Given a sequence of positive numbers $\left\{a_{n}\right\}$ satisfying $a_{n+1} \geqslant 2 a_{n}+1$, and $a_{n}<2^{n+1}$ for $n \in \mathbf{Z}_{+}$. Then the range of $a_{1}$ is $\qquad$ .

Solution: $\begin{array}{l}\text { 1. }(0,3] . \\ \text { From } a_{n+1}+1 \geqslant 2\left(a_{n}+1\right) \\ \Rightarrow a_{n}+1 \geqslant\left(a_{1}+1\right) 2^{n-1} . \\ \text { Therefore }\left(a_{1}+1\right) 2^{n-1}-1 \leqslant a_{n}<2^{n+1} \\ \Rightarrow a_{1}+1<\frac{2^{n+1}+1}{2^{n-1}}=4+\frac{1}{2^{n-1}} \\ \Rightarrow a_{1}+1 \leqslant 4 \\ \Rightarrow 0<a_{1} \leqslant 3 .\end{array}$

Answer: 0<a_{1}\leqslant3


Given that the odd function $f(x)$ is a monotonically decreasing function on $[-1,0]$, and $\alpha, \beta$ are two interior angles of an acute triangle, the relationship in size between $f(\cos \alpha)$ and $f(\sin \beta)$ is $\qquad$ .

Solution:
11. $f(\cos \alpha)>f(\sin \beta)$.

Given that $f(x)$ is a decreasing function on $[-1,0]$, it is also decreasing on $[0,1]$, and we have $\pi-(\alpha+\beta)\frac{\pi}{2}$. Therefore, $\frac{\pi}{2}>\beta>\frac{\pi}{2}-\alpha>0$, which implies $\sin \beta>\cos \alpha$, and $0<\sin \beta<1,0<\cos \alpha<1$. Hence, $f(\sin \beta)<f(\cos \alpha)$.

Answer: f(\cos\alpha)>f(\sin\beta)


The sum of four positive real numbers is 8, and the sum of their squares is 40. Determine the maximum possible value that one of these numbers can take.

Solution:
## Solution.

Let $a, b, c, d$ be the observed numbers. Without loss of generality, we can assume that $d$ is the largest of them. From $a+b+c+d=8$ and $a^{2}+b^{2}+c^{2}+d^{2}=40$ it follows that

$$
a+b+c=8-d, \quad a^{2}+b^{2}+c^{2}=40-d^{2}
$$

If we square the first equation and apply the second, we get

$$
\begin{gathered}
(a+b+c)^{2}=(8-d)^{2} \\
a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=(8-d)^{2} \\
40-d^{2}+2(a b+b c+c a)=64-16 d+d^{2} \\
a b+b c+c a=12-8 d+d^{2}
\end{gathered}
$$

It is known that

$$
a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a
$$

(easily derived from the inequality $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0$ ).

If we apply this inequality $(**)$ and $(***)$, we get:

$$
\begin{gathered}
a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a \\
40-d^{2} \geqslant 12-8 d+d^{2} \\
d^{2}-4 d-14 \leqslant 0
\end{gathered}
$$

The solutions to the equation $d^{2}-4 d-14=0$ are $d=2 \pm 3 \sqrt{2}$, and the solution to the above inequality is $d \in[2-3 \sqrt{2}, 2+3 \sqrt{2}]$. The largest real number $d$ in this interval is $2+3 \sqrt{2}$.

We need to check if there exist positive real numbers $a, b$, and $c$ that, along with $d=2+3 \sqrt{2}$, satisfy both given equations.

If we set $a=b=c=\frac{8-d}{3}=2-\sqrt{2}$, it is easy to verify that indeed $a^{2}+b^{2}+c^{2}+d^{2}=$ $3(2-\sqrt{2})^{2}+(2+3 \sqrt{2})^{2}=40$.

The maximum possible value of one of the observed numbers is $2+3 \sqrt{2}$.

Note: Instead of inequality $(***)$, we can apply the inequality between the arithmetic and quadratic means:

$$
\begin{aligned}
& \frac{a+b+c}{3} \leqslant \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}} \\
& (a+b+c)^{2} \leqslant 3\left(a^{2}+b^{2}+c^{2}\right) \\
& (8-d)^{2} \leqslant 3\left(40-d^{2}\right) \\
& d^{2}-4 d-14 \leqslant 0
\end{aligned}
$$

Answer: 2+3\sqrt{2}


The positive integer solutions $(x, y)$ of the equation $2 x^{2}-x y-3 x+y+2006=0$ are $\qquad$ pairs.

Solution:
12.4.

From $2 x^{2}-x y-3 x+y+2006=0$, we get $y=\frac{2 x^{2}-3 x+2006}{x-1}=2 x-1+\frac{2005}{x-1}$.

Therefore, $x-1$ can take the values $1,5,401,2005$. Hence, the equation has 4 pairs of positive integer solutions.

Answer: 4


If every student has watched every movie at most once, at least how many different movie collections can these students have?

Solution:
1. **Understanding the Problem:**
   We have a complete graph \( K_{23} \) where each vertex represents a student and each edge represents a movie watched by the pair of students connected by that edge. We need to determine the minimum number of different movie collections among the students.

2. **Graph Theory Application:**
   In \( K_{23} \), each vertex has a degree of \( 22 \) because each student has watched a movie with every other student. We need to color the edges such that no two edges incident to the same vertex share the same color. This is equivalent to finding an edge coloring of the graph.

3. **Edge Coloring Theorem:**
   According to Vizing's Theorem, the chromatic index (the minimum number of colors needed to color the edges of a graph such that no two edges incident to the same vertex share the same color) of a simple graph \( G \) is either \( \Delta(G) \) or \( \Delta(G) + 1 \), where \( \Delta(G) \) is the maximum degree of the graph.

4. **Applying Vizing's Theorem:**
   For \( K_{23} \), the maximum degree \( \Delta(G) \) is \( 22 \). Therefore, the chromatic index is either \( 22 \) or \( 23 \). Since \( K_{23} \) is a complete graph, it falls into the category of graphs where the chromatic index is \( \Delta(G) + 1 \). Thus, the chromatic index is \( 23 \).

5. **Interpreting the Result:**
   This means we need at least \( 23 \) different colors to color the edges of \( K_{23} \). Each color represents a different movie. Therefore, there are at least \( 23 \) different movies.

6. **Movie Collections:**
   Each student has watched \( 22 \) movies, and each movie collection is unique to each student. Since there are \( 23 \) students, and each student has a unique collection of \( 22 \) movies out of the \( 23 \) different movies, the minimum number of different movie collections is \( 23 \).

\[
\boxed{23}
\]

Answer: 23


A store has 126 boxes of apples, with each box containing at least 120 and at most 144 apples. Now, boxes with the same number of apples are grouped together. Let the largest group of boxes have $n$ boxes, then the minimum value of $n$ is $\qquad$ .

Solution:
5.6.

Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.

Answer: 6


Given $n>2$ natural numbers, among which there are no three equal, and the sum of any two of them is a prime number. What is the largest possible value of $n$?

Solution:
1. Answer: 3. Note that the triplet $1,1,2$ satisfies the condition. Suppose there are more than 3 numbers. Consider any 4 of them a, b, c, d. Among these four numbers, there cannot be two even numbers, otherwise their sum would be an even prime greater than two. Therefore, at least 3 of them must be odd. Their pairwise sums are even primes. Then all three considered numbers must be 1, which violates the condition.

Grading recommendations:

only answer - 0 points

example for three numbers provided - 2 points.

Answer: 3


Solution:
Answer: $\frac{27}{1081}$
Solution: Note that
$$
\begin{aligned}
\frac{x^{3} y^{3}}{x^{6}+3 x^{4} y^{2}+3 x^{3} y^{3}+3 x^{2} y^{4}+y^{6}} & =\frac{x^{3} y^{3}}{\left(x^{2}+y^{2}\right)^{3}+3 x^{3} y^{3}} \\
& =\frac{1}{\frac{\left(x^{2}+y^{2}\right)^{3}}{x^{3} y^{3}}+3} \\
& =\frac{1}{\left(\frac{x^{2}+y^{2}}{x y}\right)^{3}+3} \\
& =\frac{1}{\left(\frac{x}{y}+\frac{y}{x}\right)^{3}+3} .
\end{aligned}
$$

Let $u=\frac{x}{y}$. We have $\frac{x}{y}+\frac{y}{x}=u+\frac{1}{u}$. To get the minimum value of the entire expression, we need to make $u+\frac{1}{u}$ as large as possible. We can do this by setting $x=\frac{1}{2}$ and $y=\frac{3}{2}$. The function $f(u)=u+\frac{1}{u}$ is increasing on $(1,+\infty)$. Therefore $\frac{x}{y}+\frac{y}{x}=\frac{1}{3}+3=\frac{10}{3}$ and the minimum is
$$
\frac{1}{\left(\frac{x}{y}+\frac{y}{x}\right)^{3}+3}=\frac{1}{\left(\frac{10}{3}\right)^{3}+3}=\frac{27}{1081}
$$

Answer: \frac{27}{1081}


Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations

$$
\left\{\begin{array}{l}
\text { GCD }(a ; b ; c)=33, \\
\text { LCM }(a ; b ; c)=3^{19} \cdot 11^{15} .
\end{array}\right.
$$

Solution:
Answer: 9072.

Solution. Let $a=3^{\alpha_{1}} \cdot 11^{\alpha_{2}}, b=3^{\beta_{1}} \cdot 11^{\beta_{2}}, c=3^{\gamma_{1}} \cdot 11^{\gamma_{2}}$ (the numbers $a, b, c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,

$$
\operatorname{LCM}(a ; b ; c)=3^{\max \left(\alpha_{1} ; \beta_{1} ; \gamma_{1}\right)} \cdot 11^{\max \left(\alpha_{2} ; \beta_{2} ; \gamma_{2}\right)}, \quad \text { GCD }(a ; b ; c)=3^{\min \left(\alpha_{1} ; \beta_{1} ; \gamma_{1}\right)} \cdot 11^{\min \left(\alpha_{2} ; \beta_{2} ; \gamma_{2}\right)}
$$

Considering the system given in the condition, we obtain the relations

$$
\left\{\begin{array} { l }
{ \operatorname { max } ( \alpha _ { 1 } ; \beta _ { 1 } ; \gamma _ { 1 } ) = 1 9 , } \\
{ \operatorname { min } ( \alpha _ { 1 } ; \beta _ { 1 } ; \gamma _ { 1 } ) = 1 }
\end{array} \quad \text { and } \quad \left\{\begin{array}{l}
\max \left(\alpha_{2} ; \beta_{2} ; \gamma_{2}\right)=15 \\
\min \left(\alpha_{2} ; \beta_{2} ; \gamma_{2}\right)=1
\end{array}\right.\right.
$$

Consider the first system (37). The following sets of numbers $\left(\alpha_{1} ; \beta_{1} ; \gamma_{1}\right)$ are possible:

$(1 ; 1 ; 19)$ - 3 sets (due to different permutations of these numbers);

$(1 ; 19 ; 19)$ - also three sets

$(1 ; k ; 19)$, where $2 \leqslant k \leqslant 18-$ there are 17 different values of $k$, and for each of them 6 permutations - a total of 102 options.

Thus, there are $3+3+6 \cdot 17=108$ ways to choose the triplet of numbers $\left(\alpha_{1} ; \beta_{1} ; \gamma_{1}\right)$. Similarly, we establish that there are $3+3+6 \cdot 13=84$ ways to choose $\left(\alpha_{2} ; \beta_{2} ; \gamma_{2}\right)$, and since one choice is made independently of the other, the total number of ways is $108 \cdot 84=9072$.

Answer: 9072


In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=30^{\circ}, A C=2, M$ is the midpoint of $A B$. $\triangle A C M$ is folded along $C M$, making the distance between $A$ and $B$ equal to $2 \sqrt{2}$. At this point, the volume of the tetrahedron $A-B C M$ is $\qquad$ .

Solution:
Solve: Fill in $\frac{2}{3} \sqrt{2}$. Reason: To find the volume of the tetrahedron after folding, the key is to determine the height of the tetrahedron. It can be observed that after folding, $\triangle A B C$ is a right triangle $(A C=2, B C=2 \sqrt{3}, A B=2 \sqrt{2})$, which helps solve the problem.

The folded tetrahedron is shown in Figure 5-12. Take the midpoint $D$ of $C M$ and connect $A D$. In $\triangle B C M$, draw $D E \perp C M$ intersecting $B C$ at $E$, and connect $A E$. Then,
$$
A D=2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3}, D E=C D \cdot \tan 30^{\circ}=1 \cdot \frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{3}, C E=
$$
$2 D E=\frac{2 \sqrt{3}}{3}$. In $\triangle A B C$, $A C=2, A B=2 \sqrt{2}, B C=2 \sqrt{3}$, so
$A C^{2}+A B^{2}=B C^{2}$, thus $\angle B A C=90^{\circ}$. In $\triangle A C E$, $A E^{2}=A C^{2}+$
$C E^{2}-2 A C \cdot C E \cos \angle A C E=4+\frac{4}{3}-2 \cdot 2 \cdot \frac{2}{3} \cdot \sqrt{3} \cdot \frac{2}{2 \sqrt{3}}=$
$\frac{8}{3}$, so $A E^{2}+C E^{2}=\frac{8}{3}+\frac{4}{3}=4=A C^{2}$, and $A E^{2}+D E^{2}=\frac{8}{3}+\frac{1}{3}=3=A D^{2}$, therefore $A E \perp B C, A E \perp$ $D E$, hence $A E \perp$ plane $B C M$. Therefore, $V_{A-B C M}=\frac{1}{3} \cdot A E \cdot S_{\triangle H C M}=\frac{1}{3} \cdot \frac{2 \sqrt{6}}{3} \cdot \frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{\sqrt{3}}{2}=\frac{2 \sqrt{2}}{3}$.

Answer: \frac{2\sqrt{2}}{3}


The number $\text{N}$ is between $9$ and $17$. The average of $6$, $10$, and $\text{N}$ could be
$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution:
We know that $9<N<17$ and we wish to bound $\frac{6+10+N}{3}=\frac{16+N}{3}$.
From what we know, we can deduce that $25<N+16<33$, and thus \[8.\overline{3}<\frac{N+16}{3}<11\]
The only answer choice that falls in this range is choice $\boxed{\text{B}}$

Answer: B


The sides and vertices of a pentagon are labelled with the numbers $1$ through $10$ so that the sum of the numbers on every side is the same. What is the smallest possible value of this sum?

Solution:
1. Let \( a_1, a_2, \ldots, a_{10} \) be a permutation of \( 1, 2, \ldots, 10 \) where \( a_1, a_2, \ldots, a_5 \) are on the vertices and the other 5 numbers are on the sides. We need to find the smallest possible value of the sum \( S \) such that the sum of the numbers on every side of the pentagon is the same.

2. The sum of the numbers on each side of the pentagon can be expressed as:
   \[
   S = a_i + a_{i+1} + b_i
   \]
   where \( a_i \) and \( a_{i+1} \) are the numbers on the vertices and \( b_i \) is the number on the side between \( a_i \) and \( a_{i+1} \).

3. Since there are 5 sides, the total sum of the numbers on all sides is:
   \[
   5S = 2(a_1 + a_2 + a_3 + a_4 + a_5) + (a_6 + a_7 + a_8 + a_9 + a_{10})
   \]
   This equation accounts for each vertex number being counted twice and each side number being counted once.

4. The sum of all numbers from 1 to 10 is:
   \[
   1 + 2 + \cdots + 10 = \frac{10 \cdot 11}{2} = 55
   \]

5. Substituting the total sum into the equation, we get:
   \[
   5S = 2(a_1 + a_2 + a_3 + a_4 + a_5) + (55 - (a_1 + a_2 + a_3 + a_4 + a_5))
   \]
   Simplifying, we have:
   \[
   5S = 55 + (a_1 + a_2 + a_3 + a_4 + a_5)
   \]

6. To minimize \( S \), we want to minimize \( a_1 + a_2 + a_3 + a_4 + a_5 \). The smallest possible sum of 5 distinct numbers from 1 to 10 is:
   \[
   1 + 2 + 3 + 4 + 5 = 15
   \]

7. Substituting this into the equation, we get:
   \[
   5S = 55 + 15 = 70 \implies S = \frac{70}{5} = 14
   \]

8. To verify that \( S = 14 \) is achievable, we need to find a permutation of \( 1, 2, \ldots, 10 \) such that the sum of the numbers on each side is 14. One such permutation is:
   \[
   1 - 9 - 4 - 8 - 2 - 7 - 5 - 6 - 3 - 10 - 1
   \]
   Checking the sums:
   \[
   1 + 9 + 4 = 14, \quad 9 + 4 + 8 = 14, \quad 4 + 8 + 2 = 14, \quad 8 + 2 + 7 = 14, \quad 2 + 7 + 5 = 14, \quad 7 + 5 + 6 = 14, \quad 5 + 6 + 3 = 14, \quad 6 + 3 + 10 = 14, \quad 3 + 10 + 1 = 14
   \]
   All sums are 14, confirming that \( S = 14 \) is achievable.

The final answer is \( \boxed{14} \).

Answer: 14


Given that three points $A, B, C$ on the parabola $y=x^{2}$ form an isosceles right triangle with vertex $B$ being the right-angle vertex.
(1) If the slope of line $B C$ is $k$, find the coordinates of vertex $B$;
(2) When the slope $k$ of line $B C$ varies, find the minimum area of the isosceles right triangle $A B C$.

Solution:
15. (1) Let the coordinates of points $A, B, C$ be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$. Since the slope of line $BC$ is $k$, the slope of line $AB$ is $-\frac{1}{k}$. Therefore, we have $y_{3}-y_{2}=k\left(x_{3}-x_{2}\right), y_{1}-y_{2}=-\frac{1}{k}\left(x_{1}-x_{2}\right)$.

Since points $A, B, C$ are on the parabola, we have $y_{1}=x_{1}^{2}, y_{2}=x_{2}^{2}, y_{3}=x_{3}^{2}$. Substituting these into the above equations, we get $x_{3}+x_{2}=k, x_{1}+x_{2}=-\frac{1}{k}$. Thus, $x_{3}=k-x_{2}, x_{1}=-\frac{1}{k}-x_{2}$.
Given that $|AB|=|BC|$, we have $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}=\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}}$,

which simplifies to $\sqrt{1+\frac{1}{k^{2}}}\left(x_{2}-x_{1}\right)=\sqrt{1+k^{2}}\left(x_{3}-x_{2}\right), x_{2}-x_{1}=k\left(x_{3}-x_{2}\right)$.
Therefore, $x_{2}-\left(-\frac{1}{k}-x_{2}\right)=k\left(k-x_{2}-x_{2}\right)$, which simplifies to $x_{2}=\frac{k^{3}-1}{2 k(k+1)}$. Hence, the coordinates of point $B$ are $\left(\frac{k^{3}-1}{2 k(k+1)}, \frac{\left(k^{3}-1\right)^{2}}{4 k^{2}(k+1)^{2}}\right)$. Since $1 \geqslant 0$, we have $k \geqslant 1$.
The length of the legs of the isosceles right triangle is
$$
\begin{aligned}
\sqrt{k^{2}+1}\left(x_{1}-x_{2}\right) & =\sqrt{k^{2}+1}\left(k-2 x_{2}\right) \\
& =\sqrt{k^{2}+1}\left(k-\frac{k^{3}-1}{k(k+1)}\right) \\
& =\frac{k^{2}+1}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \\
& \geqslant \frac{2 k}{k} \cdot \frac{\sqrt{\frac{(k+1)^{2}}{2}}}{k+1} \\
& =\sqrt{2} .
\end{aligned}
$$

Equality holds if and only if $k=1$, i.e., when point $B$ is at the origin. Therefore, the minimum area of the isosceles right triangle is 1.

Answer: 1


A group of adventurers is showing off their loot. It is known that exactly 5 adventurers have rubies; exactly 11 have emeralds; exactly 10 have sapphires; exactly 6 have diamonds. In addition, it is known that

- if an adventurer has diamonds, then they have either emeralds or sapphires (but not both at the same time);
- if an adventurer has emeralds, then they have either rubies or diamonds (but not both at the same time).

What is the minimum number of adventurers that can be in such a group?

Solution:
Answer: 16.

Solution. Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or diamonds. Then, from the second condition, it follows that 5 adventurers have rubies and emeralds, and 6 have emeralds and diamonds. That is, every adventurer who has diamonds must also have emeralds. Then, from the first condition, there cannot be an adventurer who has both sapphires and diamonds. Therefore, there are at least $10+6=16$ adventurers.

Indeed, there can be this many adventurers: let's say we have 6 adventurers who have emeralds and diamonds, 5 adventurers who have rubies, emeralds, and sapphires, and 5 adventurers who have only sapphires. One can verify that this example meets all the conditions.

Answer: 16


The selling price of a commodity is an integer number of yuan. 100 yuan can buy at most 3 pieces. Person A and Person B each brought several hundred-yuan bills. The money A brought can buy at most 7 pieces of this commodity, and the money B brought can buy at most 14 pieces. When the two people's money is combined, they can buy 1 more piece. The price of each piece of this commodity is yuan.

Solution:
【Analysis】100 yuan can buy at most 3 items, so the price of the goods is between 26 yuan and 33 yuan (including 26 yuan and 33 yuan), because if it is less than 25 yuan, then 4 items can be bought, and if it is more than 33 yuan, then 3 items cannot be bought. Person A can buy at most 7 items with the money they have, so A has 200 yuan (100 yuan can only buy 3 items, and 300 yuan can buy at least 9 items), and the price of the goods is between 26 yuan and 28 yuan (including 26 yuan and 28 yuan), because if it is more than 29 yuan, then 7 items cannot be bought. Person B can buy at most 14 items with the money they have, so B has 400 yuan (200 yuan can buy at most 7 items, and adding another 100 yuan can buy at most 3 + 1 = 4 more items, making only 11 items, while 500 yuan can buy at least 15 items), and the price of the goods is between 27 yuan and 28 yuan (including 27 yuan and 28 yuan), because if it is less than 26 yuan, then 15 items can be bought.
Finally, when A and B's money is combined, a total of 600 yuan can buy 22 items. It is found that if the price is 28 yuan, the total amount exceeds 600 yuan, so only 27 yuan meets all the requirements.
In summary, the price of each item of this product is 27 yuan.

Answer: 27


A circle $\omega$ is circumscribed around triangle $ABC$. A line tangent to $\omega$ at point $C$ intersects the ray $BA$ at point $P$. On the ray $PC$ beyond point $C$, a point $Q$ is marked such that $PC = QC$. The segment $BQ$ intersects the circle $\omega$ again at point $K$. On the smaller arc $BK$ of the circle $\omega$, a point $L$ is marked such that $\angle LAK = \angle CQB$. Find the angle $\angle PCA$, given that $\angle ALQ = 60^{\circ}$.

Solution:
Answer: $30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-37.jpg?height=449&width=462&top_left_y=398&top_left_x=931)

Solution. Inscribed angles $L A K$ and $L B K$ subtend the arc $L K$ and are therefore equal. Given that $\angle L B Q = \angle B Q C$, it follows that $B L \| P Q$. Then the line $\ell$, passing through point $C$ and perpendicular to the tangent $P Q$, contains the diameter of the circle $\omega$ and is perpendicular to its chord $B L$. Therefore, points $B$ and $L$ are symmetric with respect to $\ell$, meaning that $\ell$ is the axis of symmetry of the trapezoid $P Q L B$. From this, it follows that $B P = L Q$ and $B C = L C$. Hence, triangles $Q L C$ and $P B C$ are congruent by the three sides, so $\angle Q L C = \angle A B C$. Additionally, $\angle A L C = \angle A B C$ as inscribed angles subtending the same chord. Therefore, $L C$ is the angle bisector of $\angle A L Q$, and

$$
\angle A B C = \angle A L C = \frac{1}{2} \angle A L Q = 30^{\circ}.
$$

Note that the angle between the chord $A C$ and the tangent $C P$ is equal to the inscribed angle subtending $A C$. Therefore, $\angle P C A = \angle A B C = 30^{\circ}$.

Answer: 30


Three beads with masses \( m_{1}=150 \) g, \( m_{3}=30 \) g, \( m_{2}=1 \) g (see figure) can slide along a horizontal rod without friction.

Determine the maximum speeds of the larger beads if at the initial moment of time they were at rest, while the small bead was moving with a speed of \( V=10 \) m/s.

Consider the collisions to be perfectly elastic.

![](https://cdn.mathpix.com/cropped/2024_05_06_dad796a08c49b07fdecag-4.jpg?height=171&width=857&top_left_y=1045&top_left_x=628)

Solution:
Solution. After each collision, the magnitude of the velocity of the bead with mass $m_{2}$ decreases. After a certain number of collisions, its velocity will be insufficient to catch up with the next bead $m_{1}$ or $m_{3}$. After such a final collision, the velocities of the beads will no longer change. Let $V_{1}$ and $V_{3}$ be the magnitudes of the velocities of the beads $m_{1}$ and $m_{3}$, respectively, and $V_{2}$ be the projection of the velocity of the bead $m_{2}$ onto the direction of motion of the bead $m_{3}$ (after the final collision).

According to the laws of conservation of momentum and energy, we get:

$$
\begin{aligned}
& -m_{1} V_{1}+m_{2} V_{2}+m_{3} V_{3}=m_{2} V \\
& \frac{m_{1} V_{1}^{2}}{2}+\frac{m_{2} V_{2}^{2}}{2}+\frac{m_{3} V_{3}^{2}}{2}=\frac{m_{2} V^{2}}{2}
\end{aligned}
$$

Since, according to the problem statement, $m_{2} \ll m_{1}$ and $m_{2} \ll m_{3}$, the terms containing $m_{2}$ can be neglected in the left-hand sides of equations (1) and (2).

Indeed, if, for example, after the final collision, the bead $m_{2}$ moves in the direction of the bead $m_{1}$, then $\left|V_{2}\right| \leq V_{1}$ and therefore $\frac{m_{2}\left|V_{2}\right|}{m_{1} V_{1}} \ll 1, \frac{m_{2} V_{2}^{2}}{m_{1} V_{1}^{2}} \ll 1$. Similarly, the situation is considered when the bead $m_{2}$ moves after the final collision in the direction of the bead $m_{3}$.

Thus, we obtain the system of equations:

$$
\left\{\begin{array}{l}
-m_{1} V_{1}+m_{3} V_{3}=m_{2} V \\
m_{1} V_{1}^{2}+m_{3} V_{3}^{2}=m_{2} V^{2}
\end{array}\right.
$$

Denoting $m_{2}=m$, then $m_{3}=n m, m_{1}=5 n m$, where $n=30$, we write the system as:

$$
\left\{\begin{array}{c}
-5 n V_{1}+n V_{3}=V \\
5 n V_{1}^{2}+n V_{3}^{2}=V^{2}
\end{array}\right.
$$

Solving this system, we find

$$
\begin{gathered}
\frac{V_{1}}{V}=-\frac{1}{6 n}+\sqrt{\frac{1}{30 n}-\frac{5}{30^{2} n^{2}}} \approx-\frac{1}{6 n}+\sqrt{\frac{1}{30 n}}=-\frac{1}{180}+\frac{1}{30}=\frac{1}{36} \\
\frac{V_{3}}{V}=\frac{1}{6 n}+\sqrt{\frac{30}{36 n}-\frac{5}{36 n^{2}}} \approx \frac{1}{6 n}+\sqrt{\frac{30}{36 n}}=\frac{1}{180}+\frac{1}{6}=\frac{31}{180}
\end{gathered}
$$

Answer: $V_{1} \approx 0.28 \mathrm{~m} / \mathrm{c} ; V_{3} \approx 1.72 \mathrm{~m} / \mathrm{c}$.

Answer to variant 212: $V_{1} \approx 0.28 \mathrm{~m} / \mathrm{c} ; V_{3} \approx 1.72 \mathrm{~m} / \mathrm{c}$.

Answer: V_{1}\approx0.28\mathrm{~}/\mathrm{};V_{3}\approx1.72\mathrm{~}/\mathrm{}


We are looking for all four-digit numbers (in decimal notation) with the digit sequence $\{a ; b ; c ; d\}$ where $a ; b ; c ; d \in N, 1 \leq a \leq 9,0 \leq b ; c ; d \leq 9$, whose sixfold has the digit sequence $\{a ; a ; c ; b ; d\}$.

Solution:
According to the problem statement,

$$
6(1000 a+100 b+10 c+d)=11000 a+100 c+10 b+d
$$

or (after appropriate rearrangement): $5 d=10(500 a+4 c-59 b)$.

It immediately follows that $d$ is an even number: $d=2 e$ with $e \in N, 0 \leq e \leq 4$; thus, the equation takes the form

$$
e=500 a+4 c-59 b \quad \text { or } \quad 59 b=500 a+4 c-e
$$

Since $a \geq 1$, it follows that $59 b \geq 496$, so $b>8$. Therefore, $b=9,531=500 a+4 c-e$.

Since $4 c-e<100$, it follows that $a=1$ and from this

$$
31=4 c-e \quad ; \quad c=7+\frac{3+e}{4}
$$

Thus, $e=1, d=2, c=8$. Indeed, $6 \cdot 1982=11892$.

Answer: 1982


Find all natural numbers \(n\) such that \(n^{2}\) does not divide \(n!\).

Solution:
\section*{Solution}

Answer: \(\mathrm{n}=4\) or prime.

If \(\mathrm{n}=\mathrm{rs}\), with \(12\), then \(r<2 r<n\), and hence \(n^{2}\) divides \(n!\). This covers all possibilities except \(n=4\) or \(n=\) prime, and it is easy to see that in these cases \(n^{2}\) does not divide \(n!\).

Answer: 4


Mom has two apples and three pears. Every day for five consecutive days, she gives out one fruit. In how many ways can this be done?

Solution:
56. Here is one way to distribute apples and pears: 000 'on the 1st and 3rd day - apples, on the 2nd, 4th, and 5th days - pears),

## 000

(on the first three days - pears, on the last two days - apples). Thus, we need to count all the tables with two light and three dark circles; but we have already done this when solving problem 44. The answer is $C_{5}^{2}=10$

Answer: 10


A taxi company's repair station has 7 taxis to be repaired. If 1 worker repairs these 7 cars, the repair times are $12, 17, 8, 18, 23, 30, 14$ minutes, respectively. Each taxi incurs an economic loss of 2 yuan for every minute it is out of service. Now, 3 workers with the same work efficiency are working separately. To minimize the economic loss, what is the minimum loss in yuan?

Solution: 【Analysis】To minimize the loss, the total time spent on repairing and waiting for each car should be minimized. With 3 workers repairing 7 cars, to minimize the total, each worker should repair as close to the same number of cars as possible, so 2 workers each repair 2 cars, and 1 worker repairs 3 cars. When the worker repairing 3 cars works on the first car, 1 car is being repaired and 2 cars are waiting, so the time is counted 3 times. When working on the second car, 1 car is being repaired and 1 car is waiting, so the time is counted 2 times. When working on the third car, 1 car is being repaired, so the time is counted 1 time. For the workers repairing 2 cars, when working on the first car, 1 car is being repaired and 1 car is waiting, so the time is counted 2 times. When working on the second car, 1 car is being repaired, so the time is counted 1 time. Therefore, the repair time for 1 car is counted 3 times, the repair time for 3 cars is counted 2 times, and the repair time for 3 cars is counted 1 time. Thus, the minimum total time is $8 \times 3+(1) 2+14+17 \times 2+18+23+30=181$ minutes, and the minimum loss is $2 \times 181=362$ yuan.

Answer: 362


A team of mechanics can complete a certain task of processing parts 15 hours faster than a team of apprentices. If the team of apprentices works for 18 hours on this task, and then the team of mechanics continues working on the task for 6 hours, only 0.6 of the entire task will be completed. How much time does the team of apprentices need to complete this task independently?

Solution:
## Solution.

Let $x$ hours be necessary for the students to complete the task; $x-15$ hours for the team of mechanics. In 18 hours, the students will complete $\frac{1}{x} \cdot 18$ of the entire task, and in 6 hours, the mechanics will complete $\frac{6}{x-15}$ of the entire task. According to the condition, $\frac{18}{x}+\frac{6}{x-15}=0.6$, from which $x=45$ hours.

Answer: 45 hours.

Answer: 45


Use five different colors to color the five vertices of the "pentagram" in Figure 1 (each vertex is colored with one color, and some colors may not be used), so that the two vertices on each line segment are of different colors. Then the number of different coloring methods is .

Solution:
12.1020.

Convert it into a coloring problem of a disk with 5 sectors, each to be colored with one of 5 colors, such that no two adjacent sectors have the same color (as shown in Figure 4).

Let the number of ways to color a disk with $k$ sectors using 5 colors be $x_{k}$. Then, we have $x_{k}+x_{k-1}=5 \times 4^{k-1}$. Therefore,
$$
\begin{array}{l}
x_{5}=\left(x_{5}+x_{4}\right)-\left(x_{4}+x_{3}\right)+ \\
\left(x_{3}+x_{2}\right)-x_{2} \\
=5\left(4^{4}-4^{3}+4^{2}-4\right)=1020 .
\end{array}
$$

Answer: 1020


To guard the object, the client agreed with the guards on the following: all of them will indicate the time intervals of their proposed shifts with the only condition that their union should form a predetermined time interval set by the client, and he will choose any set of these intervals that also satisfies the same condition, and pay for the work at a rate of 300 rubles per hour for each guard. What is the longest time interval the client can set to ensure that he definitely stays within 90000 rubles?

Solution: 6. $\frac{90000}{2 \cdot 300}=150 \text{h}$.

Answer: 150


Determine all prime numbers $p$ such that $p^{2}-6$ and $p^{2}+6$ are both prime numbers.

Solution:
Solution: If $p>5$ then the units digit of $p$ must be $1,3,7$ or 9 .
- If the units digit of $p$ is 1 or 9 then the units digit of $p^{2}$ is 1 . Therefore the units digit of $p^{2}-6$ is 5 . Since $p^{2}-6>5$ this means that $p^{2}-6$ is not prime.
- If the units digit of $p$ is 3 or 7 then the units digit of $p^{2}$ is 9 . Therefore the units digit of $p^{2}+6$ is 5 . Since $p^{2}+6>5$ this means that $p^{2}+6$ is not prime.

Therefore we must have $p \leq 5$. Hence $p$ must be 2,3 or 5 .
- If $p=2$ then $p^{2}+6=10$ is not prime.
- If $p=3$ then $p^{2}+6=15$ is not prime.
- If $p=5$ then $p^{2} \pm 6$ are 19 and 31 which are both prime.

Therefore the only answer is $p=5$.
Alternative Solution: Consider the following product modulo 5 .
$$
p\left(p^{2}-6\right)\left(p^{2}+6\right)=p^{5}-36 p \equiv p^{5}-p(\bmod 5)
$$

By Fermat's Little Theorem, this product is $0(\bmod 5)$. So if $p, p^{2}-6$ and $p^{2}+6$ are all prime numbers then at least one of them must be equal to 5 .
- If $p=5$ then $6 p^{2}-1=29$ and $6 p^{2}+1=31$. This is one solution.
- If $6 p^{2}-1=5$ then $p= \pm 1$. Neither 1 nor -1 is prime, so this case leads to no solutions.
- If $6 p^{2}+1=5$ then $p$ is not an integer. No solutions in this case.

Therefore the only solution is $p=5$.

Answer: 5


As shown in the figure, 5 identical small rectangular prisms are exactly assembled into a rectangular prism with a volume of 22.5. Then, the sum of all edge lengths of one small rectangular prism is.

Solution:
22
【Analysis】According to the graph, we can get length $=3 \times$ height, length $=2 \times$ width, length: width: height $=6: 3: 2$. Let the length of the cuboid be $6 x$, the width be $3 x$, and the height be $2 x$. Therefore, the volume of a cuboid is: $6 x \times 3 x \times 2 x=36 x^{3}$. From this, we can find the cubic value of $x$, and then the value of $x$, and finally, the sum of the lengths of the edges of the cuboid can be calculated.
【Solution】Solution: According to the graph, we can get length $=3 \times$ height, length $=2 \times$ width,
length: width: height $=6: 3: 2$, let the length of the small cuboid be $6 x$, the width be $3 x$, and the height be $2 x$.
Therefore, the volume of a small cuboid is: $6 x \times 3 x \times 2 x=36 x^{3}$
$$
\begin{aligned}
180 x^{3} & =22.5, \\
x^{3} & =0.125, \\
x & =0.5 ;
\end{aligned}
$$

length $=0.5 \times 6=3$, width $=0.5 \times 3=1.5$, height $=0.5 \times 2=1$,
sum of edge lengths $=(3+1.5+1) \times 4=22$.
Answer: The sum of all the edge lengths of a small cuboid is 22.
Therefore, the answer is: 22.

Answer: 22


Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $C:$
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

If there exists a point $P$ on the ellipse $C$ such that $P F_{1} \perp P F_{2}$, then the range of the eccentricity $e$ of the ellipse is $\qquad$

Solution:
3. $\left[\frac{\sqrt{2}}{2}, 1\right)$.

Let $A$ be the upper vertex of the ellipse $C$.
According to the problem, we have
$$
\begin{array}{l}
\angle F_{1} A F_{2} \geqslant 90^{\circ} \Rightarrow \angle F_{2} A O \geqslant 45^{\circ} \\
\Rightarrow c \geqslant b \Rightarrow a \leqslant \sqrt{2} c \Rightarrow \frac{\sqrt{2}}{2} \leqslant e<1 .
\end{array}
$$

Answer: [\frac{\sqrt{2}}{2},1)


In jar A, there are 6 balls, of which 4 are red and 2 are white; in jar B, there are 4 balls, of which 3 are red and 1 is white; in jar C, there are 5 balls, of which 2 are red and 3 are white. One ball is drawn from each of the jars A, B, and C. Find:
(1) the probability that exactly 2 of the balls are white;
(2) the probability that at least 2 of the balls are white.

Solution:
(1) $P=\frac{1}{3} \times \frac{1}{4} \times \frac{2}{5}+\frac{1}{3} \times \frac{3}{4} \times \frac{3}{5}+\frac{2}{3} \times \frac{1}{4} \times \frac{3}{5}=\frac{19}{60}$,
( 2 ) $P=\frac{17}{60}+\frac{1}{3} \times \frac{1}{4} \times \frac{3}{5}=\frac{20}{60}=\frac{1}{3}$

Answer: \frac{1}{3}


Let planar vectors $\boldsymbol{a}, \boldsymbol{b}$ satisfy $|\boldsymbol{a}+\boldsymbol{b}|=3$, then the maximum value of $\boldsymbol{a} \cdot \boldsymbol{b}$ is $\qquad$

Solution: $\boldsymbol{a} \cdot \boldsymbol{b}=\frac{1}{4}\left[(\boldsymbol{a}+\boldsymbol{b})^{2}-(\boldsymbol{a}-\boldsymbol{b})^{2}\right] \leqslant \frac{9}{4}$, equality holds when $\boldsymbol{a}=\boldsymbol{b}$.

Answer: \frac{9}{4}


Find all real numbers $x$ for which $\left(x^{2}-7 x+11\right)^{x^{2}-13 x+42}=1$.

Solution: B1. The value of the power $a^{b}$ is equal to 1 only in the case when $a=1$, $a=-1$ and $b$ is an even integer, or $b=0$ and $a \neq 0$. If $x^{2}-7 x+11=1$, it follows that $x^{2}-7 x+10=0$ or $(x-2)(x-5)=0$. From this, we get the solutions $x=2$ and $x=5$. If $x^{2}-7 x+11=-1$ or $x^{2}-7 x+12=0$, it follows that $(x-3)(x-4)=0$ and thus $x=3$ or $x=4$. In both cases, $x^{2}-13 x+42$ is an even number, so these are also solutions. If, however, $x^{2}-13 x+42=0$ or $(x-6)(x-7)=0$, it follows that $x=6$ or $x=7$ and in both cases $x^{2}-7 x+11 \neq 0$. The solutions are therefore the numbers $2,3,4,5,6$ and 7. [^0]

Answer: 2,3,4,5,6,7


Given that $(\sin t)^{\prime}=\cos t$, differentiate the function $y=\sin 2 x$.

Solution:
Solution. The complexity coefficient of the given function is 2. The order of the intermediate functions is as follows: $y=\sin u, u=2 x$. We find

$$
y^{\prime}=\cos 2 x \cdot(2 x)^{\prime}=\cos 2 x \cdot 2=2 \cos 2 x
$$

Answer: 2\cos2x


Given that $a$ and $b$ are positive integers, the quadratic equation $x^{2}-2 a x+b=0$ has two real roots $x_{1}$ and $x_{2}$, and the quadratic equation $y^{2}+2 a y+b=0$ has two real roots $y_{1}$ and $y_{2}$. It is also given that $x_{1} y_{1}-x_{2} y_{2}=2008$. Find the minimum value of $b$.

Solution:
12.B. For the equation in $x$, $x^{2}-2 a x+b=0$, the roots are $a \pm \sqrt{a^{2}-b}$. For the equation in $y$, $y^{2}+2 a y+b=0$, the roots are $-a \pm \sqrt{a^{2}-b}$.
Let $\sqrt{a^{2}-b}=t$.
Then when $x_{1}=a+t, x_{2}=a-t, y_{1}=-a+t$, $y_{2}=-a-t$, we have $x_{1} y_{1}-x_{2} y_{2}=0$, which does not satisfy the condition;

When $x_{1}=a-t, x_{2}=a+t, y_{1}=-a-t$, $y_{2}=-a+t$, we have $x_{1} y_{1}-x_{2} y_{2}=0$, which does not satisfy the condition;

When $x_{1}=a-t, x_{2}=a+t, y_{1}=-a+t$, $y_{2}=-a-t$, we get $x_{1} y_{1}-x_{2} y_{2}=4 a t$;

When $x_{1}=a+t, x_{2}=a-t, y_{1}=-a-t$, $y_{2}=-a+t$, we get $x_{1} y_{1}-x_{2} y_{2}=-4 a t$.
Since $t=\sqrt{a^{2}-b}>0$, we have $a t=502$.
Given that $a$ is a positive integer, we know that $t$ is a rational number, and thus, $t$ is an integer.

From $a t=502$, we get $a=251, t=2$, so the minimum value of $b$ is $b=a^{2}-t^{2}=251^{2}-2^{2}=62997$.
Therefore, the minimum value of $b$ is 62997.

Answer: 62997


If numbers $a_{1}, a_{2}, a_{3}$ are taken in increasing order from the set $1,2, \cdots, 14$, such that both $a_{2}-a_{1} \geqslant 3$ and $a_{3}-a_{2} \geqslant 3$ are satisfied, then the total number of different ways to choose such numbers is $\qquad$.

Solution:
If $a_{1}, a_{2}, a_{3}$ meet the conditions, then $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-2, a_{3}^{\prime}=a_{3}-4$ are distinct, so $a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}$ are selected in ascending order from the numbers $1,2, \cdots, 10$. Conversely, this is also true.
Therefore, the number of different ways to choose them is $C_{10}^{3}=120$.

Answer: 120


There are $8$ passengers boarding 6 carriages randomly. How many ways are there for the passengers to board the carriages such that exactly two carriages are empty?

Solution: Assume that carriages $5$ and $6$ are empty, and carriages $1, 2, 3, 4$ are all occupied. Let the set of all ways for 8 people to board these 4 carriages be $S$, then $|S|=4^{8}$. Let $A_{i}$ represent the set of ways where the $i$-th carriage is empty $(i=1,2,3,4)$, then $\left|A_{i}\right|=3^{8}(i=1,2,3,4),\left|A_{i} \cap A_{j}\right|=2^{8}(1 \leqslant i< j \leqslant 4),\left|A_{i} \cap A_{j} \cap A_{k}\right|=1(1 \leqslant i<j<k \leqslant 1),\left|A_{1} \cap A_{2} \cap A_{3} \cap A_{4}\right|=0$. By the principle of inclusion-exclusion: $\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3} \cap \bar{A}_{4}\right|=|S|-\mathrm{C}_{4}^{1} \cdot\left|A_{1}\right|+\mathrm{C}_{4}^{2} \cdot\left|A_{1} \cap A_{2}\right|-\mathrm{C}_{4}^{3} \cdot \left|A_{1} \cap A_{2} \cap A_{3}\right|+\left|A_{1} \cap A_{2} \cap A_{3} \cap A_{4}\right|=4^{8}-4 \cdot 3^{8}+6 \cdot 2^{8}-4=40824$, so the total number of boarding ways is $C_{6}^{2} \cdot 40824=612360$.

Answer: 612360


Find all functions $f(x)$, defined for all real $x$ and satisfying the equation $2 f(x)+f(1-x)=x^{2}$

Solution:
Substitute $1-x$ for $x$.

## Solution

In the given equation, substitute $1-x$ for $x$. We obtain the system

$$
\begin{aligned}
& 2 f(x)+f(1-x)=x^{2} \\
& f(x)+2 f(1-x)=(1-x)^{2}
\end{aligned}
$$

from which we find $f(x)=\frac{1}{3}\left(2 x^{2}-(1-x)^{2}\right)=\frac{1}{3}\left(x^{2}+2 x-1\right)$. Verification shows that the found function satisfies the condition.

## Answer

$f(x)=\frac{1}{3}\left(x^{2}+2 x-1\right)$

Answer: f(x)=\frac{1}{3}(x^{2}+2x-1)


Let $0<\theta<\pi$, the complex numbers $z_{1}=1-\cos \theta+i \sin \theta, z_{2}=a^{2}+a i, u \in \mathbf{R}, z_{1} z_{2}$ is a pure imaginary number, and $\bar{a}=z_{1}^{2}+z_{2}^{2}-2 z_{1} z_{2}$. Then $\theta=$ $\qquad$ when it is a negative real number.

Solution:
12. $\frac{\pi}{2}$.

Since $z_{1} z_{2}=\left[a^{2}(1-\cos \theta)-a \sin \theta\right]+\left[a(1-\cos \theta)+a^{2} \sin \theta\right] i$ is a pure imaginary number, we have
$\left\{\begin{array}{l}a^{2}(1-\cos \theta)-a \sin \theta=0 \\ a(1-\cos \theta)+a^{2} \sin \theta \neq 0\end{array}\right.$, so $a \neq 0$. And $a=\frac{\sin \theta}{1-\cos \theta}$.
Also, $\omega=z_{1}^{2}+z_{2}^{2}-2 z_{1} z_{2}=\left(z_{1}-z_{2}\right)^{2}$ is a non-negative real number, so $z_{1}-z_{2}$ is a real number, and $z_{1}-z_{2}=\left(1-\cos \theta-a^{2}\right)+(\sin \theta-a) i$. Therefore, $a=\sin \theta \neq 0$. Thus, $\sin \theta=\frac{\sin \theta}{1-\cos \theta}$, which gives $\cos \theta=0$. Given $0<\theta<\pi$, we finally get $\theta=\frac{\pi}{2}$.

Answer: \frac{\pi}{2}


The product of all positive divisors of a natural number $n$ is $20^{15}$. Determine $n$.

Solution:
1. The number $20^{15}=2^{30} 5^{15}$ is divisible only by the prime numbers 2 and 5, so the sought number $n$ must be of the form $n=2^{a} 5^{b}$, where $a, b$ are natural numbers. Each of its positive divisors is thus of the form $2^{\alpha} 5^{\beta}$, where $\alpha \in\{0,1, \ldots, a\}$ and $\beta \in\{0,1, \ldots, b\}$, and from the theorem of the unique factorization of a natural number into a product of primes, it follows that for different $\alpha, \beta$ we get different divisors.

For each $\beta \in\{0,1, \ldots, b\}$, consider now all divisors of the number $n$ that are divisible by 5 exactly to the power of $\beta$. They are

$$
\underbrace{2^{0} 5^{\beta}, 2^{1} 5^{\beta}, 2^{2} 5^{\beta}, \ldots, 2^{a} 5^{\beta}}_{a+1 \text { numbers }}
$$

and their product is

$$
2^{0+1+2+\ldots+a} 5^{(a+1) \beta}=2^{a(a+1) / 2} 5^{(a+1) \beta} .
$$

If we multiply all these products for $\beta=0,1, \ldots, b$, we get the product of all positive divisors of the number $n$, which is thus equal to

$$
2^{a(a+1)(b+1) / 2} 5^{(a+1) b(b+1) / 2} .
$$

It follows that to find the number $n$, it suffices to solve the system of equations in the set of natural numbers

$$
\begin{aligned}
& \frac{1}{2} a(a+1)(b+1)=30 \\
& \frac{1}{2}(a+1) b(b+1)=15 .
\end{aligned}
$$

The expressions on both sides of these equations are clearly non-zero, and by dividing them, we get $a=2 b$ after rearrangement. Substituting into the first equation of the system, we then have

$$
b(b+1)(2 b+1)=30
$$

Since $30=2 \cdot 3 \cdot 5$, we see that one solution is $b=2$. Since the function $x(x+1)(2 x+1)$ is increasing on the set of positive numbers as the product of three increasing positive functions, this is the only solution. The system (1) therefore has a unique solution $a=4, b=2$, which corresponds to the sought natural number $n=2^{4} 5^{2}=400$.

There is a unique natural number $n$ that satisfies the conditions of the problem, and that is $n=400$.

For a complete solution, award 6 points. Of these, award 1 point for noting that each divisor of the number $n$ is of the form $2^{\alpha} 5^{\beta}$, 2 points for finding their product, 1 point for setting up the system (1) (or an equivalent system), and the remaining 2 points for solving it and finding the number $n$. If it does not follow from the solution method (1) that the found solution is unique and the student does not justify it, deduct 1 point. If the solver only proves that the number $n=400$ has the required property without any indication of its derivation, award 2 points (these points cannot be added to the points according to the previous scheme).

Answer: 400


Given real numbers $x, y$ satisfy
$$
\left(2015+x^{2}\right)\left(2015+y^{2}\right)=2^{22} \text {. }
$$

Then the maximum value of $x+y$ is $\qquad$

Solution:
5. $2 \sqrt{33}$.

Notice that, $2^{22}=2048^{2}=(2015+33)^{2}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{\left[(2015-33)+33+x^{2}\right]\left[(2015-33)+y^{2}+33\right]} \\
\geqslant[(2015-33)+\sqrt{33} y+x \sqrt{33}]^{2} \\
\Rightarrow(2015+33)^{2} \\
\quad \geqslant[(2015-33)+\sqrt{33}(x+y)]^{2} \\
\Rightarrow x+y \leqslant 2 \sqrt{33} .
\end{array}
$$

When $x=y=\sqrt{33}$, the equality holds.
Therefore, the maximum value of $x+y$ is $2 \sqrt{33}$.

Answer: 2\sqrt{33}


Calculate the limit of the function:

$$
\lim _{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}
$$

Solution:
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{(\sqrt{1+\tan x}-\sqrt{1+\sin x})(\sqrt{1+\tan x}+\sqrt{1+\sin x})}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}= \\
& =\lim _{x \rightarrow 0} \frac{1+\tan x-(1+\sin x)}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}=\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}= \\
& =\lim _{x \rightarrow 0} \frac{\tan x-\frac{\sin x \cdot \cos x}{\cos x}}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}=\lim _{x \rightarrow 0} \frac{\tan x-\tan x \cdot \cos x}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}= \\
& =\lim _{x \rightarrow 0} \frac{\tan x(1-\cos x)}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}=
\end{aligned}
$$

$\tan x \sim x$, as $x \rightarrow 0$

$1-\cos x \sim \frac{x^{2}}{2}$, as $x \rightarrow 0$

We get:

$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{x\left(\frac{x^{2}}{2}\right)}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}=\lim _{x \rightarrow 0} \frac{x^{3}}{2 x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}= \\
& =\lim _{x \rightarrow 0} \frac{1}{2(\sqrt{1+\tan x}+\sqrt{1+\sin x})}=\frac{1}{2(\sqrt{1+\tan 0}+\sqrt{1+\sin 0})}= \\
& =\frac{1}{2(\sqrt{1+0}+\sqrt{1+0})}=\frac{1}{4}
\end{aligned}
$$

## Problem Kuznetsov Limits 16-5

Answer: \frac{1}{4}


The number of positive integer solutions to the equation $x^{4004}+y^{400x}=z^{2002}$ is $(\quad)$.
A. 0
B. 1
C. finite
D. infinite

Solution:
3. A

Let $x_{0}=x^{1001}, y_{0}=y^{1001}, z_{0}=z^{1001}$, then the original equation can be transformed into $x_{0}^{4}+y_{0}^{4}=z_{0}^{2}, x_{0}>0$, $y_{0}>0, z_{0}>0$. Using the method of infinite descent, it can be proven that the equation $x_{0}^{4} + y_{0}^{4}=z_{0}^{2}$ has no positive integer solutions. Therefore, the answer is A.

Answer: A


The sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}=(-1)^{n} n-a_{n}, n=1,2,3, \cdots$, and $a_{10}=a_{1}$, then the maximum value of $a_{n} a_{n+1}$ is $\qquad$ .

Solution:
According to the problem, $a_{n+1}+a_{n}=(-1)^{n} n$, let $n=1,2, \cdots, 9$ we get
$$
\left\{\begin{array} { l }
{ a _ { 2 } + a _ { 1 } = - 1 , } \\
{ a _ { 3 } + a _ { 2 } = 2 , } \\
{ a _ { 4 } + a _ { 3 } = - 3 , } \\
{ a _ { 5 } + a _ { 4 } = 4 , } \\
{ \cdots \cdots } \\
{ a _ { 1 0 } + a _ { 9 } = - 9 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a_{2}+a_{1}=-1, \\
-a_{3}-a_{2}=-2, \\
a_{4}+a_{3}=-3, \\
-a_{5}-a_{4}=-4, \\
\cdots \cdots \\
a_{10}+a_{9}=-9
\end{array} \Rightarrow a_{10}+a_{1}=-\frac{9 \times 10}{2}=-45,\right.\right.
$$

Also, $a_{10}=a_{1} \Rightarrow a_{1}=-\frac{45}{2}$. Following the above steps, we get
when $n$ is odd, $-a_{n}+a_{1}=-\frac{n(n-1)}{2} \Rightarrow a_{n}=\frac{n(n-1)-45}{2}$;
when $n$ is even, $a_{n}+a_{1}=-\frac{n(n-1)}{2} \Rightarrow a_{n}=-\frac{n(n-1)-45}{2}$.
Thus, $a_{n} a_{n+1}=-\frac{1}{4}[n(n-1)-45][n(n+1)-45]=-\frac{1}{4}\left[\left(n^{2}-45\right)-n\right]\left[\left(n^{2}-45\right)+n\right]$ $=-\frac{1}{4}\left[\left(n^{2}-45\right)^{2}-n^{2}\right]=-\frac{1}{4}\left(n^{4}-91 n^{2}+45^{2}\right)$.
By the properties of quadratic functions, when $n=7$, $a_{n} a_{n+1}$ reaches its maximum value $\frac{33}{4}$.

Answer: \frac{33}{4}


For the positive integer $n$, define $a_{n}$ as the unit digit of $n^{(n+1)^{n+2}}$. Then $\sum_{n=1}^{2010} a_{n}=$ $\qquad$ .

Solution:
4.5 829 .

When $n \equiv 0,1,5,6(\bmod 10)$, $a_{n} \equiv n^{(n+1)^{n+2}} \equiv n(\bmod 10)$;
When $n \equiv 2,4,8(\bmod 10)$,
$$
\begin{array}{l}
(n+1)^{n+2} \equiv 1(\bmod 4) \\
\Rightarrow a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k+1} \equiv n(\bmod 10) ;
\end{array}
$$

When $n \equiv 3,7,9(\bmod 10)$,
$$
\begin{array}{l}
(n+1)^{n+2} \equiv 0(\bmod 4) \\
\Rightarrow a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k} \equiv 1(\bmod 10) .
\end{array}
$$

Therefore, $a_{n}$ is sequentially $1,2,1,4,5,6,1,8,1,0,1$, $2, \cdots$ which is a periodic sequence, with the smallest positive period being 10.
Thus, $\sum_{n=1}^{2010} a_{n}=201 \times 29=5829$.

Answer: 5829


There is a certain amount of saltwater with a concentration of $3 \%$. After adding $a$ grams of pure water, the concentration of the saltwater becomes $2 \%$. After adding another $a$ grams of pure water, the concentration of the saltwater at this point is ( ).
(A) $1 \%$
(B) $1.25 \%$
(C) $1.5 \%$
(D) $1.75 \%$

Solution:
6. C.

Let's assume that in a 3% salt solution, there are 97 grams of water and 3 grams of salt. Therefore, adding 50 grams of water can turn it into a 2% salt solution, and then adding another 50 grams of water can turn it into a 1.5% salt solution.

Answer: C


There are 150 black and white chess pieces in total, divided into 50 piles, with 3 pieces in each pile. Among them, there are 15 piles with only 1 white piece, and 25 piles with no less than 2 white pieces. The number of piles with only white pieces is twice the number of piles with only black pieces. In these 150 chess pieces, the number of black pieces is $\qquad$.

Solution: Answer: 65

Answer: 65


How much does the dress cost? The worker's monthly pay, that is, for thirty days, is ten dinars and a dress. He worked for three days and earned the dress. What is the cost of the dress?

Solution: 8. Since for 30 days he will receive 10 dinars and a dress, then for 3 days he will receive a dinar and $\frac{1}{10}$ of a dress, therefore, the dress costs $1 \frac{1}{9}$ dinars. One can reason differently: since he earned a dress in three days, then in 27 days he should have received 10 dinars, and in 3 days accordingly $1 \frac{1}{9}$ dinars.

Answer: 1\frac{1}{9}


Let $p>13$ be a prime of the form $2q+1$, where $q$ is prime. Find the number of ordered pairs of integers $(m,n)$ such that $0\le m<n<p-1$ and
\[3^m+(-12)^m\equiv 3^n+(-12)^n\pmod{p}.\]

Solution:
1. **Given Conditions and Initial Setup:**
   - Let \( p > 13 \) be a prime of the form \( p = 2q + 1 \), where \( q \) is also a prime.
   - We need to find the number of ordered pairs of integers \((m, n)\) such that \( 0 \leq m < n < p-1 \) and
     \[
     3^m + (-12)^m \equiv 3^n + (-12)^n \pmod{p}.
     \]

2. **Primitive Root Analysis:**
   - We claim that exactly one of \( 3 \) and \( -12 \) is a primitive root modulo \( p \).
   - The order of any residue \( x \) modulo \( p \) must divide \( p-1 = 2q \), so it must be one of \( 1, 2, q, 2q \).
   - If the order is \( 1 \) or \( 2 \), then the residue is \( \pm 1 \), which is not the case here.
   - Thus, the order of \( x \) is either \( q \) or \( 2q \).

3. **Legendre Symbol and Primitive Root:**
   - Using the Legendre symbol, we have:
     \[
     \left( \frac{x}{p} \right) \equiv x^{\frac{p-1}{2}} \pmod{p}.
     \]
   - A residue \( x \neq \pm 1 \) is a primitive root if and only if \( \left( \frac{x}{p} \right) = -1 \).
   - Since \( p \equiv 3 \pmod{4} \), we have:
     \[
     \left( \frac{3}{p} \right) \left( \frac{-12}{p} \right) = \left( \frac{-36}{p} \right) = \left( \frac{-1}{p} \right) = -1.
     \]
   - Therefore, one of the Legendre symbols is \( -1 \) and the other is \( 1 \).

4. **Identifying the Primitive Root:**
   - Let \( g \in \{3, -12\} \) be the primitive root, and let \( h \) be the other one.
   - Since \( h \) is a square, we must have \( h = g^{2k} \) for some integer \( 1 \leq k < q \) and \( 2k - 1 \neq q \) (since \( h \not\equiv -g \pmod{p} \)).

5. **Restating the Problem:**
   - We need to find the number of ordered pairs \((m, n)\) such that:
     \[
     g^m + (g^m)^{2k} \equiv g^n + (g^n)^{2k} \pmod{p}.
     \]
   - Since \( g^m \) uniquely attains each value in the range \([1, p-1]\) modulo \( p \), we are looking for pairs \((x, y)\) such that \( 1 \leq x, y \leq p-1 \) and:
     \[
     x + x^{2k} \equiv y + y^{2k} \pmod{p}.
     \]

6. **Solving the Restated Problem:**
   - Rearrange the equation:
     \[
     x^{2k-1} \equiv \frac{(y/x)^{2k} - 1}{1 - (y/x)} \pmod{p}.
     \]
   - The right side is defined since \( x \neq y \).
   - If \( y/x \equiv -1 \pmod{p} \), then \( x = 0 \), which is not allowed.
   - Thus, \( y/x \in \{2, \ldots, p-2\} \pmod{p} \).

7. **Counting the Solutions:**
   - For each value of \( y/x \), \( x \) is uniquely determined since we may raise both sides to the power \( (2k-1)^{-1} \pmod{2q} \).
   - There are \( p-3 \) values of \( y/x \), so there are \( p-3 \) solutions.

\(\blacksquare\)

The final answer is \( \boxed{ q-1 } \).

Answer: q-1


For the number $2^{100}$, the sum of its digits was found, then the sum of the digits of the result, and so on. Eventually, a single-digit number was obtained. Find it.

Solution:
$2^{100}=\left(2^{6}\right)^{16} \cdot 2^{4} \equiv 1^{16} \cdot 16 \equiv 7(\bmod 9)$.

## Answer

7.

Answer: 7


Consider the triangle $ABC$, $AE$ the bisector of $\angle BAC$, such that $[AE] \equiv [EC]$. Find the measure of $\angle ABC$ if $AC=2AB$.

Solution:
## SUBJECT III

Draw $E F \perp A C$ (2p); $\square E A C$ is isosceles (1p); $\Rightarrow[E F] \equiv[F C] \equiv[F A]$ (1p); $A B=\frac{A C}{2} \Rightarrow[E F] \equiv[A B]$ (1p); $\square A B E \equiv \triangle A F E(L . U . L)$ (1p); $m(\square B)=90^{\circ}$ (1p).

Answer: 90


A frog is located at the origin $O$ of a number line and needs to perform 2022 jumps, each of a different length from $1, 2, \ldots, 2022$. The jumps must be performed in such an order that the following rules are respected:

- if the frog is currently at point $O$ or to the left of it, it must jump to the right,
- if the frog is to the right of point $O$, it must jump to the left.

Determine the largest natural number $k$ for which the frog can jump in such a way that it never lands on any of the numbers $1, 2, \ldots, k$.

Solution:
## Solution.

Let $n=$ 2022. Suppose there is a sequence of $n$ jumps of lengths $1,2, \ldots, n$ in some order such that the frog never lands on any of the numbers $1,2, \ldots, k$ for $k \geqslant 1$. Note that if the frog is on a number $a \geqslant k+1$, it actually stands on a number $a \geqslant k+1$ since it cannot be on any of the numbers $1,2, \ldots, k$. This means the frog can only jump left if it is on a number $a \geqslant k+1$, so it cannot reach a number less than $k+1-n$. Therefore, the frog can only land on numbers $i$ that satisfy

$$
k+1-n \leqslant i \leqslant 0 \quad \text { or } \quad k+1 \leqslant i \leqslant n
$$

Consider what happens at the moment the frog makes a jump of length exactly $k$. After this jump, it must remain on the same side of the numbers $1,2, \ldots, k$ because a jump of length at least $k+1$ is required to jump over all these numbers.

If this jump is made from a position $a>0$, it will end up at the number $a-k$, so it must be true that $a-k \geqslant k+1$ or $2 k+1 \leqslant a \leqslant n$. Thus, we have $2 k+1 \leqslant n$ in this case.

If this jump is made from a position $a \leqslant 0$, then $k+1-n \leqslant a \leqslant 0$. It will then end up at the number $a+k$, so it must be true that $a+k \leqslant 0$ and $2 k+1-n \leqslant a+k \leqslant 0$. Again, we conclude $2 k+1 \leqslant n$.

Thus, we have shown that $k \leqslant \frac{n-1}{2}=1010.5$. Since $k$ is an integer, we have $k \leqslant 1010$.

Now we show that it is possible to make a sequence of jumps such that the frog never lands on any of the numbers $1,2, \ldots, k$ for $k=1010$. First, make a jump of length $k+1$.

Then we make jumps in pairs. First, make a pair of jumps of lengths $k+2$ and 1, then a pair of jumps of lengths $k+3$ and 2. We continue making pairs of jumps: first of length $k+i+1$, and then $i$, for $i=4, \ldots, k$. Finally, make a jump of length $n=2 k+2=2022$. Note that we have made jumps of all possible lengths $1,2, \ldots, n$ exactly once.

We show that this sequence of jumps never lands on any of the numbers $1,2, \ldots, k$ for $k=1010$. After the first jump, the frog lands on the number $k+1>k$. Suppose the frog is at number 0 or $k+1$ and is about to make a pair of jumps of lengths first $k+i+1$, and then $i$. If it is at number 0, the first jump takes it to the number $k+i+1>k$, and the second to the number $k+i+1-i=k+1>k$. If it is at number $k+1$, the first jump takes it to the number $k+1-(k+i+1)=-i<0$, and the second to the number $k+1-(2 k+2)=-k-1<1$. Thus, it never lands on any of the numbers $1,2, \ldots, k$, so $k=1010$ is the desired answer.

Answer: 1010


If a store sells a certain product, which costs 100 yuan, at 120 yuan, it can sell 300 units. If the price of the product is increased by 1 yuan based on 120 yuan, it will sell 10 fewer units, and if the price is reduced by 1 yuan, it will sell 30 more units. Question: To maximize profit, what price should the store set for the product?

Solution:
Solution: (1) If sold at 120 yuan each, 300 can be sold, and the profit is $300 \times 20 = 6000$ (yuan).
(2) If the price is increased by $x (x > 0)$ yuan, then (300 $- 10x$) can be sold. Let the profit be $y$ yuan, then
$$
\begin{array}{l}
y = (20 + x)(300 - 10x) \\
= -10x^2 + 100x + 6000 \\
= -10(x - 5)^2 + 6250 .
\end{array}
$$

Because $-100)$ yuan, then (300 $+ 30x$) can be sold. Let the profit be $y$ yuan, then
$$
\begin{array}{l}
y = (20 - x)(300 + 30x) \\
= -30x^2 + 300x + 6000 \\
= -30(x - 5)^2 + 6750 .
\end{array}
$$

Because $-30 < 0$. So,
When $x = 5$, $y_{\text{max}} = 6750$.
Therefore, when the price is reduced to 115 yuan, the maximum profit of 6750 yuan can be obtained.

In conclusion, when the price is set at 115 yuan, the store can achieve the maximum profit.

Answer: 115


What is the largest factor of $130000$ that does not contain the digit $0$ or $5$?

Solution:
1. First, we factorize \(130000\):
   \[
   130000 = 2^4 \cdot 5^4 \cdot 13
   \]
2. We need to find the largest factor of \(130000\) that does not contain the digit \(0\) or \(5\). If the factor is a multiple of \(5\), it will end in \(0\) or \(5\), which is not allowed. Therefore, we exclude any factors that include \(5\) in their prime factorization.

3. We are left with the factors of \(2^4 \cdot 13\):
   \[
   2^4 \cdot 13 = 208
   \]
   However, \(208\) contains the digit \(0\), so it is not allowed.

4. Next, we check the smaller factors of \(2^4 \cdot 13\):
   \[
   2^3 \cdot 13 = 104
   \]
   \(104\) also contains the digit \(0\), so it is not allowed.

5. Continuing, we check:
   \[
   2^2 \cdot 13 = 52
   \]
   \(52\) contains the digit \(5\), so it is not allowed.

6. Finally, we check:
   \[
   2 \cdot 13 = 26
   \]
   \(26\) does not contain the digit \(0\) or \(5\), so it is allowed.

Conclusion:
\[
\boxed{26}
\]

Answer: 26


Given point $A(x, y)$ is in the fourth quadrant, and $|x|=2,|y|=3$, point $B$ is symmetric to $A$ with respect to the origin, and then point $B$ is translated downward by two units to get point $C$, then the area of $\triangle A B C$ is $\qquad$.

Answer: 4


At the mathematics olympiad in two rounds, it is necessary to solve 14 problems. For each correctly solved problem, 7 points are given, and for each incorrectly solved problem, 12 points are deducted. How many problems did the student solve correctly if he scored 60 points?

Solution:
3. Let's determine the number of problems solved by the student, given that he received 7 points for each problem: $60: 7=8$ (remainder 4). Since 12 points were deducted for each incorrectly solved problem, the number of correctly solved problems was more than 8. Further, by trial and error, we can establish that there were 12.

Answer: The student solved 12 problems correctly.

Answer: 12


A gardener is preparing to plant a row of 20 trees, with two types of trees available: maple trees or sycamore trees. The number of trees between any two maple trees (not including these two maple trees) cannot be equal to 3. How many maple trees can there be at most among the 20 trees?

Solution:
【Answer】 12
【Solution】In any continuous eight trees, once a maple tree is planted, it means that another position can only plant a sycamore tree. We use the following diagram to illustrate, using — to represent a maple tree, and ○ to represent a sycamore tree. Once the second position is planted with a maple tree, position $A$ must plant a sycamore tree. No matter where the maple tree appears, there is always a corresponding position that can only plant a sycamore tree, so in eight consecutive trees, there can be at most four maple trees.

According to the previous derivation, in the first 16 trees of 20 trees, there are at most 8 maple trees, so the total number of maple trees is at most $8+4=12$. We can plant them as follows:

Answer: 12


The average of 6 different non-zero natural numbers is 12. If one of the two-digit numbers $\overline{\mathrm{ab}}$ is replaced by $\overline{\mathrm{ba}}$ (where $a$ and $b$ are non-zero digits), then the average of these 6 numbers becomes 15. Therefore, the number of $\overline{\mathrm{ab}}$ that satisfy the condition is __.

Solution:
【Solution】Solution: The original sum of the six numbers is: $12 \times 6=72$
The sum of the 6 numbers now is: $15 \times 6=90$
That is, $\overline{b a}-\overline{a b}=90-72=18$
Then, $10 b+a-10 a-b=18$
$$
\begin{array}{r}
9(b-a)=18 \\
b-a=2
\end{array}
$$

Since, the original sum of the six numbers is 72, the smallest sum of the other 5 numbers is $1+2+3+4+5=15$, then $\overline{\mathrm{ab}}$ is at most $72-15=57$, so $a$ is at most 5, then $a$ can be $5, 4, 3, 2, 1$, at this time $b$ is 7, 6, 5, 4, 3 respectively, $\overline{\mathrm{ba}}$ is at most $90-15=75$, so $b$ is at most 7, all the above situations are consistent.

Answer: There are 5 $\overline{\mathrm{ab}}$ that meet the conditions. The answer is: 5.

Answer: 5


Let $f$ be the function defined by

$$
f(x)=x^{4}-(x+1)^{4}-(x+2)^{4}+(x+3)^{4}
$$

where the domain of $f$ is

a) the set of all integers,

b) the set of all real numbers.

Investigate, for both case a) and case b), whether the function $f$ attains a minimum value, and determine, if applicable, this minimum value in each case.

Solution:
From the representation

$$
f(x)=\left(\left(x+\frac{3}{2}\right)-\frac{3}{2}\right)^{4}-\left(\left(x+\frac{3}{2}\right)-\frac{1}{2}\right)^{4}-\left(\left(x+\frac{3}{2}\right)+\frac{1}{2}\right)^{4}+\left(\left(x+\frac{3}{2}\right)+\frac{3}{2}\right)^{4}
$$

we see that $f$ is axis-symmetric with respect to $x=-\frac{3}{2}$ and is at most a quadratic polynomial. In particular, $f$ has its global minimum or maximum at $x=-\frac{3}{2}$.

From $f(-1)=f(-2)=16$ and $f\left(-\frac{3}{2}\right)=10$, it follows that $f$ is a parabola opening upwards and thus has its real minimum at $x=-\frac{3}{2}$ and its minimum over the integers at $x=-1, x=-2$.

## Source anonymous

Answer: 10


Solution:
## Solution.

Domain of definition: $x+y>0$.

Let's write the system of equations as

$$
\left\{\begin{array} { l }
{ 3 ^ { - x } \cdot 2 ^ { y } = 1152 } \\
{ x + y = 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
3^{-x} \cdot 2^{y}=1152 \\
x=5-y
\end{array}\right.\right.
$$

Since $x=5-y$, then

$$
3^{y-5} \cdot 2^{y}=1152 \Leftrightarrow \frac{3^{y} \cdot 2^{y}}{3^{5}}=1152 \Leftrightarrow 6^{y}=279936 \Leftrightarrow 6^{y}=6^{7} \Leftrightarrow y=7
$$

Then $x=-2$.

Answer: $(-2 ; 7)$.

Answer: (-2;7)


Let set $\mathrm{A}=\{1,2,3,4,5,6,7\}$, f: $\mathrm{A} \rightarrow \mathrm{A}$ is a bijection, and satisfies $\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x})))=\mathrm{x}, \mathrm{x} \in \mathrm{A}$, try to find the number of such mappings $\mathrm{f}$.

Solution: Question 183, Answer: Construct a directed graph $G(V, E)$, where the 7 points in $V$ represent the 7 elements in $A$. If $f(a)=b$, then connect a directed edge $\overline{a b}$. Since $f$ is a bijection, each point in $V$ has an outdegree and indegree of 1, thus $G$ can be divided into several disjoint directed cycles, and the lengths of these cycles are all divisors of 3, meaning each cycle's length is either 1 or 3. The number of graphs without cycles of length 3 is 1; the number of graphs with exactly one cycle of length 3 is $\frac{A_{7}^{3}}{3}=70$; the number of graphs with exactly two cycles of length 3 is $\frac{\frac{13}{3} A_{1}^{3}}{3}=280$. Therefore, the number of $\mathrm{f}$ that satisfy the condition is $1+70+280=351$.

Answer: 351


This year, some contestants at the Memorial Contest ABC are friends with each other (friendship is always mutual). For each contestant $X$, let $t(X)$ be the total score that this contestant achieved in previous years before this contest. It is known that the following statements are true:
$1)$ For any two friends $X'$ and $X''$, we have $t(X') \neq t(X''),$
$2)$ For every contestant $X$, the set $\{ t(Y) : Y \text{ is a friend of } X \}$ consists of consecutive integers.
The organizers want to distribute the contestants into contest halls in such a way that no two friends are in the same hall. What is the minimal number of halls they need?

Solution:
1. **Graph Representation**:
   - Consider a graph \( G \) where each vertex represents a contestant.
   - There is an edge between two vertices \( A \) and \( B \) if and only if the contestants \( A \) and \( B \) are friends.
   - Each vertex \( A \) is assigned a value \( t(A) \), which is the total score that contestant \( A \) achieved in previous years before this contest.

2. **Proving the Graph is Acyclic**:
   - We need to prove that the graph \( G \) has no cycles.
   - Suppose, for the sake of contradiction, that there is a cycle in the graph. Let this cycle be \( u_1, u_2, \ldots, u_k \).
   - By the properties given, for any two friends \( X' \) and \( X'' \), \( t(X') \neq t(X'') \). This implies that all \( t(u_i) \) values in the cycle are distinct.
   - Additionally, for every contestant \( X \), the set \( \{ t(Y) : Y \text{ is a friend of } X \} \) consists of consecutive integers.

3. **Contradiction in the Cycle**:
   - In a cycle \( u_1, u_2, \ldots, u_k \), there must exist an index \( i \) such that \( t(u_{i-1}) < t(u_i) < t(u_{i+1}) \) (or the reverse order).
   - However, by condition (2), there must be a friend \( v \) of \( u_i \) such that \( t(v) = t(u_i) \), which contradicts condition (1) that \( t(X') \neq t(X'') \) for any two friends \( X' \) and \( X'' \).

4. **Conclusion on Graph Structure**:
   - Since the assumption of a cycle leads to a contradiction, the graph \( G \) must be acyclic.
   - An acyclic graph with no odd cycles is a bipartite graph. This means the graph can be colored using two colors such that no two adjacent vertices share the same color.

5. **Minimal Number of Halls**:
   - Since the graph is bipartite, we can partition the vertices into two sets where no two friends (connected vertices) are in the same set.
   - Therefore, the minimal number of halls required is 2.

The final answer is \(\boxed{2}\)

Answer: 2


If $\log _{2}\left(\log _{8} x\right)=\log _{8}\left(\log _{2} x\right)$, find $\left(\log _{2} x\right)^{2}$.

Solution:
[Solution] By the change of base formula,
$$
\begin{array}{l}
\log _{2}\left(\log _{8} x\right)=\log _{2}\left(\frac{\log _{2} x}{3}\right), \\
\log _{8}\left(\log _{2} x\right)=\frac{\log _{2}\left(\log _{2} x\right)}{3} .
\end{array}
$$

Let $y=\log _{2} x$, then
$$
\begin{array}{l}
\log _{2} \frac{y}{3}=\frac{1}{3} \log _{2} y, \\
\log _{2}\left(\frac{y}{3}\right)^{3}=\log _{2} y,
\end{array}
$$

which means $\left(\frac{y}{3}\right)^{3}=y$.
Noting that $y \neq 0$, we get
$$
y^{2}=27 \text {, }
$$

i.e., $\left(\log _{2} x\right)^{2}=27$.

Answer: 27


As shown in the figure, it is known that the three vertices of $\triangle ABC$ are on the ellipse $\frac{x^{2}}{12}+\frac{y^{2}}{4}=1$, and the coordinate origin $O$ is the centroid of $\triangle ABC$. Try to find the area of $\triangle ABC$.

Solution:
Given $A(2 \sqrt{3} \cos \theta, 2 \sin \theta)$, connect $A O$ to intersect $B C$ at point $D$,
then $D(-\sqrt{3} \cos \theta, -\sin \theta)$, and by the conclusion of the point difference method,
$$
k_{B C} \cdot \frac{y_{D}}{x_{D}}=-\frac{1}{3} \Rightarrow k_{B C}=-\frac{\sqrt{3}}{3 \tan \theta} \text {. }
$$

Thus, the line $B C: y+\sin \theta=-\frac{\sqrt{3}}{3 \tan \theta}(x+\sqrt{3} \cos \theta)$
$$
\Rightarrow x \cos \theta+\sqrt{3} y \sin \theta+\sqrt{3}=0 \text {. }
$$

Solving the system of equations $\left\{\begin{array}{l}x \cos \theta+\sqrt{3} y \sin \theta+\sqrt{3}=0, \\ x^{2}+3 y^{2}=12\end{array} \Rightarrow x^{2}+2 \sqrt{3} x \cos \theta+3-12 \sin ^{2} \theta=0\right.$.
Thus, $|B C|=\sqrt{1+\frac{1}{3 \tan ^{2} \theta}} \cdot \sqrt{12 \cos ^{2} \theta-4\left(3-12 \sin ^{2} \theta\right)}$
$=\frac{\sqrt{\cos ^{2} \theta+3 \sin ^{2} \theta}}{|\sqrt{3} \sin \theta|} \cdot|6 \sin \theta|=2 \sqrt{3} \cdot \sqrt{\cos ^{2} \theta+3 \sin ^{2} \theta}$, and the distance from point $A$ to line $B C$ is
$$
d=\frac{\left|2 \sqrt{3} \cos ^{2} \theta+2 \sqrt{3} \sin ^{2} \theta+\sqrt{3}\right|}{\sqrt{\cos ^{2} \theta+3 \sin ^{2} \theta}}=\frac{3 \sqrt{3}}{\sqrt{\cos ^{2} \theta+3 \sin ^{2} \theta}},
$$

Therefore, $S_{\triangle A B C}=\frac{1}{2}|B C| \cdot d=\frac{1}{2} \cdot 2 \sqrt{3} \cdot 3 \sqrt{3}=9$.

Answer: 9


Let $u_{0}, u_{1}, u_{2}, \ldots$ be integers such that

$\triangleright u_{0}=100$;

$\triangleright$ for all $k \geqslant 0$, the inequality $u_{k+2} \geqslant 2+u_{k}$ is satisfied;

$\triangleright$ for all $\ell \geqslant 0$, the inequality $u_{\ell+5} \leqslant 5+u_{\ell}$ is satisfied.

Find all possible values for the integer $u_{2023}$.

Solution:
Solution

For all integers $n \geqslant 0$, we know that

$$
u_{n} \geqslant u_{n+5}-5 \geqslant u_{n+10}-10 \geqslant u_{n+8}-8 \geqslant u_{n+6}-6 \geqslant u_{n+4}-4 \geqslant u_{n+2}-2 \geqslant u_{n}
$$

Thus, all our inequalities are equalities, and $u_{n}=u_{n+2}-2=u_{n+5}-5$.

In particular, for all integers $n \geqslant 0$, $k \geqslant 0$, and $\ell \geqslant 0$, we have $u_{n+5 k+2 \ell}=u_{n}+5 k+2 \ell$. Here, for $n=0$, $k=1$, and $\ell=1009$, this means that $u_{2023}=u_{0}+2023=2123$. Thus, 2023 is the only potentially possible value for the integer $u_{2023}$.

Conversely, if each number $u_{n}$ is equal to $u_{n}+100$, our sequence of numbers satisfies the conditions of the statement, and $u_{2023}=2123$. In conclusion, the only possible value is indeed 2123.

## $\square$

Answer: 2123


Find all positive integers $m$ and $n$ such that $(2^{2^{n}}+1)(2^{2^{m}}+1)$ is divisible by $m\cdot n$.

Solution:
1. **Lemma Proof**: We start by proving the lemma: If for some prime \( p \) and natural \( t \), \( p \mid 2^{2^t} + 1 \), then \( p > t \).

   - Let \( k \) be the minimal integer such that \( p \mid 2^k - 1 \). This implies \( k \mid p-1 \).
   - Since \( k \mid 2^{t+1} \), we can write \( k = 2^r \) for some \( r \leq t+1 \).
   - If \( r \leq t \), then \( 2^r \mid 2^t \), which implies \( p \mid 2^{2^r} - 1 \mid 2^{2^t} - 1 \). This would mean \( p \mid 2^{2^t} - 1 \), contradicting \( p \mid 2^{2^t} + 1 \). Hence, \( r = t+1 \) and \( k = 2^{t+1} \).
   - Therefore, \( p \geq k + 1 = 2^{t+1} + 1 > t \).

2. **Analyzing the Divisibility Condition**: We need to find all positive integers \( m \) and \( n \) such that \( (2^{2^n} + 1)(2^{2^m} + 1) \) is divisible by \( m \cdot n \).

   - Assume \( n \geq m > 1 \). Let \( p \) be a prime divisor of \( m \). Then \( p \leq m \leq n \).
   - From the lemma, if \( p \mid 2^{2^m} + 1 \), then \( p > m \), which is a contradiction since \( p \leq m \). Therefore, \( m = 1 \).

3. **Case \( m = 1 \)**: We need \( n \mid 5(2^{2^n} + 1) \).

   - For \( n = 1 \), \( 5(2^{2^1} + 1) = 5 \cdot 5 = 25 \), and \( 1 \mid 25 \), so \( (m, n) = (1, 1) \) is a solution.
   - For \( n > 1 \), let \( p \) be a prime divisor of \( n \). By the lemma, \( p > n \), which is a contradiction. Hence, \( n = 5 \).

4. **Case \( n = 5 \)**: We need to check if \( 5 \mid 5(2^{2^5} + 1) \).

   - \( 2^{2^5} + 1 = 2^{32} + 1 \equiv 2 \pmod{5} \), so \( 5 \mid 5(2^{32} + 1) \).
   - Therefore, \( (m, n) = (1, 5) \) is a solution.

5. **Case \( n = 1 \)**: We need to check if \( 1 \mid 5(2^{2^1} + 1) \).

   - \( 5(2^{2^1} + 1) = 5 \cdot 5 = 25 \), and \( 1 \mid 25 \), so \( (m, n) = (5, 1) \) is a solution.

The final answer is \(\boxed{(m, n) = (1, 1), (1, 5), (5, 1)}\).

Answer: (m, n) = (1, 1), (1, 5), (5, 1)


Determine the functions $f$ from $\mathbb{Q}_{+}^{*}$ to $\mathbb{Q}_{+}^{*}$ satisfying $f(x+1)=f(x)+1$ and $f\left(x^{3}\right)=f(x)^{3}$ for all $x \in \mathbb{Q}_{+}^{*}$

Solution:
. From $f(x+1)=f(x)+1$, we deduce $f(x+n)=f(x)+n$ for all natural integers $n$. Let $\frac{p}{q} \in \mathbb{Q}_{+}^{*}$ and let $u=f\left(\frac{p}{q}\right)$. We are given two expressions of $A=f\left(\left(\frac{p}{q}+q^{2}\right)^{3}\right)$. On one hand,

$$
A=f\left(\frac{p^{3}}{q^{3}}+3 p^{2}+3 p q^{3}+q^{6}\right)=u^{3}+3 p^{2}+3 p q^{3}+q^{6}
$$

since $f\left(\frac{p^{3}}{q^{3}}\right)=f\left(\frac{p}{q}\right)^{3}=u^{3}$ and $3 p^{2}+3 p q^{3}+q^{6}$ is a natural integer. On the other hand,

$$
A=f\left(\frac{p}{q}+q^{2}\right)^{3}=\left(u+q^{2}\right)^{3}=u^{3}+3 u^{2} q^{2}+3 u q^{4}+q^{6}
$$

By writing that the two expressions of $A$ are equal, we get

$$
3 u^{2} q^{2}+3 u q^{4}-\left(3 p^{2}+3 p q^{3}\right)=0
$$

This is a quadratic equation in $u$, of which $\frac{p}{q}$ is a root. Since the product of the roots is strictly negative, the other root is strictly negative, and we conclude that $u=p / q$. Thus, $f(x)=x$ for all $x \in \mathbb{Q}_{+}^{*}$.

Answer: f(x)=x


On the table, there is a stack of 11 magazines numbered from 1 to 11 (the top one is â1, the bottom one is â11). At 12:00, Dima and Fedia started reading the magazines, following this rule: each of them goes through the magazines from top to bottom, finds the first unread one, and takes it for themselves, while the read magazine is returned to the top of the stack; if there are no unread magazines in the stack, they wait until one appears. Dima reads each magazine for 30 minutes, while Fedia reads each for 31 minutes. In what order will the magazines be in the stack when Dima and Fedia finish reading them, if Dima was the first to take a magazine at 12:00?

Solution:
4.3. Solution. Note that in one and a half hours and one minute, Fedya and Dima will read the first three magazines, and they will be on top in reverse order. After this, they will no longer be used, so in the end, they will end up at the bottom in reverse order. The same will happen with the second and third sets of three magazines, but the 10th and 11th magazines will not switch places (which is not difficult to see with a simple examination). Thus, the final order will be $10,11,9,8,7,6,5,4,3,2,1$.

## Senior League

## First Round

Answer: 10,11,9,8,7,6,5,4,3,2,1


In rectangle $A B C D$, $A B=3, A D=4, E$ is a point on $A B$, and $A E=1$. Now, $\triangle B C E$ is folded along $C E$ so that the projection of point $B$ on plane $A E C D$ falls on $A D$. Then, the volume of the pyramid $B-A E C D$ is $\qquad$ .

Solution:
Draw $B F \perp E C$, intersecting $A D, E C$ at $F, G$ respectively.
It is easy to find that $B F=\frac{3 \sqrt{5}}{2}, B G=\frac{4 \sqrt{5}}{5}$.
By the projection theorem, we have
$$
\begin{array}{l}
B^{\prime} F^{2}=\frac{3 \sqrt{5}}{2} \cdot\left(\frac{8 \sqrt{5}}{5}-\frac{3 \sqrt{5}}{2}\right)=\frac{3}{4} \\
\Rightarrow B^{\prime} F=\frac{\sqrt{3}}{2} .
\end{array}
$$

Therefore, $V_{B-A E C D}=\frac{1}{3} \cdot \frac{1}{2} \cdot 4 \cdot 4 \cdot \frac{\sqrt{3}}{2}=\frac{4 \sqrt{3}}{3}$.

Answer: \frac{4\sqrt{3}}{3}


Given $f(x)=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x$ is an even function, $g(x)=2^{x}+\frac{a+b}{2^{x}}$ is an odd function, where $a, b \in \mathbf{C}$. Then $a+a^{2}+a^{3}+\cdots+a^{2000}+b+b^{2}+b^{3}+\cdots+b^{2000}=$

Solution:
3. -2 .

From the known $f(-x)=f(x), g(-x)=-g(x)$ we get
$$
\begin{array}{l}
\log _{\frac{1}{3}}\left(3^{-x}+1\right)-\frac{1}{2} a b x \\
=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x . \\
2^{-x}+\frac{a+b}{2^{-x}}=-\left(2^{x}+\frac{a+b}{2^{x}}\right) .
\end{array}
$$

Simplifying, from equation (1) we get $a b=1$, and from equation (2) we get $a+b=-1$. Therefore, $a$ and $b$ are a pair of conjugate complex roots of the equation $x^{2}+x+1=0$, which are the imaginary cube roots of 1. Then
$$
\begin{aligned}
a & +a^{2}+a^{3}+\cdots+a^{1999}+a^{2000}+b+b^{2} \\
& +b^{3}+\cdots+b^{1999}+b^{2000} \\
= & a^{1999}+a^{2000}+b^{1999}+b^{2000} . \\
= & a+a^{2}+b+b^{2}=-1-1=-2 .
\end{aligned}
$$

Answer: -2


Solution:
## Solution.

Let $\left\{\begin{array}{l}\sqrt[3]{x}=u, \\ \sqrt[3]{y}=v,\end{array}\left\{\begin{array}{l}x=u^{3} \\ y=v^{3}\end{array}\right.\right.$, Relative to $u$ and $v$ the system takes the form

$\left\{\begin{array}{l}u+v=4, \\ u^{3}+v^{3}=28\end{array} \Leftrightarrow\left\{\begin{array}{l}u+v=4, \\ (u+v)\left(u^{2}-u v+v^{2}\right)=28\end{array} \Leftrightarrow\left\{\begin{array}{l}u+v=4, \\ \left((u+v)^{2}-3 u v\right)=7\end{array} \Leftrightarrow\right.\right.\right.$

$\Leftrightarrow\left\{\begin{array}{l}u+v=4 \\ u v=3\end{array}\right.$

from which

$\left\{\begin{array}{l}u_{1}=1, \\ v_{1}=3 ;\end{array}\left\{\begin{array}{l}u_{2}=3 \\ v_{2}=1\end{array}\right.\right.$

Then $\left\{\begin{array}{l}\sqrt[3]{x}=1, \\ \sqrt[3]{y}=3\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1}=1, \\ y_{1}=27 ;\end{array}\left\{\begin{array}{l}\sqrt[3]{x}=3, \\ \sqrt[3]{y}=1\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{2}=27, \\ y_{2}=1 .\end{array}\right.\right.\right.\right.$

Answer: $(1 ; 27),(27 ; 1)$.

Answer: (1;27),(27;1)


There are two boxes of apples. After taking 5 apples from the first box and putting them into the second box, the first box still has 2 more apples than the second box. Originally, the first box had $\qquad$ more apples than the second box.

Solution: answer: 12

Answer: 12


We have four containers. The first three contain water, the fourth is empty. In the second one, there is twice as much water as in the first, and in the third one, there is twice as much water as in the second. We pour half of the water from the first container, a third of the water from the second container, and a quarter of the water from the third container into the fourth container. Now, the fourth container has 26 liters of water. How much water is there in total in all the containers?

Solution:
From the text, it follows that in the first container there is one part of water, in the second container there are two parts of water, and in the third container there are four parts of water. To make it easier to remove water from each container, we will divide each part into six sixths. So, in the first container there are 6 pieces, in the second 12, and in the third 24 pieces of water. Now we pour into the fourth container 3 pieces from the first, 4 pieces from the second, and 6 pieces from the third container. This is a total of 13 pieces, which is 26 liters. Therefore, one piece is two liters of water.

Originally, there were: 12 liters in the first container, 24 liters in the second container, and 48 liters in the third container, i.e., a total of 84 liters of water.

Answer: 84


Find the maximum positive integer $n$, such that there exist $n$ integer points in the coordinate plane, where any 3 points form a triangle, and the centroid of the triangle is not an integer point.

Solution:
[Solution] Let the points $\left(x_{1}, y_{1}\right) ;\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$ be 3 integer points, then the coordinates of the centroid of the triangle formed by them are
$$
\left(\frac{1}{3}\left(x_{1}+x_{2}+x_{3}\right), \frac{1}{3}\left(y_{1}+y_{2}+y_{3}\right)\right) .
$$

It is easy to verify that the following 8 integer points
$$
(0,0),(0,3),(3,1),(3,4),(1,0),(1,3),(4,1),(4,4) \text { form a triangle with any 3 }
$$
points and the centroid of the triangle is not an integer point. Therefore, the maximum positive integer $n \geqslant 8$.

On the other hand, we will prove that any 9 integer points do not satisfy the condition in the problem. If not, suppose there are 9 integer points that satisfy the condition. For an integer point $(x, y)$, let the remainders of $x, y$ when divided by 3 be $r_{1}, r_{2}$, where $r_{1}$ and $r_{2}$ are non-negative integers, then the integer point $(x, y)$ is called of type $\left(r_{1}, r_{2}\right)$. Thus, we divide all integer points on the plane into 9 categories. Since the centroid of a triangle formed by 3 points of the same type is also an integer point, any 3 points among the 9 points cannot belong to the same category. Therefore, these 9 points must belong to at least 5 of the 9 categories. Select 1 point from each of the 5 categories, we get 5 points that belong to different categories, and these 5 points have only 3 different values for their x-coordinates: $0,1,2$. If 3 points among the 5 points have the same x-coordinate, then because they belong to different categories, their y-coordinates must be $0,1,2$, thus the centroid of the triangle formed by these 3 points is an integer point, which is impossible. Therefore, the x-coordinates of the 5 points must take 3 different values, and the number of points for each value must be $2,2,1$. Since whether the centroid is an integer point is invariant under translation, we can assume that the point corresponding to the number 1 is $(0,0)$, then the x-coordinates of the other 4 points are two 1s and two 2s. The 6 categories of points with x-coordinates 1 and 2 can be divided into 3 groups
$$
\{(1,1),(2,2)\},\{(1,0),(2,0)\},\{(1,2),(2,1)\} .
$$

By the pigeonhole principle, among the other 4 points, there must be two points belonging to the same group. These two points, together with the point $(0,0)$, form a triangle whose centroid is an integer point, which is a contradiction.
In conclusion, the maximum positive integer $n=8$.

Answer: 8


How many six-digit numbers of the form $\overline{abcacv}$ are divisible by 23?

(Note: There appears to be a typo in the format provided; I've maintained the structure as given except for what seems to be a typo, changing 'b' to 'v' at the end which I assume was meant to be 'b'. If 'v' is correct, please disregard this correction.)

However, based on the instruction and assuming the intention was to repeat digits correctly, the problem should read:

How many six-digit numbers of the form $\overline{abcacb}$ are divisible by 23?

Solution:
IV /2. The six-digit number $\overline{a b c a c b}$ can be written as

$$
\begin{gathered}
x=100100 a+10001 b+1010 c=23 \cdot(4352 a+434 b+43 c)+4 a+19 b+21 c= \\
=23 \cdot(4352 a+435 b+44 c)+2(2 a-2 b-c)
\end{gathered}
$$

This number is divisible by 23 exactly when the value of the expression $2 a-2 b-c$ is divisible by 23. Since $a, b$, and $c$ are digits and $a \neq 0$, we have $-25 \leq 2 a-2 b-c \leq 18$. The value of the expression $2 a-2 b-c$ is divisible by 23 only if it is equal to 0 or -23.

In the first case, $c=2(a-b)$, so $4 \geq a-b \geq 0$ or $4+b \geq a \geq b$. For $b=0$, we have four options for $a$, for $5 \geq b \geq 1$ we can choose $a$ in 5 ways, for $b=6$ there are 4 options, for $b=7$ there are 3, for $b=8$ there are 2, and for $b=9$ there is one option. Thus, we have a total of $4+5 \cdot 5+4+3+2+1=39$ numbers.

In the second case, $2 a-2 b-c=-23$. If $7 \geq b$, then $23=2 \cdot 7+9 \geq 2 b+c=23+2 a$, so $a=0$, which is not possible. For $b=8$, we get $c=7+2 a$ and the only solution is $a=1$ and $c=9$. For $b=9$, we have $c=5+2 a$, from which we get two possibilities, namely $a=1, c=7$ and $a=2, c=9$. In the second case, we thus have 3 possibilities.

The total number of six-digit numbers with the given properties is $39+3=42$.

The conditions that the digits $a, b$, and $c$ (2a-2b-c=0 or -23 or the analogous condition) must satisfy 2 points

![](https://cdn.mathpix.com/cropped/2024_06_07_5a4348e8f3c4edef190ag-10.jpg?height=51&width=1694&top_left_y=2553&top_left_x=181)
Listing all possibilities in the case $2 a-2 b-c=-23 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots . \ldots . \ldots . \ldots . \ldots . \ldots$ points

Answer: 42


Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.

\[
\left\{
\begin{array}{l}
x=2 \ln (\operatorname{ctg} t) + \operatorname{ctg} t \\
y=\operatorname{tg} t + \operatorname{ctg} t
\end{array}
\right.
\]

$t_{0}=\frac{\pi}{4}$

Solution:
## Solution

Since $t_{0}=\frac{\pi}{4}$, then

$x_{0}=2 \ln \left(\operatorname{ctg} \frac{\pi}{4}\right)+\operatorname{ctg} \frac{\pi}{4}=2 \ln 1+1=1$

$y_{0}=\operatorname{tg} \frac{\pi}{4}+\operatorname{ctg} \frac{\pi}{4}=1+1=2$

Let's find the derivatives:

$x_{t}^{\prime}=(2 \ln (\operatorname{ctg} t)+\operatorname{ctg} t)^{\prime}=2 \cdot \frac{1}{\operatorname{ctg} t} \cdot \frac{-1}{\sin ^{2} t}+\frac{-1}{\sin ^{2} t}=-\frac{1}{\sin ^{2} t} \cdot(2 \operatorname{tg} t+1)$

$y_{t}^{\prime}=(\operatorname{tg} t+\operatorname{ctg} t)^{\prime}=\frac{1}{\cos ^{2} t}-\frac{1}{\sin ^{2} t}$

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{1}{\cos ^{2} t}-\frac{1}{\sin ^{2} t}\right) /\left(-\frac{1}{\sin ^{2} t} \cdot(2 \operatorname{tg} t+1)\right)=\left(\frac{1}{\sin ^{2} t}-\frac{1}{\cos ^{2} t}\right) \cdot \frac{\sin ^{2} t}{2 \operatorname{tg} t+1}=$

$=\frac{1-\operatorname{tg}^{2} t}{2 \operatorname{tg} t+1}$

Then:
$y_{0}^{\prime}=\frac{1-\operatorname{tg}^{2} \frac{\pi}{4}}{2 \operatorname{tg} \frac{\pi}{4}+1}=\frac{1-1^{2}}{2 \cdot 1+1}=0$

Equation of the tangent line:

$y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right)$

$y-2=0 \cdot(x-1)$

$y=2$

Since $y_{0}^{\prime}=0$, the equation of the normal line:

$x=x_{0}$

$x=1$

## Problem Kuznetsov Differentiation 17-9

Answer: 2,\;1


Find the polynomial of least degree, having integral coefficients and leading coefficient equal to 1, with $\sqrt{3}-\sqrt{2}$ as a zero.

Solution:
9. $x^{4}-10 x^{2}+1$
We let $x=\sqrt{3}-\sqrt{2}$. We find the monic polynomial equation of least degree in terms of $x$. Squaring, we get
$$
x^{2}=(\sqrt{3}-\sqrt{2})^{2}=5-2 \sqrt{6} \quad \text { or } \quad x^{2}-5=-2 \sqrt{6} .
$$

Squaring the last equation, we finally get
$$
\left(x^{2}-5\right)^{2}=(-2 \sqrt{6})^{2} \text { or } \quad x^{4}-10 x^{2}+1=0 \text {. }
$$

Answer: x^{4}-10x^{2}+1


Given a cyclic quadrilateral $ABCD$ with side lengths $AB=1, BC=3, CD=DA=2$, then the area of quadrilateral $ABCD$ is . $\qquad$

Solution: Let $\angle ABC = \alpha$, then $AC^2 = 1 + 9 - 6 \cos \alpha = 4 + 4 + 8 \cos \alpha \Rightarrow \cos \alpha = \frac{1}{7}$, so $S_{ABCD} = \frac{1}{2} \cdot (1 \cdot 3 + 2 \cdot 2) \cdot \frac{4 \sqrt{3}}{7} = 2 \sqrt{3}$.

Answer: 2\sqrt{3}


Diagonal $BD$ of quadrilateral $ABCD$ is the diameter of the circle circumscribed around this quadrilateral. Find diagonal $AC$, if $BD=2, AB=1, \angle ABD: \angle DBC=4: 3$.

Solution:
Triangle $B A D$ is a right triangle.

## Solution

Let $\angle A B D=4 \boldsymbol{\alpha}$, then $\angle D B C=3 \alpha$. From the right triangle $A B D$ (angle $A$ is a right angle, as an inscribed angle subtending the diameter $B D$), we find that

$$
\cos \angle A B D=\frac{A B}{B D}=\frac{1}{2}
$$

Therefore,

$$
\angle A B D=4 \alpha=60^{\circ}, \alpha=15^{\circ} \text {. }
$$

Then

$$
\angle C B D=3 \alpha=45^{\circ}, \angle A B C=105^{\circ}
$$

Let $R$ be the radius of the circle. Since $2 R=B D=2$, then
$A C=2 R \sin \angle A B C=2 \sin 105^{\circ}=2 \sin \left(60^{\circ}+45^{\circ}\right)=$

$=2\left(\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}\right)=\frac{\sqrt{2}+\sqrt{6}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-42.jpg?height=475&width=531&top_left_y=274&top_left_x=772)

Answer

$\frac{\sqrt{2}+\sqrt{6}}{2}$.

Answer: \frac{\sqrt{2}+\sqrt{6}}{2}


Odredite neki period funkcije $f(x)=2 \sin \left(\frac{3}{4} x\right)+\cos \left(\frac{4}{5} x-\frac{\pi}{3}\right)$.

Solution:
## Rješenje.

Temeljni period funkcije $f_{1}(x)=2 \sin \left(\frac{3}{4} x\right)$ je $P_{1}=\frac{2 \pi}{\frac{3}{4}}=\frac{8 \pi}{3}$.

Temeljni period funkcije $f_{2}(x)=\cos \left(\frac{4}{5} x-\frac{\pi}{3}\right)$ je $P_{2}=\frac{2 \pi}{\frac{4}{5}}=\frac{5 \pi}{2}$.

Uočimo da je svaki period funkcije $f_{1}$ oblika $\frac{8 \pi}{3} \cdot n, n \in \mathbb{N}$, a svaki period funkcije $f_{2}$ oblika $\frac{5 \pi}{2} \cdot m, m \in \mathbb{N}$.

Period funkcije $f=f_{1}+f_{2}$ je svaki broj koji je istovremeno oblika $\frac{8 \pi}{3} \cdot n \mathrm{i} \frac{5 \pi}{2} \cdot m$, za neke $m, n \in \mathbb{N}$.

Proširimo razlomke do istog nazivnika: $\frac{8 \pi}{3}=\frac{16 \pi}{6}, \frac{5 \pi}{2}=\frac{15 \pi}{6}$.

Najmanji zajednički višekratnik brojeva 16 i 15 je 240 . Zato je $15 \cdot \frac{8 \pi}{3}=16 \cdot \frac{5 \pi}{2}=$ $\frac{240 \pi}{6}=40 \pi$ period funkcije $f$.

Period funkcije $f$ je svaki broj oblika $40 k \pi, k \in \mathbb{N}$.

Napomena: Učeniku treba dati sve bodove za točan postupak i napisan bilo koji odgovor koji je višekratnik broja $40 \pi$.

Napomena: Ako je učenik pogodio period $P$, treba provjeriti da je $f(x+P)=f(x)$ za svaki $x \in \mathbb{R}$. U protivnom dobiva samo 1 bod.

Answer: 40\pi


If the equation
$$
x^{3}-5 x^{2}+(4+k) x-k=0
$$

has three roots that can serve as the lengths of the three sides of an isosceles triangle, then the value of the real number $k$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6

Solution:
The original equation can be transformed into
$$
(x-1)\left(x^{2}-4 x+k\right)=0 \text {. }
$$

Given that the three roots of the original equation can serve as the three side lengths of an isosceles triangle, we know that $x=1$ is a root of $x^{2}-4 x+k=0$, or $x^{2}-4 x+k=0$ has two equal roots.
Solving this, we get $k=3$ or 4.
When $k=3$, the three roots of the equation are $1,1,3$, which cannot serve as the side lengths of an isosceles triangle;

When $k=4$, the three roots of the equation are $1, 2, 2$, which can serve as the side lengths of an isosceles triangle.
Therefore, the value of $k$ is 4. Choose B.

Answer: B


Simplify the expression $\frac{\left(1-\left(\frac{x}{y}\right)^{2}\right) \cdot y^{2} \cdot(x-y)^{-1}}{(\sqrt{x}-\sqrt{y})^{2}+\sqrt{4 x y}}$, given that $x, y>0$ and $x \neq y$.

Solution:
B3. Let's correct the multiplication and consider the negative exponent $\mathrm{v}$ in the numerator $\left(1-\left(\frac{x}{y}\right)^{2} \cdot y^{2} \cdot(x-y)^{-1}=\right.$ $\left(y^{2}-x^{2}\right) \cdot \frac{1}{x-y}$. Square and partially root in the denominator $(\sqrt{x}-\sqrt{y})^{2}+\sqrt{4 x y}=x-$ $2 \sqrt{x y}+y+2 \sqrt{x y}$. Factor the numerator and simplify the denominator $\frac{(y-x)(y+x) \frac{1}{x-y}}{x+y}$. Simplify and get $(y-x) \frac{1}{x-y}=\frac{y-x}{x-y}$, simplify again and get -1.

![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-14.jpg?height=65&width=1685&top_left_y=1875&top_left_x=228)

![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-14.jpg?height=57&width=1685&top_left_y=1933&top_left_x=228)

Denominator simplification $x-2 \sqrt{x y}+y+2 \sqrt{x y}=x+y \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

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![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-14.jpg?height=57&width=1685&top_left_y=2082&top_left_x=228)

![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-14.jpg?height=54&width=1682&top_left_y=2132&top_left_x=233)

Answer: -1


Anton and Britta play a game with the set $M=\left \{ 1,2,\dots,n-1 \right \}$ where $n \geq 5$ is an odd integer. In each step Anton removes a number from $M$ and puts it in his set $A$, and Britta removes a number from $M$ and puts it in her set $B$ (both $A$ and $B$ are empty to begin with). When $M$ is empty, Anton picks two distinct numbers $x_1, x_2$ from $A$ and shows them to Britta. Britta then picks two distinct numbers $y_1, y_2$ from $B$. Britta wins if
$(x_1x_2(x_1-y_1)(x_2-y_2))^{\frac{n-1}{2}}\equiv 1\mod n$
otherwise Anton wins. Find all $n$ for which Britta has a winning strategy.

Solution:
1. **Case 1: \( n \) is not a prime number**

   If \( n \) is not a prime number, then \( n \) has a prime divisor \( q \). Anton can choose this prime divisor \( q \) on his first move. At the end of the game, Anton can choose \( x_1 = q \) and any other number \( x_2 \) from his set \( A \). Since \( q \) is a prime divisor of \( n \), we have:
   \[
   q^{\frac{n-1}{2}} \equiv 0 \pmod{n}
   \]
   This means that the expression \( (x_1 x_2 (x_1 - y_1) (x_2 - y_2))^{\frac{n-1}{2}} \) will be congruent to 0 modulo \( n \), which is not congruent to 1. Therefore, Anton wins in this case.

2. **Case 2: \( n = p \), where \( p \) is a prime number**

   If \( n \) is a prime number \( p \), Britta can follow a specific strategy to ensure her win. The strategy is as follows:
   - If Anton chooses a number \( t \) from \( M \), Britta will choose \( p - t \) from \( M \).
   - Since \( p - 1 \) is even and Anton starts the game, Britta can always follow this strategy until the end of the game.

   At the end of the game, when Anton picks \( x_1 \) and \( x_2 \) from his set \( A \), Britta will pick \( y_1 = p - x_1 \) and \( y_2 = p - x_2 \) from her set \( B \). We need to evaluate the expression:
   \[
   (x_1 x_2 (x_1 - y_1) (x_2 - y_2))^{\frac{p-1}{2}}
   \]
   Substituting \( y_1 = p - x_1 \) and \( y_2 = p - x_2 \), we get:
   \[
   (x_1 x_2 (x_1 - (p - x_1)) (x_2 - (p - x_2)))^{\frac{p-1}{2}} = (x_1 x_2 (2x_1 - p) (2x_2 - p))^{\frac{p-1}{2}}
   \]
   Since \( x_1 \) and \( x_2 \) are elements of \( \{1, 2, \ldots, p-1\} \), and \( p \) is a prime number, we have:
   \[
   (x_1 x_2 (2x_1 - p) (2x_2 - p))^{\frac{p-1}{2}} \equiv (2x_1 x_2)^{p-1} \equiv 1 \pmod{p}
   \]
   This is because \( (2x_1 x_2)^{p-1} \equiv 1 \pmod{p} \) by Fermat's Little Theorem, which states that for any integer \( a \) not divisible by a prime \( p \), \( a^{p-1} \equiv 1 \pmod{p} \).

   Therefore, Britta wins when \( n \) is a prime number.

Conclusion:
Britta has a winning strategy if and only if \( n \) is a prime number.

The final answer is \( \boxed{ n } \) is a prime number.

Answer: n


Calculate the following infinite product: $3^{\frac{1}{3}} \cdot 9^{\frac{1}{9}} \cdot 27^{\frac{1}{27}} \ldots$.

Solution:
129. Let $N=3^{\frac{1}{3}} \cdot 3^{\frac{2}{9}} \cdot 3^{\frac{3}{27}} \ldots=3^{\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\cdots+\frac{n}{3^n}+\ldots}=3^{M}$. Then

$$
\frac{M}{3}=\frac{1}{3^{2}}+\frac{2}{3^{3}}+\frac{3}{3^{4}}+\cdots+\frac{n-1}{3^{n}}+\cdots
$$

Therefore,

$$
\left(1-\frac{1}{3}\right) M=\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots+\frac{1}{3^{n}}+\ldots=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}
$$

Thus, $N=3^{\frac{3}{4}}=\sqrt[4]{27}$.

[J. F. Arena, S. S. M., 46, 678 (October 1946).]

Answer: \sqrt[4]{27}


What is the value of the product

$\log _{3} 2 \cdot \log _{4} 3 \cdot \log _{5} 4 \ldots \log _{10} 9$

if it is known that $\lg 2=0.3010$?

Solution:
11.53 Instruction. Convert all factors to base 10.

Answer: 0.3010.

Answer: 0.3010


Among the following four numbers, the largest is ( )
A: $\ln \sqrt{2}$
B: $\frac{1}{e}$
C: $\frac{\ln \pi}{\pi}$
D: $\frac{\sqrt{10} \ln 10}{20}$

Solution:
Answer B.
Analysis Consider the function $f(x)=\frac{\ln x}{x}$, then the four options are $f(2), f(e), f(\pi), f(\sqrt{10})$. Since $f^{\prime}(x)=\frac{1-\ln x}{x^{2}}$, it is clear that $f(x)$ is monotonically increasing on $(0, e)$ and monotonically decreasing on $(e,+\infty)$, so the maximum value of $f(x)$ on $(0,+\infty)$ is $f(e)$.

Answer: B


Solve the equation:
\[
\left|\begin{array}{ccc}
-4 & 2 & -4+3 i \\
3 & -5 & 2-5 i \\
-4+3 i & 2-5 i & x
\end{array}\right|=0
\]

Solution:
$$
\begin{array}{l}
\because\left|\begin{array}{lcc}
-4 & 2 & -4+3 \mathrm{i} \\
8 & -5 & 2-5 \mathrm{i} \\
-4+3 \mathrm{i} & 2-5 \mathrm{i} & \mathbf{x}
\end{array}\right| \begin{array}{l}
\text { (multiply the first row by } \\
\text { the row above) }
\end{array} \\
\left|\begin{array}{ccc}
-4 & 2 & -4+3 i \\
3 & -5 & 2-5 i \\
3 & -5 i & x+4-3 i
\end{array}\right| \text { (multiply the second row by }-\mathrm{i} \\
\left|\begin{array}{rrl}
-4 & 2 & -4+3 i \\
8 & -5 & 2-5 i \\
0 & 0 & x-1-5 i
\end{array}\right|=(x-1-5 i)\left|\begin{array}{rr}
-4 & 2 \\
3 & -5
\end{array}\right| \\
=14(x-1-5 i)=0 \text {, } \\
\therefore x=1+5 i \text {. } \\
\end{array}
$$

Answer: x=1+5i


For the quadratic trinomial $p(x)=(a+1) x^{2}-(a+1) x+2022$, it is known that $-2022 \leq p(x) \leq 2022$ for $x \in[0 ; 1]$. Find the greatest possible value of $a$.

Solution:
# Answer: 16175.

Solution. Since $p(0)=p(1)=2022$, the graph of the quadratic trinomial is a parabola symmetric about the line $x=\frac{1}{2}$. From the conditions that $-2022 \leq$ $p(x) \leq 2022$ for $x \in[0 ; 1]$ and $p(0)=p(1)=2022$, it follows that the branches of the parabola are directed upwards. Then the minimum value of $p(x)$ is $p\left(\frac{1}{2}\right)=2022-\frac{(a+1)}{4}$. The maximum possible value of $a$ will be achieved when $p\left(\frac{1}{2}\right)=-2022$. Therefore,

$$
2022-\frac{(a+1)}{4}=-2022 \Rightarrow \frac{(a+1)}{4}=4044 \Rightarrow a+1=16176 \Rightarrow a=16175
$$

Answer: 16175


On the edge $B B_{1}$ of a regular triangular prism $A B C A_{1} B_{1} C_{1}$, a point $T$ is taken such that $B T: B_{1} T=2: 3$. The point $T$ is the vertex of a right circular cone such that three vertices of the prism lie on the circumference of its base.

a) Find the ratio of the height of the prism to the edge of its base.

b) Suppose it is additionally known that $C C_{1}=5$. Find the volume of the cone.

Solution:
Answer. a) $\sqrt{5}$; b) $V=\frac{180 \pi \sqrt{3}}{23 \sqrt{23}}$.

Solution. If three vertices of the prism lie on the circumference of the base of the cone, this means that the three vertices of the prism are equidistant from point $T$, i.e., three of the segments $T A, T B, T C, T A_{1}, T B_{1}, T C_{1}$ are equal to each other. Note that $T A_{1}=T C_{1}>T A=T C$; moreover, $T A>T B, T A_{1}>T B_{1}$. From these inequalities, it follows that the segments $T A_{1}$ and $T C_{1}$ are the longest, and the segment $T B$ is the shortest. Therefore, the segments $T A, T C$, and $T B_{1}$ are equal to each other.

a) Let $T B_{1}=3 x, T B=2 x$. Then $T A=3 x$. By the Pythagorean theorem for triangle $A B T$, we find that $A B=x \sqrt{5}$. Thus, the desired ratio is $\frac{5 x}{x \sqrt{5}}=\sqrt{5}$.

b) From triangle $B B_{1} A$ by the Pythagorean theorem, we get that $A B_{1}=x \sqrt{30}$. The radius of the base of the cone is the radius of the circle circumscribed around triangle $A B_{1} C$. Its sides are: $A B_{1}=C B_{1}=x \sqrt{30}, A C=x \sqrt{5}$. Then we find its height: $B_{1} H=\frac{x \sqrt{115}}{2}$, area: $S=\frac{1}{2} \cdot x \sqrt{5} \cdot \frac{x \sqrt{115}}{2}=\frac{5 x^{2} \sqrt{23}}{4}$, radius of the circumscribed circle: $R=\frac{x \sqrt{30} \cdot x \sqrt{30} \cdot x \sqrt{5}}{4 S}=\frac{6 x \sqrt{5}}{\sqrt{23}}$.

The slant height of the cone is the segment $A T$, i.e., it is equal to $3 x$. Then the height of the cone $h=\sqrt{9 x^{2}-\frac{180 x^{2}}{23}}=\frac{3 x \sqrt{3}}{\sqrt{23}}$.

We find the volume of the cone: $V=\frac{\pi}{3} \cdot \frac{180 x^{2}}{23} \cdot \frac{3 x \sqrt{3}}{\sqrt{23}}=\frac{180 \pi x^{3} \sqrt{3}}{23 \sqrt{23}}$. Since $x=1$, we finally get $V=\frac{180 \pi \sqrt{3}}{23 \sqrt{23}}$.

Answer: \frac{180\pi\sqrt{3}}{23\sqrt{23}}