There are 6 terms in total.
Let the geometric sequence $\left\{a r^{n-1}\right\}$ satisfy $100 \leqslant a$, and it is clear that $r$ is a rational number. Let $r=\frac{p}{q}, p>q \geqsla... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Xiaohong lives on the fourth floor. She needs 2 minutes to go from the ground floor to the second floor. So, how many minutes does she need to go from the ground floor to the fourth floor? | 1. $(4-1) \times 2=6$ (minutes) | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Example 14 (Question from the 10th "Hope Cup" Invitational Competition) Given that real numbers $x, y$ satisfy the equation $(x+2)^{2}+y^{2}=1$, what is the minimum value of $\frac{y-1}{x-2}$? | Let $\frac{y-1}{x-2}=k$, then $y-1=k x-2 k, y=k x-2 k+1$.
Let $\vec{a}=(x+2, k x-2 k+1), \vec{b}=(k,-1)$,
$$
\begin{array}{l}
1=(x+2)^{2}+y^{2}=(x+2)^{2}+(k x-2 k+1)^{2} \\
=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(k x+2 k-k x+2 k-1)^{2}}{k^{2}+1^{2}}=\frac{(4 k-1)^{2}}{k^{2}+1}... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Given
$$
f(x)=\left\{\begin{array}{ll}
\left(\frac{1}{4}\right)^{x}, & x \in(-\infty, 1) ; \\
\log _{\frac{1}{2}} x, & x \in[1,+\infty) .
\end{array}\right.
$$
Then $f(f(-1))=(\quad)$.
(A) 2
(B) -2
(C) $\frac{1}{4}$
(D) $-\frac{1}{2}$ | $$
-1 \text {. B. }
$$
According to the problem,
$$
f(f(-1))=f(4)=\log _{\frac{1}{2}} 4=-2 \text {. }
$$ | -2 | Algebra | MCQ | Yes | Yes | olympiads | false |
8. From the set $\{1,2, \cdots, 63\}$, seven numbers are randomly drawn, and the smallest number among them is denoted as $x$. Then the mathematical expectation $\mathrm{E}(x)=$ $\qquad$ | 2. 2 .
It is known that, $\frac{a_{n}}{n!}=\frac{1}{\mathrm{C}_{n}^{0}}+\frac{1}{\mathrm{C}_{n}^{1}}+\cdots+\frac{1}{\mathrm{C}_{n}^{n}}$.
When $n>3$, by the monotonicity of binomial coefficients, we have
$$
\begin{array}{l}
2+\frac{2}{n}<\frac{a_{n}}{n!}<2+\frac{2}{n}+\frac{n-3}{\mathrm{C}_{n}^{2}} \\
=2+\frac{2}{n}+... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11 Given $\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=3$, then the value of $\frac{\tan \alpha}{\tan \beta}$ is | 11 From the given, we have
$$
\sin \alpha \cos \beta + \cos \alpha \sin \beta = 3(\sin \alpha \cos \beta - \cos \alpha \sin \beta),
$$
which means
$$
\sin \alpha \cos \beta = 2 \cos \alpha \sin \beta,
$$
so
$$
\frac{\tan \alpha}{\tan \beta} = \frac{\sin \alpha \cos \beta}{\cos \alpha \sin \beta} = 2.
$$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Let $a, b, c$ be distinct positive integers. Then the minimum value of $\frac{a b c}{a+b+c}$ is $\qquad$ . | 11. 1 .
Assume without loss of generality that $a>b>c$. Then $a \geqslant 3, b \geqslant 2, c \geqslant 1$.
Thus, $a b \geqslant 6, b c \geqslant 2, c a \geqslant 3$.
Therefore, $\frac{a+b+c}{a b c}=\frac{1}{b c}+\frac{1}{c a}+\frac{1}{a b}$
$\leqslant \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$
$\Rightarrow \frac{a b c}{a... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Given two points $P(\cos \alpha, \sin \alpha), Q(\cos \beta, \sin \beta)$, then the maximum value of $|P Q|$ is
A. $\sqrt{2}$
B. 2
C. 4
D. Does not exist | 1. B Hint: $|P Q|=\sqrt{(\cos \alpha-\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2}}=\sqrt{2-2 \cos (\alpha-\beta)} \leqslant \sqrt{2+2}=2$. Or use the fact that $P, Q$ are both on the unit circle. | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
7. There are 100 chess pieces, and two people take turns to take the pieces. Each time, you are allowed to take 1 or 2 pieces. The one who takes the last piece wins. If you go first, then the first time you should take ( ) pieces to ensure victory. | 【Analysis】 $100 \div(1)+2=33 \cdots 1$, first take 1, making the number of stones 99, then adopt the following strategy: if the opponent takes 2, then take 1; if the opponent takes 1, then take 2. This way, the number of stones can always be a multiple of 3. Thus, the second player always faces a multiple of 3, and can... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. (6 points) The figure is a rectangle. To divide it into 7 parts, at least $\qquad$ straight lines need to be drawn.
| 【Analysis】Two lines divide a square into 4 parts, the third line intersects with the first two lines, adding 3 more parts, making a total of 7 parts; the fourth line intersects with the first 3 lines, adding another 4 parts, making a total of 11 parts; the fifth line intersects with the first 4 lines, adding another 5 ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. For the parabola $y^{2}=4 a x(a>0)$, the focus is $A$. With $B(a+4,0)$ as the center and $|A B|$ as the radius, a semicircle is drawn above the $x$-axis, intersecting the parabola at two different points $M, N$. Let $P$ be the midpoint of the segment $M N$. (I) Find the value of $|A M|+|A N|$; (II) Does there exist ... | 3. ( I) The equation of the semicircle is $[x-(a+4)]^{2}+y^{2}=16(y \geqslant 0)$, and when combined with $y^{2}=4 a x(a>0)$, eliminating $y$ yields $x^{2}-(8-$ $2 a) x+a(a+8)=0$.
Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, then $|A M|=x_{1}+a,|A N|=x_{2}+a$, hence $|A M|+|A N|=x_{1}+x_{2}+2 a=(8-$ $... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. (5 points) Place 100 ping-pong balls into 26 boxes arranged in a row from left to right. If the leftmost box contains 4 ping-pong balls, and the sum of the number of ping-pong balls in any 4 consecutive boxes is 15, then the number of ping-pong balls in the rightmost box is $\qquad$. | 【Solution】According to the analysis, 26 boxes are divided into: $26 \div 4=6$ (groups) $\cdots 2$ (boxes).
$\because$ In any 4 adjacent boxes, the total number of ping-pong balls is 15, so the number of balls in the boxes at positions $1,5,9 \cdots 25$ are all 4.
The number of ping-pong balls in the rightmost box: $10... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$2.420 \times 814 \times 1616$ divided by 13 has a remainder of | 【Analysis】 $420 \times 814 \times 1616 \equiv 4 \times 8 \times 4 \equiv 128 \equiv 11(\bmod 13)$ | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$32 \cdot 22$ satisfies the inequality $10^{4} \leqslant A \leqslant 10^{5}$ for the number of integers $A$ is $x \cdot 10^{4}+$ 1, then the value of $x$ is
(A) 9 .
(B) 8 .
(C) 7 .
(D) 6 .
(China Shanghai Junior High School Mathematics Competition, 1986) | [Solution] Since $10^{5}-10^{4}=9 \cdot 10^{4}$, the number of integers $A$ such that $10^{4} \leqslant A \leqslant 10^{5}$ is $9 \cdot 10^{4}+1$, which means $x=9$. Therefore, the correct choice is $(A)$. | 9 | Number Theory | MCQ | Yes | Yes | olympiads | false |
3. Let $a_{0}=a, a_{n+1}=2 a_{n}-n^{2}(n=0,1,2$, $\cdots)$, if all terms of $\left\{a_{n}\right\}$ are positive integers, then the minimum value of $\left\{a_{n}\right\}$ is $\qquad$ | 3. Solution: It is easy to know that $a_{n+1}-(n+1)^{2}=2 a_{n}-2 n^{2}-2 n-1$, so $a_{n+1}-(n+1)^{2}-2(n+1)=2\left(a_{n}-n^{2}\right)-4 n-3$, which means $a_{n+1}-(n+1)^{2}-2(n+1)-3=2\left(a_{n}-n^{2}-2 n-3\right)$, after substitution, the general term of $\left\{a_{n}\right\}$ can be solved as $a_{n}=n^{2}+2 n+3+(a-3... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 4.4.5 Let $V=\{1,2,3,4\}$, find the number of non-isomorphic graphs of order 4 with $V$ as the vertex set. | Let $S_{4}$ be the set of all 4-vertex graphs with vertex set $V=\{1,2,3,4\}$. The edge set $E_{1}$ of a graph $H_{1}=\left(V, E_{1}\right)$ in $S_{4}$ is a subset of $X=\{\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\}, \{3,4\}\}$.
We can view $H_{1}$ as a coloring of the edges in set $X$ with two colors “yes” $(y)$ or “no” ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 15 Given positive numbers $a, b, c$ satisfying $a+b+c=3$, prove:
$$
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant 5 \text {. }
$$ | Proof: Let $f(x)=\frac{x^{2}+9}{3 x^{2}-6 x+9}, x \in(0,3)$, then $f^{\prime}(x)=\frac{-6\left(x^{2}+6 x-9\right)}{\left(3 x^{2}-6 x+9\right)^{2}}=$ $\frac{-2\left(x^{2}+6 x-9\right)}{3\left(x^{2}-2 x+3\right)^{2}}$. Since $f^{\prime}(x)$ is not monotonic, we cannot handle it using concavity or convexity. Noting the ge... | 5 | Inequalities | proof | Yes | Yes | olympiads | false |
6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1}+(-1)^{n} a_{n}=2 n-1 \text {, }
$$
and the sum of the first 2019 terms of the sequence $\left\{a_{n}-n\right\}$ is 2019. Then the value of $a_{2020}$ is $\qquad$ .
6.1.
$$
\begin{array}{l}
\text { From } a_{n+1}+(-1)^{n} a_{n}=2 n-1 \\
\Rightarrow\left... | Let $b_{n}=a_{n}-n$. Then
$$
b_{2 n-1}+b_{2 n}+b_{2 n+1}+b_{2 n+2}=0 \text {. }
$$
Hence, the sum of the first 2019 terms of $\left\{a_{n}-n\right\}$ is
$$
\begin{array}{l}
b_{1}+b_{2}+b_{3}+b_{4}+\cdots+b_{2018}+b_{2019} \\
=\left(b_{1}+b_{2}+b_{3}+b_{4}\right)+\left(b_{5}+b_{6}+b_{7}+b_{8}\right)+ \\
\cdots+\left(b... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given an odd function $f(x)$ defined on $\mathbf{R}$ whose graph is symmetric about the line $x=2$, when $0<x \leq 2$, $f(x)=x+1$, then $f(-100)+f(-101)=$ $\qquad$ . | Answer 2.
Analysis: Given that $f(x)$ is an odd function and its graph is symmetric about the line $x=2$, we know that $f(-x)=-f(x)$, and $f(2-x)=f(2+x)$, so
$$
f(x+4)=f(-x)=-f(x), \quad f(x+8)=-f(x+4)=f(x)
$$
$f(x)$ is a periodic function with a period of 8.
Also, $f(3)=f(1)=2, f(4)=f(0)=0$, so $f(-100)+f(-101)=f(4)+f... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. Given $A(\sqrt{\pi}, a) 、 B(\sqrt{\pi}, b)$ are two different points on the curve $y^{2}+x^{4}=2 x^{2} y+1$. Then the value of $|a-b|$ is ( ).
(A) 1
(B) $\frac{\pi}{2}$
(C) 2
(D) $\sqrt{1+\pi}$
(E) $1+\sqrt{\pi}$ | 12. C.
From the problem, substituting $x=\sqrt{\pi}$ into the equation gives
$$
\begin{array}{l}
y^{2}+(\sqrt{\pi})^{4}=2(\sqrt{\pi})^{2} y+1 \\
\Rightarrow(y-\pi)^{2}=1 \Rightarrow y=\pi \pm 1 .
\end{array}
$$
Therefore, $|a-b|=|(\pi+1)-(\pi-1)|=2$. | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
17. Given point $A(x, y)$ is in the fourth quadrant, and $|x|=2,|y|=3$, point $B$ is symmetric to $A$ with respect to the origin, and then point $B$ is translated downward by two units to get point $C$, then the area of $\triangle A B C$ is $\qquad$. | $4$ | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6、Positive real numbers $a_{1}, a_{2}, \mathrm{~L}, a_{n}$ satisfy $\sum_{i=1}^{n} a_{i}=17$, and $\sum_{i=1}^{n} \sqrt{a_{i}^{2}+(2 i-1)^{2}}$ has a minimum value that is an integer, then $n=$ | 6. 12
Analysis: Consider each term $t_{i}=\sqrt{a_{i}^{2}+(2 i-1)^{2}}$ as the hypotenuse of a right triangle with legs $2 i-1$ and $a_{i}$. By connecting these right triangles to form a "ladder" (ensuring the right-angle sides are parallel), let $A, B$ be the endpoints of the zigzag line formed by the hypotenuses. Th... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.65 Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, the sum of the first $n$ terms is $S_{n}$, 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 .
(China High School Mathematics Lea... | [Solution] 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 the common ratio $-\frac{1}{3}$.
Therefore, $S_{n}-n=\frac{8 \times\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{1+\frac{1}{3}}=6-6 \times\left... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 2 Let $x \in \mathbf{R}$, try to find the minimum value of the function $f(x)=\left(x^{2}+4 x+5\right)\left(x^{2}+4 x+2\right)+2 x^{2}+8 x+1$. | Let $u=x^{2}+4 x$, then
$$
f(x)=g(u)=(u+5)(u+2)+2 u+1=u^{2}+9 u+11 \text {. }
$$
By the properties of quadratic functions, when $u=-\frac{9}{2}$, $g(u)$ reaches its minimum value, and when $u>-\frac{9}{2}$, $g(u)$ is a strictly increasing function of $u$. Since $u=x^{2}+4 x=(x+2)^{2}-4$, we know that $u \geqslant-4$, ... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$23 \cdot 28$ On the surface of a cube, there exist skew diagonals of the square faces. If the distance between two such skew diagonals is 1, then the volume of the cube is
(A) 1.
(B) $3 \sqrt{3}$.
(C) 1 or $3 \sqrt{3}$.
(D) $3 \sqrt{3}$ or $3 \sqrt{2}$.
(2nd "Hope Cup" National Mathematics Invitational Competition, 19... | ［Solution］We discuss in two cases:
(1) If the two skew diagonals are on opposite faces of the cube (see Figure (1)), knowing that the edge length of the cube is 1, then
$$
V_{\text {cube }}=1^{3}=1 .
$$
(2) If the two skew diagonals are on adjacent faces of the cube (see Figure (2)), the edge length $a=\sqrt{3}$ can be... | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
Let $A_{1}, A_{2}, \cdots, A_{n}$, satisfy
$$
A_{1} \cup A_{2} \cup \cdots \cup A_{n}=\mathbf{Z},
$$
and for any two numbers $b>c$ in each $A_{i}$, we have
$$
b-c \geqslant a^{i} .
$$ | If $02$. That is, $2^{n-1}>a^{n}$.
Let $A_{i}=\left\{2^{i-1} m \mid m\right.$ be an odd number $\}, i=1,2, \cdots, n-1, A_{n}=\left\{2^{n-1}\right.$ multiples $\}$. Then (1) holds. Moreover, for any two numbers $b>c$ in $A_{n}, b-c \geqslant 2^{n-1}>a^{n}$. For any two numbers $b>c$ in $A_{i}(1 \leqslant i \leqslant n-... | 2 | Number Theory | proof | Yes | Yes | olympiads | false |
4. (3 points) It is said that Li Bai, the Poetic Immortal of the Tang Dynasty, went to buy wine. He walked on the street, and when he met a shop, his wine doubled; when he saw a flower, he drank 2 cups. On the way, he met shops and flowers four times, and in the end, he had 2 cups of wine left in his flask. The flask o... | 4. (3 points) It is said that during the Tang Dynasty, the poet Li Bai went to drink wine, carrying a pot on the street, doubling the wine when encountering a shop, and drinking 2 cups when seeing flowers. On the way, he met shops and flowers four times, and finally, there were 2 cups of wine left in the pot. The pot o... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
15. Scientists conduct an experiment, making a record every 5 hours. When making the twelfth record, the hour hand of the clock points exactly to 9. What number did the hour hand point to when they made the first record? | 15.【Solution】From the first record to the twelfth record, there are eleven intervals, totaling $5 \times 11=55$ (hours). The hour hand completes one full circle in 12 hours, and the remainder of 55 divided by 12 is $7, 9-7=2$
Answer: The hour hand points to 2. | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
74. In the school attendance record, 15 students were absent on Monday, 12 students were absent on Tuesday, and 9 students were absent on Wednesday. Over these three days, there were 22 students with absence records. Therefore, the maximum number of students who were absent all three days is $\qquad$ people. | answer: 7 | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. There are 1000 numbers arranged in a row, where among any three consecutive numbers, the middle number equals the sum of the numbers before and after it. If the first number and the second number are both 1, then the sum of these 1000 numbers is
A. 1000
B. 1
C. 0
D. -1 | 1. B According to the problem, the 1000 numbers are arranged as follows: $1,1,0,-1,-1,0,1,1,0,-1,-1,0, \cdots$ In other words, they cycle in a pattern of $1,1,0,-1,-1,0$.
Since $1000 \div 6=166 \cdots \cdots 4$, and the sum of the 6 numbers in each cycle is 0, the sum of these 1000 numbers is $1+1+0+(-1)=$ 1. Therefore... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. Afanti helps the tenant farmer to claim wages from the landlord. To make it difficult for Afanti, the landlord proposes a requirement: for every integer $n$ greater than 1921 that makes $\frac{n-1921}{2021-n}$ a natural number, he will give 1 gold bar as wages. How many gold bars can Afanti claim for the tenant farm... | $8$ | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Represent August 12, 2018 using eight digits as $12-08-2018$, where each digit appears exactly twice. Then including August 12, the number of days in 2018 that satisfy this property is ( ).
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9 | 1. B.
Since each digit appears twice, and the digits of 2018 are all different, the month and day must be composed of 2018. Therefore, the possible months are: 01, 02, 08, 10, and 12.
01 month can have 28 days; 02 month can have 18 days; 08 month can have 12, 21 days; 10 month can have 28 days. In total, there are 6 po... | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
4. In a triangular pyramid, two of the three faces are isosceles right triangles. The other one is an equilateral triangle with a side length of 1. Then, the number of such triangular pyramids is $\qquad$. | 4. 3 cases concerning the tetrahedron $ABCD$ are as follows:
(1) $AB=AC=BC=AD=1, CD=BD=\sqrt{2}$;
(2) $BC=BD=CD=1, AB=AC=AD=\frac{\sqrt{2}}{2}$;
(3) $AB=BC=AC=AD=BD=1, CD=\sqrt{2}$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$31 \cdot 11-$ a two-digit number, when the tens and units digits are swapped, the resulting two-digit number is $\frac{7}{4}$ times the original number. How many such two-digit numbers are there?
(A) 1.
(B) 2.
(C) 4.
(D) Infinitely many.
(E) 0.
(China Tianjin Junior High School Mathematics Competition, 1984) | ［Solution］Let the two-digit number be $10a + b$. According to the problem, we have
$$
10b + a = \frac{7}{4}(10a + b),
$$
which simplifies to $66a - 33b = 0$, or $2a - b = 0$.
Since $a, b$ are both single digits, then $\left\{\begin{array}{l}a=1,2,3,4, \\ b=2,4,6,8 .\end{array}\right.$ Therefore, the answer is $(C)$. | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
7. Given $F_{1}, F_{2}$ are the left and right foci of the hyperbola $C: \frac{x^{2}}{4}-\frac{y^{2}}{12}=1$, point $P$ is on the hyperbola $C$, and $G, I$ are the centroid and incenter of $\triangle F_{1} P F_{2}$, respectively. If $G I \| x$-axis, then the circumradius $R$ of $\triangle F_{1} P F_{2}$ is $\qquad$ ـ. | Answer 5.
Analysis Suppose $P\left(x_{0}, y_{0}\right)$ is in the first quadrant, $\left|P F_{1}\right|=r_{1},\left|P F_{2}\right|=r_{2}$. According to the problem, $r_{1}-r_{2}=4,\left|F_{1} F_{2}\right|=8$. Since $G, I$ are the centroid and incenter of $\triangle F_{1} P F_{2}$ respectively, and $G I \| x$-axis, the ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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