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Three. (50 points)
Given a positive integer $m \geqslant 17$, there are $2 m$ players participating in a round-robin tournament. Each round, the $2 m$ players are divided into $m$ pairs, with each pair competing against each other. In the next round, the players are re-paired, and this continues for a total of $2 m-1$ ... | For $m \geqslant 3$, the minimum possible $n_{\text {min }}=m-1$.
Suppose a feasible program has been carried out for $l$ rounds: $l \leqslant m-2$.
Let $k$ be the maximum positive integer such that there exist $k$ players who have all played against each other. Without loss of generality, let these $k$ players be $A_{... | -1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let real numbers $x, y$ satisfy
$$
x^{2}+\sqrt{3} y=4, y^{2}+\sqrt{3} x=4, x \neq y \text {. }
$$
Then the value of $\frac{y}{x}+\frac{x}{y}$ is . $\qquad$ | 3. -5 .
From the conditions, we have
$$
\left\{\begin{array}{l}
x^{2}-y^{2}+\sqrt{3} y-\sqrt{3} x=0, \\
x^{2}+y^{2}+\sqrt{3}(x+y)=8 .
\end{array}\right.
$$
From equation (1) and $x \neq y$, we know $x+y=\sqrt{3}$.
Substituting into equation (2) gives $x^{2}+y^{2}=5$.
$$
\begin{array}{l}
\text { Also, } 2 x y=(x+y)^{2... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Marie-Pascale solves 4 math problems per day. Kaeli solves $x$ math problems per day. Over a number of days, Marie-Pascale solves a total of 72 problems. Over the same number of days, Kaeli solves 54 more problems than Marie-Pascale. What is the value of $x$ ? | 3. Since Marie-Pascale solves 4 math problems per day and solves 72 problems in total, this takes her $72 \div 4=18$ days.
Since Kaeli solves 54 more problems than Marie-Pascale, she solves $72+54=126$ problems.
Since Kaeli solves $x$ problems per day over 18 days and solves 126 problems in total, then $x=\frac{126}{18... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Calculate: $\sqrt{9+8 \cos 20^{\circ}}-\sec 20^{\circ}=$ | 4.3.
From the triple angle formula, we get
$8 \cos ^{3} 20^{\circ}=1+6 \cos 20^{\circ}$.
Then $\sqrt{9+8 \cos 20^{\circ}}-\sec 20^{\circ}$
$$
=\frac{\sqrt{9 \cos ^{2} 20^{\circ}+8 \cos ^{3} 20^{\circ}}-1}{\cos 20^{\circ}}=3 .
$$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Given $x, y, z \in \mathbf{R}_{+}$ and $x+y+z=1$. Then the maximum value of $x+\sqrt{2 x y}+3 \sqrt[3]{x y z}$ is $\qquad$ | 3. 2 .
By the AM-GM inequality, we have
$$
\frac{x}{2}+y \geqslant \sqrt{2 x y}, \frac{x}{2}+y+2 z \geqslant 3 \sqrt[3]{x y z} \text {. }
$$
Adding the two inequalities, we get
$$
x+\sqrt{2 x y}+3 \sqrt[3]{x y z} \leqslant 2(x+y+z)=2 \text {. }
$$
The equality holds when $x: y: z=4: 2: 1$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2 \cdot 94$ for the equation $x^{2}-2|x|+2=m$. If the equation has exactly 3 real roots, then the value of $m$ is
(A) 1 .
(B) $\sqrt{3}$.
(C) 2 .
(D) 2.5 .
(China Beijing Junior High School Mathematics Competition, 1991) | [Solution] Rewrite the equation as $|x|^{2}-2|x|+2-m=0$, then $|x|=1 \pm \sqrt{m-1}$,
i.e., $x=1 \pm \sqrt{m-1}$ or $x=-1 \mp \sqrt{m-1}$. When $m=2$, $x=2,0,-2$, there are exactly 3 roots. Therefore, the answer is $(C)$. | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. Through the vertex $D$ of the tetrahedron $ABCD$, construct a sphere with radius 1. This sphere is tangent to the circumsphere of the tetrahedron $ABCD$ at point $D$, and is also tangent to the plane $ABC$. If $AD=2\sqrt{3}$, $\angle BAC=60^{\circ}$, $\angle BAD=\angle CAD=45^{\circ}$, then the radius of the circums... | 5.3.
Draw a perpendicular from point $D$ to plane $A B C$, with the foot of the perpendicular being $H$. Draw $D E \perp A B$ at point $E$, and $D F \perp A C$ at point $F$.
$$
\begin{array}{l}
\text { Then } H E \perp A B, H F \perp A C, \text { and } \\
A E=A F=A D \cos 45^{\circ}=\sqrt{6} . \\
\text { By } \triangl... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$5 、$ In a certain examination room of the "Zhonghuan Cup", there are a total of 45 students, among whom 35 are good at English, 31 are good at Chinese, and 24 are good at both subjects. How many students are not good at either subject? $\qquad$ people | 【Answer】3
Analysis: $45-(35+31-24)=45-42=3$ people | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In the sequence $\left\{a_{n}\right\}$, $a_{1}=2, a_{2}=7, a_{n+2}$ is equal to the unit digit of $a_{n+1} a_{n}$, then $a_{2009}=$ | 2. 2 Detailed Explanation: The first few terms are $2,7,4,8,2,6,2,2,4,8,2$, starting from the third term, it has a cycle of 6, so $a_{2009}=a_{6 \times 334+5}=a_{5}=2$. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. (10 points) Hongxing Primary School organized students to participate in a drill. At the beginning, only 40 boys participated. Later, the team was adjusted, each time reducing 3 boys and adding 2 girls. After $\qquad$ adjustments, the number of boys and girls became equal. | 10. (10 points) Hongxing Primary School organized students to participate in a drill. At the beginning, only 40 boys participated. Later, the team was adjusted, each time reducing 3 boys and adding 2 girls. After 8 adjustments, the number of boys and girls was equal.
【Solution】Solution: $40 \div(3+2)$
$$
\begin{array}{... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A TV station is going to broadcast a 30-episode TV series. If it is required that the number of episodes aired each day must be different, what is the maximum number of days the TV series can be broadcast?
---
The translation maintains the original format and line breaks as requested. | 2.【Solution】If it is broadcast for more than 8 days, then since the number of episodes broadcast each day is different, there must be at least $1+2+3+4+5+6+7+8=36$ episodes,
so a 30-episode series cannot be broadcast for more than 8 days as required. On the other hand, $1+2+3+4+5+6+9=30$
so it can be broadcast for a ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8・165 Given a sequence of natural numbers $\left\{x_{n}\right\}$ that satisfies
$$
x_{1}=a, x_{2}=b, x_{n+2}=x_{n}+x_{n+1}, n=1,2,3, \cdots
$$
If one of the terms in the sequence is 1000, what is the smallest possible value of $a+b$? | [Solution] Take the sequence $\left\{t_{n}\right\}$ satisfying:
$$
t_{1}=1, t_{2}=0, t_{n+2}=t_{n}+t_{n+1}, n=1,2,3, \cdots
$$
It is easy to prove that for any natural number $n$ we have
$$
x_{n}=t_{n} a+t_{n+1} b .
$$
The terms of the sequence $\left\{t_{n}\right\}$ that are less than 1000 are $1,0,1,1,2,3,5,8,13,21... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let $x_{1}, x_{2}$ be the roots of the equation for $x(x \in \mathbf{C})$
$$
x^{2}-3 \sin \theta \cdot x+\sin ^{2} \theta+1=0(\theta \in[0, \pi))
$$
and $\left|x_{1}\right|+\left|x_{2}\right|=2$. Then $\theta=$ $\qquad$ | 4. 0 .
Notice,
$$
\Delta=9 \sin ^{2} \theta-4\left(\sin ^{2} \theta+1\right)=5 \sin ^{2} \theta-4 \text {. }
$$
(1) If $\Delta \geqslant 0$.
From $x_{1} x_{2}=\sin ^{2} \theta+1>0$, we know
$$
\begin{array}{l}
\left|x_{1}\right|+\left|x_{2}\right|=\left|x_{1}+x_{2}\right| \\
=|3 \sin \theta|=3 \sin \theta=2 \\
\Right... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. As shown in Figure $1, \odot O_{1}$
intersects with $\odot O_{2}$ at points $A$ and $B$,
point $C$ is on $\odot O_{1}$,
outside $\odot O_{2}$, the extensions of $C A$ and
$C B$ intersect $\odot O_{2}$ at points $D$ and
$E, \odot O_{1}$ has a radius of
$5, A C=8, A D=12, D E=14, S_{\triangle C D E}=112$. Then the ra... | 6. A.
As shown in Figure 7, draw the diameter $C F$ of $\odot O_{1}$, extend $C F$ to intersect $D E$ at point $G$, and connect $A B, A F$.
$$
\begin{array}{l}
\text { Then } \angle A F C=\angle A B C=\angle D, \\
\angle A C F=\angle G C D .
\end{array}
$$
Therefore, $\triangle A C F \backsim \triangle G C D$
$$
\Rig... | 5 | Geometry | MCQ | Yes | Yes | olympiads | false |
9. The function $f(x)=$ $\frac{x+1}{x}+\frac{x+2}{x+1}+\cdots+\frac{x+2023}{x+2022}$ has a center of symmetry at $(a, b)$, then $2 a+b=$ $\qquad$ _. | $$
f(x)=2023+\frac{1}{x}+\frac{1}{x+1}+\cdots+\frac{1}{x+2022}=2023+g(x) \text {, }
$$
where $g(x)$ satisfies $g(-x-2022)=-g(x) \Rightarrow g(x)$ has a symmetry center at $(-1011,0)$, thus $f(x)$ has a symmetry center at $(-1011,2023)$. Therefore, $2 a+b=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
【Question 13】As shown in the figure, three line segments divide the regular hexagon into four parts. It is known that the areas of three parts are $2$, $3$, and $4$ square centimeters, respectively. What is the area of the fourth part (the shaded area in the figure)? $\qquad$ square centimeters. | Analysis:
First, analyze the regular hexagon. As shown in the left figure, the area of a regular hexagon can be expressed as $6 S$. It is easy to see that the area of triangle $D E F$ is $S$, and the area of triangle $C D F$ is $2 S$.
As shown in the right figure, connect $D F$ and $C F$. Let the area of the regular h... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$$
\begin{aligned}
f(x)= & a_{2020} x^{2020}+a_{2019} x^{2019}+\cdots+ \\
& a_{2018} x^{2018}+a_{1} x+a_{0},
\end{aligned}
$$
where, $a_{i} \in \mathbf{Z}, i=0,1, \cdots, 2020$. Find the number of ordered pairs $(n, p)$ (where $p$ is a prime) that satisfy $p^{2}<n<2020$, and for any $i=0,1, \cdots, n, \mathrm{C}_{n}^{... | Three, Lemma $1 \mathrm{C}_{n}^{0}, \mathrm{C}_{n}^{1}, \cdots, \mathrm{C}_{n}^{n}$ are all odd if and only if $n=2^{r}-1$.
Proof of Lemma 1: By a well-known conclusion, $\mathrm{C}_{n}^{i}$ is even if and only if $i+(n-i)=n$ causes a carry in binary.
By the problem statement, when $i=0,1, \cdots, n$, $i+(n-i)=n$ doe... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$29 \cdot 182^{1000}$ divided by 13, the remainder is
(A) 1.
(B) 2.
(C) 3.
(D) 7.
(E) 11.
(23rd American High School Mathematics Examination, 1972) | ［Solution］Notice the following fact:
If $N_{1}=Q_{1} D+R_{1}, N_{2}=Q_{2} D+R_{2}$, then
$$
N_{1} N_{2}=\left(Q_{1} Q_{2} D+Q_{1} R_{2}+Q_{2} R_{1}\right) D+R_{1} R_{2} \text {. }
$$
That is, if $N_{1}, N_{2}$ are divided by $D$ with remainders $R_{1}, R_{2}$, then $N_{1} N_{2}$ divided by $D$ has a remainder of $R_{1... | 3 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Example 9 In a sequence of numbers $a_{1}, a_{2}, \cdots, a_{n}$, the sum of any five consecutive terms is negative, while the sum of any nine consecutive terms is positive. What is the maximum value of $n$? Prove your conclusion. [5]
$(2015$, National Junior High School Mathematics Invitational Competition) | When there are 13 terms, arrange them as follows:
$$
\begin{array}{l}
a_{1}, a_{2}, \cdots, a_{9} ; \\
a_{2}, a_{3}, \cdots, a_{10} ; \\
a_{3}, a_{4}, \cdots, a_{11} ; \\
a_{4}, a_{5}, \cdots, a_{12} ; \\
a_{5}, a_{6}, \cdots, a_{13} .
\end{array}
$$
Consider calculating the sum $M$ of all numbers in the table.
By the... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
24. Use the digits $1, 2, 3, 4, 5$ to form a five-digit number $\overline{P Q R S T}$ without repeating any digit, and $4|\overline{P Q R}, 5| \overline{Q R S}$, $3 \mid \overline{R S T}$. Then the digit $P$ is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 | 24. A.
From $5 \mid \overline{Q R S} \Rightarrow S=5$.
From $4|\overline{P Q R} \Rightarrow 4| \overline{Q R}$
$\Rightarrow R=2$ or 4.
When $R=2$, by
$$
\begin{array}{l}
3|\overline{R S T} \Rightarrow 3| \overline{25 T} \\
\Rightarrow 2+5+T \equiv 0(\bmod 3) \\
\Rightarrow T \equiv 2(\bmod 3) .
\end{array}
$$
Thus, $... | 1 | Number Theory | MCQ | Yes | Yes | olympiads | false |
2. Let real numbers $x, y, z, w$ satisfy $x+y+z+w=x^{7}+y^{7}+z^{7}+w^{7}=0$, find the value of $w(w+x)(w+y)(w+z)$.
(IMO - 26 Shortlist) | 2. Let $x, y, z, w$ be the four roots of the quartic polynomial $f(x)=t^{4}+a_{1} t^{3}+a_{2} t^{2}+a_{3} t+a_{4}$, and let $T_{n}=x^{n}+y^{n}+z^{n}+w^{n}$. Then $T_{1}=-a_{1}=0$, i.e., $a_{1}=0$. $T_{2}=-2 a_{2}$, $T_{3}=-3 a_{3}$, $T_{4}=2 a_{2}^{2}-4 a_{4}$, $T_{5}=5 a_{2} a_{3}$, $T_{7}=7 a_{3}(a_{4}-a_{2}^{2})=0$,... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5 In a rectangle $A B C D$ with an area of 1 (including the boundary), there are 5 points, among which no three points are collinear. Find the minimum number of triangles, formed by these 5 points as vertices, whose area is not greater than $\frac{1}{4}$. (Supplied by Leng Gangsong) | The proof of this problem requires the following commonly used conclusion, which we will use as a lemma: The area of any triangle within a rectangle is no more than half the area of the rectangle.
In rectangle $ABCD$, if a triangle formed by any three points has an area no greater than $\frac{1}{4}$, then these three ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. At the initial moment, a positive integer $N$ is written on the blackboard. In each step, Misha can choose a positive integer $a>1$ that is already written on the blackboard, erase it, and write down all its positive divisors except itself. It is known that after several steps, there are exactly $N^{2}$ numbers on t... | 7. It suffices to prove: For any positive integer $N$, the numbers written on the blackboard are always no more than $N^{2}$, and when $N \geqslant 2$, the numbers written on the blackboard are fewer than $N^{2}$.
Obviously, the conclusion holds for $N=1$.
For $N \geqslant 2$, assume the conclusion holds for all positi... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. If $3 \sin ^{3} x+\cos ^{3} x=3$, then the value of $\sin ^{2018} x+\cos ^{2018} x$ is $\qquad$ | Answer 1.
Analysis First, from $3 \sin ^{3} x+\cos ^{3} x=3$ we know that $\sin x \geq 0$ must hold, otherwise $\cos ^{3} x=3-3 \sin ^{3} x \geq 3$, which is a contradiction. Also, since $\sin ^{3} x \leq \sin ^{2} x, \cos ^{3} x \leq \cos ^{2} x$, we have
$$
3=3 \sin ^{3} x+\cos ^{3} x \leq 3 \sin ^{2} x+\cos ^{2} x=1... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In $\triangle A B C$, $\angle B=60^{\circ}, \angle C=75^{\circ}, S_{\triangle A B C}=\frac{1}{2}(3+\sqrt{3}), B C=$ | 3. 2 .
$$
\angle A=180^{\circ}-60^{\circ}-75^{\circ}=45^{\circ}, S_{\triangle A B C}=\frac{1}{2} a c \sin B=\frac{1}{2} a c \sin B=\frac{1}{2} a c \sin 60^{\circ}=\frac{1}{2}(3+
$$
$\sqrt{3}) \Rightarrow a c=2(1+\sqrt{3}), \frac{a}{c}=\frac{\sin A}{\sin C}=\frac{\sin 45^{\circ}}{\sin 75^{\circ}}=\frac{1}{\sin 30^{\circ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) The sum of the ages of four people, A, B, C, and D, this year is 72 years. A few years ago (at least one year), when A was 22 years old, B was 16 years old. It is also known that when A was 19 years old, C's age was 3 times D's age (at that time, D was at least 1 year old). If the ages of A, B, C, and D ... | 【Analysis】It is known that the sum of the ages of four people this year is 72 years. A few years ago (at least one year), when A was 22 years old, B was 16 years old. When A was 19 years old, B was 13 years old, and C's age was three times D's age (at this time D is at least 1 year old). Since this was at least 4 years... | 6 | Algebra | MCQ | Yes | Yes | olympiads | false |
$29 \cdot 21$ If $n$ is any integer, $n^{2}\left(n^{2}-1\right)$ is divisible by $x$, then $x$ equals
(A) 12.
(B) 24.
(C) Any multiple of 12.
(D) $12-n$.
(E) 12 and 24.
(7th American High School Mathematics Examination, 1956) | [Solution] When $n=2$, $n^{2}\left(n^{2}-1\right)=12$, so $(B),(C),(D),(E)$ are not true. Therefore, the answer is $(A)$. | 12 | Number Theory | MCQ | Yes | Yes | olympiads | false |
11. Find the minimum value of the function
$$
y=2 x+\sqrt{4 x^{2}-8 x+3}
$$ | II. 11. From
$4 x^{2}-8 x+3 \geqslant 0 \Rightarrow x \leqslant \frac{1}{2}$ or $x \geqslant \frac{3}{2}$.
Therefore, the domain of the function is
$$
\left(-\infty, \frac{1}{2}\right] \cup\left[\frac{3}{2},+\infty\right) \text {. }
$$
Let $y=f(x)=2 x+\sqrt{4 x^{2}-8 x+3}$. Then
$$
\begin{array}{l}
f^{\prime}(x)=2+\fr... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1- 17 Let $A$ be the sum of the digits of the decimal number $4444^{4444}$, and let $B$ be the sum of the digits of $A$. Find the sum of the digits of $B$ (all numbers here are in decimal). | [Solution] First, we estimate the number of digits in $10000^{4444}$, which has $4 \times 4444 + 1 = 17777$ digits. Since each digit can be at most 9, and $4444^{4444} < 10000^{4444}$, we have
$$
A < 17777 \times 9 = 159993.
$$
Now, let's represent $A$ as
$$
A = a_{5} \cdot 10^{5} + a_{4} \cdot 10^{4} + a_{3} \cdot 10... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. $A, B, C, D, E$ five people participate in an exam, which has 7 questions, all of which are true/false questions. The scoring rule is: for each question, a correct answer earns 1 point, a wrong answer deducts 1 point, and no answer neither earns nor deducts points. Figure 15-12 records the answers of $A, B, C, D, E... | 13. Let $x_{k}=\left\{\begin{array}{l}1, \text { if the } k \text {th problem is correct, } \\ -1, \text { if the } k \text {th problem is incorrect, }\end{array}\right.$ $k=1,2, \cdots, 7$. Thus, when the $k$th problem is correct, i.e., $x_{k}=1$. If it is judged as correct (i.e., marked with a “ $\checkmark$ ”), then... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$33 \cdot 47$ Given $y=\sqrt{11+6 \sqrt{2}}, x$ represents the fractional part of $y$. Then the value of $x^{2}+2 y$ is
(A) 7.
(B) 8.
(C) 9.
(D) 10.
(China Sichuan Province Junior High School Mathematics Competition, 1991) | [Solution] From $y=\sqrt{11+6 \sqrt{2}}=\sqrt{9+6 \sqrt{2}+2}$
$$
\begin{array}{lc}
& =3+\sqrt{2}, \\
\because & 1<\sqrt{2}<2, \\
\text { then } & x=(3+\sqrt{2})-(3+1)=\sqrt{2}-1, \\
\text { thus } & x^{2}+2 y=(\sqrt{2}-1)^{2}+2(3+\sqrt{2})=9 .
\end{array}
$$
Therefore, the answer is $(C)$. | 9 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. (10 points) $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ are natural numbers satisfying $0<a_{1}<a_{2}<a_{3} \cdots<a_{n}$, and $\frac{13}{14}=\frac{1}{\mathrm{a}_{1}}, \frac{1}{\mathrm{a}_{2}}, \frac{1}{\mathrm{a}_{3}}+\cdots$ $+\frac{1}{a_{n}}$, then the minimum value of $n$ is . $\qquad$ | 【Analysis】To make $n$ the smallest, it means to minimize the number of terms, so each term should be as small as possible. $0\frac{1}{3}$,
Let $a_{2}=3$, then $\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n}}=\frac{3}{7}-\frac{1}{3}=\frac{2}{21}>\frac{1}{11}$,
Let $a_{3}=11$, then $\frac{1}{\mathrm{a}_{4}}+\cdots+\frac{1}{\mathr... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (6 points) Define new operations: $a \triangle b=(a+b)+2, a \bigcirc b=a \times 3+b$, when $(X \triangle 24) \bigcirc 18=60$, $X$ $=$ . $\qquad$ | 【Solution】Solve: $(X \triangle 24) \bigcirc 18=60$,
$$
\begin{aligned}
(X+24+2) \bigcirc 18 & =60, \\
(X+26) \times 3+18 & =60, \\
X+26 & =14, \\
X & =-12 ;
\end{aligned}
$$
Therefore, the answer is: -12. | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Given the sets
$$
\begin{array}{l}
A=\left\{(x, y) \mid x=m, y=-3 m+2, m \in \mathbf{Z}_{+}\right\}, \\
B=\left\{(x, y) \mid x=n, y=a\left(a^{2}-n+1\right), n \in \mathbf{Z}_{+}\right\} .
\end{array}
$$
Then the number of integers $a$ such that $A \cap B \neq \varnothing$ is $\qquad$. | 7. 10 .
Let $(x, y) \in A \cap B$. Then
\[
\begin{array}{l}
x=n, \\
y=-3 n+2=a\left(a^{2}-n+1\right)\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
\]
Thus, $(a-3) n=a^{3}+a-2$.
If $a=3$, then in equation (1),
the left side $=0$,
the right side $=27+3-2>0$, a contradiction.
Therefore, $a \neq 3$.
Hence, $n=\frac{a^{3... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (8 points) The unit digit of 2010 2009s is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 【Solution】Solution: The unit digit is determined only by the unit digit of a multi-digit number, that is, the unit digit is determined by the product of 2010 nines,
According to the cycle of $9$ and $1$ every 2 digits,
$$
2010 \div 2=1005
$$
There is no remainder,
Therefore, the unit digit of the product of 2010 nine... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Given that $f(x)$ is a periodic function on $\mathbf{R}$ with the smallest positive period of 2, and when $0 \leqslant x<2$, $f(x)=x^{3}-x$, then the number of intersections of the graph of the function $f(x)$ with the $x$-axis in the interval $[0,6]$ is $\qquad$ . | $f(x)=x(x+1)(x-1) \Rightarrow f(x)$ has 2 zeros in one period, so the graph of the function $f(x)$ intersects the $x$-axis 7 times in the interval $[0,6]$. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(3) Let $M$ be a set of $n$ points in the plane, satisfying:
(1) There exist 7 points in $M$ that are the 7 vertices of a convex heptagon;
(2) For any 5 points in $M$, if these 5 points are the 5 vertices of a convex pentagon, then this convex pentagon contains at least one point from $M$ inside it. Find the minimum va... | First, we prove that \( n \geqslant 11 \).
Consider a convex heptagon \( A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7} \) with vertices in \( M \). Connect \( A_{1} A_{5} \). By condition (1), there is at least one point in \( M \) within the convex pentagon \( A_{1} A_{2} A_{3} A_{4} A_{5} \), denoted as \( P_{1} \). Con... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.54 If $2 x^{3}-h x+k$ has two factors $x+2$ and $x-1$, then the value of $2 h-3 k$ is
(A) 4.
(B) 3.
(C) 2.
(D) 1.
(E) 0.
(21st American High School Mathematics Examination, 1970) | [Solution] Let $f(x)=2 x^{3}-h x+k$. According to the problem, we have
$$
\begin{array}{l}
f(-2)=2 \cdot(-2)^{3}-h(-2)+k=2 h+k-16=0, \\
f(1)=2 \cdot 1^{3}-h \cdot 1+k=k-h+2=0 .
\end{array}
$$
From the above, we solve to get $h=6, k=4$, then $|2 h-3 k|=0$. Therefore, the answer is $(E)$. | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 9 Let $M=\{1,2,3, \cdots, 40\}$, find the smallest positive integer $n$, such that $M$ can be divided into $n$ pairwise disjoint subsets and for any 3 numbers $a, b, c$ (not necessarily distinct) taken from the same subset, $a \neq b+c$.
| Solve for $n=4$, divide $M$ into the following 4 pairwise disjoint subsets: $A=\{1,4,10,13,28,31,37,40\}$, $B=\{2,3,11,12,29,30,38,39\}$, $C=\{14,15,16, \cdots, 25,26,27\}$, $D=\{5,6,7,8,9,32,33,34,35,36\}$. For any 3 numbers $a, b, c$ (not necessarily distinct) taken from the same subset, $a \neq b+c$. Therefore, the ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
31. Given $\overline{A B C D E F} \times B=\overline{E F A B C D}$, the same letter represents the same digit, and different letters represent different digits. Then the possible cases for $\overline{A B C D E F}$ are $\qquad$ kinds. | answer: 2 | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$12 \cdot 1$ If $S=i^{n}+i^{-n}$, where $i$ is the imaginary unit, and $n$ is an integer, then the number of distinct possible values of $S$ is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(E) more than 4 .
(8th American High School Mathematics Examination, 1957) | ［Solution］Consider four cases:
(1) When $n=4 k$, $\quad S=i^{4 k}+i^{-4 k}=2$;
(2) When $n=4 k+1$, $S=i^{4 k+1}+i^{-(4 k+1)}=0$;
(3) When $n=4 k+2$, $\quad S=i^{4 k+2}+i^{-(4 k+2)}=-2$;
(4) When $n=4 k+3$, $S=i^{4 k+3}+i^{-(4 k+3)}=0$. Therefore, the answer is $(C)$. | 3 | Algebra | MCQ | 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 $(\mathrm{x})$.
A. 5
B. 6
C. 7
D. 8 | 【Answer】 $\mathrm{A}$
【Solution】【Detailed】 $(x, y)=(-42,6),(6,-42),(8,56),(14,14),(56,8)$, a total of 5 pairs. When $x=7$, there is no solution; when $x \neq 7$, $y=\frac{7 x}{x-7}$, so $x-7 \mid 7 x$, thus $x-7 \mid 49$, hence $x=-42,6,8,14,56$. In summary, there are 5 pairs of solutions. Therefore, the answer is: $\m... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. Let $A=\{(\alpha, \beta, \gamma) \mid \alpha+\beta \sin x+\gamma \cos x=0, x \in(-\infty,+\infty)\}$, then the number of elements in the set $A$ is $m=$
$\qquad$ . | Take $x=0, \frac{\pi}{2}, \pi$, we have $\left\{\begin{array}{l}\alpha+\gamma=0, \\ \alpha+\beta=0, \\ \alpha-\gamma=0\end{array} \Rightarrow\left\{\begin{array}{l}\alpha=0, \\ \beta=0, \\ \gamma=0 .\end{array}\right.\right.$ Therefore, $m=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
149 The unit digit of a positive integer $m$ is denoted by $f(m)$. $a_{n}=f\left(2^{n+1}-1\right)(n=1,2, \cdots)$, then $a_{1994}$ $=$ ـ. $\qquad$ | $149 \quad 7 . \quad 1994=4 \times 498+2$,
$$
\begin{aligned}
a_{1994} & =f\left(2^{1 \times 498-3}-1\right)=f\left(2^{4 \times 498} \times 8-1\right) \\
& =f\left(16^{198} \times 8-1\right)=f(6 \times 8-1)=7 .
\end{aligned}
$$ | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. (40 points) Let real numbers $a_{1}, a_{2}, \cdots, a_{n} \in[0,2]$, and define $a_{n+1}=a_{1}$, find the maximum value of $\frac{\sum_{i=1}^{n} a_{1}^{2} a_{i+1}+8 n}{\sum_{i=1}^{n} a_{i}^{2}}$. | 【Difficulty】
【Knowledge Points】Algebra Module -- Extreme Value Problems
【Reference Answer】When $a_{1}=a_{2}=\cdots=a_{n}=2$, the original expression $=4$ $-10$
points
Next, we prove $\frac{\sum_{i=1}^{n} a_{i}^{2} a_{i+1}+8 n}{\sum_{i=1}^{n} a_{i}^{2}} \geq 4$, i.e., $\sum_{i=1}^{n} a_{i}^{2} a_{i+1}+8 n \geq 4 \sum_{... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
11. (5 points) Use a 34-meter long rope to form a rectangle, and the side lengths of the rectangle are all integer meters. There are $\qquad$ different ways to form the rectangle (rectangles with the same side lengths are considered the same way).
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note at the end is not p... | 11. (5 points) A 34-meter long rope is used to form a rectangle, and the side lengths of the rectangle are all integers in meters. There are $\qquad$ 8 different ways to form the rectangle (rectangles with the same side lengths are considered the same).
【Solution】Solution: Let the length of the rectangle be $a m$, and... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. $[a]$ represents the greatest integer not exceeding $a$, $\{a\}$ represents the fractional part of $a$, for example:
$$
[2.25]=2, \quad\{2.25\}=0.25, \quad[4]=4, \quad\{4\}=0
$$
Then the sum of all $x$ that satisfy $x+2[x]+4\{x\}=10$ is $\qquad$ | $6$ | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (3 points) The unit digit of $2^{2012}$ is $\qquad$ . (where, $2^{n}$ represents $n$ twos multiplied together) | 【Answer】Solution: $2012 \div 4=503$;
No remainder, which means the unit digit of $2^{2012}$ is 6. Therefore, the answer is: 6. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. (5 points) Arrange 6 balls in a row, where balls 1, 2, and 3 are black, and balls 4, 5, and 6 are white, as shown in Figure 1. If balls 2 and 5 are swapped, the 6 balls will be arranged in an alternating black and white pattern, as shown in Figure 2. Now, there are 20 balls arranged in a row in sequence, with balls... | 【Analysis】From the problem, we know that 6 balls (3 black, 3 white) only need to swap the middle two balls once to form a black and white alternating arrangement, i.e., the middle 2 balls remain unchanged, and the remaining 4 balls need to be swapped once. Therefore, these 4 balls can be considered as one cycle.
If we... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(8) Given a moving point $P(x, y)$ satisfies $\left\{\begin{array}{l}2 x+y \leqslant 2, \\ x \geqslant 0, \\ \left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right) \geqslant 1,\end{array}\right.$ then the area of the figure formed by the moving point $P(x, y)$ is $\qquad$ | 82 Prompt: From $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right) \geqslant 1$ we get
$$
x+\sqrt{x^{2}+1} \geqslant \sqrt{y^{2}+1}-y,
$$
Therefore,
$$
\ln \left(x+\sqrt{x^{2}+1}\right) \geqslant \ln \left(\sqrt{y^{2}+1}-y\right) .
$$
Let $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)$, then it is easy to see that ... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
11. In an arithmetic sequence with all terms being real numbers, the common difference is 4, and the square of the first term plus the sum of the remaining terms does not exceed 100. Such a sequence can have at most $\qquad$ terms. | 11. 8 Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence with a common difference of 4, then $a_{n}=a_{1}+4(n-1)$. From the given information,
$$
\begin{aligned}
a_{1}^{2}+a_{2}+\cdots+a_{n} \leqslant 100 & \Leftrightarrow a_{1}^{2}+\frac{\left(a_{2}+a_{n}\right)(n-1)}{2} \leqslant 100 \\
& \Leftrightarrow a_{... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. For any real numbers $x, y$, define the operation $x * y$ as $x * y=a x+b y+$ $c x y$, where $a, b, c$ are constants, and the operations on the right side of the equation are the usual real number addition and multiplication. It is known that $1 * 2=3,2 * 3=4$, and there is a non-zero real number $d$, such that for ... | 4. For any real number $x$, we have
$$
\begin{array}{cc}
& x * d=a x+b d+c d x=x_{1} \\
\therefore & 0 * d=b d=0 . \\
\because & d \neq 0, \quad \therefore \quad b=0 .
\end{array}
$$
(where $d \neq 0$),
Thus, from
$$
\left\{\begin{array}{l}
1 * 2=a+2 b+2 c=3, \\
2 * 3=2 a+3 b+6 c=4 .
\end{array}\right.
$$
we get
$$
... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
85. 50 chess pieces are arranged in a circle, numbered $1,2,3, \cdots \cdots, 50$ in a clockwise direction. Then, starting in a clockwise direction, every other piece is removed until only one piece remains. If the remaining piece is number 42, then the first piece removed is $\qquad$ number. | Answer: 7 | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
52. There are 2018 balls in a box, numbered $1, 2, 3, \ldots, 2018$. Now, 1616 balls are drawn from the box and their numbers are multiplied. Find the units digit of the product. | Reference answer: 0 | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The graph of the inverse function of $y=\frac{a-x}{x-a-1}$ is symmetric about the point $(-1,4)$. Then the real number $a$ equals ( ).
A. 2
B. 3
C. -2
D. -3 | 4. B. Given $y=\frac{a-x}{x-a-1}=-1-\frac{1}{x-(a+1)}$, the center of symmetry is $((a+1),-1)$. The center of symmetry of its inverse function is $(-1,(a+1)) \Rightarrow a+1=4 \Rightarrow a=3$. | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
8. Given the function
$$
f(x)=\mathrm{e}^{x}(x+a)(a \in \mathbf{R}) \text {. }
$$
Then the number of zeros of the function $g(x)=f(x-a)-x^{2}$ is $\qquad$ . | 8. When $a>1$, $g(x)$ has 3 zeros.
Notice,
$$
\begin{array}{l}
g(x)=f(x-a)-x^{2}=x \mathrm{e}^{x-a}-x^{2} \\
=x\left(\mathrm{e}^{x-a}-x\right) .
\end{array}
$$
From $g(x)=0$, we get
$x=0$ or $\mathrm{e}^{x-a}-x=0$.
Let $h(x)=\mathrm{e}^{x-a}-x$.
Also, $h(0)=\mathrm{e}^{-a} \neq 0$, which means $x=0$ is not a zero of $... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2 \cdot 48$ Try to find the remainder when $10^{10}+10^{10^{2}}+10^{10^{3}}+\cdots+10^{10^{10}}$ is divided by 7.
(5th Moscow Mathematical Olympiad, 1939) | [Solution] Let $A=10^{10}+10^{10^{2}}+10^{10^{3}}+\cdots+10^{10^{10}}$.
First, we prove that if $6 \mid n-r$, then $7 \mid 10^{n}-10^{r},(n>r)$.
In fact,
$$
\begin{aligned}
10^{n}-10^{r} & =10^{r}\left(10^{n-r}-1\right)=10^{r}\left(10^{6 k}-1\right) \\
& =\left(\left(10^{6}\right)^{k}-1\right) \cdot 10^{r} .
\end{align... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
15. (15 points) Two cars, A and B, travel from City A to City B at a speed of 55 kilometers/hour. At 10:00 AM, the distance Car A has traveled is 5 times the distance Car B has traveled; at noon 12:00 PM, the distance Car A has traveled is 3 times the distance Car B has traveled. How many hours later did Car B start co... | 【Analysis】It is known that during the 2 hours from 10:00 to 12:00, both car A and car B are driving. Therefore, the time that car B starts later than car A is the time that car B drives less than car A before 10:00. Thus, the problem can be solved by finding the time each car has driven before 10:00.
(1) The speed of b... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$13 \cdot 22$ The product of the largest and smallest roots of the equation $x|x|-5|x|+6=0$ is
(A) -6 .
(B) 3 .
(C) -3 .
(D) 6 .
(E) -8 .
(China Hubei Province Wuhan City Junior High School Mathematics Competition, 1985) | [Solution] When $x \geqslant 0$, the equation becomes $x^{2}-5 x+6=0$, solving this yields
$$
x_{1}=2, \quad x_{2}=3 .
$$
When $x<0$, the equation becomes $-x^{2}+5 x+6=0$, solving this yields
$x_{3}=-1, x_{4}=6$ (discard as it does not meet the condition).
Thus, the maximum root of the equation is 3, the minimum roo... | -3 | Algebra | MCQ | Yes | Yes | olympiads | false |
21. Given that the vertex of parabola $G$ is at the origin, and its focus is on the positive $y$-axis, the distance from point $P(m, 4)$ to its directrix is 5.
(1) Find the equation of parabola $G$;
(2) As shown in the figure, a line passing through the focus of parabola $G$ intersects the parabola $G$ and the circle $... | (1) $\frac{p}{2}=1 \Rightarrow p=2$, so the equation of the parabola $G$ is $x^{2}=4 y$.
(2) Let the line $A B: y=k x+1, A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
$C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$, combining with $x^{2}=4 y$ we get
$$
x^{2}-4 k x-4=0 \text {, }
$$
Combining with $... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 1 The number of equilateral triangles that can be formed by three vertices of a cube is
A. 4
B. 8
C. 12
D. 24 | Now, consider a vertex with three mutually perpendicular faces, each having a diagonal that together forms a unique equilateral triangle. Therefore, a one-to-one correspondence can be established between the set of equilateral triangles and the set of vertices. Since a cube has 8 vertices, there are exactly 8 equilater... | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
【Question 11】
Xiao Ling is reading an interesting storybook. Every day she reads twice the number of pages she has read in the previous days, and on the sixth day, she read $\frac{1}{9}$ of the book. On which day did Xiao Ling finish reading the book? | 【Analysis and Solution】Fraction Application Problem.
On the 6th day, she read $\frac{1}{9}$ of the book;
From day $1 \sim 5$, she read $\frac{1}{9} \div 2=\frac{1}{18}$ of the book;
From day $1 \sim 6$, she read $\frac{1}{18}+\frac{1}{9}=\frac{1}{6}$ of the book;
On the 7th day, she read $\frac{1}{6} \times 2=\frac{1}{... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
51. There is a sequence of numbers $1,2,3,5,8,13$, $\qquad$ starting from the 3rd number, each number is the sum of the two preceding ones. The remainder when the 2021st number in this sequence is divided by 3 is $\qquad$. | answer: 2 | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Let $E$ be a moving point inside square $ABCD$. It is known that the minimum value of the sum of the distances from $E$ to points $A$, $B$, and $C$ is $\sqrt{2}+\sqrt{6}$. Try to find the side length of this square. | 5. For a point $P$ inside a square, consider $P A+P B+P C$. Rotate $\triangle A P C$ by $60^{\circ}$ to $\triangle A^{\prime} P^{\prime} C$, then $\triangle A A^{\prime} C$ and $\triangle P P^{\prime} C$ are both equilateral triangles. Thus, $P C=P P^{\prime}$, and $A^{\prime}$ is a fixed point. By the invariance of ro... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. For positive integer $a$ and integers $b$, $c$, the quadratic equation $a x^{2}+b x+c=0$ has two roots $\alpha$, $\beta$. And it satisfies $0<\alpha<\beta<$ 1. Find the minimum value of $a$. | 8. Let $f(x)=a x^{2}+b x+c$, then $f(0)=c>0, f(1)=a+b+c>0, \Delta=b^{2}-4 a c>0.0c>0, 0-(a+c)+1 .(\sqrt{a}-\sqrt{c})^{2}>1, \sqrt{a}>\sqrt{c}+1 \geqslant 2$, so $a \geqslant 5$. And when the equality holds, $c=1, b=-5$. That is, $a_{\text {min }}=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. A store has 126 boxes of apples, with each box containing at least 120 and at most 144 apples. Now, boxes with the same number of apples are grouped together. Let the largest group of boxes have $n$ boxes, then the minimum value of $n$ is $\qquad$ . | 5.6.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
For what values of the parameter $a$ does the equation $\frac{\log _{a} x}{\log _{a} 2}+\frac{\log _{x}(2 a-x)}{\log _{x} 2}=\frac{1}{\log _{\left(a^{2}-1\right)} 2}$ have:
(1) solutions?
(2) exactly one solution? | The original equation is equivalent to
$$
\left\{\begin{array} { l }
{ a ^ { 2 } - 1 > 0 } \\
{ a ^ { 2 } - 1 \neq 1 } \\
{ x \neq 1 } \\
{ x > 0 } \\
{ \operatorname { log } _ { 2 } x + \operatorname { log } _ { 2 } ( 2 a - x ) = \operatorname { log } _ { 2 } ( a ^ { 2 } - 1 ) }
\end{array} \Leftrightarrow \left\{\be... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\pm$ Find all positive real numbers $a$ such that there exists a positive integer $n$ and $n$ pairwise disjoint infinite sets $A_{1}, A_{2}, \cdots, A_{n}$ satisfying $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=\mathbf{Z}$, and for any two numbers $b>c$ in each $A_{i}$, we have $b-c \geqslant a^{i}$. (Yuan Hanhui) | If $0a^{n}$, let
$A_{i}=\left\{2^{i-1} m \mid m\right.$ is odd $\}, i=1,2, \cdots, n-1$, $A_{n}=\left\{2^{n-1}\right.$ multiples $\}$,
then this partition satisfies the requirements.
If $a \geqslant 2$, suppose $A_{1}, A_{2}, \cdots, A_{n}$ satisfy the requirements, let $M=\{1,2, \cdots$,
$\left.2^{n}\right\}$, we wil... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
61. Xiao Gang is Xiao Ming's cousin. It is known that the cube of Xiao Gang's age this year is a four-digit number, and the fourth power of his age is a six-digit number; if these two numbers are combined, they exactly use all ten digits $0,1,2,3,4,5,6$, $7,8,9$ without repetition. Moreover, the sum of Xiao Ming's and ... | answer: 12 | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
32. Find the remainder when the 2017-digit number $\underbrace{7777}_{2017 \text { 7s }}$ is divided by 30. | Reference answer: 7 | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Let the integer $N>1, 1=d_{1}<d_{2}<\cdots<d_{s}=N$ be all the positive divisors of $N$. It is known that
$$
\left(d_{1}, d_{2}\right)+\left(d_{2}, d_{3}\right)+\cdots+\left(d_{s-1}, d_{s}\right)=N-2 \text {. }
$$
Find all possible values of $N$. | 3. $N=3$.
Let $r_{i}=d_{i+1}-d_{i}-\left(d_{i+1}, d_{i}\right)(1 \leqslant i \leqslant s-1)$.
Then $r_{i} \geqslant 0,\left(d_{i+1}, d_{i}\right) \mid r_{i}$.
By the condition,
$$
\sum_{i=1}^{s-1} r_{i}=\sum_{i=1}^{s-1}\left(d_{i+1}-d_{i}\right)-\sum_{i=1}^{s-1}\left(d_{i+1}, d_{i}\right)=1 \text {. }
$$
Therefore, t... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. In a lottery with 100000000 tickets, each ticket number consists of eight digits. A ticket number is called "lucky" if and only if the sum of its first four digits equals the sum of its last four digits. Then, the sum of all lucky ticket numbers, when divided by 101, leaves a remainder of | 7.0.
If the eight-digit number $x=\overline{a b c d e f g h}$ is lucky, then $y=99999999-x$ is also lucky, and $x \neq y$.
$$
\begin{array}{l}
\text { and } x+y=99999999=9999 \times 10001 \\
=99 \times 101 \times 10001,
\end{array}
$$
Therefore, the sum of all lucky numbers must be a multiple of 101. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{n+1}=a_{n}+a_{n}^{2}\left(n \in \mathbf{N}^{*}\right)$. Let
$$
S_{n}=\frac{1}{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)}, \quad T_{n}=\sum_{k=1}^{n} \frac{1}{1+a_{k}}
$$
Find the value of $S_{n}+T_{n}$. | Since $a_{n+1}=a_{n}+a_{n}^{2}$, we have $a_{n+1}=a_{n}\left(1+a_{n}\right)$, so $\frac{1}{1+a_{n}}=\frac{a_{n}}{a_{n+1}}$, thus
$$
S_{n}=\frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdot \frac{a_{3}}{a_{4}} \cdots \frac{a_{n}}{a_{n+1}}=\frac{a_{1}}{a_{n+1}}=\frac{1}{a_{n+1}}
$$
Since $a_{n+1}=a_{n}\left(1+a_{n}\rig... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. If $w$ is an imaginary root of the equation $x^{3}=1$, then the product $\left(1-w+w^{2}\right)\left(1+w-w^{2}\right)$ equals ( )
A. 4
B. $w$
C. 2
D. $w^{2}$ | 1. A $\left(1+\omega-\omega^{2}\right)\left(1-\omega+\omega^{2}\right)=(2+2 \omega)(-2 \omega)=4$ | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. As shown in the figure, square $O P Q R$ is inscribed in $\triangle A B C$. It is known that the areas of $\triangle A O R$, $\triangle B O P$, and $\triangle C R Q$ are $S_{1}=1$, $S_{2}=3$, and $S_{3}=1$, respectively. Therefore, the side length of square $O P Q R$ is
(A) $\sqrt{2}$;
(B) $\sqrt{3}$;
(C) 2;
(D) 3. | 7. (C)
Solution: Let the side length of the square $O P Q R$ be $x$, i.e., $O P=P Q=Q R=O R=x$. Draw the altitude $A D$ of $\triangle A B C$, intersecting $O R$ at $F$. In $\triangle A O R$, $A F=\frac{2 S_{1}}{O R}=\frac{2}{x}$.
Similarly, we get $B P=\frac{6}{x}$,
$$
Q C=\frac{2}{x},
$$
Thus, $S_{\triangle A B C}=\... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
5. There is a number, when divided by 3 the remainder is 2, and when divided by 4 the remainder is 1. What is the remainder when this number is divided by 12? | 5.【Solution】Let the number when divided by 12 have a remainder of a. Then a when divided by 3 has a remainder of 2; when divided by 4, the remainder is 1. Among $0,1,2, \cdots, 11$,
the only a that meets these conditions is 5, so the number when divided by 12 has a remainder of 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Calculate $\sec \frac{2 \pi}{9}+\sec \frac{4 \pi}{9}+\sec \frac{6 \pi}{9}+\sec \frac{8 \pi}{9}$. | 4. If $z=\cos \theta+i \sin \theta, n \in \mathbf{N}^{*}$, then
$\cos n \theta=\operatorname{Re}\left(z^{n}\right)=\frac{1}{2}\left(z^{n}+\bar{z}^{n}\right)=\frac{1}{2 z^{n}}\left(z^{2 n}+1\right)$, i.e., $\sec n \theta=\frac{2 z^{n}}{z^{2 n}+1}$.
Let $z=\cos \frac{2 \pi}{9}+\mathrm{i} \sin \frac{2 \pi}{9}$, then $z^{9... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
111 Let $x=\{-1,0,1\}, y=\{-2,-1,0,1,2\}$ and for all elements $x, x+f(x)$ are even, then the number of mappings $f$ from $x$ to $y$ is
A. 7
B. 10
C. 12
D. 15 | 111 C. $x$ and $f(x)$ are both even, or both odd.
$f(0)$ has three values: $-2, 0, 2$. $f(1)$ and $f(-1)$ each have two values. $f$ has a total of $3 \times 2 \times 2$ $=12$. | 12 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
3. Given $\log _{a} b+3 \log _{b} a=\frac{13}{2}$. When $a>b>1$, the value of $\frac{a+b^{4}}{a^{2}+b^{2}}$ is ( ).
(A) 13
(B) 4
(C) 2
(D) 1 | 3. D.
Let $\log _{a} b=t$. Then $t+\frac{3}{t}=\frac{13}{2} \Rightarrow t=\frac{1}{2}$ or 6 $\Rightarrow b=\sqrt{a}$ or $a^{6}$ (discard). Hence $\frac{a+b^{4}}{a^{2}+b^{2}}=\frac{a+a^{2}}{a^{2}+a}=1$. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
$10 \cdot 25$ is the sum of an arithmetic sequence with an even number of terms. The sums of the odd-numbered terms and the even-numbered terms are 24 and 30, respectively. If the last term exceeds the first term by 10.5, then the number of terms in the arithmetic sequence is
(A) 20.
(B) 18.
(C) 12.
(D) 10.
(E) 8.
(24t... | [Solution] Let $a, d$ and $2 n$ represent the first term, common difference, and number of terms of the sequence, respectively. If $S_{1}$ and $S_{2}$ represent the sums of all odd-numbered terms and all even-numbered terms, respectively, then
$$
\begin{array}{l}
S_{1}=\frac{n}{2}[2 a+(n-1) \cdot 2 d]=24, \\
S_{2}=\fra... | 8 | Algebra | MCQ | Yes | Yes | olympiads | false |
14. (15 points) Distribute 530 books to 48 students. What is the minimum number of students who will receive the same number of books? | 【Analysis】(1) If 48 students receive different numbers of books, then at least: $0+1+2+3+\cdots+47=1128>530$, which does not meet the condition;
(2) If only 2 students receive the same number of books, then at least: $(0+1+2+3+\cdots+23) \times 2=552>530$, which does not meet the condition;
(3) If only 3 students recei... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. Xiao Ming puts several chess pieces into the small squares of a $3 * 3$ grid. Each small square can be left empty or can contain one or more chess pieces. Now, by counting the total number of chess pieces in each row and each column, 6 numbers are obtained, and these 6 numbers are all different. What is the minimum... | 【Answer】 8
Analysis: Trying the smallest sum $0+1+2+3+4+5=15$, since the sum of the three rows = the sum of the three columns = the total sum, 15 is not an even number, so $16 \div 2=8,8=0+2+6=1+3+4$, after trials, it can be placed as shown in the figure, then the minimum number of chess pieces needed is 8 | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
17-108 If the three sides of $\triangle ABC$ are $BC=a, CA=b, AB=c$, and a point $P$ is taken arbitrarily inside $\triangle ABC$, and lines parallel to the three sides are drawn through $P$ to intersect the sides (as shown in the figure), if $DE=a', FG=b', HI=c'$, then the value of $\frac{a'}{a}+\frac{b'}{b}+\frac{c'}{... | [Solution 1] From $I F // B C, H E // A C$, we have $\triangle A B C \sim \triangle H I P$, thus
$$
\frac{c^{\prime}}{c}=\frac{I P}{a}=\frac{B D}{a} .
$$
Also, $\triangle A B C \backsim \triangle G P F$, so $\frac{b^{\prime}}{b}=\frac{P F}{a}=\frac{E C}{a}$.
Therefore, the answer is $(C)$.
[Solution 2] Let the altitu... | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
3. (10 points) A project needs 100 days to complete. Initially, 10 people worked for 30 days and completed $\frac{1}{5}$ of the entire project. Then, 10 more people were added to complete the project. How many days earlier can the task be completed?
Fill in the blank: $\qquad$ days. | 【Analysis】First, consider this project as a unit "1". According to the work efficiency $=$ work volume $\div$ work time, use the work volume completed by 10 people in 30 days divided by $10 \times 30$ to find out what fraction of the project each worker completes per day; then find out what fraction of the project 10 m... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. The maximum value of the function $f(x)=\sqrt{2 x-7}+\sqrt{12-x}+\sqrt{44-x}$ is | $f^{\prime}(x)=\frac{1}{\sqrt{2 x-7}}-\frac{1}{2 \sqrt{12-x}}-\frac{1}{2 \sqrt{44-x}}$ is monotonically decreasing in the interval $\left(\frac{7}{2}, 12\right)$, and $f^{\prime}(8)=\frac{1}{3}-\frac{1}{4}-\frac{1}{12}=0$, so $f(x)$ is monotonically increasing on $\left(\frac{7}{2}, 8\right)$ and monotonically decreasi... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 2 (2004 Jiangsu Summer Camp) Given the function $f(n)=k$, where $k$ is the $n$-th digit after the decimal point of the repeating decimal $0 . \dot{9} 1827364 \dot{5}$, then the value of $\underbrace{f\{f \cdots f[f(1)]\}}_{200 \uparrow f}$ is
A. 9
B. 8
C. 7
D. 3 | Solution: It is easy to get $f_{9 s+t}(1)=f_{t}(1), s, t \in \mathbf{N}^{*}$, $f_{2004}(1)=f_{6}(1)=8$, choose $(\mathrm{B})$. | 8 | Number Theory | MCQ | Yes | Yes | olympiads | false |
1. Define a new operation " $\otimes$ ": $a \otimes b=\frac{(a+1)(b-1)}{a \times b}-1$, then the calculation result of $(43 \otimes 47) \times 43 \times 47$ is $\qquad$ . | 【Match the answer】3
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
37. Find the smallest $a \in \mathbf{N}^{*}$, such that the following equation has real roots:
$$
\cos ^{2} \pi(a-x)-2 \cos \pi(a-x)+\cos \frac{3 \pi x}{2 a} \cdot \cos \left(\frac{\pi x}{2 a}+\frac{\pi}{3}\right)+2=0 .
$$ | $$
\begin{array}{l}
\text { 37. Original equation } \Leftrightarrow[\cos \pi(a-x)-1]^{2}+\left[\cos \frac{3 \pi}{2 a} x \cdot \cos \left(\frac{\pi x}{2 a}+\frac{\pi}{3}\right)+1\right]=0 \\
\Leftrightarrow[\cos \pi(a-x)-1]^{2}+\frac{1}{2}\left[\cos \left(\frac{2 \pi}{a} x+\frac{\pi}{3}\right)+\cos \left(\frac{\pi x}{a}... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\frac{1}{2} \sin 2 y+a=0$. Then the value of $\cos (x+2 y)$ is $\qquad$ . | 15.1.
Solving the system of equations and eliminating $a$ yields
$$
x=-2 y \Rightarrow \cos (x+2 y)=1 .
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The teacher fills some numbers into the circles in the figure below (each circle can and can only contain one number). The sum of the three numbers in each of the two closed loops on the left and right is 30, and the sum of the four numbers in each of the two closed loops on the top and bottom is 40. If the number i... | 【Answer】 11
【Solution】As shown in the figure, $\left\{\begin{array}{l}a+b+c+d=40 \\ X+Y+c+b=40\end{array} \Rightarrow(a+b+c+d+X+Y)+c+b=80\right.$, $\left\{\begin{array}{l}a+b+X=30 \\ c+d+Y=30\end{array} \Rightarrow a+b+c+d+X+Y=60\right.$, we deduce that $c+b=20$. Substituting $c+b=20$ into $X+Y+c+b=40 \Rightarrow X+Y=2... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Given that $n$ is a positive integer, the expansion of the binomial $\left(x^{2}+\frac{1}{x^{3}}\right)^{n}$ contains a term with $x^{7}$. Then the minimum value of $n$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | $-1 . \mathrm{C}$.
Notice that, the general term formula of the required binomial expansion is
$$
T_{r+1}=\mathrm{C}_{n}^{r}\left(x^{2}\right)^{n-r}\left(\frac{1}{x^{3}}\right)^{r}=\mathrm{C}_{n}^{r} x^{2 n-5 r} .
$$
From $2 n-5 r=7(r \in \mathbf{N})$, we know that when $r=1$, $n$ has the minimum value 6. | 6 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 5 In $\triangle A B C$, $D 、 E 、 F$ are the midpoints of $B C$ 、 $C A 、 A B$ respectively, and $G$ is the centroid. How many non-similar triangles $A B C$ are there for each value of $\angle B A C$ such that $A E G F$ is a cyclic quadrilateral? | Solution: From the fact that $A, E, G, F$ are concyclic, we get $\angle CGE = \angle BAC$,
Also, since $DE$ is the midline of $\triangle ABC$, we have $DE \parallel AB$.
Therefore, $\angle CED = \angle BAC$, hence $\angle CGE = \angle CED$.
Let $DE$ intersect $CF$ at $M$, then $\triangle CEM \sim \triangle CGE$, thu... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 7.16 Use red and blue colors to paint the 6 faces of a cube, with each face painted one color. Find the number of distinct patterns of the painted cube where the number of red faces is 3. | Let $A$ denote the set of the 6 faces of a cube, and let $a$ and $b$ represent red and blue, respectively. Let $B=\{a, b\}$. Assign the weight $w(a)=x$ and $w(b)=1$. Then, a face coloring of the cube is a mapping from $A$ to $B$, and the weight of any face-colored cube can be expressed as $x^{k}(0 \leqslant k \leqslant... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
13. Let $M$ be a finite set of numbers. If it is known that among any three elements of $M$, there are always two numbers whose sum belongs to $M$, how many elements can $M$ have at most? | 13. Let $M=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}, a_{1}>a_{2}>\cdots>a_{m}$. If $m \geqslant 8$, since multiplying each number by -1 does not change the property of $M$, we can assume without loss of generality that $a_{4}>0$. Then for the set of three numbers $a_{1}, a_{2}, a_{3}$, it is clear that only $a_{2}+a_... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
18. (12 points) Given vectors $\boldsymbol{m}=(\sin A, \cos A)$, $\boldsymbol{n}=(\cos B, \sin B)$, $\boldsymbol{m} \cdot \boldsymbol{n}=\sin 2 C$, and $\angle A$, $\angle B$, $\angle C$ are the angles opposite to sides $a$, $b$, $c$ of $\triangle ABC$ respectively.
(1) Find the size of $\angle C$;
(2) If $\sin A$, $\s... | 18. (1) From the given information,
$$
\begin{array}{l}
\sin A \cdot \cos B + \cos A \cdot \sin B = \sin 2 C \\
\Rightarrow \sin C = \sin 2 C \Rightarrow \cos C = \frac{1}{2}.
\end{array}
$$
Since $\angle C$ is an interior angle of $\triangle ABC$, we have $\angle C = \frac{\pi}{3}$.
(2) From the given information,
$$... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A positive integer $N$ is appended to the right of any positive integer (for example, appending 21 to the right of 35 yields 3521). If the new number is always divisible by $N$, then $N$ is called a “magic number”. How many “magic numbers” are there among the positive integers less than 600? | Number Theory, Divisors and Multiples.
(1) If $N$ is a one-digit number;
Then by the problem, we get $N \mid 10 m + N$, so $N \mid 10 m$;
Since $m$ is any positive integer; thus $N \mid 10$;
$10 = 2 \times 5$ has $(1+1) \times (1+1) = 4$ divisors: $1, 2, 5, 10$;
Therefore, $N = 1, 2, 5$.
(2) If $N$ is a two-digit numb... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. Let $x \in \mathbf{R}$, the inequality $2 x^{2}-a \sqrt{x^{2}+1}+3 \geqslant 0$ always holds, then the maximum value of the real number $a$ is $\qquad$ | Thus, $2 \sqrt{x^{2}+1}+\frac{1}{\sqrt{x^{2}+1}} \geqslant 3$,
so the maximum value of the real number $a$ is 3. | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
7. Let the set $A=\left\{n \left\lvert\, \frac{n}{3} \in \mathbf{N}_{+}\right.\right\}, B=\left\{y \mid y=x+4+\sqrt{5-x^{2}}\right\}$, then the number of elements in the set $A \cap B$ is
A. 1
B. 2
C. 3
D. 4 | Let $x=\sqrt{5} \sin \theta, \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then $y=\sqrt{5} \sin \theta+\sqrt{5} \cos \theta+4=\sqrt{10} \sin \left(\theta+\frac{\pi}{4}\right)+4$, since $\theta+\frac{\pi}{4} \in\left[-\frac{\pi}{4}, \frac{3 \pi}{4}\right] \Rightarrow \sin \left(\theta+\frac{\pi}{4}\right) \in\... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. A toy factory produces cubic building blocks of the same size, with each face painted one of three colors: red, yellow, or blue, and each color is used on exactly 2 faces. When two blocks can be rotated to have the same color faces in the same positions, they are considered the same type of block. Try to explain: wh... | 5. There are 6 different types of building blocks.
5.【Solution】The bottom face can always be red.
If the top face is also red, by flipping, the front can be yellow, and the left side is not yellow. At this time, the back can be yellow or blue, which gives 2 possibilities.
If the top face is not red, by rotating, the b... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(15) Find the smallest positive real number $k$, such that the inequality
$$
a b+b c+c a+k\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9,
$$
holds for all positive real numbers $a, b, c$. | (15) When $a=b=c=1$, we can get $k \geqslant 2$. Now we prove the inequality
$$
a b+b c+c a+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$.
By the AM-GM inequality, we have
$$
a b+\frac{1}{a}+\frac{1}{b} \geqslant 3 \sqrt[3]{a b \cdot \frac{1}{a} \cdot \f... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
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