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5. The speed of the car is 54 kilometers per hour. When it starts, the last three digits of the car's odometer read $\overline{a b c}$. After driving for several hours (an integer), the last three digits of the odometer read $\overline{c b a}$, with no change in the other digits. The car has driven for $\qquad$ hours. | $11$ | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$19 \cdot 28$ convex polygon, the interior angles less than $120^{\circ}$ cannot be more than
(A) 2.
(B) 3.
(C) 4.
(D) 5.
(China Zhejiang Ningbo Junior High School Mathematics Competition, 1987) | [Solution] The sum of the exterior angles of a convex polygon is $360^{\circ}$. If we set the number of its interior angles less than $120^{\circ}$ to $k$, then the corresponding exterior angles of these angles are greater than $60^{\circ}$.
But $k \cdot 60^{\circ}<360^{\circ}$, we get $k<6$.
Thus, the maximum value of... | 5 | Geometry | MCQ | Yes | Yes | olympiads | false |
$1 \cdot 27$ If $a, b, c$ are non-zero real numbers and satisfy
and
$$
\begin{array}{l}
\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}, \\
x=\frac{(a+b)(b+c)(c+a)}{a b c}, \quad \text { and } x<0,
\end{array}
$$
then $x$ equals
(A) -1 .
(B) -2 .
(C) -4 .
(D) -6 .
(E) -8 .
(29th American High School Mathematics Exam... | ［Solution］Notice that
that is
$$
\begin{array}{c}
\frac{a+b-c}{c}+2=\frac{a-b+c}{b}+2=\frac{-a+b+c}{a}+2, \\
\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a} .
\end{array}
$$
If $a+b+c=0$, these equations are satisfied.
If $a+b+c \neq 0$, then divide the latter equations by $a+b+c$, yielding
$$
\frac{1}{c}=\frac{1}{b}... | -1 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. The line $\frac{x}{4}+\frac{y}{3}=1$ intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at points $A, B$. There is a point $P$ on the ellipse such that the area of $\triangle P A B$ equals 3. How many such points $P$ are there?
A. 1
B. 2
C. 3
D. 4 | 1. B Let $P_{1}(4 \cos \alpha, 3 \sin \alpha)\left(0<\alpha<\frac{\pi}{2}\right), P_{1}$ is on the ellipse in the first quadrant. Consider the area $S$ of quadrilateral $P_{1} A O B$. $S=S_{\triangle O A P_{1}}+S_{\triangle O B P_{1}}=\frac{1}{2} \times 4 \cdot 3 \sin \alpha+\frac{1}{2} \times 3 \cdot 4 \cos \alpha=6 \... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
3. Let the function be
$$
\begin{array}{l}
f(x, y) \\
=\sqrt{x^{2}+y^{2}-6 y+9}+\sqrt{x^{2}+y^{2}+2 \sqrt{3} x+3}+ \\
\quad \sqrt{x^{2}+y^{2}-2 \sqrt{3} x+3} .
\end{array}
$$
Then the minimum value of $f(x, y)$ is ( ). | 3. C.
Let $A(0,3), B(-\sqrt{3}, 0), C(\sqrt{3}, 0)$, $D(0,1), P(x, y)$.
Then $f(x, y)=|P A|+|P B|+|P C|$.
Notice that,
$$
\angle A D B=\angle B D C=\angle C D A=120^{\circ} \text {. }
$$
Thus, $D$ is the Fermat point of $\triangle A B C$.
Therefore, the minimum value of $f(x, y)$ is achieved when point $P$ is $D$, an... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. If three angles $x, y, z$ form an arithmetic sequence with a common difference of $\frac{\pi}{3}$, then $\tan x \tan y+\tan y \tan z+\tan z \tan x=$ $\qquad$ | Given $x=y-\frac{\pi}{3}, z=y+\frac{\pi}{3}$, then $\tan z=\frac{\tan y+\sqrt{3}}{1-\sqrt{3} \tan y}, \tan x=\frac{\tan y-\sqrt{3}}{1+\sqrt{3} \tan y}$, so
$$
\tan x \tan y=\frac{\tan ^{2} y-\sqrt{3} \tan y}{1+\sqrt{3} \tan y}, \tan y \tan z=\frac{\tan ^{2} y+\sqrt{3} \tan y}{1-\sqrt{3} \tan y}, \tan z \tan x=\frac{\ta... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
29. One day, Fat Controller gave four little engines Thomas, Edward, James, and Percy each a task, and asked Thin Controller to notify the little engines. Thin Controller mixed up the tasks when assigning them, and as a result, none of the four little engines received their original task. Therefore, there are $\qquad$ ... | $9$ | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Four. (20 points) Let $x, y, z$ be the lengths of the sides of a triangle, and $x+y+z=1$. Find the minimum value of the real number $\lambda$ such that
$$
\lambda(x y+y z+z x) \geqslant 3(\lambda+1) x y z+1
$$
always holds. | From the original inequality, we have
$$
\lambda \geqslant \frac{3 x y z+1}{x y+y z+z x-3 x y z} \text {. }
$$
When $x=y=z=\frac{1}{3}$, $\lambda \geqslant 5$.
Next, we prove that the minimum value of $\lambda$ is 5, i.e., we need to prove
$$
5(x y+y z+z x) \geqslant 18 x y z+1 \text {. }
$$
Let $x=a+b, y=b+c, z=c+a$... | 5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Find all positive integers $n$, such that $n-1$ and $\frac{n(n+1)}{2}$ are both perfect numbers. | Analysis requires the following knowledge:
(1) When $n, m$ are coprime, $\sigma(m n)=\sigma(m) \sigma(n)$, i.e., the sum of the positive divisors of $n$, $\sigma(n)$, is a multiplicative function.
(2) A necessary and sufficient condition for $n$ to be an even perfect number is $n=2^{p-1}\left(2^{p}-1\right)$, and both ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11. (10 points) The teacher made 50 cards, each with numbers from 1 to 50, for the 50 students in the class. Each card has one side red and the other side blue, with the same number written on both sides. The teacher placed all 50 cards on the table with the blue side up and said to the students: "Please come to the fr... | 【Analysis】Each card, the number written on it will be flipped as many times as it has divisors. Cards that have been flipped an odd number of times will have the red side facing up, and only perfect squares have an odd number of divisors, so this problem is essentially asking how many cards have perfect squares written... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. A line with a slope of $\frac{2}{21} \sqrt{21}$ passing through the foci of the hyperbola $16 x^{2}-9 y^{2}=144$ intersects the hyperbola. Find the length of the chord intercepted. | 2. The hyperbola equation is transformed to $\frac{y^{2}}{16}-\frac{x^{2}}{9}=1$, with the foci on the $y$-axis, $a=4, b=3, c=5, e=\frac{5}{4}, e p=\frac{9}{4}$, slope $k^{\prime}=$ $\frac{2}{21} \sqrt{21}$, taking $k=\frac{1}{k^{\prime}}=\frac{\sqrt{21}}{2}$. From property 2, we find $l=\frac{2 e p\left(1+k^{2}\right)... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) Lelé placed some small squares and isosceles right triangles without overlapping on the bottom of a large square box with a side length of 7 cm. If the side lengths of the small squares are all 2 cm, and the hypotenuse lengths of the isosceles right triangles are all 3 cm, then the maximum number of each... | 【Answer】Solution: "Each put in" should mean the number of each type put in is equal,
since the sum of the area of each small square and a triangle is $2 \times 2 + 3 \times 3 \div 4 = 6.25$,
the area of the large square is $7 \times 7 = 49$,
and $49 \div 6.25 = 7.84$,
so the maximum number that can be put in is 7 each,... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
14. (15 points) Divide a large square with a side length of 7 into smaller squares with side lengths of 1, 2, or 3. What is the minimum number of squares with a side length of 1? Draw your division method in the diagram. Answer: There are at least $\qquad$ squares with a side length of 1. | 【Solution】Solution: Let the number of $3 \times 3$ squares be $x$, the number of $2 \times 2$ squares be $y$, and the number of $1 \times 1$ squares be $z$. Then we have the equation: $9 x+4 y=49-z$. Simple trials show that $x \leqslant 4, y \leqslant 9$. When $z=0$, solving $9 x+4 y=49$ gives $x=5, y=1$ (discard); $x=... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In $\triangle A B C$, prove that: $\frac{1}{\sin \frac{A}{2}}+\frac{1}{\sin \frac{B}{2}}+\frac{1}{\sin \frac{C}{2}} \geqslant 6$. | 4. $y=\frac{1}{\sin x}$ is a convex function in $\left(0, \frac{\pi}{2}\right)$, so the left side $\geqslant \frac{3}{\sin \frac{A+B+C}{6}}=6$. | 6 | Inequalities | proof | Yes | Yes | olympiads | false |
8. Given that $A$, $B$, and $C$ are three points on $\odot O$, and
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \text {. }
$$
Then the dot product $\overrightarrow{A B} \cdot \overrightarrow{A C}=$ | 8. 0 .
Let $AD$ be the diameter of $\odot O$, and connect $DB$, $DC$. Then, from the given equation, we have $\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{AC}$.
Therefore, quadrilateral $ABDC$ is an inscribed parallelogram in the circle, which means it is a rectangle, $\angle BAC=90^{\circ}$, and $\overrig... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
18. Amelia tosses a coin, with the probability of landing heads up being $\frac{1}{3}$; Brian also tosses a coin, with the probability of landing heads up being $\frac{2}{5}$. Amelia and Brian take turns tossing the coin, and the first one to get heads wins. All coin tosses are independent. Starting with Amelia, the pr... | 18. D.
Let $P_{0}$ be the probability of Amelia winning.
Notice,
$P_{0}=P($ Amelia wins in the first round $)+$ $P($ both fail to win in the first round $) \cdot P_{0}$, where, if both fail to win in the first round, it still starts with Amelia, and her probability of winning remains $P_{0}$.
In the first round, the p... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. Arrange the interior angles of a convex $n$-sided polygon in ascending order, the difference between any two adjacent angles is $5^{\circ}$, then the maximum value of $n$ is . $\qquad$ | $12$ | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Given $x=\frac{b}{a}, a, b$ are coprime positive integers, and $a \leqslant 8, \sqrt{2}-1<x<\sqrt{3}-1$, then the number of fractions $x$ that satisfy the condition is $\qquad$. | answer: 7 | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Given that the function $f(x)$ is defined on $(-1,1)$, $f\left(\frac{1}{2}\right)=-1$, and satisfies for $x, y \in(-1,1)$, $f(x)+f(y)=f\left(\frac{x+y}{1+x y}\right)$. Prove:
(1) If there is a sequence $x_{1}=\frac{1}{2}, x_{n+1}=\frac{2 x_{n}}{1+x_{n}^{2}}$, then $f\left(x_{n}\right)=-2^{n-1}$;
(2) $1+f\left... | Proof: From the given, we have $f(0)+f(0)=f\left(\frac{0+0}{1+0}\right)=f(0)$, hence $f(0)=0$.
For $x \in(-1,1)$, we have $f(x)+f(-x)=f\left(\frac{x-x}{1-x^{2}}\right)=f(0)=0$,
which implies $f(-x)=-f(x)$. Therefore, $f(x)$ is an odd function on $(-1,1)$.
(1) Taking $x=y$, we get $2 f(x)=f\left(\frac{2 x}{1+x^{2}}\righ... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
In $\triangle ABC$, $AC > AB$. $P$ is the intersection of the perpendicular bisector of $BC$ and the internal angle bisector of $\angle A$. Draw $PX \perp AB$, intersecting the extension of $AB$ at point $X$, and $PY \perp AC$, intersecting $AC$ at point $Y$. $Z$ is the intersection of $XY$ and $BC$. Find the value of ... | Solution 1: Construct the circumcircle $\odot O$ of $\triangle ABC$, and let the midpoint of arc $\overparen{BC}$ (not containing $A$) be $D$, and the midpoint of chord $BC$ be $M$.
Since $\overparen{BD} = \overparen{CD}$, we have
$$
BD = DC, \angle BAD = \angle CAD,
$$
and $D$ lies on the angle bisector of $\angle BA... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $1 \square 2 \square 3 \square 6 \square 12$, fill in the $\square$ with “ + ” or “ - ”. The total number of different natural number results that can be obtained is $\qquad$. | 【Solution】Solution: According to the problem, we have:
all are “ $+$ ”, 1 way,
contains 1 “ - ”, “ - ” can be placed in any of the 4 positions, a total of 4 ways.
contains 2 “ - ”, “ - ” cannot be placed before 12, choosing 2 out of the other three positions gives 3 ways, contains 3 “ - ”, “ - ” cannot be placed before... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
87 Given the complex sequence $\left\{a_{n}\right\}$ with the general term:
$$
a_{n}=(1+i)\left(1+\frac{i}{\sqrt{2}}\right)\left(1+\frac{i}{\sqrt{3}}\right) \cdots\left(1+\frac{i}{\sqrt{n}}\right)
$$
then $\left|a_{n}-a_{n+1}\right|=$
A. $\frac{\sqrt{2}}{2}$
B. $\sqrt{2}$
C. 1
D. 2 | 87 C. It is easy to prove by mathematical induction that $\left|a_{n}\right|=\sqrt{n+1}$. Therefore,
$$
\left|a_{n+1}-a_{n}\right|=\left|a_{n}-a_{n}\left(1+\frac{i}{\sqrt{n+1}}\right)\right|=\left|a_{n}\right| \cdot\left|\frac{i}{\sqrt{n+1}}\right|=1 .
$$ | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
8. (Mathematics for Middle School, 2004, Issue 6, Mathematical Olympiad Training Problem) Given that for every pair of real numbers $x, y$, the function $f$ satisfies $f(x)+f(y)=f(x+y)-x y-1$. If $f(1)=1$, then the number of integers $n$ that satisfy $f(n)=n$ is ( ).
A. 1
B. 2
C. 3
D. infinitely many | 8. B. Reason: Let $y=1$ to get $f(x)+f(1)=f(x+1)-x-1$, which means $f(x+1)=f(x)+x+2$. Let $x=0$ to get $f(1)=f(0)+2$.
Given $f(1)=1$, we know $f(0)=-1$. When $n \in \mathbf{N}$, $f(n)=\sum_{k=1}^{n}[f(k)-f(k-1)]+f(0)=\sum_{k=1}^{n}(k+1)+f(0)=\frac{1}{2} n(n+3)-1$.
Similarly, $f(-n)=-\frac{1}{2} n(3-n)-1$, so $f(n)=\f... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. A football team played a total of 14 games in a season, and the number of games they won is less than both the number of games they drew and the number of games they lost. Therefore, the maximum number of games this team could have won this season is $\qquad$.
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。 | 【Analysis】 $14=4+5+5$, so the maximum is 4 matches. | 4 | Number Theory | proof | Yes | Yes | olympiads | false |
1. (8 points) Calculate: $(18 \times 23-24 \times 17) \div 3+5$, the result is | 【Solution】Solve: $(18 \times 23-24 \times 17) \div 3+5$
$$
\begin{array}{l}
=(6 \times 3 \times 23-6 \times 4 \times 17) \div 3+5 \\
=6 \times(3 \times 23-4 \times 17) \div 3+5 \\
=6 \times(69-68) \div 3+5 \\
=6 \div 3+5 \\
=7
\end{array}
$$
Therefore, the answer is: 7. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let $x_{1}, x_{2}, \cdots, x_{n}$ be a sequence of integers satisfying the following conditions: (i) $-1 \leqslant x_{i} \leqslant 2$, for $i=1,2,3, \cdots, n$. (ii) $x_{1}+x_{2}+\cdots+x_{n}=19$ and (iii) $x_{1}^{2}+x_{2}^{2}+\cdots+x_{m}^{2}=99$. Let $m$ and $M$ be the minimum and maximum values of $x_{1}^{3}+x_{2... | 5. D Let's assume among the $n$ numbers, $-1, 0, 1, 2$ appear $p, q, r, s$ times respectively. Then we have $-p + r + 2s = 19, p + r + 4s = 99$; thus $r + 3s = 59, r + s = 40$, and $s_{\min} = 0, s_{\max} = 19$. Also, $x^3 + x^3 + \cdots + x_n^3 = -p + r + 8s = 19 + 6s$, so $\frac{M}{m} = \frac{19 + 6 \times 19}{19 + 6... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 7 (17th IMO problem) When $4444^{444}$ is written as a decimal number, the sum of its digits is $A, B$ is the sum of the digits of $A$. Find the sum of the digits of $B$.
| Solution: First, every positive decimal integer $n=a_{0} 10^{k}+a_{1} 10^{k-1}+\cdots+a_{k-1} \cdot 10+a_{k} \equiv a_{0}+a_{1}+\cdots+a_{k}(\bmod 9)$, which means every number is congruent to the sum of its digits modulo 9.
Secondly, since the number of digits in $4444^{4441}$ does not exceed $4 \times 4444=17776$, w... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.8 Let $m, n, p, q$ be non-negative integers, and for all $x>0$,
$$
\frac{(x+1)^{m}}{x^{n}}-1=\frac{(x+1)^{p}}{x^{q}}
$$
always holds, find the value of $\left(m^{2}+2 n+p\right)^{2 q}$. | [Solution] Since
$$
\frac{(x+1)^{m}}{x^{n}}-1=\frac{(x+1)^{p}}{x^{q}}
$$
holds for all $x>0$, we can take $x=1$, obtaining
$$
2^{m}-1=2^{p} \text {. }
$$
Since $m$ and $p$ are non-negative integers, $2^{m}-1$ is a positive odd number, so it can only be that $2^{p}=1, p=0$.
Thus,
$$
2^{m}-1=1, m=1 .
$$
Now, taking $x... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (5 points) A construction team installed 1200 streetlights in the first month, 1300 streetlights in the second month. At this point, there are still 500 streetlights left to be installed. The total number of installed streetlights is $\qquad$ times the number of uninstalled streetlights. | 【Analysis】First, calculate the total number of streetlights that have been installed, which is $1200+1300=2500$. Then, according to the meaning of division, the solution can be obtained.
【Solution】Solution: $(1200+1300) \div 500$
$$
\begin{array}{l}
=2500 \div 500 \\
=5 \text { (times); }
\end{array}
$$
Answer: The t... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The solution set of the inequality $(m-1) x<\sqrt{4 x-x^{2}}$ with respect to $x$ is $\{x \mid 0<x<2\}$. Find the value of the real number $m$.
---
Please note that the problem statement and the solution set provided are directly translated, maintaining the original format and structure. | Solve $4 x-x^{2} \geqslant 0 \Rightarrow 0 \leqslant x \leqslant 4$
The solution set of the original inequality is $\{x \mid 0<x<2\}$
Therefore, $m-1 \geqslant 0$ (if $m-1<0$, then the solution set of the inequality would include $\{x \mid 0<x<4\}$)
Squaring both sides of the inequality gives: $(m-1)^{2} x^{2}<4 x-x^{2... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9. The integers $x_{0}, x_{1}, \cdots, x_{2000}$ satisfy the conditions, $x_{0}=0,\left|x_{1}\right|=\left|x_{0}+1\right|,\left|x_{2}\right|=\left|x_{1}+1\right|, \cdots$, $\left|x_{2004}\right|=\left|x_{2003}+1\right|$, find the minimum value of $\left|x_{1}+x_{2}+\cdots+x_{2004}\right|$. | 9. From the known, we have
$$
\left\{\begin{array}{l}
x_{1}^{2}=x_{3}^{2}+2 x_{0}+1, \\
x_{2}^{2}=x_{1}^{2}+2 x_{1}+1, \\
\cdots \cdots \\
x_{2004}^{2}=x_{2003}^{2}+2 x_{2003}+1 .
\end{array}\right.
$$
Thus, $x_{2004}^{2}=x^{2}+2\left(x_{0}+x_{1}+\cdots+x_{2003}\right)+2004$. Given $x_{0}=0$, then $2\left(x_{1}+x_{2}+... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (10 points) In a $3 \times 3$ grid (each cell is a $1 \times 1$ square), place two identical pieces, with at most one piece per cell, there are $\qquad$ different ways to place them. (If two placements can be made to coincide by rotation, they are considered the same placement). | 【Analysis】We can discuss the problem by dividing it into different cases. The four vertices have the same position value, the square right in the middle has one position value, and the remaining four squares have the same position value. Therefore, we can calculate the different placement methods for each of the three ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) As shown in the figure, in the square $A B C D$ with a side length of 12 cm, an isosceles triangle $P A B$ is constructed with $A B$ as the base and a leg length of 10 cm. Then the area of triangle $P A C$ is $\qquad$ square centimeters. | 【Analysis】Draw $P E \perp A B, P F \perp B C$, the area of triangle $P A C$ = the area of triangle $P A B$ + the area of triangle $P B C$ - the area of triangle $A B C$.
【Solution】Solution: Since $10^{2}-(12 \div 2)^{2}=64$, then $P E=8, P F=6$,
Therefore, the area of triangle $P A B$ is: $12 \times 8 \div 2=48$ (squar... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Let set $A=\left\{a-1,2 \log _{2} b\right\}$ and $B=\left\{a+1, \log _{2}(16 b-64)\right\}$ have exactly one common element which is $a$, then the real number $a=$ | Ronggou 6.
Analysis: Since $a-1 \neq a, a+1 \neq a$, the common element is $2 \log _{2} b=\log _{2}(16 b-64)$, solving this gives $b=8, a=2 \log _{2} b=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Let the sequence $\left\{a_{n}\right\}$ satisfy:
$$
\begin{array}{l}
a_{1}=\frac{1}{4}, a_{n+1}=a_{n}+a_{n}^{2}\left(n \in \mathbf{Z}_{+}\right) . \\
\text {Let } T_{2020}=\frac{1}{a_{1}+1}+\frac{1}{a_{2}+1}+\cdots+\frac{1}{a_{2020}+1} .
\end{array}
$$
If the value of $T_{2020}$ lies in the interval $(k, k+1)$, the... | 8.3.
From the problem, we know
$$
\begin{array}{l}
\frac{1}{a_{n+1}}=\frac{1}{a_{n}\left(a_{n}+1\right)}=\frac{1}{a_{n}}-\frac{1}{a_{n}+1} \\
\Rightarrow \frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}} \\
\Rightarrow T_{2020}=\frac{1}{a_{1}}-\frac{1}{a_{2}}+\frac{1}{a_{2}}-\frac{1}{a_{3}}+\cdots+\frac{1}{a_{2020}}... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) There are three natural numbers, their sum is 2015, the sums of each pair are $m+1, m+2011$ and $m+2012$, then $m=$ $\qquad$ | 2. (5 points) There are three natural numbers, their sum is 2015, the sums of each pair are $m+1, m+2011$ and $m+2012$, then $m=$ $\qquad$ .
【Solution】Solution: According to the problem, we have:
$$
\begin{array}{r}
m+1+m+2011+m+2012=2015 \times 2 \\
3 m+4024=4030 \\
3 m=6 \\
m=2
\end{array}
$$
Therefore, the answer ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (1999 Henan Province Competition Team) Simplify $\frac{\sqrt{1-2 \sin 20^{\circ} \cos 20^{\circ}}}{\cos 20^{\circ}-\sqrt{1-\cos ^{2} 160^{\circ}}}$ to get ( ).
A. $\sqrt{1-\sin 40^{\circ}}$
B. $\frac{1}{\cos 20^{\circ}-\sin 20^{\circ}}$
C. 1
D. -1 | 1. Choose C. Reason: Original expression $=\frac{\sqrt{\cos ^{2} 20^{\circ}+\sin ^{2} 20^{\circ}-2 \sin 20^{\circ} \cos 20^{\circ}}}{\cos 20^{\circ}-\sqrt{1-\cos ^{2} 20^{\circ}}}=\frac{\cos 20^{\circ}-\sin 20^{\circ}}{\cos 20^{\circ}-\sin 20^{\circ}}=1$. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
11. (This question is worth 20 points) Let the ellipse $C_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have left and right foci $F_{1}$ and $F_{2}$, respectively, and the right vertex be $A$. Let $P$ be any point on the ellipse $C_{1}$, and the maximum value of $\overrightarrow{P F_{1}} \cdot \overrightarrow{... | Solution: (1) Let $P(x, y)$, then $\overrightarrow{P F_{1}}=(-c-x,-y), \overrightarrow{P F_{2}}=(c-x,-y)$.
$\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=x^{2}+y^{2}-c^{2}$. Also, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, so $y^{2}=b^{2}-\frac{b^{2} x^{2}}{a^{2}}, 0 \leq x^{2} \leq a^{2}$.
Therefore, $\ove... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Given that $f(x)$ is a linear function, $f[f(1)]=-1$, the image of $f(x)$ symmetric to the line $x-y=0$ is $C$, if the point $\left(n, \frac{a_{n+1}}{a_{n}}\right)\left(n \in \mathbf{N}^{\cdot}\right)$ lies on the curve $C$, and $a_{1}=1, \frac{a_{n+1}}{a_{n}}-\frac{a_{n}}{a_{n-1}}=1(n \geqslant 2)$.
(1) Find the an... | 5. Solution (1) Let $f(x)=k x+b(k \neq 0), \therefore f[f(1)]=k^{2}+k b+b=-1$. (1) Since the image of $f(x)$ is symmetric to the line $x-y=0$ as $C, \therefore$ the curve $C$ is: $f^{-1}(x)=\frac{x}{k}-\frac{b}{k}, \therefore f^{-1}(n)=\frac{n}{k}-\frac{b}{k}, f^{-1}(n-1)=\frac{n-1}{k}-\frac{b}{k}, f^{-1}(n)-f^{-1}(n-1... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. (10 points) In a sphere with a radius of 10 cm, there is a cube with integer (cm) edge lengths. What is the maximum edge length of the cube?
untranslated part:
设半径为 10 厘米的球中有一个棱长为整数 (厘米) 的正方体, 则该正方体的棱长最大等于多少？
translated part:
In a sphere with a radius of 10 cm, there is a cube with integer (cm) edge lengths. Wha... | 【Analysis】Draw the figure, see the figure below, the vertices of the inscribed cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ in the sphere are on the sphere's surface, its (body) diagonal $A C_{1}$ is the diameter of the sphere, i.e., $A C_{1}=2 \times 10=20$ (cm). Then, by the symmetry of the figure and the Pythagorean theor... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
19. If the equation $\frac{5}{x+1}+\frac{a x}{x-3}=\frac{7}{x-3}+1$ has an extraneous root, then $3 a+2=$ | $9$ | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
61. As shown in the figure, three equilateral triangles: $\triangle A B C, \triangle H F G, \triangle D C E$ are placed on the line $m$. It is known that $B C=\frac{1}{3} C E, F, G$ are the midpoints of $B C, C E$ respectively, $F M / / A C, G N / / D C$. Let the areas of the three parallelograms in the figure be $S_{1... | Answer: 3
Solution: According to the properties of an equilateral triangle, we have
$$
\angle A B C=\angle H F G=\angle D C E=60^{\circ} \text {, }
$$
Therefore,
$$
A B / / H F / / D C / / G N,
$$
Let $A C$ intersect $F H$ at $P$, and $C D$ intersect $H G$ at $Q$,
then $\triangle M B F, \triangle P F C, \triangle Q C... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
14. Let the complex numbers $z_{1}, z_{2}$ satisfy
$$
\begin{array}{l}
\left|z_{1}\right|=\left|z_{2}\right|=1 \text {, and } z_{1}+z_{2}-1=0 \text {. } \\
\text { Then }\left(z_{1}-z_{2}\right)^{2}=
\end{array}
$$ | 14. -3 .
From the problem, we know that the points corresponding to $z_{1}$, $z_{2}$, and $-1$ are the three equally divided points on the unit circle. Without loss of generality, let's assume the point corresponding to $z_{1}$ is in the first quadrant. Then,
$$
\begin{array}{l}
z_{1}=\cos \frac{\pi}{3}+\mathrm{i} \si... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16. (15 points) Let $x, y, z \geqslant 0$, and at most one of them is 0. Find
$$
f(x, y, z)=\sum \sqrt{\frac{x^{2}+256 y z}{y^{2}+z^{2}}}
$$
the minimum value, where $\sum$ denotes the cyclic sum. | 16. Let $x \geqslant y \geqslant z$.
(1) When $256 y^{3} \geqslant x^{2} z$, we have
$$
\begin{array}{l}
\frac{x^{2}+256 y z}{y^{2}+z^{2}}-\frac{x^{2}}{y^{2}}=\frac{z\left(256 y^{3}-x^{2} z\right)}{\left(y^{2}+z^{2}\right) y^{2}} \geqslant 0, \\
\frac{y^{2}+256 z x}{z^{2}+x^{2}}-\frac{y^{2}}{x^{2}}=\frac{z\left(256 x^{... | 12 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
10. The largest prime $p$ such that $\frac{p+1}{2}$ and $\frac{p^{2}+1}{2}$ are both perfect squares is . $\qquad$ | 10.7.
Let $\frac{p+1}{2}=x^{2} ， \frac{p^{2}+1}{2}=y^{2}\left(x, y \in \mathbf{Z}_{+}\right)$.
Obviously, $p>y>x, p>2$.
From $p+1=2 x^{2}, p^{2}+1=2 y^{2}$, subtracting the two equations gives $p(p-1)=2(y-x)(y+x)$.
Since $p(p>2)$ is a prime number and $p>y-x$, then $p \mid(y+x)$.
Because $2 p>y+x$, so $p=y+x$.
Thus, $... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. In the Cartesian coordinate system $X O Y$, given two points $M(-1,2)$ and $N(1,4)$, point $P$ moves on the $X$-axis. When $\angle M P N$ takes its maximum value, the x-coordinate of point $P$ is $\qquad$ | 1
12.【Analysis and Solution】The center of the circle passing through points $M$ and $N$ lies on the perpendicular bisector of segment $M N$, which is the line $y=3-x$. Let the center of the circle be $S(a, 3-a)$, then the equation of circle $S$ is: $(x-a)^{2}+(y-3+a)^{2}=2\left(1+a^{2}\right)$.
For a chord of fixed le... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$5 \cdot 24$ Let $M$ be the set of all polynomials of the form
$P(x)=a x^{3}+b x^{2}+c x+d, a, b, c, d \in R$, and for which $|P(x)| \leqslant 1$ when $x \in[-1,1]$. Prove: There must be some number $k$, such that for all $P(x) \in M$, we have $|a| \leqslant k$, and find the smallest $k$. | [Proof] Since the polynomial $P_{0}(x)=4 x^{3}-3 x$ satisfies $P_{0}(-1)=-1, P_{0}(1)$ $=1$, and at its critical points,
$$
P\left(-\frac{1}{2}\right)=1, P\left(\frac{1}{2}\right)=-1,
$$
we have $P_{0}(x) \in M$. Now we prove that for any $P(x) \in M$, $|a| \leqslant 4$. Otherwise, suppose there exists a polynomial
$$... | 4 | Algebra | proof | Yes | Yes | olympiads | false |
12. If $x \in \mathbf{N}^{*}$, and $x^{5}+x+1$ is a prime number, then $x=$ | 12. $x=1$
Since $\omega^{5}+\omega+1=0,\left(\omega^{2}\right)^{5}+\omega^{2}+1=0$ (where $\omega$ is a cube root of 1), it follows that $x^{2}+x+1 \mid x^{5}+x+1$, so $x=1$ | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Let the 9-element set $A=\{a+b \mathrm{i} \mid a, b \in\{1,2,3\}\}$, where $\mathrm{i}$ is the imaginary unit. $\alpha=\left(z_{1}, z_{2}, \cdots, z_{9}\right)$ is a permutation of all elements in $A$, satisfying $\left|z_{1}\right| \leqslant\left|z_{2}\right| \leqslant \cdots \leqslant\left|z_{9}\right|$, then the ... | According to the problem, $\left|z_{1}\right|<\left|z_{2}\right|=\left|z_{3}\right|<\left|z_{4}\right|<\left|z_{5}\right|=\left|z_{6}\right|<\left|z_{7}\right|=\left|z_{8}\right|<\left|z_{9}\right|$, thus the positions of $z_{2}, z_{3} ; z_{5}, z_{6} ; z_{7}, z_{8}$ can be swapped, so the number of such arrangements $\... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. "虚有其表", "表里如一", "一见如故", "故弄玄虚" four idioms, each Chinese character represents one of 11 consecutive non-zero natural numbers, the same character represents the same number, different characters represent different numbers, and "表" > "一" > "故" > "如" > "虚", and the sum of the four characters in each idiom is 21, then... | 【Answer】9
【Analysis】For convenience of expression, let these eleven numbers be $A / B / C / D / E / F / G / H / I / J / K$ respectively, and the total sum be $S$, then, "nong" is “$J$”, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
A+B+C+D=21 \\
D+E+F+G=21 \\
G+H+F+I=21 \\
I+J+K+A=21 \\
D>G>I>F>A
\end{array}\righ... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Given $x \Delta y=\frac{x+y}{x y}$, then $(1 \Delta 2) \Delta 3=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | \begin{tabular}{l}
5 \\
\hline 1
\end{tabular} | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
19 Given the parabola $y^{2}=2 p x(p>0)$ with two moving points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, and $O$ as the origin, $O A \perp O B$.
(1) Find the equation of the trajectory $C$ of the midpoint of segment $A B$;
(2) If the distance from a point on $C$ to the line $x-2 y+2 \sqrt{5}-p=0$ ... | (19) Let $y_{1}^{2}=2 p x_{1}, y_{2}^{2}=2 p x_{2}$, then $x_{1} x_{2}=\frac{1}{4 p^{2}}\left(y_{1} y_{2}\right)^{2}$. Since $O A \perp O B$, we have $x_{1} x_{2}+y_{1} y_{2}=0$. Therefore, $y_{1} y_{2}+\frac{1}{4 p^{2}}\left(y_{1} y_{2}\right)^{2}=0$, which implies $y_{1} y_{2}=-4 p^{2}$.
(1) Let the midpoint of $A B$... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9-46 Let real numbers $a, b, c, d$ satisfy $a^{2}+b^{2}+c^{2}+d^{2} \leqslant 1$, find
$$
S=(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4}+(c+d)^{4}
$$
the maximum value. | [Solution]
$$
\begin{aligned}
S \leqslant & (a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4} \\
& +(c+d)^{4}+(a-b)^{4}+(a-c)^{4}+(a-d)^{4}+(b- \\
& c)^{4}+(b-d)^{4}+(c-d)^{4} \\
= & 6\left(a^{4}+b^{4}+c^{4}+d^{4}+2 a^{2} b^{2}+2 a^{2} c^{2}+2 a^{2} d^{2}+2 b^{2} c^{2}\right. \\
& \left.+2 b^{2} d^{2}+2 c^{2} d^{2}\rig... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. The number of real solutions to the equation $x=10 \sin x$ is
The text above has been translated into English, preserving the original text's line breaks and format. | 3. 7 .
Plotting the graphs of $y=x$ and $y=10 \sin x$ reveals that the two graphs intersect at 7 points, the x-coordinates of which are the solutions to the equation. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Given that the three vertices $A, B, C$ of the right triangle $\triangle A B C$ are all on the parabola $y=x^{2}$, and the hypotenuse $A B$ is parallel to the $x$-axis, then the altitude $C D$ from $C$ to $A B$ equals $\qquad$ . | Let $A\left(-s, s^{2}\right), B\left(s, s^{2}\right), C\left(t, t^{2}\right)$, then $\overrightarrow{A C} \cdot \overrightarrow{B C}=(t+s)(t-s)+\left(t^{2}-s^{2}\right)\left(t^{2}-s^{2}\right)$ $=\left(t^{2}-s^{2}\right)\left(t^{2}-s^{2}+1\right)=0 \Rightarrow t^{2}=s^{2}-1 \Rightarrow y_{C}=y_{A}-1$.
Therefore, the al... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. There are 1993 boxes arranged in a row from left to right, each containing some small balls. If the leftmost box contains 7 balls, and every four adjacent boxes contain a total of 30 balls, then the rightmost box contains $\qquad$ balls. | 2. 7
Let the number of balls in the small boxes from left to right be
$$
\begin{array}{c}
7, a_{2}, a_{3}, \cdots, a_{1993}, \\
\because \quad 7+a_{2}+a_{3}+a_{4}=30, a_{2}+a_{3}+a_{4}+a_{5}=30, \\
\therefore \quad a_{5}=7 .
\end{array}
$$
Similarly, $a_{9}=a_{13}=a_{17}=\cdots=a_{4 k+1}=a_{1993}=7$. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 13 (2004 Western China Mathematical Olympiad) Let $n \in \mathbf{N}_{+}$, and let $d(n)$ denote the number of all positive divisors of $n$, and $\varphi(n)$ denote the number of integers in $1,2, \cdots, n$ that are coprime to $n$. Find all non-negative integers $c$ such that there exists a positive integer $n$... | Let the set of all positive divisors of $n$ be $A$, and the set of numbers in $1,2, \cdots, n$ that are coprime with $n$ be $B$. Since there is exactly one number $1 \in A \cap B$ in $1,2, \cdots, n$, we have $d(n)+\varphi(n) \leqslant n+1$, hence $c=0$ or 1.
(1) When $c=0$, from $d(n)+\varphi(n)=n$ we know that there ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 6 How many positive real numbers $x$ satisfy the equation $x[x[x[x]]]=2006$? | Assume there exists a positive real number $x$ satisfying $x[x[x[x]]]=2006$.
If $02006$.
From this, we know $6<x<7,[x]=6$.
Thus, $2006=x[x[6 x]]$.
But from $x<7$, we know $6 x<42$, so $[6 x] \leqslant 41$.
Therefore, $x[6 x]<7 \times 41=287$.
So, $[x[6 x]] \leqslant 286$.
Hence, $x[x[6 x]]<7 \times 286=2002<2006$, cont... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$10 \cdot 75$ product $\left(1+\frac{1}{1 \cdot 3}\right)\left(1+\frac{1}{2 \cdot 4}\right)\left(1+\frac{1}{3 \cdot 5}\right)\left(1+\frac{1}{4 \cdot 6}\right) \cdots$ $\left(1+\frac{1}{98 \cdot 100}\right)\left(1+\frac{1}{99 \cdot 101}\right)$ has an integer part of
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(China Guangzhou, Wu... | [Solution] Notice that $1+\frac{1}{k(k+2)}=\frac{k^{2}+2 k+1}{k(k+2)}=\frac{(k+1)^{2}}{k(k+2)}$, then $\begin{aligned} \text { original expression } & =\frac{2^{2}}{1 \cdot 3} \cdot \frac{3^{2}}{2 \cdot 4} \cdot \frac{4^{2}}{3 \cdot 5} \cdot \frac{5^{2}}{4 \cdot 6} \cdots \cdot \frac{99^{2}}{98 \cdot 100} \cdot \frac{1... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. Among the real numbers $99, \frac{\sqrt{3}}{2}, 0, \sqrt{2}-1, \sqrt{64}, \sqrt[3]{25}, \pi, \frac{7}{9}$, there are $\qquad$ irrational numbers. | $4$ | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Given the set $M=\{(a, b) \mid a \leqslant-1, b \leqslant m\}$. If for any $(a, b) \in M$, it always holds that $a \cdot 2^{b}-b-3 a \geqslant 0$, then the maximum value of the real number $m$ is $\qquad$ | 3. 1 .
Notice that,
$$
a \cdot 2^{b}-b-3 a \geqslant 0 \Leftrightarrow\left(2^{b}-3\right) a-b \geqslant 0
$$
holds for any $a \leqslant-1$.
$$
\text { Then }\left\{\begin{array}{l}
2^{b}-3 \leqslant 0, \\
2^{b}+b \leqslant 3
\end{array} \Rightarrow b \leqslant 1\right. \text {. }
$$ | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
$15 \cdot 228$ students each choose at least one of English, Mathematics, and History. Among them, the number of students who choose both Mathematics and English but not History is equal to the number of students who choose only Mathematics. There are no students who choose only English or only History, 6 students choo... | ［Solution］Let the number of people who choose both English and Math be $n_{1}$, and the number of people who choose all three subjects be $n_{2}$, where $n_{1} \geqslant 0, n_{2}>0$ and $n_{2}$ is even.
From the right diagram, we can see: $2 n_{1}+6 n_{2}+6=28$, that is,
$$
n_{1}+3 n_{2}=11 \text {. }
$$
The only solu... | 5 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
2.3.12 There are $n$ teams that need to hold a double round-robin home and away match (each pair of teams plays two matches, one home match for each team). Each team can play multiple away matches within a week (from Sunday to Saturday). However, if a team has a home match in a certain week, no away matches for that te... | (1) As shown in the table on the right: The " $\times$ " in the table indicates that the team has a home game in that week and cannot visit. It is easy to verify that, according to the schedule in the table, the six teams can complete the competition in four weeks.
(2) The following proves that seven teams cannot compl... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(6) If $7n+1$ and $8n+1$ can both be expressed as the sum of three distinct positive integers in a geometric progression, then the minimum value of the positive integer $n$ is $\qquad$. | 6 6 Hint: First, when $n=6$, $7 n+1=43=1+6+6^{2}, 8 n+1=$ $49=3^{2}+3 \cdot 5+5^{2}$ are both the sums of three positive integers in geometric progression. When $n \leqslant 5$, it is easy to see that
$$
\begin{array}{l}
7 \cdot 1+1=8,8 \cdot 2+1=17,8 \cdot 3+1=25, \\
7 \cdot 4+1=29,8 \cdot 5+1=41
\end{array}
$$
canno... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Think about Question 1 The sequence $a_{1}, a_{2}, \cdots$ is defined as follows: $a_{n}=2^{n}+3^{n}+6^{n}-1, n=1,2,3, \cdots$ Find all positive integers that are coprime to every term of this sequence. | Solution 1: We first prove the following conclusion: For any prime $p$ not less than 5, we have
$$
2^{p-2}+3^{p-2}+6^{p-2}-1 \equiv 0(\bmod p)
$$
Since $p$ is a prime not less than 5, we have $(2, p)=1,(3, p)=1,(6, p)=1$. By Fermat's Little Theorem, $2^{p-1} \equiv 1(\bmod p), 3^{p-1} \equiv 1(\bmod p), 6^{p-1} \equiv... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Suppose the positive integer $n$ can be expressed as the sum of the reciprocals of four distinct positive integers. Then the number of such $n$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(2011, Fudan University Independent Recruitment Examination) | Let $n=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$, and $a<b<c<d$ $\left(a, b, c, d \in \mathbf{Z}_{+}\right)$.
If $a \geqslant 3$, then $b \geqslant 4, c \geqslant 5, d \geqslant 6$.
Thus, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$
$$
\leqslant \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{19}{20}<... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
6. Let $A, B$ be points on the graph of the function $f(x)=3-x^{2}$ on either side of the $y$-axis, then the minimum value of the area of the region enclosed by the tangent lines of $f(x)$ at points $A, B$ and the $x$-axis is $\qquad$ . | Given $A\left(a, 3-a^{2}\right), B\left(-b, 3-b^{2}\right)$,
where $a, b>0$, then the tangent line $A D$:
$$
\begin{array}{l}
y=-2 a(x-a)+3-a^{2} \\
\Rightarrow y=-2 a x+a^{2}+3,
\end{array}
$$
Similarly, we get $B E: y=2 b x+b^{2}+3$.
Thus, $D\left(\frac{a^{2}+3}{2 a}, 0\right), E\left(-\frac{b^{2}+3}{2 b}, 0\right)$... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
4240 ** Let $a, b \in [0,1]$, find the maximum and minimum values of $S=\frac{a}{1+b}+\frac{b}{1+a}+(1-a)(1-b)$. | Parse Because
$$
\begin{aligned}
S & =\frac{a}{1+b}+\frac{b}{1+a}+(1-a)(1-b)=\frac{1+a+b+a^{2} b^{2}}{(1+a)(1+b)} \\
& =1-\frac{a b(1-a b)}{(1+a)(1+b)} \leqslant 1,
\end{aligned}
$$
When $a b=0$ or $a b=1$, the equality holds, so the maximum value of $S$ is 1.
Let $T=\frac{a b(1-a b)}{(1+a)(1+b)}, x=\sqrt{a b}$, then
... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5 Find the smallest positive integer $n$, such that there exist $n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}$ simultaneously satisfying the following 3 conditions:
(1) The least common multiple of $a_{1}, a_{2}, \cdots, a_{n}$ is 2002;
(2) For any $1 \leqslant i<j \leqslant n, a_{i}$ and $a_{j}$ ... | Since $2002=2 \times 7 \times 11 \times 13$, and $a_{i}$ is a divisor of 2002 $(i=1,2, \cdots, n)$, we have
$$
a_{i}=2^{\alpha_{i}} 7^{\beta_{i}} 11^{\gamma_{i}} 13^{\delta_{i}}(i=1,2, \cdots, n),
$$
where $\alpha_{i}, \beta_{i}, \gamma_{i}, \delta_{i} \in\{0,1\}$.
Let $a_{1} a_{2} \cdots a_{n}=k^{2}$. Since $a_{i} \m... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, the sum of the first $n$ terms is $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is
(A) 5 ;
(B) 6 ;
(C) 7 ;
(D) 8 . | 3. (C)
From $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right), a_{1}-1=8$, we know that the sequence $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$. Therefore,
$$
\begin{array}{l}
S_{n}-n=\frac{8 \times\left[1-\left(1-\frac{1}{3}\right)^{n}\right]}{1+\frac{1}{3}}=6-... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. If two congruent regular tetrahedrons are glued together by overlapping their bases, they form a hexahedron with all dihedral angles equal. If the shortest edge length of this hexahedron is 2, then the distance between the farthest two vertices is $\qquad$ . | 5. 3 As shown in the figure, the hexahedron $A-C D F-B$, let the side length of $\triangle C D F$ be $2 a$, and the length of edge $A D$ be $b$. Construct $C E \perp A D$ at $E$, connect $E F$, it is easy to know that $E F \perp A D$, thus $\angle C E F$ is the plane angle of the dihedral angle formed by faces $A D F$ ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The two sliders $\mathrm{A}, \mathrm{B}$ in the right figure are connected by a rod and can slide on vertical and horizontal tracks, respectively. Initially, slider $\mathrm{A}$ is 20 cm from point 0, and slider B is 15 cm from point 0. Question: When slider $\mathrm{A}$ slides down to point 0, how many centimeters has... | 22. Answer: 10.
Analysis: From $\mathrm{AB}^{2}=A 0^{2}+O B^{2}=20^{2}+15^{2}=25^{2}$, we know that the length of the connecting rod is 25 centimeters. When slider $\mathrm{A}$ slides down to point 0, the distance of slider $\mathrm{B}$ from point 0 is 25 centimeters, so slider B has moved $25-15=10$ (centimeters). | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$17 \cdot 85 \triangle P Q R$ has sides $P Q, P R$ of lengths 4 and 7, respectively, and the median $P M$ has a length of $3 \frac{1}{2}$. Then the length of $Q R$ is
(A) 6.
(B) 7.
(C) 8.
(D) 9.
(E) 10.
(15th American High School Mathematics Examination, 1964) | [Solution] Let $M R=y$. Draw $P H \perp Q R$ at $H$, and let $H M=x$. By the Pythagorean theorem,
$$
16-(y-x)^{2}=\left(3 \frac{1}{2}\right)^{2}-x^{2}
$$
which simplifies to $(y-x)^{2}-x^{2}=\frac{15}{4}$, or $y^{2}-2 x y=\frac{15}{4}$.
Also, $16-(y-x)^{2}=49-(x+y)^{2}$,
which simplifies to $(y+x)^{2}-(y-x)^{2}=33$, ... | 9 | Geometry | MCQ | Yes | Yes | olympiads | false |
$1 \cdot 36$ If $x^{2}+x-1=0$, then $2 x^{3}+3 x^{2}-x$ equals
(A) 0.
(B) 1.
(C) -1.
(D) Cannot be determined.
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990) | [Solution] Given $2 x^{3}+3 x^{2}-x=(2 x+1)\left(x^{2}+x-1\right)+1$,
then according to the problem, we have
$$
2 x^{3}+3 x^{2}-x=1 \text {. }
$$
Therefore, the correct choice is $(B)$. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. $a, b$ are positive integers, satisfying: $\frac{1}{a}-\frac{1}{b}=\frac{1}{2018}$, then the number of all positive integer pairs $(a, b)$ is $\qquad$ | Answer 4.
Analysis From $\frac{1}{a}-\frac{1}{b}=\frac{1}{2018}$, we know $1 \leq a2018$, since 1009 is a prime number, the possible factorizations of the number $2^{2} \cdot 1009^{2}$ are:
$$
1 \times 2018^{2}, 2 \times(2 \cdot 1009^{2}), 4 \times 1009^{2}, 1009 \times(4 \cdot 1009)
$$
Each of these factorizations co... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6 Given the sequence $\left\{a_{n}\right\}$ satisfies the recurrence relation $a_{n+1}=2 a_{n}+2^{n}-1(n \in$
$\left.\mathbf{N}^{*}\right)$, and $\left\{\frac{a_{n}+\lambda}{2^{n}}\right\}$ is an arithmetic sequence, then the value of $\lambda$ is $\qquad$. | $6-1$ Hint: Since
$$
\begin{aligned}
\frac{a_{n+1}+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} & =\frac{2 a_{n}+2^{n}-1+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
& =\frac{2^{n}-1-\lambda}{2^{n+1}},
\end{aligned}
$$
from $\frac{2^{n}-1-\lambda}{2^{n+1}}$ being a constant, we know $\lambda=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
39. As shown in the figure, $P A=P B, \angle A P B=2 \angle A C B, A C$ intersects $P B$ at point $D$, and $P B=4, P D=3$, then $A D \cdot D C=$ . $\qquad$ | answer: 7 | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. Find the maximum and minimum values of the function $y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$.
untranslated portion:
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11. Find the maximum and minimum values of the function $y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$. | Let $f(x)=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}, x \in[0,13]$,
then $f^{\prime}(x)=\frac{1}{2 \sqrt{x+27}}+\frac{1}{2 \sqrt{x}}-\frac{1}{2 \sqrt{13-x}}$. Since the function $y=\frac{1}{\sqrt{x+27}}+\frac{1}{\sqrt{x}}$ is monotonically decreasing on $(0,13)$, and the function $y=\frac{1}{\sqrt{13-x}}$ is monotonically increa... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
28.2.9 * Find the smallest positive integer $n(n \geqslant 3)$, such that in any set of $n$ points in the plane with no three points collinear, there are two points that are vertices of a non-isosceles triangle. | First, if 6 points in a plane are the 5 vertices and the center of a regular pentagon, all triangles formed by these 6 points are isosceles triangles. Therefore, the required positive integer \( n \geqslant 7 \).
Second, if there exist 7 points in a plane (where no three points are collinear), such that all triangles ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. If the positive real numbers $x, y$ satisfy $x-2 \sqrt{y}=\sqrt{2 x-y}$, then the maximum value of $x$ is $\qquad$ . | 2. 10 .
Let $a=\sqrt{2 x-y}, b=\sqrt{y}$. Then $a^{2}+b^{2}=2 x, x=a+2 b$.
Also, $(a-1)^{2}+(b-2)^{2}=5(a, b \geqslant 0)$, so $(2 x)_{\max }=(\sqrt{5}+\sqrt{5})^{2}=20$.
Therefore, the maximum value of $x$ is 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(2) Given $z=(\sqrt{3}-3 \mathrm{i})^{n}$, if $z$ is a real number, then the smallest positive integer $n$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | (2) A Hint: $z=(\sqrt{3}-3 \mathrm{i})^{n}=(-2 \sqrt{3})^{n}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)^{n}, n=3$ is the smallest positive integer that makes $z$ a real number. | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. Given $x \in\left(0, \frac{\pi}{2}\right)$, then the integer part of the number $M=3^{\cos ^{2} x}+3^{\sin ^{3} x}$ is
A. 2
B. 3
C. 4
D. Cannot be determined | 4. B Let $t=\sin ^{2} x$, then $t \in(0,1), m=3^{t}+3^{1-t}=3^{t}+\frac{3}{3^{t}} \cdot 3^{t} \in(1,3), 2 \sqrt{3} \leqslant M<4$. Therefore, $[M]=3$. | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
13. In $\triangle A B C$, $\angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively. Let
$$
\begin{array}{l}
f(x)=\boldsymbol{m} \cdot \boldsymbol{n}, \boldsymbol{m}=(2 \cos x, 1), \\
\boldsymbol{n}=(\cos x, \sqrt{3} \sin 2 x), \\
f(A)=2, b=1, S_{\triangle A B C}=\frac{\sqrt{3}}{2} . \\
\text { Then }... | II. 13.2.
It is known that $f(x)=1+2 \sin \left(2 x+\frac{\pi}{6}\right)$.
From the given conditions, we have $\angle A=\frac{\pi}{3}, c=2, a=\sqrt{3}$. Therefore, $\frac{a}{\sin A}=\frac{b+c}{\sin B+\sin C}=2$. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) In a deck of cards, $J, Q, K$ are considered as 11, 12, 13 points respectively. From the 13 cards numbered 1 to 13, what is the maximum number of cards you can pick such that there are no 2 pairs of cards where the sum of the points of one pair equals the sum of the points of the other pair? | 【Analysis】First, get the 13 cards from 1 to 13 points, which have differences of $1 \sim 12$. Further, we get that the equation $2 a_{n}=a_{m}+a_{l}$ $(m>n>l)$ appears at least 5 times, which is equivalent to selecting these 5 groups from 21 groups, leaving 16 groups. Then, according to the pigeonhole principle, we can... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
13. A die has 6 faces, each marked with the numbers $1, 2, 3, 4, 5, 6$. After each roll of the die, the number on the top face is recorded. The rolling process ends as soon as any number appears three times. Xiao Ming rolled the die 12 times, and his rolling process ended. The sum of all recorded numbers is 47. What wa... | 【Analysis】 $3+2 \times 4+1=12$ times, and because $1+2+3+4+5+6=21, 21 \times 2=42$, it indicates that the number appearing three times is 5 more than the number appearing once, which is $47-42=5$. Among these six numbers, it can only be $6-1=5$. Therefore, the number that appears three times, which is also the last num... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Hao Yu wrote three letters and three envelopes, and he needs to put each letter into the corresponding envelope, with only one letter in each envelope. The number of all possible scenarios where at least one letter is placed incorrectly is $\qquad$ kinds. | Counting.
(Method One)
Assume letters $A, B, C$ correspond to envelopes $a, b, c$;
(1) Only one letter is placed incorrectly: does not exist;
(2) Only two letters are placed incorrectly: there are 3 ways;
(1) Letter $A$ is placed in envelope $b$, letter $B$ is placed in envelope $a$, letter $C$ is placed in envelope $c... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. The smallest positive integer $n$ that satisfies $I=\left(\frac{1}{2}+\frac{1}{2 \sqrt{3}} \mathrm{i}\right)^{n}$ being a pure imaginary number is $n=$ . $\qquad$ | 9.3.
$$
\begin{array}{l}
\text { Given } I=\left(\frac{1}{2}+\frac{1}{2 \sqrt{3}} \mathrm{i}\right)^{n} \\
=\left(\frac{1}{\sqrt{3}}\right)^{n}\left(\cos \frac{n \pi}{6}+\mathrm{i} \sin \frac{n \pi}{6}\right)
\end{array}
$$
is a pure imaginary number, we know the smallest value of $n$ is 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=-2 \text {, }
$$
and $S_{n}=\frac{3}{2} a_{n}+n$ ( $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$).
$f(x)$ is an odd function defined on $\mathbf{R}$, and satisfies
$$
f(2-x)=f(x) \text {. }
$$
Then $f\left(a_{2021}\rig... | 4.0.
Given that $f(x)$ is an odd function defined on $\mathbf{R}$ and satisfies $f(2-x)=f(x)$,
we know that $f(-x)=f(x+2)=-f(x)$, $f(x+4)=-f(x+2)=f(x)$.
Thus, the smallest positive period of $f(x)$ is 4.
Since the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=-2$, and
$$
S_{n}=\frac{3}{2} a_{n}+n,
$$
Therefore, wh... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) Convert the product $0.243 \times 0.325233$ to a decimal, the digit at the 2013th position after the decimal point is $\qquad$
| 【Analysis】Since $0.2^{4} 3=\frac{243}{999}=\frac{9}{37}, 0.32^{523} 3=\frac{325233-3}{999990}=\frac{10841}{33333}$, converting the product
$0.243 \times 0.325233$ to a fraction, we get the result, and then solve it based on the repeating decimal.
【Solution】Solution: $0.2^{4} \dot{3}=\frac{243}{999}=\frac{9}{37}, 0.32_{... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The real root of the equation $\sin x=\lg x$ is
(A) 1
(B) 2
(C) 3
(D) Greater than 3 | (C)
4. 【Analysis and Solution】The number of solutions to the equation $\sin x=\lg x$ is the number of intersection points between the sine curve $\sin x$ and the logarithmic curve $\lg x$.
First, determine the range of $x$. By the definition of logarithms,
$$
\begin{array}{l}
x>0 \text {, and } \sin x \leqslant 1, \\
\... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. Given that $f(x)$ is an odd function with a period of 5, $f(-3)=1$, $\tan \alpha=2$, $f(20 \sin \alpha \cdot \cos \alpha)=$ $\qquad$ . | 7. -1 .
$$
\begin{array}{c}
\tan \alpha=2, \sin \alpha=2 \cos \alpha, \text { and } \cos \alpha= \pm \frac{\sqrt{5}}{5}, 20 \sin \alpha \cos \alpha=40 \cos ^{2} \alpha=8 . \\
f(20 \sin \alpha \cos \alpha)=f(8)=f(3+5)=f(3)=-f(-3)=-1 .
\end{array}
$$ | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$3 \cdot 14$ Solve the system of equations
$$
2 y=x+\frac{17}{x}, \quad 2 z=y+\frac{17}{y}, \quad 2 w=z+\frac{17}{z}, \quad 2 x=w+\frac{17}{w}
$$
for the number of real solutions $(x, y, z, w)$.
(A) 1 .
(B) 2 .
(C) 4 .
(D) 8 .
(E) 16 .
(37th American High School Mathematics Examination, 1986) | ［Solution 1］From the problem, we have
$$
\left\{\begin{array}{l}
x^{2}-2 x y+17=0, \\
y^{2}-2 y z+17=0, \\
z^{2}-2 z w+17=0, \\
w^{2}-2 w x+17=0 .
\end{array}\right.
$$
By symmetry, it is easy to see that $x=y=z=w$. Substituting this into the above equations, we get
$$
x=y=z=w= \pm \sqrt{17} \text {. }
$$
Therefore, ... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
10. (20 points) As shown in Figure 1, in the Cartesian coordinate system $x C y$, the tangent line $l$ of a circle with center at the origin $C$ and radius 1 intersects the $x$-axis at point $N$ and the $y$-axis at point $M$. Point $A(3,4)$, and $\overrightarrow{A C}=$
$x \overrightarrow{A M}+y \overrightarrow{A N}$. L... | 10. (1) Let point $M(0, m), N(n, 0)$.
From $\overrightarrow{A C}=x \overrightarrow{A M}+y \overrightarrow{A N}$
$\Rightarrow\left\{\begin{array}{l}3 x+(3-n) y=3, \\ (4-m) x+4 y=4\end{array}\right.$
$\Rightarrow\left\{\begin{array}{l}m=\frac{4(x+y-1)}{x}, \\ n=\frac{3(x+y-1)}{y}\end{array}\right.$,
$\Rightarrow 1=\frac... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. $\odot O_{1} 、 \odot O_{2}$ are externally tangent to each other, their radii are 7 and 14 respectively, the circle $\odot \mathrm{O}_{3}$ that contains these two circles and is tangent to them has its center on their line of centers, the radius of the fourth circle that is tangent to all three circles is $\qquad$. | 6. 6 .
It is known that $\odot O_{4}$ is externally tangent to $\odot O_{1}$ and $\odot O_{2}$, and internally tangent to $\odot O_{3}$. Let the radius of $\odot O_{4}$ be $x$, and connect $O_{1} O_{4}$, $O_{2} O_{4}$, and $O_{3} O_{4}$. Then, $O_{1} O_{3}=14$, $O_{3} O_{2}=7$, $O_{1} O_{4}=7+x$, $O_{3} O_{4}=21-x$, a... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) Use 40 yuan to buy three types of exercise books priced at 2 yuan, 5 yuan, and 11 yuan each, with at least one of each type, and spend all the money. Then the number of different purchasing methods is $\qquad$ kinds. | 【Analysis】Subtract 1 from each type, leaving $40-2-5-11=22$ yuan. Then, based on the remaining money, divide and solve the problem.
【Solution】Solution: Subtract 1 from each type, leaving $40-2-5-11=22$ yuan.
If you buy 2 more 11-yuan books, it will be exactly used up, counting 1 method;
If you buy 1 more 11-yuan book,... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. (10 points) 42 matchsticks of the same length are arranged as shown in the figure. If each matchstick is considered as a line segment of length 1, then 38 triangles can be counted in the figure. If you want to ensure that no triangles remain in the figure, then at least $\qquad$ matchsticks need to be removed. | 【Answer】Solution:
Answer: At least 12 matchsticks need to be removed. Therefore, the answer is: 12. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Given a positive number $x$ satisfies
$$
x^{10}+x^{5}+\frac{1}{x^{5}}+\frac{1}{x^{10}}=15250 \text {. }
$$
Then the value of $x+\frac{1}{x}$ is . $\qquad$ | 2.3.
Let $a=x^{5}+\frac{1}{x^{5}}$. Then $x^{10}+\frac{1}{x^{10}}=a^{2}-2$.
Thus, the original equation becomes
$$
\begin{array}{l}
a^{2}+a-15252=0 \\
\Rightarrow(a-123)(a+124)=0
\end{array}
$$
$\Rightarrow a=123$ or $a=-124$ (discard).
Therefore, $x^{5}+\frac{1}{x^{5}}=123$.
Let $x+\frac{1}{x}=b>0$. Then
$$
\begin{ar... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
20. Patrick asks SpongeBob: "What is your lucky number?" SpongeBob says: "There is a sequence of numbers: $0,1,3,8$, $21,55,144,377,987, \cdots \cdots$ Starting from the third number, each number multiplied by 3 is exactly equal to the sum of the numbers immediately before and after it. My lucky number is the remainder... | $2$ | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let 7 be the first term and the common difference of an arithmetic sequence both be non-negative integers, the number of terms is no less than 3, and the sum of all terms is $97^{2}$. How many such sequences are there? | Let the first term of the arithmetic sequence be $a$, and the common difference be $d$. According to the problem, we have $n a + \frac{1}{2} n(n-1) d = 97^2$, which means
$$
[2 a + (n-1) d] n = 2 \times 97^2
$$
Since $n$ is a positive integer no less than 3, and 97 is a prime number, the value of $n$ can only be one o... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. $a, b, m, n$ satisfy: $a m m^{2001}+b n^{2001}=3 ; a m^{20002}+b n^{2002}=7 ; a m^{2003}+$ $b n^{2003}=24 ; a m^{21004}+b m^{2004}=102$. Then the value of $m^{2}(n-1)$ is $\qquad$ . | 10.6
From $\left\{\begin{array}{l}a m^{2004}+b n^{2004}=\left(a m^{2003}+b n^{2003}\right)(m+n)-\left(a m^{20012}+b n^{20002}\right) m n \\ a m^{2003}+b m^{2003}=\left(a m^{2002}+b n^{2002}\right)(m+n)-\left(a m^{2001}+b n^{2001}\right) m n\end{array}\right.$
we have $\left\{\begin{array}{l}102=24(m+n)-7 m n \\ 24=7(m... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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