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Example 8 Let positive rational numbers $a, b$ satisfy
$$
a+b=a^{2}+b^{2}=s \text {. }
$$
If $s$ is not an integer, then denote
$$
s=\frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right) .
$$
Let $p$ be the smallest prime factor of $n$. Find the smallest possible value of $p$. ${ }^{[2]}$
(33rd Argentine Mathemati... | Let $a=\frac{x}{z}, b=\frac{y}{z}$, where $z$ is the least common multiple of the denominators of the reduced fractions $a$ and $b$.
This implies that if $a=\frac{x^{\prime}}{z^{\prime}}, b=\frac{y^{\prime}}{z^{\prime}}$, then $z^{\prime} \geqslant z$.
We now prove: $(x, y, z)=1$.
Assume there exists a prime $q$ such t... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Let $a_{0}=0, a_{1}=a_{2}=1, a_{3 n}=a_{n}, a_{3 n+1}=a_{3 n+2}=a_{n}+1(n \geqslant 1)$, then $a_{2021}=$ | $$
\begin{array}{l}
a_{2021}=a_{3 \cdot 673+2}=a_{673}+1=a_{3 \cdot 224+1}+1=a_{224}+2=a_{3 \cdot 74+2}+2=a_{74}+3 \\
=a_{3 \cdot 24+2}+3=a_{24}+4=a_{8}+4=a_{3 \cdot 2+2}+4=a_{2}+5=6 .
\end{array}
$$ | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. (10 points) On a plane, there are 5 different lines, and these 5 lines form $n$ intersection points. How many different values can $n$ have?
保留源文本的换行和格式，翻译结果如下：
9. (10 points) On a plane, there are 5 different lines, and these 5 lines form $n$ intersection points. How many different values can $n$ have? | 【Analysis】According to the problem, we can classify and discuss, and finally determine the value of $n$.
【Solution】Solution: According to the analysis, $n=0$, which means 5 lines are parallel to each other; $n=1$, which means five lines intersect at one point;
$n=2,3$ do not exist;
For $n=4,5,6,7,8,9,10$, the situation... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.32 Find all natural numbers $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square of a natural number.
(6th All-Russian Mathematical Olympiad, 1980) | ［Solution 1］(1) When $n \leqslant 8$,
$$
\begin{aligned}
N & =2^{8}+2^{11}+2^{n} \\
& =2^{n}\left(2^{8-n}+2^{11-n}+1\right) .
\end{aligned}
$$
For $N$ to be a perfect square, $n$ must be even, i.e., $n=2,4,6,8$.
Upon verification, $N$ is not a perfect square for any of these values.
(2) When $n=9$,
$$
N=2^{8}+2^{11}+2... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. In the complex plane $\left\{\begin{array}{l}0 \leqslant \arg (z-1) \leqslant \frac{\pi}{4},(R e(z) \text { represents the real part of } z) \text { the points that form the corresponding figure have an area of } \\ \operatorname{Re}(z) \leqslant 2\end{array}\right.$ . $\qquad$ | 7. $1 \quad S=\frac{1}{2} \times 2 \times 1=1$ | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. In $\triangle A B C$, $D$ is a point on side $B C$. It is known that $A B=13, A D=12, A C=15, B D=5$, find $D C$. | 3. By Stewart's Theorem, we have $A D^{2}=A B^{2} \cdot \frac{C D}{B C}+A C^{2} \cdot \frac{B D}{B C}-B D \cdot D C$. Let $D C=x$, then $B C=5+x$, then $12^{2}=$ $13^{2} \cdot \frac{x}{5+x}+15^{2} \cdot \frac{5}{5+x}-5 x$, solving for $x$ yields $x_{1}=9\left(\right.$ discard $\left.x_{2}=-9\right)$. | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Stephanie has 1000 eggs to pack into cartons of 12 eggs. While packing the eggs, she breaks $n$ eggs. The unbroken eggs completely fill a collection of cartons with no eggs left over. If $n<12$, what is the value of $n$ ? | 1. Solution 1
Since $\frac{1000}{12} \approx 83.33$, then the largest multiple of 12 less than 1000 is $83 \times 12=996$.
Therefore, if Stephanie filled 83 cartons, she would have broken 4 eggs.
If Stephanie filled fewer than 83 cartons, she would have broken more than 12 eggs. If Stephanie filled more than 83 cartons... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
32. Given real numbers $a, b, x, y$ satisfy $a+b=x+y=2, a x+b y=5$, then $\left(a^{2}+b^{2}\right) x y+a b\left(x^{2}+y^{2}\right)=$ $\qquad$ | answer: -5 | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5 Find all positive integers that are coprime with all terms of the sequence $\left\{a_{n}=2^{n}+3^{n}+6^{n}-1, n \geqslant\right.$ $1\}$. | Solution: Clearly, $\left(1, a_{n}\right)=1$. Let $m>1$ be a positive integer that is coprime with all terms in $\left\{a_{n}\right\}$, and let $p$ be a prime factor of $m$. If $p>3$, then by Fermat's Little Theorem, we have:
$$
2^{p-1} \equiv 1(\bmod p), 3^{p-1} \equiv 1(\bmod p), 6^{p-1} \equiv
$$
$1(\bmod p)$.
$$
\t... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4、In the Fibonacci sequence, the first term and the second term are both 1, starting from the 3rd term, each term is equal to the sum of the two preceding ones. Numbers that appear in the Fibonacci sequence are called Fibonacci numbers. Express 121 as the sum of several different Fibonacci numbers, there are $\qquad$ d... | 【Answer】 8
【Analysis】The Fibonacci numbers less than 121 are $1,1,2,3,5,8,13,21,34,55,89$. Starting from the largest 89; (1) The largest is 89, then 121-89=32, the sum of the remaining numbers is 32, $32=21+8+3=21+8+2+1=21+5+3+2+1$, there are 4 ways; (2) The largest is 54, then $89=34+55$, there are 4 more ways; In tot... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 14 (1998 Shanghai Competition Problem) Given that there are three vertices $A, B, C$ of a square on the parabola $y=x^{2}$, find the minimum value of the area of such a square.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result dir... | Let's assume $A$ and $B$ are on the right side of the $y$-axis (including the $y$-axis), and the coordinates of $A, B, C$ are $\left(x_{1}, x_{1}^{2}\right)$, $\left(x_{2}, x_{2}^{2}\right), \left(x_{3}, x_{3}^{2}\right) \left(x_{3} \geq 0\right)$, then
$$
\left\{\begin{array}{l}
x_{1}^{2}-x_{2}^{2}=k\left(x_{1}-x_{2}\... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) Today is January 30, we first write down 130; the rule for writing the next number is: if the number just written is even, divide it by 2 and add 2 to write it down, if the number just written is odd, multiply it by 2 and subtract 2 to write it down, thus we get: $130, 67, 132, 68 \cdots$, so the 2016th ... | 【Answer】Solution: According to the problem, we have:
The number pattern is $130, 67, 132, 68, 36, 20, 12, 8, 6, 5, 8, 6, 5, 8, 6, 5$,
Removing the first 7 terms, we get a cyclic sequence $2016-7=2009$. Each 3 numbers form a cycle $2009 \div 3=667 \cdots 2$
The second number in the cyclic sequence is 6.
Therefore, the a... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5・17 Let $a<b<c<d$. If the variables $x, y, z, t$ are some permutation of the numbers $a, b, c, d$, how many different values can the expression
$$
n=(x-y)^{2}+(y-z)^{2}+(z-t)^{2}+(t-x)^{2}
$$
take? | 【Sol】Let $n(x, y, z, t)=(x-y)^{2}+(y-z)^{2}+(z-t)^{2}+(t-$ $x)^{2}$. Since the expression
$$
\begin{aligned}
& n(x, y, z, t)+(x-z)^{2}+(y-z)^{2} \\
= & n(x, y, z, t)+\left(x^{2}+y^{2}+z^{2}+t^{2}\right)-2(x z+y t)
\end{aligned}
$$
is independent of the permutation of the numbers $a, b, c, d$ chosen as the values of th... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
22. (12 points) Given the function
$$
f(x)=\frac{m x-n}{x}-\ln x \quad (m, n \in \mathbf{R}) \text{.}
$$
(1) If the tangent line to the function $f(x)$ at the point $(2, f(2))$ is parallel to the line $x-y=0$, find the value of the real number $n$;
(2) Discuss the maximum value of the function $f(x)$ on the interval $[... | 22. (1) Notice,
$$
f^{\prime}(x)=\frac{n-x}{x^{2}}, f^{\prime}(2)=\frac{n-2}{4} \text {. }
$$
Since the tangent line to the function $f(x)$ at the point $(2, f(2))$ is parallel to the line $x-y=0$, we have,
$$
\frac{n-2}{4}=1 \Rightarrow n=6 \text {. }
$$
(2) It is easy to see that, $f^{\prime}(x)=\frac{n-x}{x^{2}}(x>... | 6 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
5. (2007 Anhui Province College Entrance Examination Question) The function $f(x)$ defined on $\mathbf{R}$ is both an odd function and a periodic function, with $T$ being one of its positive periods. If the number of roots of the equation $f(x)=0$ in the closed interval $[-T, T]$ is denoted as $n$, then $n$ could be
A.... | 5. D Since $f(x)$ is an odd function on $\mathbf{R}$, for $x \in \mathbf{R}, f(-x)=-f(x)$. Let $x=0$. We get $f(0)=-f(0)$, hence $f(0)=0$. Also, $f(x)$ has a period of $T$, so $f(-T)=f(T)=f(0)=0$.
$f(x)$ has at least one more root in $[0, T]$, otherwise $f(x)$ would be positive (or negative) throughout $[0, T]$. By the... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
15. Given $k$ as a positive integer, the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=3, a_{n+1}=\left(3^{\frac{2}{2 x-1}}-1\right) S_{n}+3\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$.
Let $b_{n}=\frac{1}{n} \log _{3} a_{1}... | 15. From the problem, we have
$$
\begin{array}{l}
a_{2}=\left(3^{\frac{2}{2 k-1}}-1\right) \times 3+3=3^{\frac{2 k+1}{2 k-1}} . \\
\text { Also, } a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3, \\
a_{n}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n-1}+3(n \geqslant 2),
\end{array}
$$
Thus, \(a_{n+1}-a_{n}=\left(3^{\fra... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the given figure, $A B C D$ is a square with sides of length 4 , and $Q$ is the midpoint of $C D$. $A B C D$ is reflected along the line $A Q$ to give the square $A B^{\prime} C^{\prime} D^{\prime}$. The two squares overlap in the quadrilateral $A D Q D^{\prime}$. Determine the area of quadrilateral $A D Q D^{\pr... | Solution: Since the side length of $A B C D$ is 4 , so $D Q=2$. The area of $A D Q$ is $4 \times 2 / 2=4$ and by symmety the area of $A D^{\prime} Q$ is the same. Thus, the area of $A D Q D^{\prime}$ is $4+4=8$. | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Find all positive integers $k$ such that the indeterminate equation $x^{2}+y^{2}=k x y-1$ has positive integer solutions for $x$ and $y$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove that among the positive integer solutions of the given indeterminate equation, the one that minimizes $x+y$ is $\left(x_{0}, y_{0}\right)$.
If $x_{0}=y_{0}$, then $x_{0}^{2} \backslash\left(2 x_{0}^{2}+1\right)$.
Therefore, $x_{0}^{2} \mid 1 \Rightarrow x_{0}=1$.
Thus, $k=\frac{x_{0}^{2}+y_{0}^{2}+1}{x_{0} y_{0}}... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
89. The blackboard is written with a number $\underbrace{222 \cdots 2}_{9 \uparrow 2}$, and the following operation is performed: erase the last digit, multiply the number on the blackboard by 4, and then add the last digit that was just erased. Continue performing such operations until a single-digit number appears on... | answer: 6 | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (11th "Hope Cup" Test Question) Given $x, y, z \in \mathbf{R}^{+}$, and $\frac{1}{x}+\frac{2}{y}+\frac{3}{z}=1$, then the minimum value of $x+\frac{y}{2}+\frac{z}{3}$ is
A. 5
B. 6
C. 8
D. 9 | 5. Solution 1 Since $x, y, z \in \mathbf{R}^{+}$, and $\frac{1}{x}+\frac{2}{y}+\frac{3}{z}=1$,
then $x+\frac{y}{2}+\frac{z}{3}=\left(x+\frac{y}{2}+\frac{z}{3}\right)\left(\frac{1}{x}+\frac{2}{y}+\frac{3}{z}\right)$
$$
=3+\left(\frac{y}{2 x}+\frac{2 x}{y}\right)+\left(\frac{z}{3 x}+\frac{3 x}{z}\right)+\left(\frac{2 z}... | 9 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. (6 points) 2 fen and 5 fen coins total 15, with a total value of 51 fen. The 2 fen coins are $\qquad$ more than the 5 fen coins. | 【Answer】Solution: Assume all are 5-cent coins,
$$
\begin{array}{l}
2 \text { cents: }(5 \times 15-51) \div(5-2), \\
=24 \div 3, \\
=8 \text { (coins); } \\
5 \text { cents: } 15-8=7 \text { (coins); } \\
8-7=1 \text { (coin); }
\end{array}
$$
Answer: The 2-cent coins are 1 more than the 5-cent coins. Therefore, the an... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. If $a^{2}+b^{2}=4$, then
$$
\begin{array}{l}
\sqrt[3]{a(b-4)}+\sqrt{a b-3 a+2 b-6} \\
=(\quad)
\end{array}
$$
(A) -2
(B) 0
(C) 2
(D) 4 | - 1. C.
From the condition $a^{2}+b^{2}=4$, we know $-2 \leqslant a, b \leqslant 2$. Therefore, $a+2 \geqslant 0$, and $b-3<0$.
Thus, $a b-3 a+2 b-6$
$$
=(a+2)(b-3) \leqslant 0 \text{. }
$$
Also, from the meaningfulness of the square root, we have
$$
a b-3 a+2 b-6 \geqslant 0 \text{. }
$$
Therefore, $a=-2, b=0$.
Hen... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
6. There are 99 bags, each containing $1,2,3,4, \cdots \cdots, 99$ gold coins respectively. Each time the fairy wields her magic wand over some of the bags, the number of gold coins in these bags increases by the same amount. To make the number of gold coins in each bag exactly 100, the fairy needs to wield her magic w... | $7$ | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In a convex quadrilateral $ABCD$, $\angle B=\angle D=90^{\circ}, AB=4, AD=5, \angle A=60^{\circ}$, $\frac{BC}{CD}=$ $\qquad$ . | 1. 2 .
Connect $A C$, let $A C=R$, then $\sin \angle D A C=\frac{D C}{A C}=\frac{\sqrt{R^{2}-25}}{R}, \cos \angle D A C=\frac{5}{R}, \sin \angle B A C=$
$$
\begin{array}{l}
\frac{B C}{A C}=\frac{\sqrt{R^{2}-16}}{R}, \cos \angle B A C=\frac{A B}{A C}=\frac{4}{R}, \frac{1}{2}=\cos 60^{\circ}=\cos (\angle D A C+\angle B ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
52. As shown in the figure, the area of $\triangle ABC$ is 24, point $D$ is on line segment $AC$, point $F$ is on the extension of line segment $BC$, and $BC=4CF$. Quadrilateral $DCFE$ is a parallelogram. The area of the shaded part in the figure is . $\qquad$ | answer: 6 | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
96. A car travels from location A to location B. If the car increases its speed by $20 \%$, it can arrive 1 hour earlier than the original time; if it travels 200 kilometers at the original speed and then increases its speed by $25 \%$, it can arrive 40 minutes earlier than the original time. If the car travels at 45 k... | Reference answer: 10 | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Roll a die 6 times, let the number obtained on the $i$-th roll be $a_{i}$, if there exists a positive integer $k$ such that $\sum_{i=1}^{k} a_{i}=6$ the probability $p=\frac{n}{m}$, where $m, n$ are coprime positive integers, then $\log _{6} m-\log _{7} n=$ $\qquad$ | 7. 1 .
Solution: When $k=1$, the probability is $\frac{1}{6}$;
When $k=2$, $6=1+5=2+4=3+3$, the probability is $5 \cdot$ $\left(\frac{1}{6}\right)^{2}$;
When $k=3$, $6=1+1+4=1+2+3=2+2+2$, the probability is $(3+6+1) \cdot\left(\frac{1}{6}\right)^{3}=10 \cdot\left(\frac{1}{6}\right)^{3}$;
When $k=4$, $6=1+1+1+3=1+1+2+... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (20 points) Xiaohong and Xiaoming both brought money to buy the book "Fun Mathematics". When they arrived at the bookstore, Xiaohong found that she was short of 2 yuan and 2 jiao, and Xiaoming was short of 1 yuan and 8 jiao. But the money they brought together was just enough to buy one book. Then the price of "Fun ... | 【Analysis】According to "The two of them put their money together, and it is just enough to buy a book." This indicates: The money each person lacks can be made up by the other, so the price of the book is the sum of the two people's money. The amount one person lacks is the amount the other person has, based on this to... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3 Consider a square on the complex plane, whose four vertices correspond to the four roots of a monic quartic equation with integer coefficients $x^{4}+p x^{3}+q x^{2}+r x+s=0$. Find the minimum value of the area of such a square. (Xiong Bin) | Let the complex number corresponding to the center of the square be $a$. Then, after translating the origin of the complex plane to $a$, the vertices of the square are uniformly distributed on a circle, i.e., they are the solutions to the equation $(x-a)^{4}=b$, where $b$ is some complex number. Thus,
$$
\begin{aligned... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
17. As shown in the figure, it is known that the three vertices of $\triangle A B C$ are on the ellipse $\frac{x^{2}}{12}+\frac{y^{2}}{4}=1$, and the coordinate origin $O$ is the centroid of $\triangle A B C$. Try to find the area of $\triangle A B C$. | Given $A(2 \sqrt{3} \cos \theta, 2 \sin \theta)$, connect $A O$ to intersect $B C$ at point $D$,
then $D(-\sqrt{3} \cos \theta, -\sin \theta)$, and by the conclusion of the point difference method,
$$
k_{B C} \cdot \frac{y_{D}}{x_{D}}=-\frac{1}{3} \Rightarrow k_{B C}=-\frac{\sqrt{3}}{3 \tan \theta} \text {. }
$$
Thus,... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. For every pair of real numbers $x, y$, the function $f$ satisfies the functional equation $f(x)+f(y)=f(x+y)-x y-1$. If $f(1)=1$. Then the number of integers $n$ that satisfy $f(n)=n$ (where $n \neq 1$) is
A. 0
B. 1
C. 2
D. 3 | 3. $\mathrm{B}$ Hint: Let $x=1, f(y+1)=f(y)+y+2(*)$. For $y$ being a positive integer, $f(y+1)>y+2>y+1$; also from * we have $f(y)=f(y+1)-(y+2)$, it is easy to see that for $y>0$. The original functional equation $f(n)=n$ has only the solutions $n=1,-2$. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. (10 points) On the table, there are 20 cards numbered $1 \sim 20$. Each time, Xiao Ming takes out 2 cards, with the requirement that one card's number is 2 more than twice the number of the other card. What is the maximum number of cards Xiao Ming can take out?
A. 12
B. 14
C. 16
D. 18 | 【Analysis】Since 2 cards are drawn each time, and it is required that one card's number is 2 more than twice the other, the larger number of the drawn cards cannot exceed 20. Let the other card number be $x$, then $2 x+2 \leqslant 20$, and $x \geqslant 1$. Solve accordingly.
【Solution】Let the other card number be $x$, ... | 12 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
15. (6 points) 4 workers can produce 108 parts in 3 hours. Now, to produce 504 parts in 8 hours, the number of additional workers needed is $\qquad$.
| 【Solution】Solution: $504 \div 8 \div(108 \div 3 \div 4)-4$,
$$
\begin{array}{l}
=504 \div 8 \div 9-4, \\
=63 \div 9-4, \\
=7-4, \\
=3 \text { (people), }
\end{array}
$$
Answer: Need to add 3 people, so the answer should be: 3 . | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. A line $l$ is drawn through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ intersecting the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$ such that there are exactly 3 such lines $l$, then $\lambda=$ $\qquad$. | By the symmetry of the hyperbola, when $|A B|=\lambda$ of the line $l$ is exactly 3, $\lambda=\max \left\{2 a, \frac{2 b^{2}}{a}\right\}$ $=\max \{2,4\}=4$, where $\frac{2 b^{2}}{a}$ is the length of the latus rectum of the hyperbola. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. In the convex quadrilateral $ABCD$,
$$
\angle BAD=\angle BCD=120^{\circ}, BC=CD=10.
$$
Then $AC=$ . $\qquad$ | 2. 10 .
From the problem and the sum of the interior angles of a convex quadrilateral being $360^{\circ}$, we know
$$
\begin{array}{l}
\angle A B C+\angle A D C=120^{\circ} . \\
\text { If } A C>B C=C D=10 \text {, then } \\
\angle A B C+\angle A D C \\
>\angle B A C+\angle D A C=\angle B A D=120^{\circ} \text {, }
\e... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
B. If real numbers $x, y$ satisfy
$$
x^{3}+y^{3}=26 \text {, and } x y(x+y)=-6 \text {, }
$$
then $|x-y|=$ $\qquad$ | B. 4 .
From the given information, we have
$$
\begin{array}{l}
(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)=8 \\
\Rightarrow x+y=2 \Rightarrow x y=-3 . \\
\text { Therefore, }|x-y|=\sqrt{(x+y)^{2}-4 x y}=4 .
\end{array}
$$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Expression: $1^{2015}+2^{2015}+3^{2015}+\cdots+2013^{2015}+2014^{2015}$, the unit digit of the calculation result is | 【Analysis and Solution】
Number Theory, Congruence.
The last digit of power operations repeats every four:
1's last digit is always 1;
2's last digits are $2, 4, 8, 6$;
3's last digits are $3, 9, 7, 1$;
4's last digits are $4, 6$;
5's last digit is always 5;
6's last digit is always 6;
7's last digits are $7, 9, 3, 1$;
... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 7.17 Use red, blue, and yellow to paint the 6 faces of a cube, so that 2 faces are painted red, 2 faces are painted blue, and 2 faces are painted yellow. Find the number of distinct colored cubes that can be formed. | Solution: Let the number of differently styled colored cubes be $N$. Let $A$ denote the set of the 6 faces of the cube, and let $b_{1}, b_{1}^{\prime}, b_{2}, b_{2}^{\prime}, b_{3}, b_{3}^{\prime}$ denote 6 different colors, and let $B=\left\{b_{1}, b_{1}^{\prime}, b_{2}, b_{2}^{\prime}, b_{3}, b_{3}^{\prime}\right\}$.... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. In the Cartesian coordinate system, if the circle with center $(r+1,0)$ and radius $r$ has a point $(a, b)$ satisfying $b^{2} \geqslant 4 a$, then the minimum value of $r$ is $\qquad$
In the Cartesian coordinate system, if the circle with center $(r+1,0)$ and radius $r$ has a point $(a, b)$ satisfying $b^{2} \geqsl... | $$
\begin{array}{l}
\text { From }(a-r-1)^{2}+b^{2}=r^{2} \Rightarrow a^{2}-2(r+1) a+2 r+1+b^{2}=0 \\
\Rightarrow a^{2}-2(r+1) a+2 r+1+4 a \leqslant 0 \Rightarrow a^{2}-2(r-1) a+2 r+1 \leqslant 0,
\end{array}
$$
Notice that the above inequality has a solution, thus $\Delta=4(r-1)^{2}-4(2 r+1) \geqslant 0$ $\Rightarrow... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16. The following calculation that equals 9 is ( )
A. $3 \times 3 \div 3+3$
B. $3 \div 3+3 \times 3$
C. $3 \times 3-3+3$
D. $3 \div 3+3 \div 3$ | 【Solution】Solution: $A 、 3 \times 3 \div 3+3$
$$
\begin{array}{l}
=3+3 \\
=6 ;
\end{array}
$$
$$
\begin{array}{l}
\text { B. } 3 \div 3+3 \times 3 \\
=1+9 \\
=10
\end{array}
$$
$$
\begin{array}{l}
\text { C. } 3 \times 3-3+3 \\
=9-3+3 \\
=9 ;
\end{array}
$$
$$
\begin{array}{l}
\text { D. } 3 \div 3+3 \div 3 \\
=1+1 \\
... | 9 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 4 Given that $\alpha^{2005}+\beta^{2005}$ can be expressed as a bivariate polynomial in terms of $\alpha+\beta, \alpha \beta$. Find the sum of the coefficients of this polynomial.
(2005 China Western Olympiad) | Since we are asked to find the sum of the coefficients of this polynomial, we should set the variables to 1, i.e., let $\alpha+\beta=1$, $\alpha \beta=1$, so $\alpha, \beta$ are the two roots of the polynomial $f(x)=x^{2}-x+1$. Let $T_{n}=\alpha^{n}+\beta^{n}$, then by Newton's formula, we have
$T_{n}=T_{n-1}-T_{n-2}$ ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (3 points) Xiaohui started from point $A$ and walked 15 meters west to reach point $B$, then walked 23 meters east from point $B$ to reach point $C$. The distance from point $C$ to point $A$ is $\qquad$ meters. | 【Analysis】We solve this by drawing a diagram. Walking 15 meters west, and then walking 23 meters east, the distance from point $C$ to point $A$ is actually the difference between 23 meters and 15 meters.
【Solution】Solution: The diagram is as follows:
The distance from point $C$ to point $A$ is:
$$
23-15=8 \text { (met... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
60. On the blackboard, there is a sequence of numbers: $1,2,3, \cdots, 2018,2019$, and you can arbitrarily erase several numbers and write down the remainder of the sum of the erased numbers divided by 17 (write 0 if it is exactly divisible). Continue this operation until only one number remains on the blackboard, this... | Reference answer: 6
Exam point: Operations and strategies | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. Let $d$ be a positive divisor of 2015. Then the maximum value of the unit digit of $d^{\frac{2005}{d}}$ is $\qquad$ . | 9.7.
Notice that, $2015=5 \times 13 \times 31$.
Therefore, 2015 has eight positive divisors.
Let $G(n)$ denote the unit digit of $n$, then
$$
\begin{array}{l}
G\left(1^{2015}\right)=G\left(31^{165}\right)=1, \\
G\left(2015^{1}\right)=G\left(5^{403}\right)=G\left(65^{31}\right) \\
=G\left(155^{13}\right)=5 .
\end{array... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The number of zeros of the function $f(x)=x^{2} \ln x+x^{2}-2$ is $\qquad$ . | 3. 1 .
From the condition, we have
$$
f^{\prime}(x)=2 x \ln x+x+2 x=x(2 \ln x+3) \text {. }
$$
When $0 < x < \mathrm{e}^{-\frac{3}{2}}$, $f^{\prime}(x)<0$; when $x > \mathrm{e}^{-\frac{3}{2}}$, $f^{\prime}(x)>0$.
Thus, $f(x)$ is a decreasing function on the interval $\left(0, \mathrm{e}^{-\frac{3}{2}}\right)$ and an ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let $p(x)$ be a polynomial of degree $3n$, satisfying
$$
\begin{array}{c}
p(0)=p(3)=\cdots=p(3 n)=2, \\
p(1)=p(4)=\cdots=p(3 n-2)=1, \\
p(2)=p(5)=\cdots=p(3 n-1)=0,
\end{array}
$$
and $p(3 n+1)=730$, find $n$. | 5. We have $\Delta^{3 n+1} p(0)=\sum_{k=0}^{3 n+1}(-1)^{k} C_{3 n+1}^{k} p(3 n+1-k)$
$$
\begin{array}{l}
=730+\sum_{j=0}^{n}(-1)^{3 j+1} C_{3 n+1}^{3 j+1} \cdot 2+\sum_{j=1}^{n}(-1)^{3 j} C_{3 n+1}^{3 j} \\
=729-2 \sum_{j=0}^{n}(-1)^{j} C_{3 m-1}^{3,+1}+\sum_{j=0}^{n}(-1)^{j} C_{3 j-1}^{3 j}=0 .
\end{array}
$$
To find... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let $x$, $y$, and $z$ all be prime numbers, and $x \leqslant y \leqslant z$. Then the number of positive integer solutions to the indeterminate equation $x^{2}+y^{2}+z^{2}=2019$ is ( ) groups.
(A) 3
(B) 4
(C) 5
(D) 6 | 5. C.
Notice that, prime numbers $x \leqslant y \leqslant z$.
From $3 x^{2} \leqslant 2019$
$$
\Rightarrow 2 \leqslant x \leqslant 25 \text {. }
$$
Therefore, all possible values of $x$ and $y$ are:
$$
2, 3, 5, 7, 11, 13, 17, 19, 23 \text {. }
$$
From $3 z^{2} \geqslant 2019, z^{2}<2019$
$$
\Rightarrow 26 \leqslant ... | 5 | Number Theory | MCQ | Yes | Yes | olympiads | false |
8. (10 points) King Arthur summoned 6 knights to the palace. Each knight has exactly 2 friends. They sat around a round table (the names and seats of the knights are shown in the figure), and it turned out that in this arrangement, any two adjacent knights are friends. King Arthur wants to rearrange the seats, so how m... | 【Analysis】First, according to the problem requirements, rotations that are the same are considered the same method, so we only need to consider the case where one person is in the first position, and then sort according to the conditions given in the problem.
【Solution】Solution: For convenience, let's use the numbers ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$1 \cdot 6$ Let $n$ be an integer. If the tens digit of $n^{2}$ is 7, what is the units digit of $n^{2}$? | [Solution] Let $n=10 x+y$, where $x$ and $y$ are integers, and $0 \leqslant y \leqslant 9$. Thus, we have
$$
\begin{aligned}
n^{2} & =100 x^{2}+20 x y+y^{2} \\
& =20\left(5 x^{2}+x y\right)+y^{2}
\end{aligned}
$$
If the tens digit of $n^{2}$ is the odd number 7, then the tens digit of $y^{2}$ is odd, which leads to
$$... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$24 \cdot 6$ The area of the upper base of a frustum of a cone is 1, and the area of the lower base is 16. A plane parallel to the bases cuts this frustum, and the distance from this plane to the upper base is twice the distance from it to the lower base. The area of this section is
(A) 4.
(B) 9.
(C) $\frac{5}{2}$.
(D... | ［Solution］Let the areas of the upper base, lower base, and the section of the frustum be $S_{\text {upper }}, S_{\text {lower }}$, and $S_{\text {middle, then }}$ thus
$$
\begin{array}{c}
\sqrt{S_{\text {middle }}}=\frac{\sqrt{S_{\text {upper }}}+2 \sqrt{S_{\text {lower }}}}{3}=\frac{1+2 \cdot 4}{3}=3, \\
S_{\text {mid... | 9 | Geometry | MCQ | Yes | Yes | olympiads | false |
$1 \cdot 71 \sqrt{3+2 \sqrt{2}}-\sqrt{3-2 \sqrt{2}}$ equals
(A) 2.
(B) $2 \sqrt{3}$.
(C) $4 \sqrt{2}$.
(D) $\sqrt{6}$.
(E) $2 \sqrt{2}$.
(21st American High School Mathematics Examination, 1970) | $$
\begin{array}{l}
\text { [Solution] Original expression }=\sqrt{2+2 \sqrt{2}+1}-\sqrt{2-2 \sqrt{2}+1} \\
=\sqrt{(1+\sqrt{2})^{2}}-\sqrt{(\sqrt{2}-1)^{2}} \\
=1+\sqrt{2}-\sqrt{2}+1=2 \text {. } \\
\end{array}
$$
Therefore, the answer is $(A)$. | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. Given the sequence $\left\{a_{n}\right\}\left(n \in \mathbb{R}^{*}\right)$ satisfies $a_{n}=\left\{\begin{array}{ll}n, & n=1,2,3,4,5,6, \\ -a_{n-3}, & n \geq 7, n \in \mathbb{N}^{*},\end{array}\right.$ then $a_{2023}=$ $\qquad$. | answer: -4 | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. As shown in the figure, $M, P, Q$ are moving points on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with foci $F_{1}$ and $F_{2}$, $B$ is one endpoint of the minor axis of the ellipse and satisfies $\left|\overrightarrow{B F_{1}}+\overrightarrow{B F_{2}}\right|=2 c$. If $\overrightarrow{M F_{1}}=\... | 11. Solution: From $\left|\overrightarrow{B F_{1}}+\overrightarrow{B F_{2}}\right|=2 c$ we know $2 b=2 c$, so $b=c$,
Therefore, the eccentricity of the ellipse $e=\frac{\sqrt{2}}{2}$. That is, $b^{2}=c^{2}=\frac{1}{2} a^{2}$,
So the equation of the ellipse can be simplified to $x^{2}+2 y^{2}=2 c^{2}$.
When point $M$ i... | 6 | Geometry | proof | Yes | Yes | olympiads | false |
5. (10 points) A store sold some pens on the first day. On the second day, after reducing the price of each pen by 1 yuan, it sold 100 more pens. On the third day, after increasing the price of each pen by 3 yuan compared to the previous day, it sold 200 fewer pens than the previous day. If the store made the same amou... | 【Analysis】Let the price of each pen on the first day be $x$ yuan, and the number sold be $y$ pens. According to the formula total price = unit price $\times$ quantity, the amount of money made on the first day is $x y$ yuan. The unit price on the second day is $x-1$ yuan, and the number of pens sold is $y+100$ pens, so... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
【Question 5】
Subtract 101011 from 10000000000, and the digit 9 appears $\qquad$ times in the resulting answer. | 【Analysis and Solution】
Number problem.
$$
1 \underbrace{000000000}_{10 \uparrow 0}-101011=\underbrace{9999999999}_{10 \uparrow 9}-101010 \text {; }
$$
$\underbrace{999999999}_{10 \text { nines }}-101010$ will not require borrowing in any digit, and only the same digit position of $9-0$ will result in 9; $\underbrace{9... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) Three positive integers greater than 1000 satisfy: the unit digit of the sum of any two of these numbers is equal to the unit digit of the third number. Then, the last three digits of the product of these 3 numbers have $\qquad$ possible values. | 【Analysis】Assuming the units of the three numbers are $a$, $b$, and $c$, according to the condition "the units digit of the sum of any two numbers is equal to the units digit of the third number," the three numbers can be divided into three cases: (1) If $a$, $b$, and $c$ are all equal, then they can only all be $0$; (... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. For $n \in \mathbf{N}$, let $S_{n}$ be the minimum value of $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$, where $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ are positive real numbers whose sum is 17. If there exists a unique $n$ such that $S_{n}$ is also an integer, find $n$. | $\begin{array}{l}\text { 10. Solution: } \because \sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} \mathrm{i}\right|, k=1,2, \cdots, n, \\ \therefore \sum_{k=1}^{n}\left|(2 k-1)+a_{k} \mathrm{i}\right| \geqslant\left|1+a_{1} \mathrm{i}\right|+\left|3+a_{2} \mathrm{i}\right|+\cdots+\left|(2 ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. There are a total of 10040 numbers written on the blackboard, including 2006 ones, 2007 twos, 2008 threes, 2009 fours, and 2010 fives. Each operation involves erasing 4 different numbers and writing the fifth type of number (for example, erasing 1, 2, 3, and 4 each once and writing one 5; or erasing 2, 3, 4, and 5 e... | 【Solution】Solution: Since after each operation, the parity of the number of each digit changes,
what was originally an odd number becomes an even number, and what was originally an even number becomes an odd number.
That is, no matter how the operation is performed, the parity of the original 2 odd numbers remains the... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. If $M=\left\{z \left\lvert\, z=\frac{t}{1+t}+i \frac{l+t}{t}\right., t \in R, t \neq-1, t \neq 0\right\}$,
$$
N=\{z|z=\sqrt{2}[\cos (\arcsin t)+i \cos (\arccos t)], t \in R,| t \mid \leqslant 1\} \text {, }
$$
then the number of elements in $M \cap N$ is
(A) 0 ;
(B) 1 ;
(C) 2 ;
(D) 4 . | 5. (A)
The points in $M$ lie on the curve
$$
M:\left\{\begin{array}{l}
x=\frac{t}{1+t} \\
y=\frac{1+t}{t}
\end{array} \quad(t \in \mathbb{R}, t \neq 0,-1)\right.
$$
The points in $N$ lie on the curve
$$
N:\left\{\begin{array}{l}
x=\sqrt{2\left(1-x^{2}\right)} \\
y=\sqrt{2} t
\end{array} \quad(t \in \mathbb{R},|t| \leq... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
12. In the Cartesian coordinate system $x O y$, it is known that points $A_{1}(-2,0), A_{2}(2,0)$, and a moving point $P(x, y)$ satisfies the product of the slopes of lines $A_{1} P$ and $A_{2} P$ is $-\frac{3}{4}$. Let the trajectory of point $P$ be curve $C$.
(1) Find the equation of $C$;
(2) Suppose point $M$ is on ... | (1) $k_{A_{1} P} \cdot k_{A_{2} P}=\frac{y^{2}}{x^{2}-4}=-\frac{3}{4} \Rightarrow \frac{y^{2}}{3}+\frac{x^{2}-4}{4}=0 \Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) Let $M(4, s), A B:\left\{\begin{array}{l}x=4+t \cos \alpha, \\ y=s+t \sin \alpha,\end{array} P Q:\left\{\begin{array}{l}x=4+t \cos \beta, \\ y=s+t \si... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
88. As shown in the figure, two people, A and B, are playing hide-and-seek outside a square wall with a side length of 100 meters. They start from two opposite corners at the same time and walk in a clockwise direction. A walks 50 meters per minute, and B walks 30 meters per minute. Both will stop for 1 minute when the... | Reference answer: 8
Key point: Travel problem - Stop and go problem | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (3 points) The left figure below is the recently discovered Archimedes' "Stomachion" puzzle, dividing a square into 14 polygonal pieces: After research, experts found that these 14 pieces can be correctly drawn on a square with a side length of $12 \mathrm{~cm}$, as shown in the figure. Question: The area of the gra... | 【Analysis】According to the characteristics of the gray quadrilateral, reasonably divide the figure into two triangles to find the area.
【Solution】Solution: As shown in the figure, the gray part is quadrilateral $A B C D$. Connect $B D$, then $S$ quadrilateral $A B C D=S_{\triangle A B D}+S_{\triangle B C D}=\frac{1}{2}... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. In $\triangle A B C$, the side lengths opposite to $\angle A 、 \angle B 、 \angle C$ are $a 、 b 、 c$, respectively, and
$$
\begin{array}{l}
\sin C \cdot \cos \frac{A}{2}=(2-\cos C) \sin \frac{A}{2}, \\
\cos A=\frac{3}{5}, a=4 .
\end{array}
$$
Then the area of $\triangle A B C$ is $\qquad$ | 7.6.
From equation (1) we know
$$
\begin{array}{l}
2 \sin \frac{A}{2}=\sin \left(C+\frac{A}{2}\right) \\
\Rightarrow 2 \sin A=2 \sin \left(C+\frac{A}{2}\right) \cdot \cos \frac{A}{2} \\
=\sin C+\sin B .
\end{array}
$$
Thus, $c+b=2a$.
Also, $a^{2}=b^{2}+c^{2}-2 b c \cos A$, which means $4^{2}=b^{2}+(8-b)^{2}-2 b(8-b) ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. For a finite set $A$, there exists a function $f: \mathbf{N}^{*} \rightarrow A$, with the following property: if $i, j \in \mathbf{N}^{*}$, and $|i-j|$ is a prime number, then $f(i) \neq f(j)$. How many elements must the set $A$ have at least? | 1. Since the absolute value of the difference between any two of the numbers $1,3,6,8$ is a prime number, according to the problem, $f(1)$, $f(3)$, $f(6)$, $f(8)$ are four distinct elements in $A$. Therefore, $|A| \geqslant 4$. On the other hand, if we let $A=\{0,1,2,3\}$, and the mapping $f: \mathbf{N}^{*} \rightarrow... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. Given $a, b, c \in R^{+}$, satisfying $a b c(a+b+c)=1$,
(I) Find the minimum value of $S=(a+c)(b+c)$;
(II) When $S$ takes the minimum value, find the maximum value of $c$.
保留了源文本的换行和格式。 | 14. Solution: (I) Since $(a+c)(b+c)=a b+a c+b c+c^{2}=a b+(a+b+c) c=a b+\frac{1}{a b}$ $\geq 2 \sqrt{a b \cdot \frac{1}{a b}}=2$, the equality holds when $a b=1$,
When $a=b=1, c=\sqrt{2}-1$, $S$ can achieve the minimum value of 2.
(II) When $S$ takes the minimum value, $a b=1$, thus $c(a+b+c)=1$,
which means $c^{2}+(a+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.76 Students solve a set of quadratic equations in numerical order: when an equation has two distinct roots, the next equation is formed as follows: the constant term is its larger root, the coefficient of the linear term is the smaller root, and the coefficient of the quadratic term $x^{2}$ is always 1. Prove that th... | [Proof] Let the initial quadratic equation be $x^{2}+p x+q=0$, and the subsequent equations be
$$
f_{j}(x)=x^{2}+p_{j} x+q_{j}=0, p_{j}0 \text {. }
$$
By Vieta's formulas:
$$
\begin{array}{l}
p_{j}+q_{j}=-p_{j-1}, \\
p_{j} q_{j}=q_{j-1} .
\end{array}
$$
Therefore, given $p_{j-1}0 .
$$
The above can be transformed in... | 5 | Algebra | proof | Yes | Yes | olympiads | false |
5.38 $P(x)$ is a polynomial of degree $3n$, such that
$$
\begin{array}{l}
P(0)=P(3)=\cdots=P(3 n)=2, \\
P(1)=P(4)=\cdots=P(3 n-2)=1, \\
P(2)=P(5)=\cdots=P(3 n-1)=0, \\
P(3 n+1)=730 .
\end{array}
$$
Determine $n$. | [Solution] Notice that when $x$ takes the values $0,1,2, \cdots, 3 n$, $P(x)-1$ cyclically takes the values $1,0,-1,1,0,-1, \cdots, 1$. Therefore, we can consider the sequence
$$
\left\{w^{n}\right\}=\left\{1, w, w^{2}, 1, w, w^{2}, \cdots\right\} \text {. }
$$
where $w=\frac{1}{2}(-1+i \sqrt{3})$ is a cube root of 1.... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) Two identical rectangular pieces of paper, each with an area of 48 square centimeters. As shown in the figure, when they are stacked together, the covered area is 72 square centimeters. Given that one diagonal $B D$ of the overlapping quadrilateral $A B C D$ is 6 centimeters, what is the length of each r... | 【Analysis】It is known that $A B C D$ is a parallelogram, and its area is $48+48-72=24$ square centimeters.
Let the width of the rectangular paper be $h$, then $h$ is also the distance between the two sets of parallel lines of the parallelogram, so $A D \times h=$ $\square A B C D$'s area $=D C \times h$,
we get $A D=D... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) On Beta Planet, there are seven countries, each of which has exactly four friendly countries and two enemy countries. There are no three countries that are all enemies with each other. For such a planetary situation, a total of $\qquad$ three-country alliances, where all countries are friends with each o... | 【Analysis】Use $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ to represent seven countries, and use dashed lines to connect to indicate enemy relations, and solid lines to connect to indicate friendly relations. Then each country has 2 dashed lines and 4 solid lines. There are a total of $7 \times 2 \div 2=7$ (lines)... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
80. The school organized students to participate in a formation drill. At the beginning, there were 40 boys participating, and no girls participated. Later, the team was adjusted, each time reducing 3 boys and adding 2 girls, so after adjusting $\qquad$ times, the number of boys and girls would be equal. | Answer: 8 | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(11) (15 points) Divide each side of the equilateral $\triangle A B C$ with side length 3 into three equal parts, and draw lines parallel to the other two sides through each division point. The 10 points formed by the intersections of the sides of $\triangle A B C$ and these parallel lines are called grid points. If $n... | (11) Let the two equal division points on side $AB$ from $A$ to $B$ be $D$ and $E$; the two equal division points on side $BC$ from $B$ to $C$ be $F$ and $G$; and the two equal division points on side $CA$ from $C$ to $A$ be $H$ and $I$. The central grid point is $K$.
If the minimum value of $n$ is 4, taking the grid ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$7 \cdot 115$ Let $S=\{1,2,3,4\}, a_{1}, a_{2}, \cdots, a_{k}$ be a sequence composed of numbers from $S$, and it includes all permutations of $S$ that do not end with 1, i.e., for any permutation $\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ of the 4 numbers in $S$, where $b_{4} \neq 1$, there exist $i_{1}, i_{2}, i_{3}, ... | [Solution] (1) For a sequence composed of $1,2,3$, if it contains all permutations of $\{1,2,3\}$, it must have at least 7 terms.
Assume that in $1,2,3$, 3 appears last, then the first 3 must be at least the 3rd term. To include the permutations $(3,2,1)$ and $(3,1,2)$, after 3 there should be $2,1,2$ or $1,2,1$. If t... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$30 \cdot 9$ There are 18 teams participating in the opening ceremony of the competition. When entering, the 1st team has 27 people, the 2nd team has 26 people, ..., the 18th team has 10 people. If they all enter in a single-file line, and all representatives of the 18 teams are numbered $1,2, \cdots, 333$ in the order... | [Solution] The last representative of the 1st team is numbered 27, the last representative of the 2nd team is numbered $27+26=53$, the last representative of the 3rd team is numbered $27+26+25=78$, and the last representative of the 4th team is numbered $27+26+25+24=102$,
Thus, the numbers of the representatives follo... | 10 | Number Theory | MCQ | Yes | Yes | olympiads | false |
1. Given the set $M=\{2,0,1,9\}, A$ is a subset of $M$, and the sum of the elements in $A$ is a multiple of 3. Then the number of subsets $A$ that satisfy this condition is ( ).
(A) 8
(B) 7
(C) 6
(D) 5 | - 1. B.
After enumeration, the subsets whose elements sum up to a multiple of 3 are $\{0\}, \{9\}, \{2,1\}, \{0,9\}, \{2,0,1\}, \{2,1,9\}, \{2,0,1,9\}$, totaling 7 subsets. | 7 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
10. Let $A$ and $B$ be two points on the parabola $y=x^{2}$ on opposite sides of the $y$-axis, distinct from $O$, such that $AB=4$. Denote the area of $\triangle AOB$ by $S$, where $O$ is the origin. Then the maximum value of $S$ is $\qquad$. | 10.8.
Let $A\left(-m, m^{2}\right), B\left(n, n^{2}\right)(n, m>0)$. Then $16=(n+m)^{2}+\left(n^{2}-m^{2}\right)^{2} \geqslant(n+m)^{2}$.
Therefore, $m+n \leqslant 4$.
$$
\begin{array}{l}
\text { Hence } S=\frac{1}{2}\left(m^{2}+n^{2}\right)(m+n)-\left(\frac{1}{2} m m^{2}+\frac{1}{2} n n^{2}\right) \\
=\frac{1}{2}(m+n... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. The value of the real number $a$ that satisfies the equation $\sqrt{1+\sqrt{1+a}}=\sqrt[3]{a}$ is $\qquad$ . | 2, 1.8.
From the given equation, we have
$$
(1+\sqrt{1+a})^{3}=a^{2}(a>1) \text {. }
$$
Let $x=\sqrt{1+a}$. Then $x \geqslant 0$, and $a=x^{2}-1$. Thus, $(1+x)^{3}=\left(x^{2}-1\right)^{2}$.
After rearranging and factoring, we get
$$
\begin{array}{l}
x(x-3)(x+1)^{2}=0 \\
\Rightarrow x_{1}=0, x_{2}=3, x_{3}=-1 \text { ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16. Let $x, y, z \geq 0$, and at most one of them is 0, find
$$
f(x, y, z)=\sqrt{\frac{x^{2}+256 y z}{y^{2}+z^{2}}}+\sqrt{\frac{y^{2}+256 z x}{z^{2}+x^{2}}}+\sqrt{\frac{z^{2}+256 x y}{x^{2}+y^{2}}}
$$
the minimum value | Let's assume $x \geq y \geq z$.
Case 1: When $256 y^{3} \geq x^{2} z$, we have
$$
\begin{array}{c}
\frac{x^{2}+256 y z}{y^{2}+z^{2}}-\frac{x^{2}}{y^{2}}=\frac{z\left(256 y^{3}-x^{2} z\right)}{\left(y^{2}+z^{2}\right) y^{2}} \geq 0 \\
\frac{y^{2}+256 z x}{z^{2}+x^{2}}-\frac{y^{2}}{x^{2}}=\frac{z\left(256 x^{3}-y^{2} z\r... | 12 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
8. A square is divided into 8 smaller squares, of which exactly 7 have a side length of 2. Then the side length of the original square is $\qquad$ . | \begin{tabular}{l}
\hline $\mathbf{8}$ \\
\hline 8
\end{tabular} | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Four (50 points) Let $k \in N^{*}$, if it is possible to color all positive integers with $k$ colors, and there exists a function $f: Z^{+} \rightarrow Z^{+}$ satisfying:
(1) For any same-colored positive integers $m, n$ (which can be the same), $f(m+n)=f(m)+f(n)$;
(2) There exist $m, n \in N^{*}$ (which can be the sam... | Four, Solution: The minimum value of $k$ is 3.
First, we provide a construction for $k=3$. Let $f(n)=\left\{\begin{array}{ll}2 n, & 3 \mid n \\ n, & 3 \nmid n\end{array}\right.$, then $f(1)+f(2)=3 \neq f(3)$, satisfying condition (2). If we color the numbers that are congruent to $0, 1, 2 \pmod{3}$ with three different... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
15. Blu is a smart blue macaw, and its young owner Linda not only teaches it to speak but also arithmetic. One day, Linda gave it an arithmetic problem:
$$
1-2+3-4+5-6+7-6+5-4+3-2+1
$$
Then the result of this expression is $\qquad$ | $1$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$15 \cdot 1$ A school's wind band consists of 100 boys and 80 girls, and the string band consists of 80 boys and 100 girls. In total, 230 students participated in these two bands, with 60 boys participating in both. Therefore, the number of girls who only participated in the wind band and not in the string band is
(A)... | ［Solution］Let the number of people who participate in both bands be $x$, as shown in the shaded area in the figure. According to the problem,
we have
$180+180-x=230$, then $x=130$.
$\because$ The number of boys participating in both bands is 60, then the number of girls participating in both bands is 70.
$\therefore$ ... | 10 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
1. The imaginary part of the complex number $z=(1+\mathrm{i})^{2}(2+\mathrm{i})$ is ( ).
(A) $-2 \mathrm{i}$
(B) -2
(C) 4 i
(D) 4 | \begin{array}{l}\text {-.1. D. } \\ z=(1+\mathrm{i})^{2}(2+\mathrm{i})=2 \mathrm{i}(2+\mathrm{i})=4 \mathrm{i}-2 .\end{array} | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
6. If $a>b>c>d$, and $\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-d} \geqslant \frac{n}{a-d}$, then the maximum value of $n$ is ( ).
A. 12
B. 9
C. 6
D. 3 | $6 . B$
Obviously, $n>0$. From the condition, we get
$$
\begin{aligned}
n & \leqslant \frac{a-d}{a-b}+\frac{a-d}{b-c}+\frac{a-d}{c-d} \\
& =\frac{a-b+b-c+c-d}{a-b}+\frac{a-b+b-c+c-d}{b-c}+\frac{a-b+b-c+c-d}{c-d} \\
& =3+\frac{b-c}{a-b}+\frac{c-d}{a-b}+\frac{a-b}{b-c}+\frac{c-d}{b-c}+\frac{a-b}{c-d}+\frac{b-c}{c-d} \geq... | 9 | Inequalities | MCQ | Yes | Yes | olympiads | false |
4. (8 points) Count the number of triangles in the figure: $\qquad$ triangles.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4. (8 points) Count the number of triangles in the figure: $\qquad$ triangles.
【Solution】Solution: 4 triangles made up of 1 piece;
2 triangles made up of 2 pieces;
2 triangles made up of 3 pieces;
Total: $4+2+2=8$ (triangles);
Answer: There are 8 triangles in the figure.
Therefore, the answer is: 8. | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. As shown in the figure, quadrilaterals $A B C D$ and $C E F G$ are two squares. Given that $\frac{S_{\triangle A B I}}{S_{\triangle E F I}}=27$, find $\frac{A B}{E F}$. | 【Analysis】Let $A B=x, E F=y$,
$$
\begin{array}{l}
\frac{A I}{I E}=\frac{S_{\triangle A B F}}{S_{\triangle B E F}}=\frac{\frac{1}{2} \times x \times(x+y)}{\frac{1}{2} \times y \times y}=\frac{x(x+y)}{y^{2}} \\
\frac{I F}{I B}=\frac{S_{\triangle A E F}}{S_{\triangle A B E}}=\frac{\frac{1}{2} \times y \times(x+y)}{\frac{1... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$10 \cdot 771^{2}, 2^{2}, 3^{2}, \cdots, 123456789^{2}$ What is the digit in the units place of the sum?
(China Junior High School Mathematics League, 1990) | [Solution] Since $123456789=10 \cdot 12345678+9$,
then
$$
\begin{array}{c}
1^{2}+2^{2}+3^{2}+\cdots+123456789^{2} \text {, the unit digit is } \\
=(1+4+9+6+5+6+9+4+1+0) \cdot 12345678+(1+4+
\end{array}
$$
$9+6+5+6+9+4+1)$, the unit digit is $=5 \cdot 8+5$, the unit digit
$$
=5 \text {. }
$$
Thus, the required unit dig... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. The integer sequence $\left\{a_{i, j}\right\}(i, j \in \mathbf{N})$, where,
$$
\begin{array}{l}
a_{1, n}=n^{n}\left(n \in \mathbf{Z}_{+}\right), \\
a_{i, j}=a_{i-1, j}+a_{i-1, j+1}(i 、 j \geqslant 1) .
\end{array}
$$
Then the unit digit of $a_{128,1}$ is $\qquad$ | 8. 4.
By the recursive relation, we have
$$
\begin{array}{l}
a_{1,1}=1, a_{2, n}=n^{n}+(n+1)^{n+1}, \\
a_{3, n}=n^{n}+2(n+1)^{n+1}+(n+2)^{n+2} .
\end{array}
$$
Accordingly, by induction, we get
$$
\begin{array}{l}
a_{n, m}=\sum_{k=0}^{m-1} \mathrm{C}_{m-1}^{k}(n+k)^{n+k} \\
=\sum_{k \geqslant 0} \mathrm{C}_{m-1}^{k}(... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
61. A shape made of building blocks, when viewed from the front is $\square$, and when viewed from the side is $\square$. If this shape is made using $n$ identical cubic blocks, then how many possible values can $n$ have? | Reference answer: 7 | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. If $a b c d=1$, then
$$
\frac{1}{1+a+a b+a b c}+\frac{1}{1+b+b c+b c d}+\frac{1}{1+c+c d+c d a}+\frac{1}{1+d+d a+d a b}=
$$
$\qquad$ | $1$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Determine all positive integers $n$, such that the equation $x^{n}+(2+x)^{n}+(2-x)^{n}=0$ has integer solutions. | 2. Solution: Clearly, if $n$ is even, then the equation has no real roots. If $n=1$, then $x=-4$.
When $n \geqslant 3$ and is odd, since the left side of the equation is a polynomial with the leading coefficient 1 and the constant term $2^{n+1}$, therefore, its integer solutions can only be of the form $-2^{t}$ (where... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
33-11 Given that $a$ is a root of the equation $x^{2}-5 x+1=0$, then the last digit of $a^{4}+a^{-4}$ is
(A) 3.
(B) 5.
(C) 7.
(D) 9.
(6th "Jinyun Cup" Junior High School Mathematics Invitational, 1989) | [Solution] Let $\alpha$ be a root of the equation, then $\alpha^{2}-5 \alpha+1=0$. That is, $\alpha^{2}+1=5 \alpha$, or $\frac{\alpha^{2}+1}{\alpha}=5$.
Also, $\alpha^{4}+\alpha^{-4}=\left(\alpha^{2}+\alpha^{-2}\right)^{2}-2$
\[
\begin{array}{l}
=\left[\left(\alpha+\alpha^{-1}\right)^{2}-2\right]^{2}-2 \\
=\left[\left(... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
7・81 Suppose 20 telephones are connected by wires, each wire connecting two telephones, and at most 1 wire connects any pair of telephones; from each telephone, at most 3 wires can extend out. Now, we want to paint each of these wires with 1 color, such that the colors of the wires extending from each telephone are all... | [Solution] When the connections are as shown in the right figure, at least 4 different colors are required.
The following proves that using 4 different colors is sufficient. Represent each telephone with a point, and when two telephones are connected by a wire, draw a line segment between the corresponding points. Thus... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A figure $\Psi$ composed of 4 unit squares is called a $T$-shaped block, and the figure $\square$ is called an $S$-shaped block. Cover an $8 \times 10$ rectangular chessboard with several $S$-shaped blocks and several $T$-shaped blocks without overlapping, and both shapes can be rotated or flipped. Then, what is the... | Analysis: In a tiling method, we categorize the blocks used into the following 4 types: horizontally placed T-shaped blocks (or their flips) are called type $A$, vertically placed T-shaped blocks (or their rotations) are called type $B$, horizontally placed S-shaped blocks (or their flips) are called type $C$, and vert... | 6 | Combinatorics | 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 . $\qquad$ | 2. Let $y=f(x), z=f(y), w=f(z)$, analyze the change of $y, z, w$ when $x$ changes from 0 to 1:
Since “$\rightarrow$” indicates a linear change, and the change is within the non-negative real number range, the graph of $w=f(f(f(x)))$ is as shown in the figure.
Obviously, the graph of $w=f(f(f(x)))$ intersects the line ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
25. In Chinese class, the teacher gave a matching question, requiring to connect the four great classical novels, "Romance of the Three Kingdoms," "Journey to the West," "Water Margin," and "Dream of the Red Chamber," with their respective authors. Careless Xiaoma connected all of them incorrectly. How many possible wa... | $9$ | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
30. In square $ABCD$ with side length $4$, $M$ and $N$ are moving points on $AB$ and $CD$ respectively. If trapezoid $BCNM$ is folded along $MN$, point $B$ lands exactly on side $AD$. Then the minimum value of the area of trapezoid $BCNM$ is $\qquad$ | $6$ | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 9 For any positive integer $n$, let $x_{n}=\mathrm{C}_{2 n}^{n}$. Find all positive integers $h(h>1)$, such that the sequence of remainders of $\left\{x_{n}\right\}$ modulo $h$ is eventually periodic. | 【Analysis】 $h=2$ is the solution we are looking for.
If $h$ is not a power of 2, let $p$ be a prime factor of $h$, and $p$ is odd. Then there exist positive integers $N$ and $T$ such that the sequence $\left\{x_{n}\right\}$ modulo $p$ has a periodic remainder sequence after the $N$-th term with period $T$.
Take a posit... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12.57 For a positive integer $n$, let $S_{n}$ be the minimum value of $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers whose sum is 17. There is a unique $n$ for which $S_{n}$ is also an integer. Find $n$.
(9th American Invitational Mathematics Examination, 19... | [Solution] Consider each term $t_{k}=\sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the hypotenuse of a right-angled triangle, with the two legs being $2 k-1$ and $a_{k}$.
By placing these right-angled triangles together to form a ladder, let $A, B$ be the starting and ending points, respectively.
The distance from $A$ to $B$ is
$$... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. How many four-digit numbers can be formed using the digits $1$, $9$, $8$, $8$ that leave a remainder of 8 when divided by 11? | Can form 4 numbers that leave a remainder of 8 when divided by 11
14.【Solution】Using $1, 9, 8, 8$, 12 four-digit numbers can be formed, namely $1988, 1898, 1889$, $9188, 9818, 9881, 8198, 8189, 8918, 8981, 8819, 8891$.
Subtracting 8 from them results in 1980, 1890, 1881, 9180, 9810, 9873, 8190, 8181, $8910, 8973, 8811... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
18 4 \% 6 Solve the equation for positive integer solutions:
$$
5^{x}-3^{y}=2
$$ | The solution $x=y=1$ is clearly a solution to (1).
Assume $x, y$ are not both 1, then neither of them is 1. Taking (1) modulo 4, we get $1-(-1)^{y}=2(\bmod 4)$, so $y$ is odd and greater than 1. Taking (1) modulo 9, we get $5^{x} \equiv 2$ (mod 9), so $x=6k+5$. Finally, taking (1) modulo 7, we have $3^{y} \equiv 3,-1,-... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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