On the Island of Misfortune with a population of 96 people, the government decided to carry out five reforms. Each reform is opposed by exactly half of all citizens. A citizen will go to a rally if they are dissatisfied with more than half of all the reforms. What is the maximum number of people the government can expect at the rally?

Solution:
Let $x$ be the number of people who came out to the rally. Consider the total number of "dissatisfactions." On one hand, each reform is opposed by exactly 48 residents, so the total number of dissatisfactions is $48 \cdot 5 = 240$. On the other hand, each person who came out to the rally is dissatisfied with at least three reforms. Therefore, the total number of dissatisfactions is no less than $3x$. Thus, $240 \geq 3x$, from which $x \leq 80$.

Let's provide an example where exactly 80 people come out to the square. Choose 80 people among the residents and divide them into five groups of 16 people each (the remaining 16 people form the government). Let the first reform be opposed by people from the first three groups, the second reform by people from the second, third, and fourth groups, the third reform by people from the third, fourth, and fifth groups, the fourth reform by people from the fourth, fifth, and first groups, and the fifth reform by people from the fifth, first, and second groups. Then, each reform is opposed by exactly $3 \cdot 16 = 48$ people, and the chosen 80 people will come out to the rally.

## Answer

80 people.

Answer: 80


A school plans to arrange 3 math lectures in September, with at least 7 days between any two lectures. How many ways are there to arrange the lectures?

Solution:
This is a limited distance combination problem, where $n=30, k=3, m=8$, so there are different arrangement methods
$$
C_{30-7 \times 2}^{3}=C_{16}^{3}=560 \text { (ways) }
$$

Answer: 560


In a chess club, 90 children attend. During the session, they divided into 30 groups of 3 people, and in each group, everyone played one game with each other. No other games were played. In total, there were 30 games of "boy+boy" and 14 games of "girl+girl". How many "mixed" groups were there, that is, groups where there were both a boy and a girl?

Solution:
Answer: 23.

Solution: There were a total of 90 games, so the number of games "boy+girl" was $90-30-14=46$. In each mixed group, two "boy+girl" games are played, while in non-mixed groups, there are no such games. In total, there were exactly $46 / 2=23$ mixed groups.

## Grade 5

Answer: 23


The rules of a "level-up game" stipulate: On the $n$-th level, a die must be rolled $n$ times. If the sum of the points obtained from these $n$ rolls is greater than $2^{n}$, the level is considered passed. Questions:
(1) What is the maximum number of levels a person can pass in this game?
(2) What is the probability that he can pass the first three levels consecutively?
(Note: A die is a uniform cube with points numbered $1,2,3,4,5,6$ on its faces. The number of points on the face that lands up after rolling the die is the result of the roll.)

Solution:
Solution: Since the dice is a uniform cube, the possibility of each number appearing after a roll is equal.
(1) Because the maximum number that can appear on a single dice is 6, and $6 \times 1 > 2^4, 6 \times 5 < 2$.
Therefore, when $n \geqslant 5$, the sum of the numbers that appear after rolling the dice $n$ times cannot be greater than 2, which means this is an impossible event, and the probability of passing is 0.
So, the maximum number of consecutive levels that can be passed is 4.
(2) Let event $A_{n}$ be "failing the $n$-th level", then the complementary event $\overline{A_{n}}$ is "passing the $n$-th level". In the $n$-th level game, the total number of basic events is $6^n$.
Level 1: The number of basic events contained in event $A_{1}$ is 2 (i.e., the cases where the number is 1 or 2),
So the probability of passing this level is: $P\left(\overline{A_{1}}\right)=1-P\left(A_{1}\right)=1-\frac{2}{6}=\frac{2}{3}$
Level 2: The number of basic events contained in event $A_{2}$ is the sum of the number of positive integer solutions to the equation $x+y=a$ when $a$ takes the values $2,3,4$. That is, $C_{1}^{3}+C_{3}^{1}+C_{3}^{1}=1+2+3=6$ (cases).
So the probability of passing this level is $P\left(\bar{A}_{2}\right)=1-P\left(A_{2}\right)=1-\frac{6}{6^{2}}=\frac{5}{6}$
Level 3: The number of basic events contained in event $A_{3}$ is the sum of the number of positive integer solutions to the equation $x+y+z=a$ when $a$ takes the values $3,4,5,6,7,8$. That is, $C_{2}^{3}+C_{3}^{3}+C_{4}^{2}+C_{5}^{3}+C_{6}^{3}+C_{5}^{2}=1+3+6+10+15+21=56$ (cases).
So the probability of passing this level is: $P\left(\bar{A}_{3}\right)=1-P\left(A_{3}\right)=1-\frac{56}{6^{3}}=\frac{20}{27}$
Therefore, the probability of passing the first 3 levels consecutively is: $P\left(\bar{A}_{1}\right) \cdot P\left(\bar{A}_{2}\right) \cdot P\left(\bar{A}_{3}\right)=\frac{2}{3} \times \frac{5}{6} \times \frac{20}{27}=\frac{100}{243}$

Answer: \frac{100}{243}


The function $f(x)=\log _{\frac{1}{3}}\left(2 x^{2}+2 x \sqrt{x^{2}+1}+1\right)^{x}$ is $(\quad$ ).
A. Even function
B. Odd function
C. Both odd and even function
D. Neither odd nor even function

Solution: 5. A. Reason: It is easy to see that $f(x)=\log _{\frac{1}{3}}\left(x+\sqrt{x^{2}+1}\right)^{2 x}$, thus $f(-x)=\log _{\frac{1}{3}}\left(-x+\sqrt{x^{2}+1}\right)^{-2 x}=\log _{\frac{1}{3}}\left(\frac{1}{-x+\sqrt{x^{2}+1}}\right)^{2 x}=\log _{\frac{1}{3}}\left(x+\sqrt{x^{2}+1}\right)^{2 x}$, hence $f(x)=f(-x)(x \in \mathbf{R})$, so $f(x)$ is an even function.

Answer: A


Mr. Kutil wanted to paint star ornaments on 20 tiles in the bathroom. On the can of paint, it was written that the paint would cover $750 \mathrm{~cm}^{2}$. How many cans of paint did Mr. Kutil have to buy at least, if one square of the grid has an area of $1 \mathrm{~cm}^{2}$? The ornament on one tile is shown in the picture.

![](https://cdn.mathpix.com/cropped/2024_04_17_151e4384192b9f5c88c8g-1.jpg?height=454&width=448&top_left_y=752&top_left_x=838)

Solution:
The tile is a square with a side of $12 \mathrm{~cm}$, its area is $(12 \mathrm{~cm})^{2}=144 \mathrm{~cm}^{2}$.

Let's calculate the area of the parts that will not be colored:

4 pentagons in the corners have a total area of $4 \cdot 15 \mathrm{~cm}^{2}=60 \mathrm{~cm}^{2}$,

4 isosceles triangles in the middle of the sides have a total area of $4 \cdot \frac{4 \cdot 6}{2} \mathrm{~cm}^{2}=$ $=48 \mathrm{~cm}^{2}$.

The total uncolored area of the tile is $(60+48) \mathrm{~cm}^{2}=108 \mathrm{~cm}^{2}$.

The colored part of the tile has an area of $(144-108) \mathrm{~cm}^{2}=36 \mathrm{~cm}^{2}$.

The colored area of 20 tiles is $(36 \cdot 20) \mathrm{~cm}^{2}=720 \mathrm{~cm}^{2}$,

$$
720 \mathrm{~cm}^{2}<750 \mathrm{~cm}^{2} \quad \text { (1 can). }
$$

Mr. Kutil will have enough with one can of paint to color the ornaments on 20 tiles.

Answer: 1


Given the sequence $\left\{a_{n}\right\}$ with the general term formula $a_{n}=\log _{3}\left(1+\frac{2}{n^{2}+3 n}\right)$, then $\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\cdots+a_{n}\right)=$

Solution:
1. 1 .

From $a_{n}=\log _{3} \frac{n^{2}+3 n+2}{n^{2}+3 n}=\log _{3} \frac{(n+1)(n+2)}{n(n+3)}$ we get
$$
\begin{aligned}
a_{1}+a_{2}+\cdots+a_{n} & =\log _{3} \frac{2 \times 3}{1 \times 4}+\log _{3} \frac{3 \times 4}{2 \times 5}+\cdots+\log _{3} \frac{(n+1)(n+2)}{n(n+3)} \\
& =\log _{3}\left[\frac{2 \times 3}{1 \times 4} \cdot \frac{3 \times 4}{2 \times 5} \cdots \cdots \frac{(n+1)(n+2)}{n(n+3)}\right] \\
& =\log _{3} \frac{3(n+1)}{n+3} .
\end{aligned}
$$

Therefore
$$
\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\cdots+a_{n}\right)=\lim _{n \rightarrow \infty} \log _{3} \frac{3(n+1)}{n+3}=\log _{3}\left(\lim _{n \rightarrow \infty} \frac{3(n+1)}{n+3}\right)=1
$$

Answer: 1


Let positive real numbers $x, y$ satisfy
$$
x^{2}+y^{2}+\frac{1}{x}+\frac{1}{y}=\frac{27}{4} \text {. }
$$

Then the minimum value of $P=\frac{15}{x}-\frac{3}{4 y}$ is

Solution:
7.6.

By the AM-GM inequality for three terms, we have
$$
\begin{array}{l}
x^{2}+\frac{1}{x}=\left(x^{2}+\frac{8}{x}+\frac{8}{x}\right)-\frac{15}{x} \geqslant 12-\frac{15}{x} \\
y^{2}+\frac{1}{y}=\left(y^{2}+\frac{1}{8 y}+\frac{1}{8 y}\right)+\frac{3}{4 y} \geqslant \frac{3}{4}+\frac{3}{4 y} .
\end{array}
$$

Adding the two inequalities, we get
$$
\begin{array}{l}
\frac{27}{4}=x^{2}+y^{2}+\frac{1}{x}+\frac{1}{y} \geqslant \frac{51}{4}+\left(\frac{3}{4 y}-\frac{15}{x}\right) \\
\Rightarrow P=\frac{15}{x}-\frac{3}{4 y} \geqslant 6 .
\end{array}
$$

Equality holds if and only if $x=2, y=\frac{1}{2}$.
Therefore, the minimum value of $P$ is 6.

Answer: 6


How many 9-digit numbers are there in the decimal system that are divisible by 11 and in which every digit except zero occurs?

Solution:
Solution. The digits are: $1,2,3,4,5,6,7,8,9$, and we know that 11 | $\overline{a b c d e f g h i}$. Due to divisibility by 11, the digits of $\overline{a b c d e f g h i}$, when summed with alternating signs, result in a number divisible by 11. Therefore, from the given digits, we need to find sums that are divisible by 11 by taking 4 digits as positive and 5 as negative (or 5 as positive and 4 as negative).

In the largest case: $9+8+7+6+5-4-3-2-1=25$.

In the smallest case: $1+2+3+4-5-6-7-8-9=-25$.

Thus, the following numbers divisible by 11 can be considered as sums: $-22,-11,0,11,22$. Since an odd number of odd numbers cannot result in an even number through addition and subtraction, only two sums need to be examined: -11 and 11.

If all digit signs are +, the sum is: $1+2+\ldots+9=45$, from which the sum of the negative-signed digits must be subtracted twice to get -11 or 11.

If the sum of the negative-signed digits is $\frac{45-(-11)}{2}=28$, then for four numbers, the possibilities are:

$$
9+8+7+4, \quad 9+8+6+5
$$

These are the only possibilities because smaller numbers would result in a sum less than 28.

For five numbers, the possibilities are:

$$
\begin{array}{lll}
9+8+7+3+1, & 9+8+6+4+1, & 9+8+6+3+2 \\
9+8+5+4+2, & 9+7+6+5+1, & 9+7+6+4+2 \\
9+7+5+4+3, & 8+7+6+5+2, & 8+7+6+4+3
\end{array}
$$

These are the only possibilities because smaller numbers would result in a sum less than 28.

If the sum of the positive-signed digits is 28, we would get the same 11 arrangements. Therefore, there are 11 ways to assign the digits to positive or negative positions, but they also need to be arranged in order. The five identical signs can be arranged in 5! ways, and the remaining four different signs in 4! ways. Thus, the total number of cases is 11$\cdot$5!$\cdot$4! = 31680.

Therefore, there are a total of 31680 nine-digit numbers in the decimal system that are divisible by 11, where every digit except 0 appears.

Answer: 31680


Let $n$ be a natural number, $a, b$ be positive real numbers, and satisfy $a+b=2$, then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is

Solution:
1
II. Fill-in-the-blank Questions
1. 【Analysis and Solution $\because a b>0$, $\therefore a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, a^{n} b^{n} \leqslant 1$. Therefore, $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}} \geqslant 1$. When $a=b=1$, the above expression $=1$, hence the minimum value is 1.

Answer: 1


In the aquarium, there are 1000 neon tetras, clownfish, and guppies. Among them, the neon tetras are the most numerous, and the guppies are the least, with their numbers in the ratio of $5: 3$. Therefore, the maximum number of clownfish is $\qquad$.

Answer: 384


Numbers $1, 2, \cdots$ are filled into a rectangular table with 8 columns in a certain order (filled from left to right, and when a row is full, move to the next row, still filling from left to right). A student first colors the cell with the number 1 black, then skips 1 cell, and colors the cell with the number 3 black; then skips 2 cells, and colors the cell with the number 6 black; then skips 3 cells, and colors the cell with the number 10 black. This continues until every column contains at least one black cell (no more coloring after that). Therefore, the number in the last black cell he colored is $\qquad$

Solution:
(Method One)
The 1st blackened cell is 1;
The 2nd blackened cell is $3=1+2$;
The 3rd blackened cell is $6=1+2+3$;
The 4th blackened cell is $10=1+2+3+4$;
$\qquad$
And so on, the $n$-th blackened cell is $1+2+\cdots+n$.
Thus, we can get the blackened cells in sequence as $1 \equiv 1(\bmod 8), 1+2=3 \equiv 3(\bmod 8)$,
$$
\begin{array}{l}
1+2+3=6 \equiv 6(\bmod 8), 1+2+3+4=10 \equiv 2(\bmod 8) \\
1+2+\cdots+5=15 \equiv 7(\bmod 8), \quad 1+2+\cdots+6=21 \equiv 5(\bmod 8) \\
1+2+\cdots+7=28 \equiv 4(\bmod 8), \quad 1+2+\cdots+8=36 \equiv 4(\bmod 8) \\
1+2+\cdots+9=45 \equiv 5(\bmod 8), \quad 1+2+\cdots+10=55 \equiv 7(\bmod 8) \\
1+2+\cdots+11=66 \equiv 2(\bmod 8), \quad 1+2+\cdots+12=78 \equiv 6(\bmod 8) \\
1+2+\cdots+13=91 \equiv 3(\bmod 8), \quad 1+2+\cdots+14=105 \equiv 1(\bmod 8) \\
1+2+\cdots+15=120 \equiv 0(\bmod 8)
\end{array}
$$

Up to this point, the numbers that are $\bmod 8$ with remainders $0 \sim 7$ have all appeared; that is, every column contains at least one black cell;
The number in the last blackened cell is 120.
(Method Two)
The 1st blackened cell is 1;
The 2nd blackened cell is $3=1+2$;
The 3rd blackened cell is $6=1+2+3$;
The 4th blackened cell is $10=1+2+3+4$;
$\qquad$
And so on, the $n$-th blackened cell is $1+2+\cdots+n$.
The blackened cells $\bmod 8$ have remainders in sequence as $1,3,6,2, 7, 5, 4, \ldots$
Up to this point, except for $\bmod 8$ with remainder 0, all other $\bmod 8$ with remainders $1 \sim 7$ have already appeared;
Therefore, we only need to consider $1+2+\cdots+n \equiv 0(\bmod 8)$;
$$
\begin{array}{l}
8 \left\lvert\, 1+2+\cdots+n=\frac{1}{2} n(n+1)\right. \\
16 \mid n(n+1)
\end{array}
$$

Since $n$ and $n+1$ are coprime;
Thus, $16 \mid n$ or $16 \mid n+1$;
We hope that $n$ is as small as possible;
Then take $n+1=16$;
$$
n=15
$$

Thus, the number in the last blackened cell is $1+2+\cdots+15=(1+15) \times 15 \div 2=120$.

Answer: 120


Determine the minimum value of the quantity $|x+y|$ given that the numbers $x$ and $y$ satisfy the relation $5 \cos (x+4 y)-3 \cos (x-4 y)-4 \sin (x-4 y)=10$.

Solution:
Solution. Let $\varphi=\arccos \frac{3}{5}$ and we get $\cos (x+4 y)-\cos (x-4 y-\varphi)=2 \Leftrightarrow$ $\left\{\begin{array}{c}\cos (x+4 y)=1, \\ \cos (x-4 y-\varphi)=-1 .\end{array}\right.$ Therefore, $\left\{\begin{array}{c}x=\frac{\varphi+\pi}{2}+\pi(n+k), \\ y=-\frac{\varphi+\pi}{8}+\frac{\pi(n-k)}{4}\end{array} \Rightarrow\right.$ $x+y=\frac{3 \varphi+\pi}{8}+\frac{\pi(5 n+3 k+1)}{4}$ for any integers $n$ and $k$. Hence, $|x+y|=\left|\frac{3 \varphi+\pi}{8}-\frac{\pi m}{4}\right|$, where $m-$ is any integer. Since $\frac{\pi}{4}<\varphi<\frac{\pi}{3}$, then $\frac{\pi}{8}<\frac{3 \varphi+\pi}{8}<\frac{\pi}{4}$ and therefore the minimum of the expression $|x+y|$ is achieved at $m=1$.

Answers in different versions: $\frac{\pi}{8}-\frac{3}{8} \arccos \frac{3}{5} ; \frac{\pi}{6}-\frac{1}{3} \arccos \frac{4}{5} ; \frac{3}{8} \arccos \frac{3}{5}-\frac{\pi}{16}$; $\frac{\pi}{3}-\frac{1}{3} \arccos \frac{4}{5}$

Answer: \frac{\pi}{8}-\frac{3}{8}\arccos\frac{3}{5}


Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and then count how many dots he got. He got 10 dots.

![](https://cdn.mathpix.com/cropped/2024_05_06_3890a5f9667fd1ab5160g-03.jpg?height=331&width=329&top_left_y=196&top_left_x=562)

Then the teacher decided to complicate the task and asked to count the number of dots, but for the letter "P" obtained in the same way from a square with a side of 10 cm. How many dots will Zhenya have this time?

Solution:
Answer: 31.

Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$.

Answer: 31


The lengths of the sides of an isosceles triangle are expressed as natural numbers (in centimeters). The product of the lengths of all sides of this triangle is 2016. What is the perimeter of this triangle?

Solution:
First method: Factorize the number 2016 into prime factors:

$2016=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 7$.

$1 \text{ POINT}$

Since we are dealing with an isosceles triangle, we need to find two sides of equal length and check if all three sides form a triangle (i.e., whether the sum of the lengths of any two sides is strictly greater than the length of the third side). It is sufficient to check whether the sum of the lengths of the legs is greater than the length of the base.

| Leg $b$ | Leg $b$ | Base $a$ | Triangle existence condition |  |
| :--- | :--- | :--- | :---: | :---: |
|  |  |  | $a+b>b$ | $b+b>a$ |
| 1 | 1 | $2016$ | yes | no |
| 2 | 2 | $2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 7=504$ | yes | no |
| $2 \cdot 2=4$ | $2 \cdot 2=4$ | $2 \cdot 3 \cdot 3 \cdot 7=126$ | yes | no |
| $2 \cdot 2 \cdot 3=12$ | $2 \cdot 2 \cdot 3=12$ | $2 \cdot 7=14$ | yes | yes |
| 3 | 3 | $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 7=224$ | yes | no |
| $3 \cdot 2=6$ | $3 \cdot 2=6$ | $2 \cdot 2 \cdot 2 \cdot 7=56$ | yes | no |

(table)

We can conclude that there is only one solution - the triplet 12, 12, 14

NOTE: If all possible combinations of side lengths are listed, but there is no visible or described verification of the triangle inequality, the number of points awarded is reduced by 2. If one or two of the possible combinations of side lengths are missing (with verification of the triangle inequality), the number of points awarded is reduced by 1.

If only the triplet 12, 12, 14 is listed after factorization into prime factors and a range is specified (with verification of the triangle inequality), the task is scored with 3 points, and without verification of the triangle inequality, with 2 points.

Answer: 38


In the set of complex numbers, solve the equation

$$
(x-1)(x-2)(x-4)(x-5)=40
$$

Solution:
Solution. Since

$$
(x-1)(x-5)=x^{2}-6 x+5,(x-2)(x-4)=x^{2}-6 x+8
$$

the equation takes the form

$$
\left(x^{2}-6 x+5\right)\left(x^{2}-6 x+8\right)=40
$$

Setting $x^{2}-6 x+5=y$, we get the equation $y(y+3)=40$ whose solutions are $y_{1}=5$ and $y_{2}=-8$. Solving now the equations

$$
x^{2}-6 x+5=5, \quad x^{2}-6 x+5=-8
$$

we obtain the roots of the given equation: $x_{1}=0, x_{2}=6, x_{3}=3+2 i, x_{4}=3-2 i$.

Answer: x_{1}=0,x_{2}=6,x_{3}=3+2i,x_{4}=3-2i


In a sequence of English letter strings, the first string $a_{1}=A$, the second string $a_{2}=B$, and each subsequent string $a_{n}(n \geq 3)$ is formed by appending the reverse of $a_{n-2}$ to $a_{n-1}$. For example, $a_{3}=a_{2} \overline{a_{1}}=B A$ (we use $\overline{a_{i}}$ to denote the reverse of $a_{i}$, which is the string read from right to left, such as $\overline{A B B}=B B A$, $\overline{A A B A}=A B A A$), $a_{4}=a_{3} \bar{a}_{2}=B A B$, $a_{5}=a_{4} \overline{a_{3}}=B A B A B$, $a_{6}=a_{5} \overline{a_{4}}=B A B A B B A B$. Therefore, among the first 1000 strings in this sequence, there are $\qquad$ palindromic strings (a palindromic string is one that reads the same from left to right as from right to left, such as $A B A$, $A A B A A$).

Solution:
【Answer】 667

【Solution】Through trial and error, we find that only $a_{3}, a_{6}, a_{9}, \cdots \cdots, a_{999}$ are not palindromic strings, the rest are. Therefore, we can directly get the answer: there are only 333 non-palindromic strings, and the remaining $1000-333=667$ are palindromic strings.

Next, we will provide a rigorous proof (this part can be ignored if there is no time during the exam):
Assume $a_{n}=P, a_{n+1}=Q$, then $a_{n+2}=Q \bar{P}, a_{n+3}=Q \bar{P} \bar{Q}$. Since $\bar{P}$ is surrounded by $Q$ and $\bar{Q}$, which can ensure the palindromic property, whether $a_{n+3}$ is a palindromic string depends on the situation of $\bar{P}$. If $a_{n}=P$ is a palindromic string, then $a_{n+3}=Q \bar{P} \bar{Q}$ is also a palindromic string; if $a_{n}=P$ is not a palindromic string, then $a_{n+3}=Q \bar{P} \bar{Q}$ is also not a palindromic string. Considering that $a_{1}$ and $a_{2}$ are both palindromic strings, $a_{4}, a_{7}, \cdots, a_{1000}$ and $a_{5}, a_{8}, \cdots, a_{998}$ are all palindromic strings. However, $a_{3}=B A$ is not a palindromic string, so $a_{6}, a_{9}, \cdots, a_{999}$ are not palindromic strings. Thus, we have proven the initial guess.

Answer: 667


Given two moving points $A$ and $B$, and a fixed point $M\left(x_{0}, y_{0}\right)$, all on the parabola $y^{2}=2 p x (p>0)$ ($A$ and $B$ do not coincide with $M$). Let $F$ be the focus of the parabola, and $Q$ be a point on the axis of symmetry. It is given that $\left(\overrightarrow{Q A}+\frac{1}{2} \overrightarrow{A B}\right) \cdot \overrightarrow{A B}=0$, and $|\overrightarrow{F A}|, |\overrightarrow{F M}|, |\overrightarrow{F B}|$ form an arithmetic sequence.
(1) Find the coordinates of $\overrightarrow{O Q}$;
(2) If $|\overrightarrow{O Q}|=3, |\overrightarrow{F M}|=\frac{5}{2}$, and the projections of points $A$ and $B$ on the directrix of the parabola are $A_{1}$ and $B_{1}$ respectively, find the range of the area of quadrilateral $A B B_{1} A_{1}$.

Solution:
21. (1) Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), Q(a, 0)$. Then, from $|\overrightarrow{F A}|, |\overrightarrow{F M}|, |\overrightarrow{F B}|$ forming an arithmetic sequence, we get $x_{0}=\frac{x_{1}+x_{2}}{2}$.
And $y_{1}^{2}=2 p x_{1}$,
$y_{2}^{2}=2 p x_{2}$.
(1) - (2) gives $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{2 p}{y_{1}+y_{2}}$.

From $\left(\overrightarrow{Q A}+\frac{1}{2} \overrightarrow{A B}\right) \cdot \overrightarrow{A B}=0$, we get $a=x_{0}+p$. Hence, $\overrightarrow{O Q}=\left(x_{0}+p, 0\right)$.
(2) From $|\overrightarrow{O Q}|=3,|\overrightarrow{F M}|=\frac{5}{2}$, we get
$$
p=1, x_{0}=2 \text {. }
$$

Then $S_{\text {quadrilateral } A B B_{1} A_{1}}$
$$
\begin{array}{l}
=\frac{\left[\left(x_{1}+\frac{1}{2}\right)+\left(x_{2}+\frac{1}{2}\right)\right]\left|y_{1}-y_{2}\right|}{2} \\
=\frac{5}{2}\left|y_{1}-y_{2}\right| .
\end{array}
$$

Since $y_{1}= \pm \sqrt{2 x_{1}}, y_{2}= \pm \sqrt{2 x_{2}}$, we have
$$
y_{1} y_{2}= \pm 2 \sqrt{x_{1} x_{2}} \text {. }
$$

And $0 \leqslant 2 \sqrt{x_{1} x_{2}} \leqslant x_{1}+x_{2}=4$, thus,
$$
-4 \leqslant y_{1} y_{2} \leqslant 4 \text {. }
$$

Therefore, $0 \leqslant\left(y_{1}-y_{2}\right)^{2}=2\left(x_{1}+x_{2}\right)-2 y_{1} y_{2}$
$$
=8-2 y_{1} y_{2} \leqslant 16 \text {. }
$$

Then $S_{\text {quadrilateral } A B B_{1} A_{1}}=\frac{5}{2}\left|y_{1}-y_{2}\right| \in(0,10]$.
Thus, the range of the area of quadrilateral $A B B_{1} A_{1}$ is $(0,10]$.

Answer: (0,10]


Given the radius of the base of a cylinder is $r$, the height is $h$, the volume is 2, and the surface area is 24. Then $\frac{1}{r}+\frac{1}{h}=(\quad)$.
(A) 6
(B) 8
(C) 12
(D) 24

Solution: 2. A. From the conditions, we know $\pi r^{2} h=2, 2 \pi r^{2}+2 \pi r h=24$. Dividing the two equations gives $\frac{1}{r}+\frac{1}{h}=6$.

Answer: A


A rod is broken into two parts at a randomly chosen point; then the larger of the two resulting parts is again broken into two parts at a randomly chosen point. What is the probability that a triangle can be formed from the three resulting pieces?

Solution:
99. For simplicity, let's assume that the length of our rod is 2 (this is legitimate, as we can always take half the length

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-331.jpg?height=559&width=378&top_left_y=968&top_left_x=149)

Fig. 87. of the rod as the unit of length). Let \( AB \) be our rod, \( K \) and \( L \) the points of the first and second break, respectively. The outcome of the experiment in this problem is completely determined by the numbers \( x = AK \) and \( z = KL \), where \( x \) is chosen randomly between 0 and 1 ( \( x \) is the length of the smaller of the two parts formed after the first break, and thus \( x \leq 1 - x \)).

Since \( y = \frac{z}{2 - x} \), dividing the written inequalities by \( 2 - x \), we get that favorable outcomes will correspond to pairs of numbers \( (x, y) \) such that \( y \geq \frac{1 - x}{2 - x} \), i.e.,

\[
y \geq 1 - \frac{1}{2 - x}
\]

We now need to visualize the region on Fig. 88 defined by these inequalities.

Introduce instead of coordinates \( (x, y) \) coordinates \( \left(x', y'\right) \), where

\[
y' = y, \quad \text{and} \quad x' = 2 - x
\]

Such a change of coordinates is equivalent to moving the origin to point \( O' \), located on the x-axis at a distance of 2 from the old origin and changing the direction of the x-axis (see Fig. 88). In the new coordinates, the inequalities (*) will transform as follows:

\[
y' \geq 1 - \frac{1}{x'}
\]

The first of these inequalities indicates that all points corresponding to favorable outcomes will be located below the hyperbola

\[
y' = \frac{1}{x'}
\]

Now let's construct the curve

\[
\begin{gathered}
y' = 1 - \frac{1}{x'} = \frac{1}{2} - \left(\frac{1}{x'} - \frac{1}{2}\right) \\
y' - \frac{1}{2} = -\left(\frac{1}{x'} - \frac{1}{2}\right)
\end{gathered}
\]

This curve will obviously be symmetric to the curve \( y' = \frac{1}{x'} \) relative to the line \( y' = \frac{1}{2} \) (see Fig. 88). The second inequality (**) shows that all points corresponding to favorable outcomes are located above the curve

\[
y' = 1 - \frac{1}{x'}
\]

Thus, favorable outcomes correspond to points filling the shaded area on Fig. 88. We need to determine the area of this region.

First, note that due to the symmetry of the curves

\[
y' = \frac{1}{x'} \text{ and } y' = 1 - \frac{1}{x'}
\]

relative to the line \( y' = \frac{1}{2} \), the area of the curvilinear triangle \( OTW \) is equal to the area of the curvilinear triangle \( TUV \); thus, the desired area is \( 1 - 2 \) area \( TUV \). But due to problem 152, the area of the curvilinear trapezoid \( OTVW \) is \( \ln 2 \) (where \( \ln \) denotes the natural logarithm, i.e., the logarithm with base \( e = 2.718 \ldots \)). From this, it follows that the area \( TUV = 1 - \ln 2 \) and, consequently, the area of the curvilinear triangle \( TVW \) is

\[
1 - 2(1 - \ln 2) = 2 \ln 2 - 1 = 0.388 \ldots
\]

This expression is also equal to the probability that the segments \( x, z \), and \( 2 - x - z \) can form a triangle.

Answer: 2\ln2-1


Find all solutions of the equation $x^{2 y}+(x+1)^{2 y}=(x+2)^{2 y}$ with $x, y \in \mathbb{N}$.

Solution:
One can easily see that neither $x$ nor $y$ can be zero.
For $y=1$, from $x^{2}+(x+1)^{2}=(x+2)^{2}$, we obtain the equation $x^{2}-2 x-3=0$, of which only the solution $x=3$ is valid.
Now let $y>1$.
Since $x$ and $x+2$ have the same parity, $x+1$ is an even number and thus $x$ is an odd number.
With $x=2 k-1(k \in \mathbb{N})$, we get the equation
(\#) $\quad(2 k-1)^{2 y}+(2 k)^{2 y}=(2 k+1)^{2 y}$
from which, by expanding, we obtain:

$$
(2 k)^{2 y}-2 y(2 k)^{2 y-1}+\ldots-2 y 2 k+1+(2 k)^{2 y}=(2 k)^{2 y}+2 y(2 k)^{2 y-1}+\ldots+2 y 2 k+1
$$

Since $y>1$, it follows that $2 y \geq 3$. If we now move all terms in $yk$ to one side and factor out $(2k)^{3}$ on the other side, we get:

$$
8 yk=(2 k)^{3}\left[2\binom{2 y}{3}+2\binom{2 y}{5}(2 k)^{2}+\ldots-(2 k)^{2 y-3}\right]
$$

from which it follows that $y$ is a multiple of $k$.
By dividing equation (\#) by $(2k)^{2 y}$, we get:

$$
\left(1-\frac{1}{2 k}\right)^{2 y}+1=\left(1+\frac{1}{2 k}\right)^{2 y}
$$

where the left side is less than 2. The right side, however, is greater than $1+\frac{2 y}{2 k} \geq 2$, which cannot be, since $y$ is a multiple of $k$.

Therefore, the given equation has only the solution $x=3$ and $y=1$.
That there are no solutions for $y > 1$ was clear to most participants, as it represents a special case of Fermat's Last Theorem, which has since been proven. Unfortunately, only a few managed to provide a complete proof of this special case.

Answer: x=3, y=1


Represent the number 36 as the product of three integer factors, the sum of which is 4. What is the smallest of the factors?

Solution:
Answer: -4.

Example: $36=(-4) \cdot(-1) \cdot 9$.

Solution. The given factorization is unique. This can be proven.

If all three factors are positive, then the largest of them is not less than 4 (since $3^{3}<36$), and the sum is greater than 4, which contradicts the condition. Therefore, two of the factors are negative, and the third is positive.

Then the positive factor is 9. Indeed, if it is not more than 6, then the sum of the absolute values of the other two is greater than 2; if it is not less than 12, then the sum of the absolute values of the other two is less than 8. Then the two negative factors are either -4 and -1, or -2 and -2. Only the first variant satisfies the condition.

Answer: -4


Find all four-digit numbers $n$ that satisfy the following conditions:
(1) The first and third digits of $n$ are the same;
(2) The second and fourth digits of $n$ are the same;
(3) The product of the digits of $n$ is a divisor of $n^2$.

Solution:
【Analysis】First, set the form of the unknowns, then discuss the relationship between the two numbers, while paying attention to the fact that this four-digit number is a multiple of 101 and also a prime number. Find the numerical relationship and enumerate accordingly.

【Solution】Solution: According to the problem, we have:
Let $n=\overline{\mathrm{ab} \mathrm{ab}}=101 \overline{\mathrm{ab}}$, and $a^{2} b^{2}$ is a divisor of $n^{2}$, so $a b$ is a divisor of $n$. Since 101 is a prime number, i.e., $a b$ is a divisor of $101 \overline{\mathrm{ab}}$, and because $a b$ is coprime with 101, $a b$ is a divisor of $\overline{\mathrm{ab}}$. Therefore, $a b \mid(10 a+b)$, which implies $a \mid b$ and $b \mid 10 a$, so $b=a, 2 a$ or $5 a$,

Thus, the value of $\overline{\mathrm{ab}}$ can be $11, 12, 15, 24, 36$.
Answer: The four-digit numbers are: $1111, 1212, 1515, 2424, 3636$.

Answer: 1111,1212,1515,2424,3636


The largest odd number that cannot be written as the sum of three distinct composite numbers is

Solution:
【Analysis】Among the positive integers, the three smallest composite numbers are $4, 6, 8$. First, calculate their sum, then compare it with the nearest odd number, and finally prove the correctness of this conclusion.

【Solution】Solution: Among the positive integers, the three smallest composite numbers are $4, 6, 8$. Their sum is $4+6+8=18$, so 17 is an odd number that cannot be expressed as the sum of three different composite numbers. Next, we will prove that any odd number $n$ greater than or equal to 19 can be expressed as the sum of three different composite numbers.

Since when $k \geqslant 3$, $4, 9, 2k$ are three different composite numbers, and $4+9+2k \geqslant 19$, by appropriately choosing $k$, any odd number $n$ greater than or equal to 19 can be expressed as the sum of $4, 9, 2k\left(k=\frac{n-13}{2}\right)$. In summary, the largest odd number that cannot be expressed as the sum of three different composite numbers is 17. Therefore, the answer is: 17.

Answer: 17


Someone chooses a natural number \(n\), adds the natural numbers from 1 to \(n\) together, and obtains the sum \(1+2+\ldots+n\) as a three-digit number, which (like 777) consists entirely of the same digits.

Determine all possibilities for choosing a number \(n\) for which this is true!

Solution:
}

Assuming \(n\) is a natural number that satisfies the conditions of the problem, then on the one hand,

\[
1+2+3+\ldots+(n-1)+n=\frac{1}{2} n(n+1)
\]

On the other hand, the sum that arises according to the problem can be written in the form \(111 x\), where \(x\) is a natural number such that \(1 \leq x \leq 9\). Therefore,

\[
n(n+1)=2 \cdot 111 x=2 \cdot 3 \cdot 37 x
\]

Since 37 is a prime number, it follows that either \(n\) or \(n+1\) is divisible by 37; Furthermore, since the sum \(1+2+\ldots+n\) is supposed to be a three-digit number, \(n(n+1)<2000\), and thus \(n^2<2000\) and therefore \(n<45\). Consequently, \(n\) can only be one of the numbers 36, 37.

Since \(\frac{36 \cdot 37}{2}=666\) and \(\frac{37 \cdot 38}{2}=703\), we obtain a three-digit number with three identical digits as the sum of the natural numbers from 1 to \(n\) exactly when \(n=36\).

Answer: 36


Let $F$ be the set of points with coordinates $(x, y)$ such that $||x|-|y||+|x|+|y|=2$.

(a) Draw $F$.

(b) Find the number of points in $F$ such that $2y=|2x-1|-3$.

Solution:
Solution: (a) If $|x| \geq|y|$, then ||$x|-| y||+|x|+|y|=|x|-|y|+$ $|x|+|y|=2|x|=2$, thus $|x|=1$ and therefore $1 \geq|y|$, so $-1 \leq y \leq 1$. We conclude that the segments $-1 \leq y \leq 1$ on the lines $x=1$ and $x=-1$ belong to $F$.

If $|x| \leq|y|$, then ||$x|-| y||+|x|+|y|=-|x|+|y|+|x|+|y|=$ $2|y|=2$, thus $|y|=1$ and therefore $1 \geq|x|$, so $-1 \leq x \leq 1$. We conclude that the segments $-1 \leq x \leq 1$ on the lines $y=1$ and $y=-1$ also belong to $F$.

Thus we have determined that $F$ consists of the sides of a square with vertices $A(-1,-1), B(1,-1), C(1,1), D(-1,1)$.

![](https://cdn.mathpix.com/cropped/2024_06_03_e679746950d7d3832835g-213.jpg?height=366&width=458&top_left_y=126&top_left_x=779)

(b) We find the number of solutions of $2 y=|2 x-1|-3$ on each of the segments $A B, B C, C D, D A$.

The segment $C D$ consists of all points $(x, y)$ such that $-1 \leq x \leq$ 1, $y=1$. The equation $2=|2 x-1|-3$ has no solution $x$ when $-1 \leq x \leq 1$. Therefore $2 y=|2 x-1|-3$ has no solution on $C D$.

The segment $A B$ consists of all points $(x, y)$ such that $-1 \leq x \leq$ $1, y=-1$. The equation $-2=|2 x-1|-3$ has two solutions: $x=0$ and $x=1$. Therefore $2 y=|2 x-1|-3$ has two solutions on $A B$.

As above we get that $2 y=|2 x-1|-3$ has a unique solution on $A D:(x, y)=(-1,0)$ and a unique solution on $B C:(x, y)=$ $(1,-1)$. Note that the last one has already been obtained as a point on $A B$. Thus there are three solutions of $2 y=|2 x-1|-3$ in $F$ : $(x, y)=(-1,0),(0,-1),(1,-1)$.

Answer: 3


Solution:
Solution.

$$
\begin{aligned}
& f^{\prime}(x)=\left(4 x^{3}-5 x^{-7}-8 x^{\frac{5}{6}}+10\right)^{\prime}=\left[\left((x)^{n}\right)^{\prime}=n x^{n-1},(\text { const })^{\prime}=0\right]= \\
& =4\left(x^{3}\right)^{\prime}-5\left(x^{-7}\right)^{\prime}-8\left(x^{\frac{5}{6}}\right)^{\prime}+(10)^{\prime}=4 \cdot 3 x^{3-1}-5 \cdot(-7) x^{-7-1}- \\
& -8 \cdot \frac{5}{6} x^{\frac{5}{6}-1}+0=12 x^{2}+35 x^{-8}-\frac{20}{3} x^{-\frac{1}{6}}=12 x^{2}+\frac{35}{x^{8}}-\frac{20}{3 \sqrt[6]{x}} . \\
& \quad \text { The derivative is defined for } x \in(0 ;+\infty) .
\end{aligned}
$$

Answer: 12x^{2}+\frac{35}{x^{8}}-\frac{20}{3\sqrt[6]{x}}


On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $-1,2,6,7,8,11,13,14,16,20$. Determine which numbers are written on the board. In your answer, write their product.

Solution: Answer: -2970 (numbers on the board: $-3,2,5,9,11$).

Answer: -2970


I was walking with my mother-in-law. We were slowly, at a speed of 3 km/h, walking down Rue Sainte-Catherine in Bordeaux, which, as everyone knows (or maybe doesn't know), is straight. Suddenly, I remembered that I needed to drop a letter in a mailbox located a bit further down the street. Leaving my mother-in-law, who continued her walk calmly, I hurried ahead, and after dropping the letter, I walked back to meet her at the same speed (5 km/h). I clearly remember that I left her alone for only 3 minutes.

How far from the mailbox were we when we parted?

Solution:
21. Let's choose meters and minutes as units of measurement. Then my speed after I left my mother-in-law is $5000 / 60 = 250 / 3 \, \text{m} /$ min, and my mother-in-law's speed (and our common speed during the walk) is 3000/60 = $50$ m $/$ min. Let $S$ be the point where we parted, $R$ the point where we met again, and $B$ the point where the mailbox is located. The distance my mother-in-law walked in 3 minutes is

$$
S R = 50 \cdot 3 = 150
$$

In this time, I walked the distance

$$
S B + B R = S R + 2 B R = (250 / 3) \cdot 3 = 250
$$

Therefore, $2 B R = 100$ and $B R = 50$. Thus,

$$
S B = S R + R B = 200
$$

So, at the moment we parted, we were 200 meters away from the mailbox.

Answer: 200


Solution:
Solution. By direct calculation, we get

$$
\begin{aligned}
20130+2 \cdot(480 \cdot 4 \cdot 14 & +30 \cdot 44 \cdot 16)-(5 \cdot 80 \cdot 43+19 \cdot 400 \cdot 3) \cdot 2= \\
& =20130+2 \cdot(480 \cdot 56+44 \cdot 480)-(400 \cdot 43+400 \cdot 57) \cdot 2 \\
& =20130+2 \cdot 480 \cdot 100-2 \cdot 400 \cdot 100 \\
& =20130+96000-80000 \\
& =20130+16000=36130
\end{aligned}
$$

Answer: 36130


Calculate: $\frac{\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} \times \frac{1}{5}+\frac{3}{2} \times \frac{3}{4} \times \frac{3}{5}}{\frac{1}{2} \times \frac{2}{3} \times \frac{2}{5}}=$

Solution: $5 \frac{1}{8}$

Answer: 5\frac{1}{8}


Let $n, b_{0} \in \mathbf{N}^{*}, n \geqslant 2, 2 \leqslant b_{0} \leqslant 2 n-1$. The sequence $\left\{b_{i}\right\}$ is defined as follows:
$$b_{i+1}=\left\{\begin{array}{ll}
2 b_{i}-1, & b_{i} \leqslant n, \\
2 b_{i}-2 n, & b_{i}>n,
\end{array} \quad i=0,1,2, \cdots\right.$$

Let $p\left(b_{0}, n\right)$ denote the smallest index $p$ such that $b_{p}=b_{0}$.
(1) For $k \in \mathbf{N}^{*}$, find the values of $p\left(2,2^{k}\right)$ and $p\left(2,2^{k}+1\right)$;
(2) Prove that for any $n$ and $b_{0}$, $p\left(b_{0}, n\right) \mid p(2, n)$.

Solution:
4. Let $m=n-1, a_{i}=b_{i}-1$, then $1 \leqslant a_{0} \leqslant 2 m$, and
$$a_{i+1}=\left\{\begin{array}{ll}
2 a_{i}, & a_{i} \leqslant m, \\
2 a_{i}-(2 m+1), & a_{i}>m .
\end{array}\right.$$

This indicates: $a_{i+1} \equiv 2 a_{i}(\bmod 2 m+1)$, and for $i \in \mathbf{N}$, we have $1 \leqslant a_{i} \leqslant 2 m$.
(1) The required values are equivalent to finding $p\left(1,2^{k}-1\right)$ and $p\left(1,2^{k}\right)$ for $\left\{a_{i}\right\}$. The former is equivalent to finding the smallest $l \in \mathbf{N}^{*}$ such that
$$2^{l} \equiv 1\left(\bmod 2\left(2^{k}-1\right)+1\right) ;$$

The latter is equivalent to finding the smallest $t \in \mathbf{N}^{*}$ such that
$$2^{t} \equiv 1\left(\bmod 2^{k+1}+1\right)$$

Since $2\left(2^{k}-1\right)+1=2^{k+1}-1$, and for $1 \leqslant l \leqslant k$, it is clear that $2^{l} \neq 1\left(\bmod 2^{k+1}-1\right)$, hence $p\left(1,2^{k}-1\right)=k+1$. Also, $2^{2(k+1)} \equiv 1\left(\bmod 2^{k+1}+1\right)$, so $\delta_{2^{k+1}+1}(2) \mid 2(k+1)$. But for $1 \leqslant t \leqslant k+1$, we have $2^{t} \neq 1\left(\bmod 2^{k+1}+1\right)$, thus $p\left(1,2^{k}\right)=\delta_{2^{k+1}+1}(2)=2(k+1)$.

Therefore, for $\left\{b_{i}\right\}$, we have $p\left(2,2^{k}\right)=k+1, p\left(2,2^{k}+1\right)=2(k+1)$.
(2) We still discuss in terms of $\left\{a_{i}\right\}$, and need to prove: $p\left(a_{0}, m\right) \mid p(1, m)$.
First, let $p(1, m)=t$, then $2^{t} \equiv 1(\bmod 2 m+1)$, and thus $2^{t} a_{0} \equiv a_{0}(\bmod 2 m+1)$, so $p\left(a_{0}, m\right) \leqslant p(1, m)$.

Second, if $p\left(a_{0}, m\right) \times p(1, m)$, then we can set
$$p(1, m)=p\left(a_{0}, m\right) q+r, 0<r<p\left(a_{0}, m\right)$$

Let $s=p\left(a_{0}, m\right), t=p(1, m)$, then combining $2^{s} a_{0} \equiv a_{0}(\bmod 2 m+1)$, we know
$$a_{0} \equiv 2^{t} a_{0}=2^{n+r} a_{0} \equiv 2^{s(q-1)+r} a_{0} \equiv \cdots \equiv 2^{r} a_{0}(\bmod 2 m+1),$$

which contradicts the minimality of $s$.
Therefore, $p\left(a_{0}, m\right) \mid p(1, m)$, and the proposition is proved.

Answer: p\left(2,2^{k}\right)=k+1, p\left(2,2^{k}+1\right)=2(k+1)


Among the natural numbers less than 5000, the numbers that are divisible by 11 and have a digit sum of 13, there are $\qquad$ in total.

Solution:
【Solution】Solution: (1) Sum of digits in odd positions $=12$, sum of digits in even positions $=1$, resulting in $3190, 3091, 4180, 4081$, a total of 4 possibilities.
(2) Sum of digits in odd positions $=1$, sum of digits in even positions $=12$. Resulting in $1309, 1408, 1507, 1606, 1705, 1804$, 1903; 319, 418, 517, 616, 715, 814, 913, a total of 14 possibilities.

Total $4+14=18$ possibilities.
Therefore, the answer is: 18.

Answer: 18


Consider the set $A=\{100,101,102, \ldots, 200\}$.

a) Determine the number of elements in $A$ that are divisible by 10.

b) Determine the elements of $A$ that, when divided by 21 give a remainder of 12, and when divided by 28 give a remainder of 5.

Solution:
a) A contains 11 numbers divisible by 10

b) If $n=21 k+12$ and $n=28 p+5, k, p \in \mathbb{N}$, then $4 n=84 k+48,3 n=84 p+15$ and $4 n-3 n=84(k-p)+33$, that is, $n=84 q+33,100 \leq n \leq 200$, thus $n=117$. $.4 \mathrm{p}$

Answer: 117


Given that $f(x)$ is a monotonic function on $(0,+\infty)$, and for any $x \in(0,+\infty)$, we have $f\left(f(x)-\log _{2} x\right)=6$. If $x_{0}$ is a solution to the equation $f(x)-f^{\prime}(x)=4$, and $x_{0} \in (a-1, a)\left(a \in \mathbf{Z}_{+}\right)$, then $a=(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4

Solution:
4. B.

From the monotonicity of $f(x)$ and $f\left(f(x)-\log _{2} x\right)=6$, we know that $f(x)-\log _{2} x=c$ (constant) $\Rightarrow f(c)=6$.
$$
\begin{array}{l}
\text { Hence } f(c)-\log _{2} c=c \Rightarrow \log _{2} c=6-c \\
\Rightarrow c=4 \Rightarrow f(x)=\log _{2} x+4 . \\
\text { From } f(x)-f^{\prime}(x)=\log _{2} x+4-\frac{1}{x \ln 2}=4 \\
\Rightarrow \log _{2} x_{0}=\frac{1}{x_{0} \ln 2}=\frac{\log _{2} \mathrm{e}}{x_{0}} \\
\Rightarrow x_{0}^{x_{0}}=\mathrm{e} \Rightarrow 1<x_{0}<2 .
\end{array}
$$

Answer: B


As shown in Figure 6, point $A$ is on the line segment $B G$, quadrilaterals $A B C D$ and $D E F G$ are both squares, with areas of $7 \mathrm{~cm}^{2}$ and $11 \mathrm{~cm}^{2}$, respectively. Find the area of $\triangle C D E$.

Solution:
Analysis: Since $CD = \sqrt{7}$, we only need to find the height $EH$ on side $CD$ (as shown in Figure 6).

From the given conditions, it is easy to see that $\triangle EDH$ is obtained by rotating $\triangle DAG$ around point $D$ by $90^{\circ}$, so,
$$
EH = AG = \sqrt{DG^2 - AD^2} = 2.
$$

Therefore, $S_{\triangle ECD} = \sqrt{7} \, \text{cm}^2$.

Answer: \sqrt{7} \, \text{cm}^2


There is a sequence, the first number is 6, the second number is 3, starting from the second number, each number is 5 less than the sum of the number before it and the number after it. What is the sum of the first 200 numbers in this sequence, from the first number to the 200th number?

Solution:
【Answer】 999
【Solution】The first number is 6, the second number is 3. According to the rule "from the second number, each number is 5 less than the sum of the number before it and the number after it," the third number is $3+5-6=2$; similarly, the subsequent numbers are: $6, 3, 2, 4, 7, 8, 6, 3, 2, 4 \ldots \ldots$. It is easy to see that every 6 numbers form a cycle. $200 \div 6=33 \ldots \ldots 2$. Therefore, the sum of the first 200 numbers is $33 \times(6+3+2+4+7+8)+6+3=999$.

Answer: 999


In $\triangle A B C$, $A B=A C$. If $\sin A=\frac{3}{5}$, then $\frac{B C}{A B}=$ .

Solution:
$=.1 . \frac{\sqrt{10}}{5}$.
As shown in Figure 7, draw $BH \perp AC$, with $H$ as the foot of the perpendicular.
$$
\begin{array}{l}
\text { Let } BH=3k, \\
AB=AC=5k .
\end{array}
$$

Then $AH=4k$,
$$
\begin{array}{c}
CH=k, \\
BC=\sqrt{BH^{2}+CH^{2}} \\
=\sqrt{10} k .
\end{array}
$$

Therefore, $\frac{BC}{AB}=\frac{\sqrt{10}}{5}$.

Answer: \frac{\sqrt{10}}{5}


Let $n$ be a positive integer. Let $(a, b, c)$ be a random ordered triple of nonnegative integers such that $a + b + c = n$, chosen uniformly at random from among all such triples. Let $M_n$ be the expected value (average value) of the largest of $a$, $b$, and $c$. As $n$ approaches infinity, what value does $\frac{M_n}{n}$ approach?

Solution:
1. **Define the problem and variables:**
   We are given a positive integer \( n \) and need to find the expected value \( M_n \) of the largest of three nonnegative integers \( a, b, \) and \( c \) such that \( a + b + c = n \). We are interested in the limit of \( \frac{M_n}{n} \) as \( n \) approaches infinity.

2. **Count the number of ordered triples:**
   The number of ordered triples \((a, b, c)\) such that \( a + b + c = n \) is given by the stars and bars theorem:
   \[
   \binom{n+2}{2}
   \]

3. **Define \( W_k(n) \):**
   Let \( W_k(n) \) denote the number of ordered triples \((a, b, c)\) such that \( a + b + c = n \) and \(\max(a, b, c) = k\).

4. **Symmetry and counting \( W_k(n) \):**
   By symmetry and the properties of the triangular array, we can determine \( W_k(n) \) for different ranges of \( k \):
   - For \( k \geq \lceil \frac{n}{3} \rceil \), the values of \( W_k(n) \) depend on the residue of \( n \mod 3 \):
     \[
     W_{\lceil \frac{n}{3} \rceil + k}(n) =
     \begin{cases}
     3(3k + 2) & \text{if } n \equiv 1 \mod 3 \\
     3(3k + 1) & \text{if } n \equiv 2 \mod 3 \\
     9k & \text{if } 3 \mid n
     \end{cases}
     \]

5. **Expected value calculation:**
   To compute \( M_n \), we need to average the values of \( k \cdot W_k(n) \) over all \( k \) from \( \lceil \frac{n}{3} \rceil \) to \( n \):
   \[
   M_n = \frac{\sum_{i=\lceil \frac{n}{3} \rceil}^{n} (i \cdot W_i(n))}{\binom{n+2}{2}}
   \]

6. **Simplify the sum:**
   Using the properties of binomial coefficients and the Hockey Stick Identity, we can simplify the sum:
   \[
   \sum_{i=1 + \lfloor \frac{n}{2} \rfloor}^{n} (i \cdot W_i(n)) = 3 \lfloor \frac{n}{2} \rfloor \binom{\lfloor \frac{n}{2} \rfloor + 2}{2} + 3 \binom{\lfloor \frac{n}{2} \rfloor + 3}{3}
   \]

7. **Asymptotic behavior:**
   As \( n \) approaches infinity, the floors and ceilings become negligible. We focus on the leading terms of the numerator and denominator:
   \[
   \frac{\sum_{i=\lceil \frac{n}{3} \rceil}^{n} (i \cdot W_i(n))}{n \binom{n+2}{2}}
   \]

8. **Leading coefficients:**
   The leading third-degree coefficients of the numerator and the denominator determine the limit. After simplification, we find:
   \[
   \lim_{n \to \infty} \frac{M_n}{n} = \frac{11}{18}
   \]

The final answer is \( \boxed{\frac{11}{18}} \)

Answer: \frac{11}{18}


Let $\left(x_{n}\right)$ be a sequence such that $x_{0}=1, x_{1}=2$, with the property that the sequence $\left(y_{n}\right)$ is given by the relation

$$
y_{n}=\binom{n}{0} x_{0}+\binom{n}{1} x_{1}+\cdots+\binom{n}{n} x_{n}, \quad \text { for } n \in \mathbb{N}_{0}
$$

is a geometric sequence. Determine $x_{2020}$.

Solution:
## Solution.

By substituting into the given equation, we get $y_{0}=1$, and $y_{1}=1+2=3$. Since $\left(y_{n}\right)$ is a geometric sequence, we conclude that $y_{n}=3^{n}$ for all $n \in \mathbb{N}_{0}$.

We will prove by mathematical induction that $x_{n}=2^{n}$ for all $n \in \mathbb{N}_{0}$.

For $n=0$ and $n=1$, we have $x_{0}=1=2^{0}$ and $x_{1}=2=2^{1}$, so the base case of induction is satisfied.

Assume that for all $k \leqslant n$ it holds that $x_{k}=2^{k}$. We will prove that $x_{n+1}=2^{n+1}$. According to the induction hypothesis, we have

$$
\begin{aligned}
& y_{n+1}=\binom{n+1}{0} x_{0}+\binom{n+1}{1} x_{1}+\cdots+\binom{n+1}{n} x_{n}+\binom{n+1}{n+1} x_{n+1} \\
&=\binom{n+1}{0} 2^{0}+\binom{n+1}{1} 2^{1}+\cdots+\binom{n+1}{n} 2^{n}+\binom{n+1}{n+1} x_{n+1} \\
&=\binom{n+1}{0} 2^{0}+\binom{n+1}{1} 2^{1}+\cdots+\binom{n+1}{n} 2^{n}+\binom{n+1}{n+1} 2^{n+1} \\
& \quad-\binom{n+1}{n+1}\left(2^{n+1}-x_{n+1}\right) \\
&=(1+2)^{n+1}-2^{n+1}+x_{n+1} \\
&=3^{n+1}-2^{n+1}+x_{n+1}
\end{aligned}
$$

where the second-to-last equality holds by the binomial theorem. Since $y_{n+1}=3^{n+1}$, it follows that $x_{n+1}=2^{n+1}$, thus completing the induction step.

Therefore, for every $n \in \mathbb{N}_{0}$, we have $x_{n}=2^{n}$, and in particular, $x_{2020}=2^{2020}$.

Answer: 2^{2020}


From the 100 positive integers $1,2,3, \cdots, 100$, if $n$ numbers are taken, among these $n$ numbers, there are always 4 numbers that are pairwise coprime. Find the minimum value of $n$.

Solution:
Let $S=\{1,2,3, \cdots, n\}, A_{i}=\{k \mid k \in S$ and $k$ is divisible by $i\}(i=2,3,5)$, then by the principle of inclusion-exclusion, we have
$$
\begin{aligned}
\left|A_{2} \cup A_{3} \cup A_{5}\right| & =\left|A_{2}\right|+\left|A_{3}\right|+\left|A_{5}\right|-\left|A_{2} \cap A_{3}\right|-\left|A_{2} \cap A_{5}\right|-\left|A_{3} \cap A_{5}\right|+\left|A_{2} \cap A_{3} \cap A_{5}\right| \\
& =\left[\frac{100}{2}\right]+\left[\frac{100}{3}\right]+\left[\frac{100}{5}\right]-\left[\frac{100}{2 \times 3}\right]-\left[\frac{100}{2 \times 5}\right]-\left[\frac{100}{3 \times 5}\right]+\left[\frac{100}{2 \times 3 \times 5}\right] \\
& =50+33+20-16-10-6+3=74 .
\end{aligned}
$$

In $A_{2} \cup A_{3} \cup A_{5}$, if we choose 4 numbers, at least 2 of them belong to the same set among $A_{2}, A_{3}, A_{5}$, and they are not coprime.
Therefore, the minimum value of $n$ is $\geqslant 75$.
Next, we prove that the minimum value of $n$ is 75.
Let $B_{1}=\{1\} \cup\{100$ prime numbers $\}, B_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}\right\}$,
$$
B_{3}=\{2 \times 47,3 \times 31,5 \times 19,7 \times 13\}, B_{4}=\{2 \times 43,3 \times 29,5 \times 17,7 \times 11\},
$$

and let $B=B_{1} \cup B_{2} \cup B_{3} \cup B_{4}$. Then $|B|=26+3 \times 4=38,|S \backslash B|=100-38=62$. Thus, when choosing 75 numbers from $S$, at least $75-62=13$ of them belong to $B=B_{1} \cup B_{2} \cup B_{3} \cup B_{4}$. By the pigeonhole principle, among these 13 numbers, at least $\left[\frac{13-1}{4}\right]+1=4$ numbers belong to the same subset among $B_{1}, B_{2}, B_{3}, B_{4}$, and they are pairwise coprime.

In conclusion, the minimum value of $n$ is 75.

Answer: 75


Given a circle with its center at the origin and radius $R$ intersects with the sides of $\triangle A B C$, where $A(4,0), B(6,8)$, $C(2,4)$. Then the range of values for $R$ is $\qquad$.

Solution:
5. $\left[\frac{8 \sqrt{5}}{5}, 10\right]$.

Solution: Draw a perpendicular from the origin $O$ to $AC$, with the foot of the perpendicular being $D$. Then point $D$ lies on side $AC$. It is easy to see that
$l_{AC}: y = -2x + 8 \Rightarrow 2x + y - 8 = 0$, so $OD = \frac{|2 \times 0 + 0 - 8|}{\sqrt{2^2 + 1}} = \frac{8 \sqrt{5}}{5}$,

Also, $OA = 4$, $OB = 10$, $OC = 2 \sqrt{5}$, hence $OD < OA < OC < OB$,

When and only when $OD \leqslant R \leqslant OB$, the circle intersects with $\triangle ABC$, so, $R \in \left[\frac{8 \sqrt{5}}{5}, 10\right]$.

Answer: [\frac{8\sqrt{5}}{5},10]


In a triangle, three lines parallel to its sides and tangent to the inscribed circle have been drawn. They cut off three triangles from the given one. The radii of the circumcircles of these triangles are $R_{1}, R_{2}, R_{3}$. Find the radius of the circumcircle of the given triangle.

Solution:
79. Prove that if $a, b, c$ are the lengths of the sides of a triangle, then the perimeters of the cut-off triangles will be $2(p-a), 2(p-b)$, $2(p-c)$. Consequently, if $R$ is the radius of the circumscribed circle, then $R_{1}+R_{2}+R_{3}=\left(\frac{p-a}{p}+\frac{p-b}{p}+\frac{p-c}{p}\right) R=R$.

Answer: $R=R_{1}+R_{2}+R_{3}$.

Answer: R_{1}+R_{2}+R_{3}


Two numbers are written on the board in the laboratory. Every day, the senior researcher Petya erases both numbers on the board and writes down their arithmetic mean and harmonic mean instead. On the morning of the first day, the numbers 1 and 2 were written on the board. Find the product of the numbers written on the board in the evening of the 1999th day.

Solution:
The product of the numbers on the board does not change. Indeed, $\frac{a+b}{2} \cdot \frac{2 a b}{a+b}=a b$. Therefore, the desired product is 2.

## Answer

2.

Send a comment

Answer: 2


Calculate the area of the figure bounded by the graphs of the functions:

$$
y=2 x-x^{2}+3, y=x^{2}-4 x+3
$$

Solution:
## Solution

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-23.jpg?height=756&width=917&top_left_y=730&top_left_x=107)

Let's find the points of intersection of the graphs of the functions:

$$
2 x - x^{2} + 3 = x^{2} - 4 x + 3 \Rightarrow \left[\begin{array}{l}
x = 0 \\
x = 3
\end{array}\right]
$$

Now let's find the area of the resulting figure:

$$
S = \int_{0}^{3} \left( \left( -x^{2} + 2 x + 3 \right) - \left( x^{2} - 4 x + 3 \right) \right) d x = \int_{0}^{3} \left( -2 x^{2} + 6 x \right) d x = -\left. \frac{2}{3} x^{3} \right|_{0}^{3} + \left. 3 x^{2} \right|_{0}^{3} = -18 + 27 = 9
$$

Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\� \%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%B8\%D0\%B2\%D0\%B5\%D0\%B9\%D0\%B8\%D0\%BD\%D0\%B8\%D0\%B5\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8\%D0\%B8

Answer: 9


Find all triplets of non-zero natural numbers $(a, b, c)$ such that

$$
2021^{a}+4=3^{b} \times 5^{c}
$$

Solution:
In a Diophantine equation, one always starts by looking at small values of the parameters. Here, it suffices to factorize $2021^{a}+4$ into prime factors, so we look at the small values of $a$. Here, when $a=1$, the equation becomes $2025=3^{b} \times 5^{c}$, so $(a, b, c)=(1,4,2)$ is a solution. Moreover, if $a \geqslant 2$, no one wants to factorize $2021^{a}+4$, so we stop here our exploration of small cases.
In a second step, one should either proceed with factorizations or look at the equation modulo a well-chosen number $n$ to reduce the set of possible values of $a, b$, and $c$. The numbers 3 and 5 are clearly prime, and our study of the case $a=1$ indicates that $2021=45^{2}-4=43 \times 47$. We note this relation in a corner of our scratch paper for later use.
Finally, using the Chinese Remainder Theorem, we reduce our equation modulo $n$ when $n$ is a power of a prime. For example:
$\triangleright n=1, n=2$ and $n=5$ do not provide any information;
$\triangleright n=3$ indicates that $a$ is odd;
$\triangleright n=4$ indicates that $b$ is even;
$\triangleright n=7$ does not provide any easy-to-exploit information;
$\triangleright n=8$ indicates, since $a$ is odd and $b$ is even, that $c$ is even.
We then set $\beta=b / 2$ and $\gamma=c / 2$. Moreover, two of the three terms in our equation are squares, which allows us to obtain the factorization

$$
43^{a} \times 47^{a}=2021^{a}=\left(3^{\beta} \times 5^{\gamma}\right)^{2}-2^{2}=\left(3^{\beta} \times 5^{\gamma}+2\right)\left(3^{\beta} \times 5^{\gamma}-2\right) .
$$

Since $3^{\beta} \times 5^{\gamma}+2$ and $3^{\beta} \times 5^{\gamma}-2$ are two odd integers whose difference is 4, they are coprime.
Only one of these two factors is divisible by 43, and only one is divisible by 47. Several cases are possible a priori:
$\triangleright$ if 47 divides $3^{\beta} \times 5^{\gamma}-2$, then $47^{a}$ divides $3^{\beta} \times 5^{\gamma}-2$, so $3^{\beta} \times 5^{\gamma}-2 \geqslant 47^{a} \geqslant 43^{a} \geqslant 3^{\beta} \times 5^{\gamma}+2$, which is impossible;
$\triangleright$ if 43 and 47 divide $3^{\beta} \times 5^{\gamma}+2$, then $3^{\beta} \times 5^{\gamma}+2=43^{a} \times 47^{a}$ and $3^{\beta} \times 5^{\gamma}-2=1$, which is impossible since $3^{\beta} \times 5^{\gamma}-2 \geqslant 3 \times 5-2=13$.
Thus, 47 divides $3^{\beta} \times 5^{\gamma}+2$ and 43 divides $3^{\beta} \times 5^{\gamma}-2$, so that $3^{\beta} \times 5^{\gamma}+2=47^{a}$ and $3^{\beta} \times 5^{\gamma}-2=43^{a}$. But then

$$
4=\left(3^{\beta} \times 5^{\gamma}+2\right)-\left(3^{\beta} \times 5^{\gamma}-2\right)=47^{a}-43^{a} \geqslant 47^{a}-43 \times 47^{a-1}=4 \times 47^{a-1}
$$

which shows that $a \leqslant 1$.
In conclusion, the only solution to the problem is $(a, b, c)=(1,4,2)$.
Comment from the graders The problem was difficult, but many students managed to score points thanks to intelligent reflections. Finding a particular solution is always a step forward, as it allows one to see what can or cannot be proven: here $(1,4,2)$ was a solution, so one could expect to prove that $b$ or $c$ was even, and it was futile to prove that $b$ or $c$ was odd without additional assumptions on $a$.

Let's recall that looking modulo a composite number is rarely intelligent: looking modulo 15, for example, serves no purpose, since the same observations can be made modulo 3 and 5, and since the calculations are simpler, it allows one to avoid making mistakes.
Here, the key to the problem was to question the parity of $b, c$ to obtain a square: obtaining squares in Diophantine equations is crucial, as it allows one to factorize and thus simplify the problem. However, once the equation is factorized, some forget to specify that the two terms are positive and even strictly greater than 1, which means one cannot immediately conclude that one of the factors is $47^{a}$ and the other is $43^{a}$. Moreover, it would be good to rigorously justify at the end that $47^{a}=43^{a}+4$ implies $a=1$. For example, one can invoke that $47^{2}>43^{2}+4$ and say that by immediate induction $47^{k}>43^{k}+4$ if $k \geqslant 2$.

## Senior Problems

Answer: (1,4,2)


For any sequence of real numbers $A=\left\{a_{1}, a_{2}, \ldots\right\}$, we define $\Delta A$ as the sequence $\left\{a_{2}-a_{1}, a_{3}-a_{2}, \ldots\right\}$.

We assume that all terms of the sequence $\Delta(\Delta A)$ are 1 and that $a_{19}=a_{92}=0$.

Determine $a_{1}$.

Solution:
We denote $d$ as the first term of $\Delta A$. Therefore, we have: $\Delta A=\{d, d+1, d+2, \ldots\}$ where the $n^{\text{th}}$ term is written as $d+n-1$.

It follows that $A=\left\{a_{1}, a_{1}+d, a_{1}+d+(d+1), a_{1}+d+(d+1)+(d+2), \ldots\right\}$ where the $n^{\text{th}}$ term is written as $a_{n}=a_{1}+(n-1)d+\frac{(n-1)(n-2)}{2}$.

$a_{n}$ is therefore a polynomial of the second degree in $n$ with a leading coefficient of $\frac{1}{2}$.

Since $a_{19}=a_{92}=0$, we necessarily have:

$$
a_{n}=\frac{1}{2}(n-19)(n-92)
$$

Thus, $a_{1}=819$.

## Intermediate Exercises

Answer: 819


Let $p$ be a prime number, and the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=1$, and for any non-negative integer $n$,
$$
a_{n+2}=2 a_{n+1}-p a_{n} \text {. }
$$

If -1 is a term in the sequence $\left\{a_{n}\right\}$, find all possible values of $p$.

Solution:
The only $p$ that satisfies the condition is $p=5$.
It is easy to see that when $p=5$, $a_{3}=-1$.
Next, we prove that it is the only solution.
Assume $a_{m}=-1\left(m \in \mathbf{N}_{+}\right)$.
Clearly, $p \neq 2$, otherwise, from $a_{n+2}=2 a_{n+1}-2 a_{n}$, we know that for $n \geqslant 2$, $a_{n}$ is always even.
Thus, the sequence cannot contain -1.
Therefore, $(2, p)=1$.
Taking the recurrence relation $a_{n+2}=2 a_{n+1}-p a_{n}$ modulo $p$ gives $a_{n+2} \equiv 2 a_{n+1}(\bmod p)$,

which means the sequence $\left\{a_{n}\right\}$ is a geometric sequence modulo $p$, and we have
$$
a_{n+1} \equiv 2^{n} a_{1}(\bmod p) \text {. }
$$

In particular, we have
$$
-1 \equiv a_{m} \equiv 2^{m-1} a_{1} \equiv 2^{m-1}(\bmod p) .
$$

Taking $a_{n+2}=2 a_{n+1}-p a_{n}$ modulo $p-1$ gives
$$
\begin{array}{l}
a_{n+2} \equiv 2 a_{n+1}-a_{n}(\bmod (p-1)) \\
\Rightarrow a_{n+2}-a_{n+1} \equiv a_{n+1}-a_{n}(\bmod (p-1)),
\end{array}
$$

which means the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence modulo $p-1$, and we have
$$
\begin{array}{l}
a_{n+1} \equiv(n+1)\left(a_{1}-a_{0}\right)+a_{0} \\
\equiv n+1(\bmod (p-1)) .
\end{array}
$$

In particular, we have
$$
-1 \equiv a_{m} \equiv m(\bmod (p-1)),
$$

which implies $m+1 \equiv 0(\bmod (p-1))$.
By $(2, p)=1$ and Fermat's Little Theorem, we get
$$
2^{p-1} \equiv 1(\bmod p) \text {. }
$$

Combining with equation (1), we get
$$
1 \equiv 2^{p-1} \equiv 2^{m+1} \equiv 4 \times 2^{m-1} \equiv-4(\bmod p) \text {. }
$$

Thus, $5 \equiv 0(\bmod p)$.
Therefore, $p=5$ is the only solution that satisfies the requirement.

Answer: p=5


In triangle $ABC$, $AB = 13$, $BC = 14$, and $CA = 15$. Let $M$ be the midpoint of side $AB$, $G$ be the centroid of $\triangle ABC$, and $E$ be the foot of the altitude from $A$ to $BC$. Compute the area of quadrilateral $GAME$.

Solution:
1. **Assign coordinates to the vertices of the triangle:**
   We can place the triangle in the coordinate plane for easier calculation. Let:
   \[
   A = (0, 12), \quad B = (5, 0), \quad C = (-9, 0)
   \]
   These coordinates satisfy the given side lengths:
   \[
   AB = \sqrt{(5-0)^2 + (0-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13
   \]
   \[
   BC = \sqrt{(5 - (-9))^2 + (0-0)^2} = \sqrt{14^2} = 14
   \]
   \[
   CA = \sqrt{(0 - (-9))^2 + (12-0)^2} = \sqrt{81 + 144} = \sqrt{225} = 15
   \]

2. **Find the coordinates of the midpoint \( M \) of \( AB \):**
   The midpoint formula gives:
   \[
   M = \left( \frac{0+5}{2}, \frac{12+0}{2} \right) = \left( \frac{5}{2}, 6 \right)
   \]

3. **Find the coordinates of the foot of the altitude \( E \) from \( A \) to \( BC \):**
   Since \( BC \) is horizontal (both \( B \) and \( C \) have the same y-coordinate), the altitude from \( A \) to \( BC \) is vertical. Thus, \( E \) has the same x-coordinate as \( A \) and the same y-coordinate as \( B \) and \( C \):
   \[
   E = (0, 0)
   \]

4. **Find the coordinates of the centroid \( G \) of \( \triangle ABC \):**
   The centroid \( G \) is the average of the coordinates of the vertices:
   \[
   G = \left( \frac{0 + 5 - 9}{3}, \frac{12 + 0 + 0}{3} \right) = \left( \frac{-4}{3}, 4 \right)
   \]

5. **Calculate the area of quadrilateral \( GAME \) using the Shoelace Theorem:**
   The Shoelace Theorem states that for a polygon with vertices \((x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)\), the area is:
   \[
   \text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) + (x_n y_1 - y_n x_1) \right|
   \]
   For quadrilateral \( GAME \) with vertices \( G \left( \frac{-4}{3}, 4 \right) \), \( A (0, 12) \), \( M \left( \frac{5}{2}, 6 \right) \), and \( E (0, 0) \), we have:
   \[
   \text{Area} = \frac{1}{2} \left| \left( \frac{-4}{3} \cdot 12 + 0 \cdot 6 + \frac{5}{2} \cdot 0 + 0 \cdot 4 \right) - \left( 4 \cdot 0 + 12 \cdot \frac{5}{2} + 6 \cdot 0 + 0 \cdot \frac{-4}{3} \right) \right|
   \]
   Simplifying inside the absolute value:
   \[
   = \frac{1}{2} \left| \left( -16 + 0 + 0 + 0 \right) - \left( 0 + 30 + 0 + 0 \right) \right|
   \]
   \[
   = \frac{1}{2} \left| -16 - 30 \right|
   \]
   \[
   = \frac{1}{2} \left| -46 \right|
   \]
   \[
   = \frac{1}{2} \times 46 = 23
   \]

The final answer is \(\boxed{23}\).

Answer: 23


Find the mass of the body bounded by the cylindrical surface $x^{2}=2 y$ and the planes $y+z=1, 2 y+z=2$, if at each point of the body the volumetric density is numerically equal to the ordinate of that point.

Solution:
Solution. According to the condition, at point $M(x, y, z)$ of the body, the volume density $\delta(M)=y$. By formula (2), the mass of this body is

$$
m=\iint_{G} \delta \delta(M) d v=\iiint_{G} y d x d y d z
$$

where $G$ is the region occupied by the given body (Fig. 182).[^31]

Evaluating the triple integral using formula (*), we obtain:

$$
\begin{aligned}
& m=\iint_{i_{x y}} y d x d y \int_{1-y}^{2(1-y)} d z=\iint_{A(/ / i} y(1--y) d x d y=\int_{n}^{1}\left(y-y^{2}\right) d y \int_{-V^{-\overline{2 y}}}^{\sqrt{-1-1}} d x= \\
& =\int_{0}^{1}\left(y-y^{2}\right) 2 \sqrt{2 y} d y=\left.2 \sqrt{2}\left(\frac{2}{5} y^{\frac{5}{2}}-\frac{2}{7} y^{\frac{7}{2}}\right)\right|_{0} ^{1}=\frac{8 \sqrt{\bar{z}}}{35} .
\end{aligned}
$$

Answer: \frac{8\sqrt{2}}{35}


Given a positive integer $ n\geq 2$, let $ B_{1}$, $ B_{2}$, ..., $ B_{n}$ denote $ n$ subsets of a set $ X$ such that each $ B_{i}$ contains exactly two elements. Find the minimum value of $ \left|X\right|$ such that for any such choice of subsets $ B_{1}$, $ B_{2}$, ..., $ B_{n}$, there exists a subset $ Y$ of $ X$ such that:
 (1) $ \left|Y\right| = n$;
 (2) $ \left|Y \cap B_{i}\right|\leq 1$ for every $ i\in\{1,2,...,n\}$.

Solution:
1. **Initial Setup and Problem Restatement**:
   We are given a set \( X \) and \( n \) subsets \( B_1, B_2, \ldots, B_n \) of \( X \), each containing exactly two elements. We need to find the minimum value of \( |X| \) such that there exists a subset \( Y \subseteq X \) with \( |Y| = n \) and \( |Y \cap B_i| \leq 1 \) for all \( i \in \{1, 2, \ldots, n\} \).

2. **Lower Bound Argument**:
   Consider \( |X| = 2n - 2 \). We construct the subsets \( B_i \) as follows:
   \[
   B_1 = \{1, 2\}, B_2 = \{3, 4\}, \ldots, B_{n-1} = \{2n-3, 2n-2\}, B_n = \{1, 3\}
   \]
   In this case, any subset \( Y \) of \( X \) with \( |Y| = n \) will have at least one \( B_i \) such that \( |Y \cap B_i| \geq 2 \). Therefore, \( |X| = 2n - 2 \) is not sufficient.

3. **Proving \( |X| \geq 2n - 1 \)**:
   We need to show that \( |X| \geq 2n - 1 \) is necessary. Assume \( |X| = 2n - 1 \). We will prove that there exists a subset \( Y \subseteq X \) with \( |Y| = n \) and \( |Y \cap B_i| \leq 1 \) for all \( i \).

4. **Constructing the Subset \( Y \)**:
   Choose \( Y \) such that \( |Y| \) is maximized and \( |Y \cap B_i| \leq 1 \) for all \( i \). Suppose \( |Y| \leq n - 1 \). Then \( |X \setminus Y| \geq n \).

5. **Contradiction Argument**:
   For each \( a \in X \setminus Y \), there exists \( 1 \leq t(a) \leq n \) and \( x_a \in Y \) such that \( (a, x_a) \in B_{t(a)} \). If not, we could add \( a \) to \( Y \) and get \( |Y'| > |Y| \), contradicting the maximality of \( |Y| \).

6. **Distinct Subsets**:
   If \( a \neq b \), then \( B_{t(a)} \neq B_{t(b)} \). Since \( |X \setminus Y| \geq n \), we have at least \( n \) distinct subsets \( B_{t(a)} \). Thus, \( |X \setminus Y| = n \).

7. **Final Contradiction**:
   The subset \( X \setminus Y \) satisfies \( |(X \setminus Y) \cap B_i| = 1 \) for all \( i \) and \( |X \setminus Y| = n \), which is a contradiction. Therefore, \( |Y| \) must be exactly \( n \).

8. **Conclusion**:
   Hence, the minimum value of \( |X| \) such that there exists a subset \( Y \subseteq X \) with \( |Y| = n \) and \( |Y \cap B_i| \leq 1 \) for all \( i \) is \( 2n - 1 \).

The final answer is \( \boxed{ 2n - 1 } \)

Answer: 2n - 1


The director of a certain school decided to take a photo of the 2008 graduates. He arranged the students in parallel rows, all with the same number of students, but this arrangement was too wide for the camera's field of view. To solve this problem, the director decided to remove one student from each row, placing them in a new row. This arrangement did not please the director because the new row had four students fewer than the others. He then decided to remove one more student from each original row, placing them in the newly created row, and found that now all the rows had the same number of students, and he finally took his photo. How many students appeared in the photo?

Solution:
The diagrams below represent the situation of the problem, where the students who were initially removed are represented in black and the students removed the second time, in gray.
![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-099.jpg?height=260&width=1538&top_left_y=2460&top_left_x=292)

Let $m$ and $n$ be the number of rows (horizontal lines) and columns of the initial formation, respectively. By taking one student from each of the $m$ rows, a new incomplete row is formed: four students are missing to complete the current $n-1$ columns, i.e., $m+4=n-1$ and therefore, $n=m+5$. Now we have $m$ rows of $n-1$ students, plus one incomplete row, where four students are missing.

By taking one student from each of the $m$ complete rows, we form a rectangle with one column less, thus filling the current three vacancies in the new row. Therefore, $m=3$ and thus, $n=8$. The total number of students in the photo is given by $n \times m=3 \times 8=24$.

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-100.jpg?height=233&width=322&top_left_y=546&top_left_x=1532)

Answer: 24


Filipa plays a game. She starts with a row of 15 squares and a coin on the centre square. Filipa then rolls a die. If she rolls an even number, she moves the coin that many squares to the right; if she rolls an odd number, she moves the coin that many squares to the left. If the results of six rolls were $1,2,3,4,5,6$, where would her coin be located?

(A) On the square where it started

(B) 1 square to the right of where it started

(C) 2 squares to the right of where it started

(D) 2 squares to the left of where it started

(E) 3 squares to the right of where it started

Solution:
The coin starts on square number 8 , counting from left to right.

After the first roll, it moves 1 square to the left, since 1 is odd.

After the second roll, it moves 2 squares to the right, since 2 is even.

The coin continues to move, ending on square $8-1+2-3+4-5+6=11$, and so is 3 squares to the right of where it started.

ANsWER: (E)

Answer: E


A tourist who lives in Magdeburg (M) wants to visit each of the cities Schwerin (S), Neubrandenburg (N), and Berlin (B) exactly once on a round trip and then return to his place of residence.

One possible route would be from Magdeburg via Berlin, Schwerin, and Neubrandenburg back to Magdeburg (see illustration).

List all the travel routes the tourist can choose under the given conditions!

How many travel routes are there in total?

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0117.jpg?height=254&width=254&top_left_y=975&top_left_x=1506)

A justification is not required.

Solution:
The required specification of travel routes can be made in graphical form or by indicating the respective sequence of cities to be chosen.

An example of a complete specification is, for instance: $M B N S M, M B S N M, M N B S M, M N S B M, M S B N M$, $M S N B M$.

The number of travel routes is 6.

Answer: 6


One side of the parallelogram is three times larger than the other side. Find the sides of the parallelogram if its perimeter is 24.

Solution:
Answer

$3,9,3,9$.

Answer: 3,9,3,9


There are several small sticks of lengths 1 cm, 2 cm, 3 cm, 4 cm, and 5 cm each. If you randomly pick 3 sticks to form a triangle, what is the maximum number of different triangles that can be formed?

Solution:
【Answer】Solution: (1) 1 cm, 1 cm, 1 cm;
(2) 1 cm, 2 cm, 2 cm;
(3) 1 cm, 3 cm, 3 cm;
(4) 1 cm, 4 cm, 4 cm;
(5) 1 cm, 5 cm, 5 cm;
(6) 5 cm, 5 cm, 5 cm;
(7) 2 cm, 2 cm, 2 cm;
(8) 2 cm, 2 cm, 3 cm;
(9) 2 cm, 3 cm, 3 cm;
(10) 2 cm, 3 cm, 4 cm;
(11) 2 cm, 4 cm, 4 cm;
(12) 2 cm, 4 cm, 5 cm;
(13) 2 cm, 5 cm, 5 cm;
(14) 3 cm, 3 cm, 3 cm;
(15) 3 cm, 3 cm, 4 cm;
(16) 3 cm, 3 cm, 5 cm;
(17) 3 cm, 4 cm, 4 cm;
(18) 3 cm, 4 cm, 5 cm;
(19) 3 cm, 5 cm, 5 cm;
(20) 4 cm, 4 cm, 4 cm;
(21) 4 cm, 4 cm, 5 cm;
(22) 4 cm, 5 cm, 5 cm.

Answer: A maximum of 22 different triangles can be formed.

Answer: 22


Solution: $\begin{array}{l}\text { 1. }-1 \text { Detailed Explanation: } \because f(\lg 20)=a \sin (\lg 20+1) \pi+b \sqrt[3]{\lg 20-1}+2=a \sin (2-\lg 5+1) \pi+b \sqrt[3]{\lg 2}+2 \\ =a \sin (\lg 5) \pi+b \sqrt[3]{\lg 2}+2 \text {. Also, } f(\lg 5)=a \sin (\lg 5+1) \pi+b \sqrt[3]{\lg 5-1}+2=-a \sin (\lg 5) \pi-b \sqrt[3]{\lg 2}+2 \text {, } \\ \therefore \quad f(\lg 20)+f(\lg 5)=4, f(\lg 20)=4-f(\lg 5)=-1 .\end{array}$

Answer: -1


The odometer of a bicycle reads $3733 \mathrm{~km}$. The first time it will again display a number with three equal digits will occur

$\begin{array}{llll}\text { (A) before } 50 \mathrm{~km} & \text { (B) between } 50 \mathrm{~km} \text { and } 100 \mathrm{~km} & \text { (C) between } 100 \mathrm{~km} \text { and } 500 \mathrm{~km}\end{array}$ (D) between $500 \mathrm{~km}$ and $1000 \mathrm{~km}$ (E) between $1000 \mathrm{~km}$ and $5000 \mathrm{~km}$.

Solution:
2) The answer is $(\mathrm{A})$

The first event will occur at km 3777. There are no intermediate events, because as long as the first two digits of the odometer are 3 and 7, the only possibility is that the other two are both equal to 3 or both equal to 7.

Answer: A


$a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ are five different real numbers. The number of different values that the sum $a_{i}+a_{j}$ can take for all $i, j$ with $1 \leq i<j \leq 5$ we call $m$.

Determine the smallest possible value of $m$.

Solution: 3. We order the five numbers from smallest to largest and call them $a, b, c, d$, and $e$, so $a<b<c<d<e$. Then it holds that: $a+b<a+c<a+d<a+e<b+e<c+e<d+e$. Therefore, the smallest value of $m$ satisfies: $m \geq 7$. With the example $a=1, b=2, c=3, d=4$ and $e=5$, we find the sums to be $3,4,5,6,7,8$ and 9. Thus, 7 is the smallest possible value of $m$.

Answer: 7


Calculate the limit of the numerical sequence:

$\lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}+(3 n+2)^{3}}{(2 n+3)^{3}-(n-7)^{3}}$

Solution:
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}+(3 n+2)^{3}}{(2 n+3)^{3}-(n-7)^{3}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((2 n+1)^{3}+(3 n+2)^{3}\right)}{\frac{1}{n^{3}}\left((2 n+3)^{3}-(n-7)^{3}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(2+\frac{1}{n}\right)^{3}+\left(3+\frac{2}{n}\right)^{3}}{\left(2+\frac{3}{n}\right)^{3}-\left(1-\frac{7}{n}\right)^{3}}=\frac{2^{3}+3^{3}}{2^{3}-1^{3}}=\frac{35}{7}=5
\end{aligned}
$$

## Problem Kuznetsov Limits 3-19

Answer: 5


As shown in Figure $1, P$ is a point on the chord $AB$ of $\odot O$, $AP=m, PB=n$, and $m>n$.
When $AB$ moves around $\odot O$ for one complete revolution, the trajectory of point $P$ is curve $C$. If the area between $\odot O$ and curve $C$ is $\left(m^{2}-n^{2}\right) \pi$, then the value of $\frac{m}{n}$ is

Solution:
5. $\frac{1+\sqrt{5}}{2}$.

As shown in Figure 5, let the midpoint of $A B$ be $M$, connect $O M$, $O P$, and $O B$, and let the radius of $\odot O$ be $R$.
$$
\begin{array}{l}
\text { By } O M \perp A B, M B=\frac{1}{2}(m+n), \\
M P=\frac{1}{2}(m+n)-n=\frac{1}{2}(m-n),
\end{array}
$$

we get $R^{2}=O B^{2}=O M^{2}+\frac{1}{4}(m+n)^{2}$,
$$
O P^{2}=O M^{2}+\frac{1}{4}(m-n)^{2} \text {. }
$$

From equation (1), we know that $O M^{2}$ is a constant. Then from equation (2), we know that $O P$ is a constant. Therefore, curve $C$ is a circle with $O$ as the center and $O P$ as the radius.
From the problem, we get
$$
\begin{array}{l}
\left(R^{2}-O P^{2}\right) \pi=\left(m^{2}-n^{2}\right) \pi \\
\Rightarrow \frac{1}{4}\left((m+n)^{2}-(m-n)^{2}\right)=m^{2}-n^{2} \\
\Rightarrow m^{2}-n^{2}=m n \\
\Rightarrow \frac{m}{n}=\frac{1+\sqrt{5}}{2} \text { (negative root discarded). }
\end{array}
$$

Answer: \frac{1+\sqrt{5}}{2}


In December 1903, the first airplane "Flyer 1" made by Wright Brothers finally succeeded in its test flight. This marked the first time humans achieved powered flight with a pilot on board.
A. Bell Brothers
B. Haier Brothers
C. Wright Brothers
D. Leon Brothers

Solution:
【Answer】C
【Analysis】In 1810, the British G. Cayley first proposed the basic principles of flight for heavier-than-air aircraft and the structural layout of airplanes, laying the foundation for modern aviation theory of fixed-wing aircraft and rotary-wing aircraft.

In the history of aviation, the person who contributed the most to gliding flight was the German O. Lilienthal. Starting from 1867, he and his brother studied bird gliding flight for more than 20 years, clarifying many theories related to flight, which laid the foundation for modern aerodynamics.

American scientist S. P. Langley made significant achievements in many scientific fields and was renowned in the global scientific community. In 1896, Langley built a powered aircraft model that flew to a height of 150m and remained airborne for nearly 3 hours. This was the first time a heavier-than-air powered aircraft achieved stable and sustained flight, which was of great significance in the history of world aviation.
At the end of the 19th century, the American Wright brothers, building on the experiences and lessons of their predecessors, established a small wind tunnel to measure the lift produced by airflow on a plate. They also built three gliders and conducted thousands of flight tests, meticulously recording lift, drag, and speed, and repeatedly modifying and improving longitudinal and lateral control. Later, they designed and manufactured a 12-horsepower, 77.2 kg piston gasoline engine, which they installed on their third glider to drive two pusher propellers. This was the "Flyer" 1. On December 17, 1903, Orville Wright piloted the "Flyer" 1 for its first test flight, covering a distance of 36m and remaining airborne for 12s. As their piloting skills improved, by the final flight, piloted by Wilbur Wright, the distance covered was 260m and the flight time was 59s. This was the first sustained and controlled powered flight in human history, turning the human dream of flying into reality and ushering in a new era of modern aviation.

Answer: C


Find all positive integers $n>1$ such that
\[\tau(n)+\phi(n)=n+1\]
Which in this case, $\tau(n)$ represents the amount of positive divisors of $n$, and $\phi(n)$ represents the amount of positive integers which are less than $n$ and relatively prime with $n$.

Solution:
To find all positive integers \( n > 1 \) such that
\[
\tau(n) + \phi(n) = n + 1,
\]
we need to analyze the properties of the divisor function \(\tau(n)\) and the Euler's totient function \(\phi(n)\).

1. **Case \( n = p \) where \( p \) is a prime:**

   - For a prime number \( p \), the number of divisors \(\tau(p)\) is 2 (since the divisors are 1 and \( p \)).
   - The Euler's totient function \(\phi(p)\) is \( p-1 \) (since all numbers less than \( p \) are relatively prime to \( p \)).

   Therefore, for \( n = p \):
   \[
   \tau(p) + \phi(p) = 2 + (p - 1) = p + 1.
   \]
   This satisfies the given equation \( \tau(n) + \phi(n) = n + 1 \).

2. **Case \( n = 4 \):**

   - For \( n = 4 \), the number of divisors \(\tau(4)\) is 3 (since the divisors are 1, 2, and 4).
   - The Euler's totient function \(\phi(4)\) is 2 (since the numbers less than 4 that are relatively prime to 4 are 1 and 3).

   Therefore, for \( n = 4 \):
   \[
   \tau(4) + \phi(4) = 3 + 2 = 5.
   \]
   This satisfies the given equation \( \tau(n) + \phi(n) = n + 1 \).

3. **Case \( n \neq 4 \) and \( n \neq p \):**

   - Suppose \( n \) is a composite number that is not equal to 4. Let \( n = pq \) where \( p \) and \( q \) are distinct primes.
   - The number of divisors \(\tau(n)\) for \( n = pq \) is 4 (since the divisors are 1, \( p \), \( q \), and \( pq \)).
   - The Euler's totient function \(\phi(n)\) for \( n = pq \) is \((p-1)(q-1)\) (since the numbers less than \( pq \) that are relatively prime to \( pq \) are those that are not divisible by either \( p \) or \( q \)).

   Therefore, for \( n = pq \):
   \[
   \tau(pq) + \phi(pq) = 4 + (p-1)(q-1).
   \]
   We need to check if this can ever equal \( pq + 1 \):
   \[
   4 + (p-1)(q-1) = pq + 1.
   \]
   Simplifying, we get:
   \[
   4 + pq - p - q + 1 = pq + 1,
   \]
   \[
   5 - p - q = 0,
   \]
   \[
   p + q = 5.
   \]
   The only pairs of primes that sum to 5 are \( (2, 3) \) and \( (3, 2) \). However, if \( p = 2 \) and \( q = 3 \), then \( n = 6 \):
   \[
   \tau(6) = 4 \quad \text{and} \quad \phi(6) = 2.
   \]
   \[
   \tau(6) + \phi(6) = 4 + 2 = 6 \neq 7.
   \]
   Thus, \( n = 6 \) does not satisfy the equation. Therefore, there are no other solutions.

Conclusion:
The only solutions are \( n = 4 \) and \( n = p \) where \( p \) is a prime.

The final answer is \( \boxed{ n = 4 } \) and \( n = p \) where \( p \) is a prime.

Answer: n = 4


In the Cartesian coordinate system $x O y$, the area of the figure enclosed by the curve
$$
2|x|+3|y|=5
$$

is ( ).
(A) $\frac{5}{3}$
(B) 5
(C) $\frac{20}{3}$
(D) $\frac{25}{3}$

Solution:
2. D.

As shown in Figure 1, it is clear that the figure enclosed by the curve $2|x|+3|y|=5$ is symmetric about the $x$-axis and $y$-axis. Therefore, we only need to consider the part of the graph in the first quadrant. In this case, $x>0, y>0$, so the curve is $2 x+3 y=5$.

Thus, the figure enclosed by the curve $2|x|+3|y|=5$ is a rhombus, and its area is
$$
2 \times \frac{5}{2} \times \frac{5}{3}=\frac{25}{3} \text {. }
$$

Answer: D


Let $a_{n}=6^{n}+8^{n}$. Then the remainder when $a_{2018}$ is divided by 49 is $\qquad$ .

Solution: \begin{array}{l}\text { 4.2. } \\ a_{2018}=(7-1)^{2018}+(7+1)^{2018} \\ =2 \sum_{k=0}^{1009} \mathrm{C}_{2018}^{2 k} 7^{2 k} \equiv 2\left(\bmod 7^{2}\right)\end{array}

Answer: 2


In a basketball tournament, the teams "Wolves" and "Bears" played the first quarter to a draw. The points scored by the Wolves in each of the 4 quarters form an increasing geometric sequence, while the points scored by the Bears in each quarter form an increasing arithmetic sequence. In the end, the Wolves won by a single point. Neither team scored more than 100 points. Determine the total number of points both teams scored together at the end of the first half.

Solution:
## Solution.

Let $a$ be the number of points both teams scored in the first quarter. Then the sequence of numbers, or points, that the Wolves scored in each quarter is $a, a q, a q^{2}, a q^{3}$, where $q$ is the common ratio of the geometric sequence and $q>1$ because the sequence is increasing.

The sequence of points that the Bears scored in each quarter is the following increasing arithmetic sequence: $a, a+d, a+2 d, a+3 d, d>0$.

The sum of the geometric sequence is the total number of points the Wolves scored at the end, and the sum of the arithmetic sequence is the total number of points the Bears scored at the end. Therefore,

$$
\begin{aligned}
& a+a q+a q^{2}+a q^{3} \leqslant 100 \\
& a+a+d+a+2 d+a+3 d \leqslant 100
\end{aligned}
$$

or

$$
\begin{aligned}
& a\left(1+q+q^{2}+q^{3}\right) \leqslant 100 \\
& 4 a+6 d \leqslant 100 \\
& a\left(1+q+q^{2}+q^{3}\right)=4 a+6 d+1(*)
\end{aligned}
$$

From the last equation, considering the parity of the left and right sides, we conclude that $a$ is an odd number, and $q$ is an even number. From the first inequality, we conclude that $q \leqslant 4$.

We have the following possibilities: $q=4$ or $q=2$.

If $q=4$, then $85 a \leqslant 100$, which means $a=1$. From $(*)$ for $q=4$ and $a=1$, we get $d \notin \mathbb{N}$.

If $q=2$, then $15 a \leqslant 100$, which means $a \leqslant 6, a \in\{1,3,5\}$. Now we use $(*)$ to check which of the numbers $a \in\{1,3,5\}$ give an integer solution for $d$.

$$
\begin{aligned}
& 15 a=4 a+6 d+1 \\
& 6 d=11 a-1
\end{aligned}
$$

The last equation has an integer solution only for $a=5, d=9$.

For these values, the Wolves had the following points: 5, 10, 20, 40, and the Bears had 5, 14, 23, 32.

We need to calculate the total sum of points that both teams scored at halftime, i.e., $a+a q+a+a+d=3 a+a q+d$.

The total sum of points is $5+10+5+14=34$.

Answer: 34


Man ermittle alle diejenigen Zahlenfolgen $\left(a_{n}\right)$ mit $n=1,2,3, \ldots$, die die folgenden Bedingungen (1), (2) erfÃ¼llen:

(1) FÃ¼r alle ganzen Zahlen $m, n$ mit $n>m>0$ gilt $a_{n+m} \cdot a_{n-m}=a_{n}^{2}-a_{m}^{2}$.

(2) Es gilt $a_{1}=1$ und $a_{2}=\frac{5}{2}$.

Solution:
I. Wenn eine Zahlenfolge $\left(a_{n}\right)$ die Bedingungen erfüllt, so folgt durch vollständige Induktion, dass

$$
a_{n}=\frac{2}{3}\left(2^{n}-2^{-n}\right)
$$

for $n=1,2,3, \ldots$ gilt:

(a) Es gilt $a_{1}=1=\frac{2}{3}\left(2-\frac{1}{2}\right.$ und $a_{2}=\frac{5}{2}=\frac{2}{3}\left(4-\frac{1}{4}\right.$ also die Behauptung für $n=1$ und $n=2$.

(b) Wenn $k \geq 2$ ist und die Behauptung für $n=k$ und $n=k-1$ gilt, d.h. wenn

$$
a_{k}=\frac{2}{3}\left(2^{k}-2^{-k}\right) \quad, \quad a_{k-1}=\frac{2}{3}\left(2^{k-1}-2^{-k+1}\right)
$$

ist, so folgt: Wegen $k \geq 2$ ist $a_{k-1} \neq 0$, also ergibt sich aus (1) und (2)

$$
\begin{aligned}
a_{k+1} & =\frac{a_{k}^{2}-a_{1}^{2}}{a_{k-1}}=\frac{\frac{4}{9}\left(\left(2^{k}-2^{-k}\right)^{2}-\frac{9}{4}\right)}{\frac{2}{3}\left(2^{k-1}-2^{-k+1}\right)} \\
& =\frac{2}{3} \cdot \frac{2^{2 k}-4-\left(\frac{1}{4}-2^{-2 k}\right)}{2^{k-1}-2^{-k+1}}=\frac{2}{3}\left(2^{k+1}-2^{-k-1}\right)
\end{aligned}
$$

d.h. die Behauptung für $n=k+1$. Daher kann nur die durch (3) gegebene Folge die Bedingungen (1), (2) erfüllen.

II. Sie erfüllte sich auch, denn (2) und in I(a) bestätigt, und für alle ganzen $n>m>0$ folgt auch (3)

$$
\begin{aligned}
a_{n+m} \cdot a_{n-m} & =\frac{2}{3}\left(2^{n+m}-2^{-n-m}\right) \cdot \frac{2}{3}\left(2^{n-m}-2^{-n+m}\right) \\
& =\frac{4}{9}\left(2^{2 n}+2^{-2 n}-2+2-2^{2 m}-2^{-2 m}\right)=a_{n}^{2}-a_{m}^{2}
\end{aligned}
$$

Somit erfüllt genau die Folge (3) die Bedingungen der Aufgabe.

Übernommen von [5]

Answer: a_{n}=\frac{2}{3}(2^{n}-2^{-n})


If the surface areas of three cubes with integer edge lengths (unit: $\mathrm{cm}$) sum up to $564 \mathrm{~cm}^{2}$, then the sum of the volumes of these three cubes is ( ).
(A) $764 \mathrm{~cm}^{3}$ or $586 \mathrm{~cm}^{3}$
(B) $764 \mathrm{~cm}^{3}$
(C) $586 \mathrm{~cm}^{3}$ or $564 \mathrm{~cm}^{3}$
(D) $586 \mathrm{~cm}^{3}$

Solution:
4.A.

Let the edge lengths of the three cubes be $a$, $b$, and $c$. Then $6\left(a^{2}+b^{2}+c^{2}\right)=564 \Rightarrow a^{2}+b^{2}+c^{2}=94$.
Assume without loss of generality that $1 \leqslant a \leqslant b \leqslant c \leqslant 31$.
Thus, $6 \leqslant c < 10$. Therefore, $c$ can only be $9, 8, 7, 6$.
If $c=9$, then $a^{2}+b^{2}=94-9^{2}=13$. It is easy to see that $a=2, b=3$, yielding the solution $(a, b, c)=(2,3,9)$.

If $c=8$, then $a^{2}+b^{2}=94-64=30, b \leqslant 5$. But $2 b^{2} \geqslant 30, b \geqslant 4$, thus, $b=4$ or $5$. If $b=5$, then $a^{2}=5$ has no solution; if $b=4$, then $a^{2}=14$ has no solution. Thus, there is no solution in this case.

If $c=7$, then $a^{2}+b^{2}=94-49=45$, with the unique solution $a=3, b=6$.

If $c=6$, then $a^{2}+b^{2}=94-36=58$. In this case, $2 b^{2} \geqslant a^{2}+b^{2}=58, b^{2} \geqslant 29$. Hence, $b \geqslant 6$, but $b \leqslant c = 6$, so $b=6$. At this point, $a^{2}=58-36=22$, which has no solution.
In summary, there are two solutions:
$$
(a, b, c)=(2,3,9),(3,6,7).
$$

The volumes are $V_{1}=764 \mathrm{~cm}^{3}$ or $V_{2}=586 \mathrm{~cm}^{3}$.

Answer: A


A set $S$ of natural numbers is called good, if for each element $x \in S, x$ does not divide the sum of the remaining numbers in $S$. Find the maximal possible number of elements of a good set which is a subset of the set $A=\{1,2,3, \ldots, 63\}$.

Solution: Let set $B$ be the good subset of $A$ which have the maximum number of elements. We can easily see that the number 1 does not belong to $B$ since 1 divides all natural numbers. Based on the property of divisibility, we know that $x$ divides the sum of the remaining numbers if and only if $x$ divides the sum of all numbers in the set $B$. If $B$ has exactly 62 elements, then $B=\{2,3,4, \ldots, 62\}$, but this set can't be good since the sum of its elements is 2015 which is divisible by 5. Therefore $B$ has at most 61 elements. Now we are looking for the set, whose elements do not divide their sum, so the best way to do that is making a sum of elements be a prime number. $2+3+4+\ldots+63=2015$ and if we remove the number 4, we will obtain the prime number 2011. Hence the set $B=\{2,3,5,6,7, \ldots, 63\}$ is a good one. We conclude that our number is 61.

Answer: 61


During the construction of the Berlin city center, the "House of Teachers" is being built at Alexanderplatz. First, the excavation pit was dug out. About $7100 \mathrm{~m}^{3}$ of soil had to be removed:

a) How many dump truck loads would this have been, if one truck can transport $4 \mathrm{~m}^{3}$ of soil?

b) How long would the "dump truck convoy" have been for the entire transport, if each dump truck is $3 \mathrm{~m}$ long?

Solution:
a) The total volume of $7100 \mathrm{~m}^{3}$ is to be divided by the loading capacity of a dump truck of $4 \mathrm{~m}^{3}$: $\frac{7100 m^{3}}{4 m^{3}}=1775$.

There were 1775 dump truck loads.

b) Since each truck is $3 \mathrm{~m}$ long, the total length of all trucks is $3 \mathrm{~m} \cdot 1775=5325 \mathrm{~m}$.

Answer: 5325\mathrm{~}


In a positive non-constant geometric progression, the arithmetic mean of the second, seventh, and ninth terms is equal to some term of this progression. What is the minimum possible number of this term?

Solution:
Answer: 3

Solution:

The second element is either the larger or the smaller of the three specified, so it cannot be equal to the arithmetic mean of all three. The first element is even less suitable.

To prove that the answer "3" is possible, let's introduce the notation: let $b_{n}=$ $b q^{n-1}$. Then we need to solve the equation $3 b q^{2}=b q+b q^{6}+b q^{8}$. Simplifying it, we get $q^{7}+q^{5}-$ $3 q+1=0$.

This equation has a root $q=1$, but it does not suit us because the progression is not constant. $q^{7}+q^{5}-3 q+1=(q-1)\left(q^{6}+q^{5}+2 q^{4}+2 q^{3}+q^{2}+2 q-1\right)$. The second factor is negative at $q=0$ and positive at $q=1$, so this expression has a root between zero and one. Thus, we can choose the common ratio of the progression and take any positive number as the first term for the answer 3.

Answer: 3


In square $ABCD$, points $M$ and $N$ are on sides $AD$ and $CD$ respectively, and $MD=2AM, ND=3CN$. Then $\sin \angle MBN=$

Solution:
2.1. $\frac{11 \sqrt{170}}{170}$.
Connect $M N$. Without loss of generality, let $A D=12$. Then
$$
A M=4, D M=8, C N=3, D N=9 \text {. }
$$

Thus, $B M=\sqrt{A B^{2}+A M^{2}}=4 \sqrt{10}$,
$$
\begin{array}{l}
B N=\sqrt{B C^{2}+C N^{2}}=3 \sqrt{17} . \\
\text { Also, } S_{\triangle B B M}=24, S_{\triangle B C N}=18, S_{\triangle O M N}=36,
\end{array}
$$
$S_{\text {square } A B C D}=144$, hence,
$$
\begin{array}{l}
S_{\triangle B M N}=S_{\text {square } A B C D}-S_{\triangle A B M}-S_{\triangle B C N}-S_{\triangle D M N} \\
=144-24-18-36=66 .
\end{array}
$$

Therefore, $S_{\triangle B M N}=\frac{1}{2} B M \cdot B N \sin \angle M B N$
$$
\begin{array}{l}
\Rightarrow 66=\frac{1}{2} \times 4 \sqrt{10} \times 3 \sqrt{17} \sin \angle M B N \\
\Rightarrow \sin \angle M B N=\frac{11}{\sqrt{170}}=\frac{11 \sqrt{170}}{170} .
\end{array}
$$

Answer: \frac{11 \sqrt{170}}{170}


Suppose $n$ is a positive integer and $d$ is a single digit in base 10. Find $n$ if

$\frac{n}{810}=0.d25d25d25\ldots$

Solution:
Repeating decimals represent [rational numbers](https://artofproblemsolving.com/wiki/index.php/Rational_number).  To figure out which rational number, we sum an [infinite](https://artofproblemsolving.com/wiki/index.php/Infinite) [geometric series](https://artofproblemsolving.com/wiki/index.php/Geometric_series), $0.d25d25d25\ldots = \sum_{n = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}$.  Thus $\frac{n}{810} = \frac{100d + 25}{999}$ so $n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}$.  Since 750 and 37 are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime), $4d + 1$ must be [divisible](https://artofproblemsolving.com/wiki/index.php/Divisible) by 37, and the only digit for which this is possible is $d = 9$.  Thus $4d + 1 = 37$ and $n = \boxed{750}$.

(Note: Any repeating sequence of $n$ digits that looks like $0.a_1a_2a_3...a_{n-1}a_na_1a_2...$ can be written as $\frac{a_1a_2...a_n}{10^n-1}$, where $a_1a_2...a_n$ represents an $n$ digit number.)

Answer: 750


The hypotenuse $AB$ of the right triangle $ABC$ is a chord of a circle with radius 10. The vertex $C$ lies on the diameter of the circle, which is parallel to the hypotenuse, $\angle A=75^{\circ}$. Find the area of triangle $ABC$.

Solution:
From the center $O$ of the given circle, drop a perpendicular $OM$ to the hypotenuse $AB$. Then $MC = MA = MB$.

Therefore, $\angle MCB = \angle B = 15^{\circ}$, $\angle BCO = \angle B = 15^{\circ}$. Consequently, $\angle MCO = 30^{\circ}$.

Let $OM = x$. From the right triangle $MCO$, we find that $MC = 2x$. By the Pythagorean theorem, $OB^2 = OM^2 + MB^2$, or $100 = x^2 + 4x^2$, from which

$x^2 = 20$. Therefore, $S_{ABC} = \frac{1}{2} AB \cdot OM = 2x^2 = 40$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-01.jpg?height=301&width=509&top_left_y=890&top_left_x=777)

## Answer

40.

## [ Projections of the bases, sides, or vertices of a trapezoid Problem 54302 Topics: Area of a trapezoid [ $\underline{\text{ T }}$ trigonometric ratios in a right triangle ]

Difficulty: $3+$

Grades: 8,9

Given the larger base $a$ of an isosceles trapezoid, its height $h$, and the angle $\alpha$ at the base, find the area of the trapezoid.

## Hint

The projection of the diagonal of an isosceles trapezoid on the base is equal to the midline of the trapezoid.

## Solution

The projection of the lateral side on the larger base is $h \operatorname{ctg} \alpha$, and the midline of the trapezoid is $a - h \operatorname{ctg} \alpha$.

Therefore, the area of the trapezoid is $h(a - h \operatorname{ctg} \alpha)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-01.jpg?height=349&width=592&top_left_y=2072&top_left_x=733)

## Answer

$h(a - h \operatorname{ctg} \alpha)$

Answer: 40


Let $p_{1}, p_{2}, \ldots, p_{97}$ be prime numbers (not necessarily distinct). What is the largest integer value that the expression

$$
\sum_{i=1}^{97} \frac{p_{i}}{p_{i}^{2}+1}=\frac{p_{1}}{p_{1}^{2}+1}+\frac{p_{2}}{p_{2}^{2}+1}+\ldots+\frac{p_{97}}{p_{97}^{2}+1}
$$

can take?

Solution:
# Answer: 38

Solution: Note that the function $\frac{x}{x^{2}+1}$ decreases when $x \geqslant 2$, which means $\sum_{i=1}^{97} \frac{p_{i}}{p_{i}^{2}+1} \leqslant 97 \cdot \frac{2}{2^{2}+1}=$ $97 \cdot 0.4=38.8$. Therefore, the answer cannot be greater than 38.

At the same time, $\frac{3}{3^{2}+1}=0.3$, which is 0.1 less than the value for $p=2$. Therefore, if in the above maximum sum, eight twos are replaced by threes, the result will be exactly 38.

Answer: 38