There is a troop of soldiers. If they form a column of five, there is one person left over at the end. If they form a column of six, there are five people left over at the end. If they form a column of seven, there are four people left over at the end. If they form a column of eleven, there are ten people left over at the end. Find the number of soldiers.

Solution:
Let $x$ be the number of soldiers we are looking for. According to the problem,

In Theorem 1, take $m_{1}=5, m_{2}=6, m_{3}=7, m_{4}=11, b_{1}=1$, $b_{2}=5, b_{3}=4, b_{4}=10$. Then we have
$$\begin{array}{l}
M=5 \times 6 \times 7 \times 11=2310 \\
M_{1}=\frac{2310}{5}=462 \\
M_{2}=\frac{2310}{6}=385 \\
M_{3}=\frac{2310}{7}=330 \\
M_{4}=\frac{2310}{11}=210
\end{array}$$

Let $M_{1}^{\prime}$ be a positive integer that satisfies $M_{1}^{\prime} M_{1} \equiv 1(\bmod 5)$, then $1 \equiv$ $M_{1}^{\prime} M_{1} \equiv 462 M_{1}^{\prime} \equiv 2 M_{1}^{\prime}(\bmod 5)$, so we get $M_{1}^{\prime}=3$. Let $M_{2}^{\prime}$ be a positive integer that satisfies $M_{2}^{\prime} M_{2} \equiv 1(\bmod 6)$, then $1 \equiv M_{2}^{\prime} M_{2} \equiv$ $385 M_{2}^{\prime} \equiv M_{2}^{\prime}(\bmod 6)$, so we get $M_{2}^{\prime}=1$. Let $M_{3}^{\prime}$ be a positive integer that satisfies $M_{3}^{\prime} M_{3} \equiv 1(\bmod 7)$, then $1 \equiv M_{3}^{\prime} M_{3} \equiv 330 M_{3}^{\prime} \equiv$ $M_{3}^{\prime}(\bmod 7)$, so we get $M_{3}^{\prime}=1$. Let $M_{4}^{\prime}$ be a positive integer that satisfies $M_{4}^{\prime} M_{4} \equiv 1 \quad(\bmod 11)$, then $1 \equiv M_{4}^{\prime} M_{4} \equiv 210 M_{4}^{\prime} \equiv M_{4}^{\prime} (\bmod$ 11). Therefore, by (46) we get
$$\begin{aligned}
x & \equiv 3 \times 462+5 \times 385+4 \times 330+10 \times 210 \\
& \equiv 6731 \equiv 2111(\bmod 2310)
\end{aligned}$$

Thus, we have
$$x=2111+2310 k, \quad k=0,1,2, \cdots$$
$$\begin{array}{l}
x \equiv 1(\bmod 5), x \equiv 5(\bmod 6), \\
x \equiv 4 \quad(\bmod 7), \quad x \equiv 10 \quad(\bmod 11).
\end{array}$$

Answer: 2111


Determine the values that the expression $V=a b+b c+c d+d a$ can take, given that the real numbers $a, b, c, d$ satisfy the following conditions:

$$
\begin{aligned}
& 2 a-5 b+2 c-5 d=4, \\
& 3 a+4 b+3 c+4 d=6 .
\end{aligned}
$$

Solution:
1. For the given expression $V$, we have

$$
V=a(b+d)+c(b+d)=(a+c)(b+d).
$$

Similarly, we can modify both given conditions:

$$
2(a+c)-5(b+d)=4 \quad \text { and } \quad 3(a+c)+4(b+d)=6
$$

If we choose the substitution $m=a+c$ and $n=b+d$, we get the solution of the system (1) as $m=2$ and $n=0$. For the given expression, we then have $V=m n=0$.

Conclusion. Under the given conditions, the expression $V$ takes only the value 0.

Alternative solution. We can interpret the conditions of the problem as a system of equations with unknowns $a, b$ and parameters $c, d$. By solving this system (using the addition or substitution method), we find $a=2-c, b=-d$ (where $c, d \in \mathbb{R}$), and substituting into the expression $V$ gives

$$
V=(2-c)(-d)-d c+c d+d(2-c)=0.
$$

For a complete solution, award 6 points. For the first solution, award 2 points for factoring the expression $V$ into a product, 2 points for transforming the conditions into the system (1), 1 point for solving the system, and 1 point for calculating the value of $V$.

Answer: 0


Let $1 = d_1 < d_2 < ...< d_k = n$ be all natural divisors of the natural number $n$. Find all possible values of the number $k$ if $n=d_2d_3 + d_2d_5+d_3d_5$.

Solution:
We are given a natural number \( n \) with divisors \( 1 = d_1 < d_2 < \cdots < d_k = n \) satisfying the equation \( n = d_2d_3 + d_2d_5 + d_3d_5 \).

We need to find all possible values of \( k \).

### Case 1: \( (d_2, d_3) = (p, p^2) \)

1. Let \( d_2 = p \), where \( p \) is the smallest prime divisor of \( n \).
2. Then \( d_3 = p^2 \).
3. From the given equation \( n = d_2d_3 + d_2d_5 + d_3d_5 \), we have:
   \[
   n = p \cdot p^2 + p \cdot d_5 + p^2 \cdot d_5
   \]
   \[
   n = p^3 + p \cdot d_5 + p^2 \cdot d_5
   \]
   \[
   n = p^3 + d_5(p + p^2)
   \]
4. Since \( d_5 \mid p^3 \) and \( d_5 > p^2 \), we get \( d_5 = p^3 \).
5. Substituting \( d_5 = p^3 \) into the equation, we get:
   \[
   n = p^3 + p \cdot p^3 + p^2 \cdot p^3
   \]
   \[
   n = p^3 + p^4 + p^5
   \]
   \[
   n = p^3(p^2 + p + 1)
   \]
6. We know \( d_5 = p^3 \) and \( d_5 > \sqrt{n} \), so:
   \[
   p^3 > \sqrt{p^3(p^2 + p + 1)}
   \]
   \[
   p^3 > p^{3/2} \sqrt{p^2 + p + 1}
   \]
   \[
   p^{3/2} > \sqrt{p^2 + p + 1}
   \]
   \[
   p^3 > p^2 + p + 1
   \]
7. This implies \( k < 10 \).

### Case 2: \( (d_2, d_3) = (p, q) \)

1. Let \( d_2 = p \) and \( d_3 = q \), where \( p \) and \( q \) are the smallest and second smallest prime divisors of \( n \), respectively.
2. From the given equation \( n = d_2d_3 + d_2d_5 + d_3d_5 \), we have:
   \[
   n = p \cdot q + p \cdot d_5 + q \cdot d_5
   \]
   \[
   n = pq + d_5(p + q)
   \]
3. Since \( d_5 \mid pq \) and \( d_5 > q \), we get \( d_5 = pq \).
4. Substituting \( d_5 = pq \) into the equation, we get:
   \[
   n = pq + pq(p + q)
   \]
   \[
   n = pq(1 + p + q)
   \]
5. We know \( d_3 < d_4 < d_5 \), so:
   \[
   q < d_4 < pq
   \]
6. This implies \( d_4 = p + q + 1 \) when \( (p, q) \neq (2, 3) \).
7. Consequently, \( n = d_5d_4 \), which implies \( k = 4 + 5 - 1 = 8 \).

### Special Case: \( (p, q) = (2, 3) \)

1. Setting \( (p, q) = (2, 3) \) in the equation, we get:
   \[
   n = 2 \cdot 3 + 2 \cdot 6 + 3 \cdot 6
   \]
   \[
   n = 6 + 12 + 18
   \]
   \[
   n = 36
   \]
2. The divisors of 36 are \( 1, 2, 3, 4, 6, 9, 12, 18, 36 \), so \( k = 9 \).

### Conclusion

The natural numbers \( n \) satisfying the given condition have 8 or 9 divisors, i.e., \( k \in \{8, 9\} \).

The final answer is \( \boxed{ k \in \{8, 9\} } \).

Answer: k \in \{8, 9\}


The least common multiple of natural numbers $a$ and $b$ is 140, and their greatest common divisor is 5. Then the maximum value of $a+b$ is $\qquad$

Solution: 【Answer】Solution: According to the analysis, one of $a$ and $b$ is 140, and the other is 5, so: the maximum value of $a+b$ is $5+140=145$; hence, the answer is: 145.

Answer: 145


Solution:
## Solution

Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$ :

$$
\left|\begin{array}{ccc}
x-1 & y-3 & z-0 \\
4-1 & -1-3 & 2-0 \\
3-1 & 0-3 & 1-0
\end{array}\right|=0
$$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-3 & z \\
3 & -4 & 2 \\
2 & -3 & 1
\end{array}\right|=0 \\
& (x-1) \cdot\left|\begin{array}{cc}
-4 & 2 \\
-3 & 1
\end{array}\right|-(y-3) \cdot\left|\begin{array}{ll}
3 & 2 \\
2 & 1
\end{array}\right|+z \cdot\left|\begin{array}{cc}
3 & -4 \\
2 & -3
\end{array}\right|=0 \\
& (x-1) \cdot 2-(y-3) \cdot(-1)+z \cdot(-1)=0 \\
& 2 x-2+y-3-z+0=0 \\
& 2 x+y-z-5=0
\end{aligned}
$$

The distance $d$ from a point $M_{0}\left(x_{0} ; y_{0} ; z_{0}\right)$ to the plane $A x+B y+C z+D=0$ :

$$
d=\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}
$$

Find:

$d=\frac{|2 \cdot 4+3-0-5|}{\sqrt{2^{2}+1^{2}+(-1)^{2}}}=\frac{|8+3-5|}{\sqrt{4+1+1}}=\frac{6}{\sqrt{6}}=\sqrt{6}$

## Problem Kuznetsov Analytic Geometry $8-11$

Answer: \sqrt{6}


Without using any unit of length or other auxiliary means, how can you cut exactly half a meter from a piece of fabric that is $\frac{8}{15}$ m?

Solution: 43. Fold the piece in half (we get $\frac{4}{15}$ m), fold the resulting piece in half again (we get $\frac{2}{15}$ m), then fold it in half once more (we get $\frac{1}{15}$ m), and finally, fold the resulting piece in half again (we get $\frac{1}{30}$ m). If we cut off this part from the piece, i.e., $\frac{8}{15}-\frac{1}{30}$, we will get $\frac{1}{2}$ m $\left(\frac{8}{15}-\frac{1}{30}=\frac{15}{30}=\frac{1}{2}\right)$.

Answer: \frac{1}{2}


A merchant bought several bags of salt in Tver and sold them in Moscow with a profit of 100 rubles. With all the money earned, he again bought salt in Tver (at the Tver price) and sold it in Moscow (at the Moscow price). This time the profit was 120 rubles. How much money did he spend on the first purchase?

Solution:
Solution. Let the merchant pay $x$ rubles for the salt during the first purchase in Tver. Then he sold it in Moscow for $x+100$ rubles. The second time, he spent $x+100$ rubles in Tver and received $x+100+120=x+220$ rubles in Moscow. Since the ratio of Moscow and Tver prices did not change, we can set up the proportion $\frac{x}{x+100}=\frac{x+100}{x+220}$, from which we get $x^{2}+220 x=x^{2}+200 x+10000, 20 x=10000, x=500$.

Answer: 500 rubles.

Answer: 500


Determine all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for every $n \in \mathbb{N}$,

$$
2 n+2001 \leqslant f(f(n))+f(n) \leqslant 2 n+2002
$$

(Romania)

Solution:
4. The function $f(n)=n+667$ satisfies the conditions of the problem.

For a given $n \in \mathbb{N}$, we define the sequence $(a_{k})$ with $a_{0}=n$ and $a_{k+1}=f(a_{k})$ for $k \geqslant 0$. If we denote $b_{k}=a_{k+1}-a_{k}-667-\frac{1}{6}$, by the condition of the problem, $-\frac{1}{2} \leqslant c_{k}=b_{k+1}+2 b_{k} \leqslant \frac{1}{2}$. Since

$$
\begin{aligned}
b_{m} & =(-2)^{m} b_{0}+\sum_{k=0}^{m-1}(-2)^{m-1-k} c_{k} \quad \text { and } \\
a_{n} & =a_{0}+667 n+\sum_{m=0}^{n-1} b_{m}=a_{0}+667 n+\sum_{m=0}^{n-1}(-2)^{m} b_{0}+\sum_{k=0}^{n-2} \sum_{m=k+1}^{n-1}(-2)^{m-1-k} c_{k} \\
& =a_{0}+667 n+\frac{1-(-2)^{n}}{3} b_{0}+\sum_{k=0}^{n-2} \frac{1-(-2)^{n-k-1}}{3} c_{k} \\
& \leqslant a_{0}+667 n+\frac{1-(-2)^{n}}{3}\left(b_{0}+\frac{(-1)^{n-1}}{2}\right)
\end{aligned}
$$

it must be true that $a_{n}\frac{1}{2}$. Since by the condition of the problem $a_{n}>0$ for all $n$, it follows that $-\frac{1}{2} \leqslant b_{0}=f(n)-n-667-\frac{1}{6} \leqslant \frac{1}{2}$, from which $f(n)=n+667$.

Answer: f(n)=n+667


From 3 boys and $n$ girls, any 3 people are selected to participate in a competition, given that the probability of having at least 1 girl among the 3 people is $\frac{34}{35}$. Then $n=$ $\qquad$ .

Solution: 13.4. From the condition, $1-\frac{\mathrm{C}_{3}^{3}}{\mathrm{C}_{n+3}^{3}}=\frac{34}{35}$, solving for $n$ yields $n=4$.

Answer: 4


What is the greatest value that the area of a right triangle can take, one vertex of which coincides with the origin, another lies on the curve $x^{2}+y^{2}=2(x+y)$, and the vertex of the right angle is located on the line $y=x$? In the answer, write the square of the found area.

Solution:
Solution. $\quad x^{2}+y^{2}-2 x-2 y=0, \quad(x-1)^{2}+(y-1)^{2}=2$.

We have the equation of a circle with center at point $(1 ; 1)$ and radius $R=\sqrt{2}$. Let's transition to the coordinate system Ouv while maintaining the scale (see figure). The equation of the circle in this coordinate system is $(u-\sqrt{2})^{2}+v^{2}=2$. The area of a right triangle, one vertex of which coincides with the origin, another lies on the circle, and the vertex of the right angle is located on the line

![](https://cdn.mathpix.com/cropped/2024_05_06_d66465e084edd0e7e96fg-10.jpg?height=622&width=697&top_left_y=1191&top_left_x=1296)
$v=0, \quad$ is calculated by the formula $S_{\text {OAB }}=\frac{1}{2} u \cdot|v|=\frac{1}{2} u \sqrt{2 \sqrt{2} u-u^{2}}$, where vertex $A(u, v)$ lies on the circle, and the vertex of the right angle $B(u, 0)$. We have $S^{2}=\frac{1}{4}\left(2 \sqrt{2} u^{3}-u^{4}\right)$. We find the zeros of the derivative of this function $\quad\left(S^{2}\right)^{\prime}(u)=\frac{u^{2}}{2}(3 \sqrt{2}-2 u)$. The only extremum point, and specifically, the point of maximum for this function is the point $u=3 \sqrt{2} / 2$, $S_{\max }=S(3 / \sqrt{2})=\frac{3 \sqrt{3}}{4}, \quad S_{\max }^{2}==\frac{27}{16}=1.6875$,

Answer: 1.6875.

Answer: 1.6875


The equation $\sum_{k=1}^{2 n+1} \cos ^{2} k x=n\left(n \in \mathbf{N}_{+}\right)$ has $\qquad$ solutions in $[0,2 \pi)$.

Solution:
$$
\begin{array}{l}
7.8 n+2 . \\
\sum_{k=1}^{2 n+1} \cos ^{2} k x=n \\
\Leftrightarrow \sum_{k=1}^{2 n+1}(\cos 2 k x+1)=2 n \\
\Leftrightarrow \sum_{k=1}^{2 n+1} \cos 2 k x=-1 .
\end{array}
$$

Obviously, $x \neq t \pi(\iota \in \mathbf{Z})$. Then
$$
\begin{array}{l}
\sum_{k=1}^{2 n+1} \cos 2 k x=\operatorname{Re}\left(\sum_{k=1}^{2 n+1} \mathrm{e}^{\mathrm{i} 2 k x}\right) \\
=\operatorname{Re}\left(\frac{\mathrm{e}^{\mathrm{i} 2(2 n+2) x}-\mathrm{e}^{\mathrm{i} 2 x}}{\mathrm{e}^{\mathrm{i} 2 x}-1}\right) \\
=\operatorname{Re}\left\{\frac{[\cos (4 n+4) x-\cos 2 x]+\mathrm{i}[\sin (4 n+4) x-\sin 2 x]}{(\cos 2 x-1)+\mathrm{i} \sin 2 x}\right\} \\
=\operatorname{Re}\left\{\frac{\sin (2 n+1) x}{\sin x} \cdot \frac{-\sin (2 n+3) x+\mathrm{i} \cos (2 n+3) x}{-\sin x+\mathrm{i} \cos x}\right\} \\
=\operatorname{Re}\left[\frac{\sin (2 n+1) x}{\sin x} \cdot \mathrm{e}^{\mathrm{i}(2 n+2) x}\right] \\
=\frac{\sin (2 n+1) x}{\sin x} \cdot \cos (2 n+2) x \\
=\frac{\sin (4 n+3) x-\sin x}{2 \sin x}=-1 .
\end{array}
$$

Thus, $\sin (4 n+3) x+\sin x=0$, which means
$$
\sin (2 n+2) x \cdot \cos (2 n+1) x=0 .
$$

Solving this, we get $x=\frac{s \pi}{2 n+2}(s=1,2, \cdots, 2 n+1,2 n+3,2 n+4, \cdots, 4 n+3)$ or $x=\frac{\left(t+\frac{1}{2}\right) \pi}{2 n+1}(t=0, 1, \cdots, 4 n+1)$. Among these, $\frac{s \pi}{2 n+2}=\frac{\left(t+\frac{1}{2}\right) \pi}{2 n+1} \Leftrightarrow \frac{s}{n+1}=\frac{2 t+1}{2 n+1}$ $\Leftrightarrow(s, t)=(n+1, n)$ or $(3 n+3,3 n+1)$. Therefore, equation (1) has $8 n+2$ solutions in $[0,2 \pi)$.

Answer: 8n+2


There is a prime number $p$ such that $16p+1$ is the cube of a positive integer.  Find $p$.

Solution:
Let the positive integer mentioned be $a$, so that $a^3 = 16p+1$. Note that $a$ must be odd, because $16p+1$ is odd.
Rearrange this expression and factor the left side (this factoring can be done using $(a^3-b^3) = (a-b)(a^2+a b+b^2)$ or synthetic divison once it is realized that $a = 1$ is a root):
\begin{align*} a^3-1 &= 16p\\ (a-1)(a^2+a+1) &= 16p\\ \end{align*}
Because $a$ is odd, $a-1$ is even and $a^2+a+1$ is odd. If $a^2+a+1$ is odd, $a-1$ must be some multiple of $16$. However, for $a-1$ to be any multiple of $16$ other than $16$ would mean $p$ is not a prime. Therefore, $a-1 = 16$ and $a = 17$.
Then our other factor, $a^2+a+1$, is the prime $p$:
\begin{align*} (a-1)(a^2+a+1) &= 16p\\ (17-1)(17^2+17+1) &=16p\\ p = 289+17+1 &= \boxed{307} \end{align*}

Answer: 307


In the set $\left\{-3,-\frac{5}{4},-\frac{1}{2}, 0, \frac{1}{3}, 1, \frac{4}{5}, 2\right\}$, two numbers are drawn without replacement. Find the probability that the two numbers drawn are the slopes of a pair of perpendicular lines.

Solution:
Two numbers form the slopes of a pair of perpendicular lines, meaning the product of the two numbers is -1. Pairs that satisfy this condition are: -3 and $\frac{1}{3}$, $-\frac{5}{4}$ and $\frac{4}{5}$, $-\frac{1}{2}$ and 2, for a total of 3 pairs.
Choosing 2 numbers from 8, there are $C_{8}^{2}=\frac{8 \times 7}{2}=28$ ways, so the required probability is $\frac{3}{28}$.

Answer: \frac{3}{28}


If Xingyou's current age in years is also his current height in feet, and in three years, Xingyou's age in years will be twice his current height in feet, what is Xingyou's age in years right now?

Solution:
1. Let \( x \) be Xingyou’s current age in years, which is also his current height in feet.
2. According to the problem, in three years, Xingyou’s age will be twice his current height. This can be expressed as:
   \[
   x + 3 = 2x
   \]
3. To solve for \( x \), we subtract \( x \) from both sides of the equation:
   \[
   x + 3 - x = 2x - x
   \]
   Simplifying this, we get:
   \[
   3 = x
   \]

The final answer is \( \boxed{3} \).

Answer: 3


Into how many parts can 7 different tangents to a given circle divide the plane? Provide examples for all answers and prove that no others exist.

Solution:
Solution. 26,27,28,29. To get the correct answer, one should notice that it is possible to consider cases of pairwise parallelism of several lines. For example, no parallel lines, one pair of parallel lines, two pairs of parallel lines, three pairs of parallel lines.

Criteria. -+ - incomplete enumeration, partially correct answer obtained.

$\pm$ - correct answer, but there is an error in the analysis of cases.

Answer: 26,27,28,29


At 8:00 AM, Xiao Cheng and Xiao Chen set off from locations A and B, respectively, heading towards each other. They met at 9:40 AM.
Xiao Cheng said: "If I had traveled 10 kilometers more per hour, we would have met 10 minutes earlier."
Xiao Chen said: "If I had set off half an hour earlier, we would have met 20 minutes earlier."
If both of them are correct, then the distance between A and B is $\qquad$ kilometers.

Solution:
【Solution】Solution: 9: $40-8: 00=1$ hour 40 minutes $=\frac{5}{3}$ hours 1 hour 40 minutes -10 minutes $=1$ hour 30 minutes $=\frac{3}{2}$ hours
$$
10 \times \frac{3}{2}=15 \text { (km) }
$$

Let the sum of the two people's speeds be $x$ km per hour
$$
\begin{array}{c}
\frac{5}{3} x-\frac{3}{2} x=15 \\
\frac{1}{6} x=15 \\
x=15 \times 6 \\
x=90 \\
90 \times \frac{5}{3}=150 \text { (km) }
\end{array}
$$

Answer: The distance between location A and location B is 150 km.
The answer is: 150 .

Answer: 150


A sample of size $n=50$ has been drawn from the population:

| variant | $x_{i}$ | 2 | 5 | 7 | 10 |
| :--- | :--- | ---: | ---: | ---: | ---: |
| frequency | $n_{i}$ | 16 | 12 | 8 | 14 |

Find the unbiased estimate of the population mean.

Solution:
The problem is solved. An unbiased estimate of the population mean is the sample mean

$$
\bar{x}_{\mathrm{B}}=\left(\sum n_{l} x_{i}\right) / n=(16.2+12.5+8.7+14 \cdot 10) / 50=5.76
$$

Answer: 5.76


Given that $\triangle ABC$ is an equilateral triangle with side length 1, $D$ is a point on side $BC$ such that $BD=p$, and $r_{1}, r_{2}$ are the radii of the inscribed circles of $\triangle ABD$ and $\triangle ADC$ respectively. Express $r_{1} r_{2}$ in terms of $p$, and find the maximum value of $r_{1} r_{2}$.

Solution:
Solution: As shown in Figure 4, in $\triangle A B D$, by the cosine rule we have
$$
\begin{array}{l}
A D^{2}=p^{2}-p+1, \\
A D=\sqrt{p^{2}-p+1} .
\end{array}
$$

The area of a triangle is equal to the product of its semiperimeter and the inradius, so we have
$$
\begin{array}{l}
\frac{1+p+\sqrt{p^{2}-p+1}}{2} \cdot r_{1}=S_{\triangle A B D} \\
=\frac{1}{2} \times 1 \times p \sin 60^{\circ} .
\end{array}
$$

Thus, $r_{1}=\frac{\sqrt{3}\left(1+p-\sqrt{p^{2}-p+1}\right)}{6}$.
Similarly, from $\triangle A D C$ we get
$$
r_{2}=\frac{\sqrt{3}\left(2-p-\sqrt{p^{2}-p+1}\right)}{6} .
$$

Therefore, $r_{1} r_{2}=\frac{1-\sqrt{p^{2}-p+1}}{4}$.
When $p=\frac{1}{2}$, the maximum value of $r_{1} r_{2}$ is $\frac{2-\sqrt{3}}{8}$.

Answer: \frac{2-\sqrt{3}}{8}


$(\tan \alpha+\cot \alpha)^{2}-(\tan \alpha-\cot \alpha)^{2}$.

Solution:
Solution. We will use the formulas for the square of the sum and difference of two numbers:

$$
\begin{gathered}
(\operatorname{tg} \alpha+\operatorname{ctg} \alpha)^{2}-(\operatorname{tg} \alpha-\operatorname{ctg} \alpha)^{2}=\operatorname{tg}^{2} \alpha+2 \operatorname{tg} \alpha \operatorname{ctg} \alpha+\operatorname{ctg}^{2} \alpha-\operatorname{tg}^{2} \alpha+ \\
+2 \operatorname{tg} \alpha \operatorname{ctg} \alpha-\operatorname{ctg}^{2} \alpha=4 \operatorname{tg} \alpha \operatorname{ctg} \alpha=4
\end{gathered}
$$

(after combining like terms, we applied identity IV).

Answer: 4


A black bishop and a white king are placed randomly on a $2000 \times 2000$ chessboard (in distinct squares). Let $p$ be the probability that the bishop attacks the king (that is, the bishop and king lie on some common diagonal of the board). Then $p$ can be expressed in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m$.

Solution:
1. **Determine the total number of ways to place the bishop and the king:**
   Since the bishop and the king must be placed on distinct squares, the total number of ways to place them is:
   \[
   2000^2 \times (2000^2 - 1)
   \]

2. **Calculate the number of ways the bishop can attack the king:**
   The bishop attacks the king if they are on the same diagonal. There are two types of diagonals on a chessboard:
   - Diagonals with a positive slope (from bottom-left to top-right)
   - Diagonals with a negative slope (from top-left to bottom-right)

3. **Count the number of squares on each diagonal:**
   - For diagonals with a positive slope, the number of squares on each diagonal ranges from 1 to 2000, then back down to 1. Specifically, there are:
     \[
     1, 2, 3, \ldots, 2000, 1999, 1998, \ldots, 1
     \]
   - The same pattern holds for diagonals with a negative slope.

4. **Calculate the number of ways to place the bishop and the king on each diagonal:**
   For a diagonal with \( k \) squares, the number of ways to place the bishop and the king is:
   \[
   k \times (k - 1)
   \]
   This is because there are \( k \) choices for the bishop and \( k-1 \) remaining choices for the king.

5. **Sum the number of ways for all diagonals:**
   - For diagonals with a positive slope:
     \[
     \sum_{k=1}^{2000} k(k-1) + \sum_{k=1}^{1999} k(k-1)
     \]
   - For diagonals with a negative slope, the sum is the same.

6. **Simplify the sum:**
   \[
   \sum_{k=1}^{2000} k(k-1) = \sum_{k=1}^{2000} (k^2 - k) = \sum_{k=1}^{2000} k^2 - \sum_{k=1}^{2000} k
   \]
   Using the formulas for the sum of squares and the sum of the first \( n \) natural numbers:
   \[
   \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{k=1}^{n} k = \frac{n(n+1)}{2}
   \]
   For \( n = 2000 \):
   \[
   \sum_{k=1}^{2000} k^2 = \frac{2000 \cdot 2001 \cdot 4001}{6}, \quad \sum_{k=1}^{2000} k = \frac{2000 \cdot 2001}{2}
   \]
   Therefore:
   \[
   \sum_{k=1}^{2000} k(k-1) = \frac{2000 \cdot 2001 \cdot 4001}{6} - \frac{2000 \cdot 2001}{2}
   \]
   Simplifying further:
   \[
   \sum_{k=1}^{2000} k(k-1) = \frac{2000 \cdot 2001}{6} (4001 - 3) = \frac{2000 \cdot 2001 \cdot 3998}{6}
   \]

7. **Double the sum for both types of diagonals:**
   \[
   2 \left( \frac{2000 \cdot 2001 \cdot 3998}{6} + \frac{1999 \cdot 2000 \cdot 3997}{6} \right)
   \]

8. **Calculate the probability \( p \):**
   \[
   p = \frac{2 \left( \frac{2000 \cdot 2001 \cdot 3998}{6} + \frac{1999 \cdot 2000 \cdot 3997}{6} \right)}{2000^2 \times (2000^2 - 1)}
   \]

9. **Simplify the fraction to find \( m \) and \( n \):**
   After simplification, we find that:
   \[
   p = \frac{1333}{2001000}
   \]
   Here, \( m = 1333 \) and \( n = 2001000 \).

The final answer is \( \boxed{1333} \)

Answer: 1333


Thirty ones are written on the board. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 30 minutes?

Solution:
Answer: 435.

Solution: Let's represent 30 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line segments. This is the same number of candies that Karlson eats. After 30 minutes, all points will be connected. In total, $\frac{30 \cdot(30-1)}{2}=435$ line segments will be drawn. Therefore, Karlson will eat 435 candies.

Answer: 435


Given prime numbers $p$ and $q$ satisfy $q^{5}-2 p^{2}=1$. Then $p+q=$ $\qquad$

Solution:
7. 14 .
$$
\begin{aligned}
\text { Given } & q^{5}-2 p^{2}=1 \\
\Rightarrow & 2 p^{2}=q^{5}-1 \\
& =(q-1)\left(q^{4}+q^{3}+q^{2}+q+1\right) .
\end{aligned}
$$

It is easy to see that \( q^{4}+q^{3}+q^{2}+q+1 \) is an odd number greater than 1, and \( p^{2} \) is also an odd number greater than 1.
$$
\begin{array}{l}
\text { Then } q-1=2 \Rightarrow q=3 \\
\Rightarrow 2 p^{2}=3^{5}-1=242 \Rightarrow p=11 \\
\Rightarrow p+q=14 .
\end{array}
$$

Answer: 14


In $\triangle ABC$, $AB=37, AC=58$. With $A$ as the center and $AB$ as the radius, an arc is drawn intersecting $BC$ at point $D$, and $D$ is between $B$ and $C$. If the lengths of $BD$ and $DC$ are both integers, find the length of $BC$.

Solution:
Solve as shown in the figure, first prove
$$
\begin{array}{l}
B C \cdot C D \\
=A C^{2}-A B^{2} .
\end{array}
$$

Draw $A H \perp B C$ at $H$, then
$$
\begin{array}{l}
A C^{2}=A H^{2}+C H^{2}, \\
A B^{2}=A H^{2}+B H^{2},
\end{array}
$$

we have
$$
\begin{array}{l}
A C^{2}-A B^{2}=C H^{2}-B H^{2} \\
\quad=(C H+B H)(C H-B H)=B C \cdot C D .
\end{array}
$$

Also, $\because A B=37, A C=58$.
$$
\therefore B C \cdot C D=58^{2}-37^{2}=3 \times 5 \times 7 \times 19 \text {. }
$$

Since $A C - A B < B C, D$ is not between $B, C$, so it should be discarded, hence we should take $B C=57$, at this time $C D$ $=35, B D=24$.

Answer: 57


In an isosceles trapezoid \(ABCD\) with bases \(AD\) and \(BC\), perpendiculars \(BH\) and \(DK\) are drawn from vertices \(B\) and \(D\) to the diagonal \(AC\). It is known that the feet of the perpendiculars lie on the segment \(AC\) and \(AC=20\), \(AK=19\), \(AH=3\). Find the area of trapezoid \(ABCD\).

Solution:
Solution. Note that right triangles $D K A$ and $B H C$ are similar, since

$\angle B C H=\angle D A K$. Let $D K=x, B H=y$. Due to similarity $\frac{D K}{K A}=\frac{B H}{H C}, \frac{x}{19}=\frac{y}{17}$. On the other hand, $C D$

$=\mathrm{AB}$ and by the Pythagorean theorem

$$
C D^{2}=D K^{2}+K C^{2}=x^{2}+1, A B^{2}=B H^{2}+H A^{2}=y^{2}+9
$$

From this, $x^{2}-y^{2}=8$. Since $y=\frac{17 x}{19}$, substituting $y$ into the last equation, we get $\mathrm{x}=19 / 3, \mathrm{y}$ $=17 / 3, S_{A B C D}=A C \cdot(D K+B H) / 2=20 \cdot(19 / 3+17 / 3)=120$.

Answer: 120.

Answer: 120


Given the function

$$
f(x)=\frac{(x+a)^{2}}{(a-b)(a-c)}+\frac{(x+b)^{2}}{(b-a)(b-c)}+\frac{(x+c)^{2}}{(c-a)(c-b)}
$$

where $a, b$, and $c$ are distinct real numbers. Determine the range of the function.

Solution:
I. solution. After bringing to a common denominator $(a \neq b, a \neq c, b \neq c)$:

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

Thus, the range of the function $f$ is: $\{1\}$.

II. solution. Due to the quadratic expressions in the function assignment, $f(x)$ can be at most quadratic $(a \neq b, a \neq c, b \neq c)$. Let's examine the values of $f(x)$ at three different points, specifically at $x=-a$, $x=-b$, and $x=-c$:

$$
\begin{aligned}
f(-a) & =0+\frac{(b-a)^{2}}{(b-a)(b-c)}+\frac{(c-a)^{2}}{(c-a)(c-b)}= \\
& =\frac{b-a}{b-c}+\frac{c-a}{c-b}=\frac{b-a}{b-c}+\frac{a-c}{b-c}=\frac{b-c}{b-c}=1 \\
f(-b) & =\frac{(a-b)^{2}}{(a-b)(a-c)}+0+\frac{(c-b)^{2}}{(c-a)(c-b)}= \\
& =\frac{a-b}{a-c}+\frac{c-b}{c-a}=\frac{a-b}{a-c}+\frac{b-c}{a-c}=\frac{a-c}{a-c}=1 \\
f(-c) & =\frac{(a-c)^{2}}{(a-b)(a-c)}+\frac{(b-c)^{2}}{(b-a)(b-c)}+0= \\
& =\frac{a-c}{a-b}+\frac{b-c}{b-a}=\frac{a-c}{a-b}+\frac{c-b}{a-b}=\frac{a-b}{a-b}=1
\end{aligned}
$$

We found that the function $f(x)$ takes the same value at three different points. Quadratic functions can take the same value at most at two points because quadratic equations have at most two solutions. Linear functions can take the same value at only one point because linear equations have at most one solution.

Thus, our function $f(x)$ cannot be quadratic, it cannot be linear, so it must be a constant function, and its range consists of a single number, 1: $R_{f}=\{1\}$.

Answer: 1


Solution:
Question 45, Solution Method One: Let $M=\frac{\sin x(2-\cos x)}{5-4 \cos x}$, and let $a=5-4 \cos x$, then $a \in[1, 9]$,
$$
\begin{array}{l}
M^{2}=\frac{\left(1-\cos ^{2} x\right)(2-\cos x)^{2}}{(5-4 \cos x)^{2}}=\frac{\left(1-\cos ^{2} x\right)(2-\cos x)^{2}}{(5-4 \cos x)^{2}}=\frac{\left[1-\left(\frac{5-a}{4}\right)^{2}\right]\left(2 \frac{5-a}{4}\right)^{2}}{a^{2}}=\frac{1}{16^{2}} \cdot \frac{(9-a)(a-1)(3+a)^{2}}{a^{2}} \\
\leq \frac{1}{16} \cdot \frac{\left.\left[\frac{(9-a)+(3+a)}{2}\right]^{2} \times \frac{1}{3} \frac{3(a-1)+(3+a)}{2}\right]^{2}}{a^{2}}=\frac{1}{16^{2}} \cdot \frac{6^{2} \times \frac{1}{3}(2 a)^{2}}{a^{2}}=\frac{3}{16}
\end{array}
$$

Therefore, $M \leq \frac{\sqrt{3}}{4}$, with equality when $a=3 \Leftrightarrow \cos x=\frac{1}{2}$.
Given $x \in \mathbb{R}$, find the maximum value of $\frac{\sin x(2-\cos x)}{5-4 \cos x}$.
Solution Method Two: Let $M=\frac{\sin x(2-\cos x)}{5-4 \cos x}$, and let $a=5-4 \cos x$, then $a \in[1, 9]$, and
$$
\begin{array}{l}
\text { Solution Method Two: Let } M=\frac{\sin x(2-\cos x)}{5-4 \cos x}, \text { and let } a=5-4 \cos x, \text { then } a \in[1, 9], \text { and } \\
M^{2}=\frac{\left(1-\cos ^{2} x\right)(2-\cos x)^{2}}{(5-4 \cos x)^{2}}=\frac{\left(1-\cos ^{2} x\right)(2-\cos x)^{2}}{(5-4 \cos x)^{2}}=\frac{\left[1-\left(\frac{5-a}{4}\right)^{2}\right]\left(2-\frac{5-2}{4}\right)^{2}}{a^{2}}=\frac{1}{16^{2}} \cdot \frac{(9-a)(a-1)(3+a)^{2}}{a^{2}} \\
=\frac{1}{16^{2}} \cdot\left(10-\left(a+\frac{9}{a}\right)\right)\left(a+\frac{9}{a}+6\right) \leq \frac{3}{16}
\end{array}
$$

Therefore, $M \leq \frac{\sqrt{3}}{4}$, with equality when $a=3 \Leftrightarrow \cos x=\frac{1}{2}$.
Solution Method Three: Let $M=\frac{\sin x(2-\cos x)}{5-4 \cos x}$, then
$$
M^{2}=\frac{\sin ^{2} x(2-\cos x)^{2}}{\left(\sin ^{2} x+(2-\cos x)^{2}\right)^{2}}=\frac{\left(\frac{\sin x}{2-\cos x}\right)^{2}}{\left(\left(\frac{\sin x}{2-\cos x}\right)^{2}+1\right)^{2}}=\frac{1}{\left(\frac{\sin x}{2-\cos x}\right)^{2}+2+\frac{1}{\left(\frac{\sin x}{2-\cos x}\right)^{2}}}
$$

Let $t=\frac{\sin x}{2-\cos x}$, then $2 t=t \cdot \cos x+\sin x \leq \sqrt{t^{2}+1} \Rightarrow t^{2} \leq \frac{1}{3}$, hence
$$
M^{2}=\frac{1}{t^{2}+2+\frac{1}{t^{2}}} \leq \frac{1}{\frac{1}{3}+2+3}=\frac{3}{16}
$$

Therefore, $M \leq \frac{\sqrt{3}}{4}$, with equality when $a=3 \Leftrightarrow \cos x=\frac{1}{2}$.

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


When $a, b$ take what real numbers, the equation
$$
x^{2}+2(1+a) x+\left(3 a^{2}+4 a b+4 b^{2}+2\right)=0
$$

has real roots?

Solution:
Solution: When $\Delta \geqslant 0$, the equation has real roots
$$
\begin{array}{l}
\Leftrightarrow[2(1+a)]^{2}-4\left(3 a^{2}+4 a b+4 b^{2}+2\right) \geqslant 0 \\
\Leftrightarrow 2 a^{2}+4 a b+4 b^{2}-2 a+1 \leqslant 0 \\
\Leftrightarrow(a+2 b)^{2}+(a-1)^{2} \leqslant 0 . \\
\text { Also, }(a+2 b)^{2}+(a-1)^{2} \geqslant 0, \text { therefore, only }
\end{array}
$$

Also, $(a+2 b)^{2}+(a-1)^{2} \geqslant 0$, therefore, only $(a+2 b)^{2}=0$ and $(a-1)^{2}=0$, that is
$$
a=1, b=-\frac{1}{2} \text {. }
$$

Thus, when $a=1, b=-\frac{1}{2}$, the given equation has real roots.

Answer: a=1, b=-\frac{1}{2}


In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?

![](https://cdn.mathpix.com/cropped/2024_05_06_c47c8a04f76ef67ab4e5g-22.jpg?height=285&width=321&top_left_y=685&top_left_x=566)

Solution:
Answer: 6.

Solution. Let $a, b, c$ be the natural numbers in the three lower circles, from left to right. According to the condition, $14 \cdot 4 \cdot a = 14 \cdot 6 \cdot c$, i.e., $2a = 3c$. From this, it follows that $3c$ is even, and therefore $c$ is even. Thus, $c = 2k$ for some natural number $k$, and from the equation $2a = 3c$ it follows that $a = 3k$.

It must also hold that $14 \cdot 4 \cdot 3k = 3k \cdot b \cdot 2k$, which means $b \cdot k = 28$. Note that by choosing the number $k$, which is a natural divisor of 28, the natural numbers $a, b, c$ are uniquely determined. The number 28 has exactly 6 natural divisors: $1, 2, 4, 7, 14, 28$. Therefore, there are also 6 ways to place the numbers in the circles.

Answer: 6


Solution:
1.D.

From the given, we have
$$
\frac{1}{a}+\frac{1}{b}=15, \frac{1}{b}+\frac{1}{c}=17, \frac{1}{c}+\frac{1}{a}=16 \text {. }
$$

Adding the three equations yields $2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=48$.
Therefore, the original expression $=\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{1}{24}$.

Answer: D


Equation $\sqrt{x+4}-\sqrt{x-3}+1=0$
(A) No roots.
(B) One real root.
(C) One real root and one imaginary root.
(D) Two imaginary roots.
(E) Two real roots.

Solution:
[Solution] Given $\sqrt{x+4}+1=\sqrt{x-3}$, squaring both sides and simplifying yields
$$
8+2 \sqrt{x+4}=0 \text {, }
$$

this equation cannot hold.
Therefore, the answer is $(A)$.

Answer: A


Let the integer sequence $a_{1}, a_{2}, \cdots, a_{k}$ have length $k$, where $1 \leqslant a_{i} \leqslant 4(i=1,2, \cdots$, $k)$. If the sequence satisfies the condition: Ã¢ÂÂIf $a_{i}=a_{j}$ and $a_{i+1}=a_{j+1}$, then $i=j$. Ã¢ÂÂthen the maximum value of $k$ is $\qquad$.

Solution:
12. 17

Given that the different combinations of $\left(a_{i}, a_{i+1}\right)$ are $4 \times 4=16$. By the condition, the same combination cannot appear twice in the sequence (since $i \neq j$, then $a_{i} \neq a_{j}$ or $a_{i+1} \neq a_{j+1}$, i.e., $\left(a_{1}, a_{i+1}\right)$ and $\left(a_{3}, a_{s+1}\right)$ are different). Therefore, $k \leqslant 32-15=17$. For example, 11213142232433441 satisfies the requirement.

Answer: 17


Compute the sum of all positive integers whose digits form either a strictly increasing or a strictly decreasing sequence.

Solution:
Denote by $I$ and $D$ the sets of all positive integers with strictly increasing (respectively, decreasing) sequence of digits. Let $D_{0}, D_{1}, D_{2}$ and $D_{3}$ be the subsets of $D$ consisting of all numbers starting with 9 , not starting with 9 , ending in 0 and not ending in 0 , respectively. Let $S(A)$ denote the sum of all numbers belonging to a set $A$. All numbers in $I$ are obtained from the number 123456789 by deleting some of its digits. Thus, for any $k=0,1, \ldots, 9$ there are $\left(\begin{array}{l}9 \\ k\end{array}\right) k$-digit numbers in $I$ (here we consider 0 a 0 -digit number). Every $k$-digit number $a \in I$ can be associated with a unique number $b_{0} \in D_{0}, b_{1} \in D_{1}$ and $b_{3} \in D_{3}$ such that

$$
\begin{aligned}
& a+b_{0}=999 \ldots 9=10^{k+1}-1 \\
& a+b_{1}=99 \ldots 9=10^{k}-1 \\
& a+b_{3}=111 \ldots 10=\frac{10}{9}\left(10^{k}-1\right)
\end{aligned}
$$

Hence we have

$$
\begin{aligned}
& S(I)+S\left(D_{0}\right)=\sum_{k=0}^{9}\left(\begin{array}{l}
9 \\
k
\end{array}\right)\left(10^{k+1}-1\right)=10 \cdot 11^{9}-2^{9} \\
& S(I)+S\left(D_{1}\right)=\sum_{k=0}^{9}\left(\begin{array}{l}
9 \\
k
\end{array}\right)\left(10^{k}-1\right)=11^{9}-2^{9} \\
& S(I)+S\left(D_{3}\right)=\frac{10}{9}\left(11^{9}-2^{9}\right)
\end{aligned}
$$

Noting that $S\left(D_{0}\right)+S\left(D_{1}\right)=S\left(D_{2}\right)+S\left(D_{3}\right)=S(D)$ and $S\left(D_{2}\right)=10 S\left(D_{3}\right)$ we obtain the system of equations

$$
\left\{\begin{aligned}
2 S(I)+S(D) & =11^{10}-2^{10} \\
S(I)+\frac{1}{11} S(D) & =\frac{10}{9}\left(11^{9}-2^{9}\right)
\end{aligned}\right.
$$

which yields

$$
S(I)+S(D)=\frac{80}{81} \cdot 11^{10}-\frac{35}{81} \cdot 2^{10} .
$$

This sum contains all one-digit numbers twice, so the final answer is

$$
\frac{80}{81} \cdot 11^{10}-\frac{35}{81} \cdot 2^{10}-45=25617208995
$$

Answer: 25617208995


Given that for all real numbers $x$, the inequality $\left[\left(\log _{3} m\right)^{2}-\log _{3}\left(27 m^{2}\right)\right] x^{2}-\left(\log _{3} m-3\right) x-1<0$ always holds, find the range of real numbers $m$.

Solution:
(1) If $\left(\log _{3} m\right)^{2}-\log _{3}\left(27 m^{2}\right)=0, \log _{3} m=3$ or -1
If $\log _{3} m=-1$, the original inequality is $4 x-1<0$, which does not meet the requirement;
If $\log _{3} m=3$, the original inequality is $-1<0$, which is obviously true for any $x \in \mathbf{R}$, solving gives $m=27$
(2) If $\left(\log _{3} m\right)^{2}-\log _{3} 27 m^{2} \neq 0$, to make the original inequality true for any $x \in \mathbf{R}$, then it should be:
$$
\left\{\begin{array}{l}
\left(\log _{3} m\right)^{2}-\log _{3}\left(27 m^{2}\right)<0 \\
\Delta=\left(\log _{3} m-3\right)^{2}+4\left[\left(\log _{3} m\right)^{2}-\log _{3}\left(27 m^{2}\right)\right]<0
\end{array}\right.
$$

Solving gives $-\frac{1}{5}<\log _{3} m<3,3^{-\frac{1}{5}}<m<27$
In summary, the range of real number $m$ is $\left(3^{-\frac{1}{5}}, 27\right]$

Answer: (3^{-\frac{1}{5}},27]


Hixi and Shanshan each have several points cards.
Hixi says to Shanshan: "If you give me 3 cards, my number of cards will be 3 times yours."
Shanshan says to Hixi: "If you give me 4 cards, my number of cards will be 4 times yours."
Hixi says to Shanshan: "If you give me 5 cards, my number of cards will be 5 times yours."
It is known that exactly one of the above three statements is incorrect. Therefore, originally Hixi had $\qquad$ points cards.

Solution:
【Answer】Solution: According to the analysis, assuming the first and second sentences are correct, then the total should be a multiple of 20. According to the first sentence,

the ratio of Xixi to Shanshan's points cards should be 15:5, and according to the second sentence, the ratio of Xixi to Shanshan's card numbers should be 4:16, the difference of 11 times for each person corresponds to 7 cards, which is not an integer, so it is discarded.

Assuming the first and third sentences are correct, the total should be a multiple of 12. According to the first sentence, the ratio of the two people's points cards is $9:3$, and according to the second sentence, the ratio of the two people's points cards is 10:2, the difference of 1 part is the extra 2 cards, which is valid, so the ratio of Xixi and Shanshan's points cards is 6:24, according to the third sentence, the ratio of Xixi and Shanshan's points cards is 25:5, the difference of 19 parts is 9 cards, which is not an integer, so it is not valid, and is discarded.

In summary, the first and third sentences are correct, Xixi has 15 points cards.
The answer is: 15.

Answer: 15


Sasha and Vanya are playing a game. Sasha asks Vanya questions. If Vanya answers correctly, Sasha gives him 7 candies. If Vanya answers incorrectly, he gives Sasha 3 candies. After Sasha asked 50 questions, it turned out that each of them had as many candies as they had at the beginning. How many questions did Vanya answer correctly?

Solution:
Answer: 15.

Solution. Let Vanya answer correctly to $x$ questions, then he answered incorrectly to $50-x$ questions. This means that during the game, Vanya received $7 x$ candies and gave away $3(50-x)$ candies. Since the total number of candies for both boys did not change after the game, we get $7 x=3(50-x)$. Solving this, we find $x=15$.

Remark. The same answer could have been obtained by understanding that to "balance the candies," Vanya should "compensate" every 3 correct answers with 7 incorrect ones.

Answer: 15


A cylinder has height 4. Each of its circular faces has a circumference of $10\pi$. What is the volume of the cylinder?

Solution:
Suppose that the cylinder has radius $r$.

Since the circumference of its circular faces is $10 \pi$, then $2 \pi r=10 \pi$ or $r=5$.

Therefore, the volume of the cylinder is $\pi r^{2} h=\pi \cdot 5^{2} \cdot 4=100 \pi$.

ANSWER: $100 \pi$

Answer: 100\pi


Let $z_{1}, z_{2}, \cdots, z_{12}$ be the roots of the equation $z^{12}=2^{36}$. Let $w_{j} (j=1,2, \cdots, 12)$ be one of $z_{j}$ and i $z_{j}$. Then the maximum value of $\operatorname{Re} \sum_{j=1}^{12} w_{j}$ is . $\qquad$

Solution:
3. $16(1+\sqrt{3})$.

As shown in Figure 2, using the geometric meaning, $y=x$, $y=-x$ and the coordinate axes divide the circle $x^{2}+y^{2}=8$ into eight parts.

When $z_{j}$ is above the line $y=-x$, $w_{j}=z_{j}$; otherwise, $w_{j}=\mathrm{i} z_{j}$.
$$
\begin{array}{l}
\text { Hence Re } \sum_{j=1}^{12} w_{j} \\
=2^{3} \sum_{j=-1}^{4} \cos \frac{2 j \pi}{12}-2^{3} \sum_{j=5}^{10} \cos \frac{2 j \pi}{12} \\
=16(1+\sqrt{3}) .
\end{array}
$$

Answer: 16(1+\sqrt{3})


The numbers $a$ and $b$ are such that the polynomial $x^{4}+x^{3}+2 x^{2}+a x+b$ is the square of some other polynomial. Find $b$.

Solution:
Answer: $\frac{49}{64} \approx 0.77$.

Solution. The polynomial whose square is the given one is a quadratic trinomial with a coefficient of 1 or -1 for $x^{2}$. We can assume that the leading coefficient is 1 (otherwise, factor out -1 with a change in the signs of the other coefficients). Let $\left(x^{2}+A x+B\right)^{2}=x^{4}+x^{3}+2 x^{2}+a x+b$. Expanding the brackets, we get

$$
x^{4}+2 A x^{3}+\left(A^{2}+2 B\right) x^{2}+2 A B x+B^{2}=x^{4}+x^{3}+2 x^{2}+a x+b,
$$

therefore, equating the coefficients of the powers, we find $2 A=1, A^{2}+2 B=2,2 A B=a$, $B^{2}=b$. Hence, $A=\frac{1}{2}, B=1-\frac{A^{2}}{2}=\frac{7}{8}, b=B^{2}=\frac{49}{64} \approx 0.77$.

Answer: \frac{49}{64}


Mr. Chrt had five dogs in his dog team - AlÃk, Brok, Muk, Raf, and Punt. He was thinking about how he could harness the dogs in a row behind him so that AlÃk would be in front of Punt.

In how many ways could Mr. Chrt do this?

Hint. On which positions could AlÃk and Punt be harnessed?

Solution:
If Alík was first, Punta could be second, third, fourth, or fifth:

$$
A P * * * \quad A * P * * \quad A * * P * \quad A * * * P
$$

If Alík was second, Punta could be third, fourth, or fifth:

$$
* A P * * \quad * A * P * \quad * A * * P
$$

If Alík was third, Punta could be fourth or fifth:

$$
* * A P * \quad * * A * P
$$

If Alík was fourth, Punta had to be fifth:

$$
* * * A P
$$

Thus, Mr. Chrt had 10 options for harnessing Alík and Punta in the required manner.

In each of these cases, the remaining three dogs could be harnessed in the unoccupied positions (marked with *) in any order. This can be done in 6 ways: one dog can go to any of the three free positions, the second dog to any of the two remaining positions, and the third dog has no choice and must go to the last free position (thus $6=3 \cdot 2$).

Mr. Chrt could harness his dogs in a total of 60 ways (since $10 \cdot 6=60$).

Note. The previous calculation of all possible orders of three elements can be generalized for any number; the resulting number is called a factorial and is denoted by an exclamation mark (here $3 !=3 \cdot 2 \cdot 1=6$).

All five of Mr. Chrt's dogs can be harnessed in a total of $5 !=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120$ ways. In half of the cases, Alík stands in front of Punta, and in the other half, it is the opposite (it is enough to swap these two dogs and leave the others in their places). Thus, Alík can be in front of Punta in 60 cases.

Answer: 60


Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ for which

$$
x+f(x f(y))=f(y)+y f(x)
$$

holds for all real numbers $x$ and $y$.

Solution:
IV/4. By substituting $x=0$ into the functional equation, we get $f(0)=f(y)+y f(0)$. Therefore, $f$ is a linear function of the form $f(x)=a-a x$ for some real number $a$. We substitute this form into the functional equation and get that for all $x, y \in \mathbb{R}$, it holds that $x+a-a x(a-a y)=a-a y+y(a-a x)$, thus

$$
x-a^{2} x+a^{2} x y=-a x y
$$

If we substitute $y=0$ into this equation, we get $x=a^{2} x$ for all $x$, hence $a^{2}=1$. Considering this in the above equation, it simplifies to $x y=-a x y$ or $(1+a) x y=0$, from which it follows that $a=-1$, since the equation must hold for all $x$ and $y$.

We get $f(x)=x-1$. If we substitute this form into both sides of the original equation, we get

$$
x+f(x f(y))=x+f(x y-x)=x+x y-x-1=x y-1
$$

and

$$
f(y)+y f(x)=y-1+y(x-1)=x y-1
$$

which is equal for all $x$ and $y$, so $f(x)=x-1$ is the only function that satisfies the given equation.

Substituting $x=0$ or $y=0$ into the functional equation 1 point Substituting the form $f(x)=a-a x$ into the original equation for $x$ and $y$ (can also be written as $f(x)=f(0)-x f(0))$

1 point

Concluding that for all $x$ and $y$ the equation $x-a^{2} x+a^{2} x y=-a x y$ holds (or equivalently,

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-17.jpg?height=48&width=1625&top_left_y=2346&top_left_x=221)

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-17.jpg?height=49&width=1625&top_left_y=2391&top_left_x=221)

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-17.jpg?height=46&width=1627&top_left_y=2444&top_left_x=223)

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-17.jpg?height=51&width=1627&top_left_y=2490&top_left_x=223)

Concluding that $f(x)=x-1$ is the solution .......................................................................

Answer: f(x)=x-1


Given a regular triangular prism with a base edge length of 1, the diagonals of the two lateral faces are perpendicular to each other. Then the length of the lateral edge of the prism is $\qquad$

Solution:
6. $\frac{\sqrt{2}}{2}$.

Let the triangular prism $A B C-A_{1} B_{1} C_{1}$, with side edge length $a$, and the skew diagonals of the side faces $A B_{1}$ and $B C_{1}$ are perpendicular to each other. Then
$$
\begin{array}{l}
\overrightarrow{A B_{1}} \cdot \overrightarrow{B C_{1}}=0 \\
\Rightarrow\left(\overrightarrow{B_{1} B}+\overrightarrow{B A}\right) \cdot\left(\overrightarrow{B B_{1}}+\overrightarrow{B_{1} C_{1}}\right)=0 \\
\Rightarrow \overrightarrow{B_{1} B} \cdot \overrightarrow{B B_{1}}+\overrightarrow{B_{1} B} \cdot \overrightarrow{B_{1} C_{1}}+\overrightarrow{B A} \cdot \overrightarrow{B B_{1}}+\overrightarrow{B A} \cdot \overrightarrow{B_{1} C_{1}}=0 \\
\Rightarrow-a^{2}+\cos 60^{\circ}=0 \Rightarrow a=\frac{\sqrt{2}}{2}
\end{array}
$$

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


In $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$ respectively. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array}
$$

then $m+n=$

Solution:
3. 2 .

Solution 1 Notice that,
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} .
$$

Since points $M, O, N$ are collinear, we have,
$$
\frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 \text {. }
$$

Solution 2 Since points $M, O, N$ are on the sides or their extensions of $\triangle A B C$, and the three points are collinear, by Menelaus' theorem we get
$$
\begin{array}{l}
\frac{A M}{M B} \cdot \frac{B O}{O C} \cdot \frac{C N}{N A}=1 . \\
\text { Hence } \frac{n-1}{1-m}=1 \Rightarrow m+n=2 .
\end{array}
$$

Answer: 2


Fill $n^{2}(n \geq 3)$ integers into an $n \times n$ grid, with one number in each cell, such that the product of the numbers in each row and each column is 30. How many ways are there to do this?

Solution:
Question 184, Solution: First, consider the case where all integers in the grid are positive integers. In this case, the prime factors of the numbers in each row and each column exactly include one $2$, one $3$, and one $5$. Among these, the $\mathrm{n}$ twos are in different rows and different columns, with $\mathrm{n}$! ways to fill them. Similarly, the threes and fives each have $\mathrm{n}$! ways to fill them. Therefore, in this case, there are $(\mathrm{n}!)^3$ ways to fill the grid.

Furthermore, the numbers in the $(n-1) \times (n-1)$ subgrid in the top-left corner can arbitrarily choose their positive or negative signs. At this point, the numbers in the $n$-th row and the $n$-th column have a unique way to choose their positive or negative signs. Hence, there are $2^{(n-1)^2} \cdot (\mathrm{n}!)^3$ ways to fill the grid that satisfy the conditions.

Answer: 2^{(n-1)^2}\cdot(\mathrm{n}!)^3


From $1,2,3, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, such that $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways are there to select the numbers that meet the above requirements?

Solution:
Solution: Obviously, $a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+(14-$ $\left.a_{3}\right)=14$, where $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3,14-$ $a_{3} \geqslant 0$.
Transform the above equation to
$$
\left(a_{1}-1\right)+\left(a_{2}-a_{1}-3\right)+\left(a_{3}-a_{2}-3\right)+(14
$$
$\left.-a_{3}\right)=7$.

At this point, $a_{1}-1, a_{2}-a_{1}-3, a_{3}-a_{2}-3,14-c_{3}$ are all non-negative integers. It is known that this indeterminate equation has $C_{i+4-1}^{\gamma}=$ $C_{10}^{?}$ different sets of non-negative integer solutions, therefore, the number of different ways that meet the requirements is $C_{10}^{3}=120$ (using the combination formula).

Answer: 120


The graph of the function $f(x)=A \cos (\omega x+\phi)$ is symmetric with respect to the origin if and only if ( )(where $k \in \mathbf{Z}$ ).
A. $\phi=2 k \pi+\frac{\pi}{2}$
B. $\frac{\varphi}{\omega}=2 k \pi \pm \frac{\pi}{2}$
C. $\phi=k \pi+\frac{\pi}{2}$
D. $\phi=k \pi+\frac{\pi}{2}$ or $\omega x=k \pi$

Solution:
3. C
$f(x)=\operatorname{Aoox}(\omega x$ ' $\varphi)$ 的图像关 $f \cdot$ 拍点对称 $\Leftrightarrow f(x)$ 图像过原点 $\Leftarrow \cos \phi$ $=0$

3. C
The graph of $f(x)=\operatorname{Aoox}(\omega x$ ' $\varphi)$ is symmetric with respect to the origin $\Leftrightarrow$ the graph of $f(x)$ passes through the origin $\Leftarrow \cos \phi$ $=0$

Answer: C


For what real number $x$ does the function $f(x)=x^{2}-x+1+\sqrt{2 x^{4}-18 x^{2}+12 x+68}$ have a minimum value? What is the minimum value?

Solution:
Given $f(x)=x^{2}-x+1+\sqrt{2 x^{4}-18 x^{2}+12 x+68}$
$$
\begin{array}{l}
=x^{2}-x+1+\sqrt{2(x+3)^{2}+2\left(x^{2}-5\right)^{2}} \\
=x^{2}-x+1+\sqrt{2} \cdot \sqrt{(x+3)^{2}+\left(x^{2}-5\right)^{2}} \\
=\sqrt{2}\left[\frac{x^{2}-x+1}{\sqrt{2}}+\sqrt{(x+3)^{2}+\left(x^{2}-5\right)^{2}}\right] .
\end{array}
$$

Thus, the above expression can be seen as the sum of two distances:
The distance from a moving point $P\left(x, x^{2}\right)$ on the parabola $y=x^{2}$ to the line $y=x-1$
$$
|P Q|=\frac{\left|x-x^{2}-1\right|}{\sqrt{1^{2}+(-1)^{2}}}=\frac{\left|x^{2}-x+1\right|}{\sqrt{2}} \text {, }
$$

and the distance from the moving point $P\left(x, x^{2}\right)$ to the fixed point $(-3,5)$
$$
|P M|=\sqrt{[x-(-3)]^{2}+\left(x^{2}-5\right)^{2}}=\sqrt{(x+3)^{2}+\left(x^{2}-5\right)^{2}} \text {. }
$$

As shown in Figure 5-2, when $|P Q|+| P M|$ is minimized, points $M, P, Q$ are collinear, and this line is the line $l: y=-x+2$ passing through point $M$ and perpendicular to the line $y=x-1$.

Therefore, the minimum value of $|P Q|+|P M|$ is the distance from point $M$ to the line $y=x-1$, which is $\left|\frac{-3-5-1}{\sqrt{1^{2}+(-1)^{2}}}\right|=\frac{9}{2} \sqrt{2}$.

At this time, the intersection points of the line $y=-x+2$ and the parabola are $P_{1}(-2,4)$ and $P_{2}(1,1)$. Hence, when $x=-2$ or $x=1$, the minimum value of the function $f(x)$ is
$$
f(x)_{\min }=\sqrt{2} \cdot \frac{9}{2} \sqrt{2}=9 .
$$

Answer: 9


Let $z, w$ be complex numbers satisfying $\left\{\begin{array}{c}z+\frac{20 i}{w}=5+i \\ w+\frac{12 i}{z}=-4+10 i\end{array}\right.$, find the minimum possible value of $|z w|^{2}$.

Solution:
Question 103, Solution: Multiplying the two expressions yields $\mathrm{zw}-\frac{240}{z \mathrm{w}}+12 \mathrm{i}+20 \mathrm{i}=(5+\mathrm{i})(-4+10 \mathrm{i})=-30+$ $46 \mathrm{i}$, rearranging gives
$$
\begin{array}{l}
(\mathrm{zw})^{2}-(-30+14 \mathrm{i}) \cdot \mathrm{zw}-240=0 \\
\Rightarrow[\mathrm{zw}-(6+2 \mathrm{i})][\mathrm{zw}-(-36+12 \mathrm{i})]=0
\end{array}
$$

Thus, $\mathrm{zw}=6+2 \mathrm{i}$ or $-36+12 \mathrm{i}$. Therefore, the minimum possible value of $|\mathrm{zw}|^{2}$ is $6^{2}+2^{2}=40$.

Answer: 40


What is the probability that the colony never goes extinct?

Solution:
The answer is $1-\left(\frac{1}{2}\right)^{100}$.

Let $p$ be the probability that a colony consisting of just one cell will survive. Then

$$
p=\frac{1}{3} \cdot 0+\frac{2}{3}\left(1-(1-p)^{2}\right)
$$

owing to the fact that when the cell splits in two, the probability both of them go extinct is $(1-p)^{2}$. Solving for $p$, we obtain $p=\frac{1}{2}$.

The colony initially has 100 cells; if we treat these as 100 distinct colonies, we obtain the claimed answer.

Answer: 1-(\frac{1}{2})^{100}


Determine all triples of prime numbers $(p, q, r)$ for which

$$
p^{q}=r-1
$$

Solution:
Scoring: 1 point is awarded for concluding that $r$ is not 2, and another 2 points for determining that $p=2$. 1 point is awarded for the case $q=2$, and 3 points for eliminating the case $q>2$.

A written solution of $(2,2,5)$ earns 1 point even if it is not justified.

Answer: (2,2,5)


The number of elements in set $S$ is denoted as $|S|$, and the number of subsets of set $S$ is denoted as $n(S)$. Given three non-empty finite sets $A$, $B$, and $C$ that satisfy the condition:
$$
\begin{array}{l}
|A|=|B|=2019, \\
n(A)+n(B)+n(C)=n(A \cup B \cup C) .
\end{array}
$$

Determine the maximum value of $|A \cap B \cap C|$, and briefly describe the reasoning process.

Solution:
$$
2^{2019}+2^{2019}+2^{|C|}=2^{|A \cup B \cup C|} \text {, }
$$

that is, $2^{2020}+2^{|C|}=2^{|A \cup B \cup C|}$.
Thus, $|A \cup B \cup C|>|C|$,
$|A \cup B \cup C|>2020$.
If $|C|>2020$, then
$$
1+2^{|C|-2020}=2^{|A \cup B \cup C|-2020} \text {. }
$$

The left side of the above equation is odd, while the right side is even, so
$$
|C|>2020
$$

does not hold.
Similarly, $|C|<2020$ does not hold either.
Therefore, $|C|=2020$.
In this case, from equation (1) we know $|A \cup B \cup C|=2021$.
Clearly,
$|A \cap B \cap C| \leqslant \min \{|A|,|B|,|C|\}=2019$.
If $|A \cap B \cap C|=2019$, then $A=B \subseteq C$.
Thus, $|A \cup B \cup C|=|C|=2020$. This is a contradiction.
Therefore, $|A \cap B \cap C| \leqslant 2018$, meaning the maximum possible value of $|A \cap B \cap C|$ is 2018.
$$
\begin{array}{l}
\text { When } A=\{1,2, \cdots, 2019\}, \\
B=\{1,2, \cdots, 2018,2020\}, \\
C=\{1,2, \cdots, 2018,2019,2021\}
\end{array}
$$

then, $|A \cap B \cap C|=2018$, meaning the maximum value of $|A \cap B \cap C|$ that satisfies the given conditions is 2018.

Answer: 2018


Solution:
2. D

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

Answer: D


The organizers of the exhibition "To the Moon and Beyond" felt that they had low attendance on the first day, so they reduced the admission fee by 12 KÄ. As a result, the number of visitors increased by $10\%$ on the second day, but the total daily revenue decreased by $5\%$. How many korunas did the admission cost after the discount?

Solution:
We will organize the information from the assignment into the following table:

|  | first day | second day |
| :---: | :---: | :---: |
| number of visitors | $n$ | $1.1 n$ |
| admission fee (Kč per person) | $x+12$ | $x$ |
| total daily revenue (Kč) | $n(x+12)$ | $1.1 n x$, or $0.95 n(x+12)$ |

From the last cell of the table, we will set up an equation, which we will solve (assuming $n>0$):

$$
\begin{aligned}
1.1 n x & =0.95 n(x+12) \\
1.1 x & =0.95 x+11.4 \\
0.15 x & =11.4 \\
x & =76
\end{aligned}
$$

The admission fee after the discount was 76 Kč.

Evaluation. 3 points for setting up the table or its equivalent, of which 1 point for the information in the first two rows under the header and 2 points for the information in the last row; 2 points for setting up and solving the equation; 1 point for the result.

If the solver includes in their work a variant that no one arrived on the first day, i.e., $n=0$, and that the admission fee after the discount could therefore be anything, this cannot be accepted as a solution to the problem. (The assignment states that the number of visitors increased by 10% on the second day.) However, if the solver adds this variant to the correct solution of the problem, do not deduct points for it.

Answer: 76


If $x, y, z>0$, and $x^{2}+y^{2}+z^{2}=1$, then $S=\frac{(z+1)^{2}}{2 x y z}$ reaches its minimum value when $x$ equals

Solution:
$31 \sqrt{\sqrt{2}-1}$.
$$
S \geqslant \frac{(z+1)^{2}}{\left(x^{2}+y^{2}\right) z}=\frac{(z+1)^{2}}{\left(1-z^{2}\right) z}=\frac{z+1}{(1-z) z} .
$$

Let $u=\frac{z+1}{(1-z) z}, 0<z<1$ and
$$
u z^{2}+(1-u) z+1=0
$$

has real roots, so
$$
(1-u)^{2} \geqslant 4 u
$$

which means $u^{2}-6 u+1 \geqslant 0$
thus $u \geqslant 3+2 \sqrt{2}$.
When $z^{2}=\frac{3+2 \sqrt{2}-1}{2(3+2 \sqrt{2})}=\sqrt{2}-1$, $u=\frac{z+1}{(1-z) z}$ reaches its minimum value $3+2 \sqrt{2}$. Therefore, when $S$ reaches its minimum value,
$$
x=\sqrt{\frac{1-(\sqrt{2}-1)^{2}}{2}}=\sqrt{\sqrt{2}-1} .
$$

Answer: \sqrt{\sqrt{2}-1}


Find the maximum value of the expression $\cos (x+y)$, given that $\cos x - \cos y = \frac{1}{4}$.

Solution: Answer: $\frac{31}{32} \approx 0.97$.

Answer: \frac{31}{32}


Given real numbers $a, b, c$ satisfy the inequalities
$$
|a| \geqslant|b+c|,|b| \geqslant|c+a|,|c| \geqslant |a+b|
$$

then the value of $a+b+c$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3

Solution:
3. A.

From the given conditions, we have
$$
\begin{array}{l}
a^{2} \geqslant(b+c)^{2}=b^{2}+2 b c+c^{2}, \\
b^{2} \geqslant(c+a)^{2}=c^{2}+2 c a+a^{2}, \\
c^{2} \geqslant(a+b)^{2}=a^{2}+2 a b+b^{2} .
\end{array}
$$

Adding the three inequalities, we get
$$
a^{2}+b^{2}+c^{2} \geqslant 2\left(a^{2}+b^{2}+c^{2}\right)+2 a b+2 b c+2 c a \text {. }
$$

Thus, we have $(a+b+c)^{2} \leqslant 0$.
Therefore, $a+b+c=0$.

Answer: A


Find the GCD of all numbers of the form $n^{13}-n$.

Solution: We will find all primes that divide $n^{13}-n$ for all $n$, i.e., $p \mid\left(n^{12}-1\right)$ for all $n$ coprime with $p$. Fermat's little theorem states that $p \mid\left(n^{p-1}-1\right)$, so if $(p-1) \mid 12$, then $p \mid\left(n^{12}-1\right)$ for all $n$ coprime with $p$. We have at least $p=2,3,5,7$ and 13. Now let's show that $p^{2}$ does not divide $n^{13}-n$ for all $n$ with a specific case: $p^{2}$ divides $p^{13}$ but not $p$, so $p^{2}$ does not divide $p^{13}-p$. The last step is to show that there is no other prime factor. The ideal solution would be to prove the converse of Fermat's theorem, but here it suffices to observe that $2^{13}-2=8190=2 \cdot 3^{2} \cdot 5 \cdot 7 \cdot 13$. The greatest common divisor is therefore $2 \cdot 3 \cdot 5 \cdot 7 \cdot 13=2730$.

Answer: 2730


Solution:
Solve by first comparing the quotient of $-\frac{x+1}{23}$ and $\frac{23 x+1}{23^{2} x+1}$.
$$
\begin{array}{l}
23 \frac{x+1}{x+1} \div \frac{23 x+1}{23^{2} x+1} \\
=\frac{(x+1)\left(23^{2} x+1\right)}{(23 x+1)^{2}} \\
=\frac{23^{2} x^{2}+\left(23^{2}+1\right) x+1}{23^{2} x^{2}+2 \times 23 x+1}>1 \text {. } \\
\end{array}
$$

When $x=23^{1988}$,
$$
\frac{23^{1988}+1}{23^{1989}+1}>\frac{23^{1989}+1}{23^{1980}+1}
$$

Answer: \frac{23^{1988}+1}{23^{1989}+1}>\frac{23^{1989}+1}{23^{1980}+1}


The quadratic function $f(x)=x^{2}+m x+n$ has real roots, the inequality $s \leq(m-1)^{2}+(n-1)^{2}+(m-n)^{2}$ holds for any quadratic function satisfying the above condition, then the maximum value of $s$ is _____ .

Solution:
Question 69, Solution Method One: In fact, the maximum value of the desired $s$ is $\min _{m^{2} \geq 4 n}\left\{(m-1)^{2}+(n-1)^{2}+\right.$ $\left.(m-n)^{2}\right\}$.

Notice that $(m-1)^{2}+(n-1)^{2}+(m-n)^{2}=\frac{9}{8}+\frac{(2 m-4 n-1)^{2}+6(m-1)^{2}+6\left(m^{2}-4 n\right)}{8} \geq \frac{9}{8}$, with equality when $m=$ $1, n=\frac{1}{4}$. Therefore, the maximum value of the desired $s$ is $\frac{9}{8}$.
Solution Method Two: Let $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+\mathrm{mx}+\mathrm{n}$ have two real roots $\mathrm{p}$ and $\mathrm{q}$, then $\left\{\begin{array}{c}p+q=-m \\ p q=n\end{array}\right.$, hence
$$
\begin{array}{l}
(m-1)^{2}+(n-1)^{2}+(m-n)^{2} \\
=(1+p+q)^{2}+(p q-1)^{2}+(p q+p+q)^{2} \\
=2\left(p^{2}+p+1\right)\left(q^{2}+q+1\right) \geq \frac{9}{8}
\end{array}
$$

Equality holds if and only if $p=q=-\frac{1}{2}$. Therefore, the maximum value of the desired $s$ is $\frac{9}{8}$.

Answer: \frac{9}{8}


AB is the diameter of a circle. C is a point not on the line AB. The line AC cuts the circle again at X and the line BC cuts the circle again at Y. Find cos ACB in terms of CX/CA and CY/CB.

Solution: : cos 2 ACB = (CX/CA).(CY/CB). It is the positive root for C outside the circle and the negative root for C inside the circle. For C outside the circle, we have cos ACB = CX/CB = CY/CA (one should check that this is true in all configurations). Hence result. For C on the circle it is still true because cos ACB = 0 and CX = CY = 0. For C inside the circle, cos ACB = - cos ACY = - CY/CA and also = - cos BCX = - CX/CB. 26th CanMO 1994 © John Scholes jscholes@kalva.demon.co.uk 18 Aug 2002

Answer: \cos\angleACB=\\sqrt{(CX/CA)\cdot(CY/CB)}


In the tetrahedron $ABCD$, there is a point $O$ inside such that the lines $AO, BO, CO, DO$ intersect the faces $BCD, ACD, ABD, ABC$ at points $A_1, B_1, C_1, D_1$ respectively, and $\frac{AO}{OA_1}=\frac{BO}{OB_1}=\frac{CO}{OC_1}=\frac{DO}{OD_1}=k$. Then $k=$ $\qquad$ .

Solution:
12. $3 \frac{V_{A B C D}}{V_{O B C D}}=\frac{A A_{1}}{O A_{1}}=\frac{A O}{O A_{1}}+\frac{O A_{1}}{O A_{1}}=1+k$, thus $\frac{V_{O B C D}}{V_{A B C D}}=\frac{1}{1+k}$.

Similarly, $\frac{V_{(X \cdot D A}}{V_{A B C D}}=\frac{V_{O D A B}}{V_{A B C D}}=\frac{V_{O A B C}}{V_{A B C D}}=\frac{1}{1+k}$.
Since $V_{O B C D}+V_{\triangle C D A}+V_{O D A B}+V_{O A B C}=V_{A B C D}$,
we have $\frac{4}{1+k}=1$, hence $k=3$.

Answer: 3


At the cross-country race, 8 runners wore white sports shirts. 4 runners wore blue sports shirts. How many runners started the cross-country race?

Solution: $8+4=12, 12$ runners started.

Answer: 12


Find all functions $f: \mathbf{Z} \rightarrow \mathbf{Z}$, such that
$$
f(-f(x)-f(y))=1-x-y \quad (x, y \in \mathbf{Z}) .
$$

Solution:
2. $f(x)=x-1$.

It is easy to verify that $f(x)=x-1$ satisfies the functional equation.
Assume the function $f: \mathbf{Z} \rightarrow \mathbf{Z}$ satisfies
$$
f(-f(x)-f(y))=1-x-y \text {. }
$$

Let $y=1$. Then $f(-f(x)-f(1))=-x$.
Let $x=1$. Then $f(-2 f(1))=-1$.
Let $n \in \mathbf{Z}$.
Let $x=n$. Then $f(-f(n)-f(1))=-n$.
Take $x=-f(n)-f(1), y=-2 f(1)$. We get
$$
\begin{array}{l}
f(n+1)=1+(f(n)+f(1))+2 f(1) \\
\Rightarrow f(n+1)=f(n)+3 f(1)+1 .
\end{array}
$$

Starting from $n=1$, whether using forward induction or backward induction, we can prove
$$
f(n)=(3 n-2) f(1)+n-1 \text {. }
$$

Let $n=0$, then $f(0)=-2 f(1)-1$.
Take $x=0, y=1$. We get
$$
\begin{array}{l}
f(-f(0)-f(1))=0 \\
\Rightarrow f(f(1)+1)=0 .
\end{array}
$$

If $n=f(1)+1$, then
$$
\begin{array}{l}
(3(f(1)+1)-2) f(1)+(f(1)+1)-1=0 \\
\Rightarrow f(1)(3 f(1)+2)=0 .
\end{array}
$$

Since $f(1)$ is an integer, therefore, $f(1)=0$.
Thus, $f(n)=n-1$.

Answer: f(x)=x-1


If $x$ is a real number satisfying the equation $$9\log_3 x - 10\log_9 x =18 \log_{27} 45,$$ then the value of $x$ is equal to $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.

Solution:
1. We start with the given equation:
   \[
   9\log_3 x - 10\log_9 x = 18 \log_{27} 45
   \]

2. We use the change of base formula for logarithms. Recall that:
   \[
   \log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2}
   \]
   and
   \[
   \log_{27} x = \frac{\log_3 x}{\log_3 27} = \frac{\log_3 x}{3}
   \]

3. Substitute these into the original equation:
   \[
   9\log_3 x - 10 \left(\frac{\log_3 x}{2}\right) = 18 \left(\frac{\log_3 45}{3}\right)
   \]

4. Simplify the equation:
   \[
   9\log_3 x - 5\log_3 x = 6\log_3 45
   \]
   \[
   4\log_3 x = 6\log_3 45
   \]

5. Divide both sides by 4:
   \[
   \log_3 x = \frac{6}{4} \log_3 45
   \]
   \[
   \log_3 x = \frac{3}{2} \log_3 45
   \]

6. This implies:
   \[
   x = 45^{\frac{3}{2}}
   \]

7. Simplify \(45^{\frac{3}{2}}\):
   \[
   45^{\frac{3}{2}} = (45^1) \cdot (45^{\frac{1}{2}}) = 45 \cdot \sqrt{45}
   \]

8. Express \(45\) in terms of its prime factors:
   \[
   45 = 3^2 \cdot 5
   \]
   \[
   \sqrt{45} = \sqrt{3^2 \cdot 5} = 3 \sqrt{5}
   \]

9. Therefore:
   \[
   x = 45 \cdot 3 \sqrt{5} = 135 \sqrt{5}
   \]

10. Identify \(m\) and \(n\):
    \[
    m = 135, \quad n = 5
    \]

11. Calculate \(m + n\):
    \[
    m + n = 135 + 5 = 140
    \]

The final answer is \(\boxed{140}\)

Answer: 140


If the three sides of $\triangle A B C$ are all unequal, the area is $\frac{\sqrt{15}}{3}$, and the lengths of the medians $A D$ and $B E$ are 1 and 2, respectively, then the length of the median $C F$ is $\qquad$.

Solution: 7. $\sqrt{6}$

Answer: \sqrt{6}


If $\frac{z}{x+y}=2$ and $\frac{y}{x+z}=3$, what is the value of $\frac{z}{y+z}$?

Solution:
Solution. From the assumptions, we get

$$
\begin{aligned}
& z=2 x+2 y \\
& y=3 x+3 z
\end{aligned}
$$

If we express $y$ and $z$ in terms of $x$, we get $y=-\frac{9}{5} x, z=-\frac{8}{5} x$.

Substituting this into the expression $\frac{z}{y+z}$, we get

$$
\frac{z}{y+z}=\frac{-\frac{8}{5} x}{-\frac{9}{5} x-\frac{8}{5} x}=\frac{8}{17}
$$

Note that $x$ cannot be $0$, otherwise the conditions of the problem would state $\frac{z}{y}=2$ and $\frac{y}{z}=3$, which cannot both be true. Therefore, the above calculation is valid.

(2 points)

Answer: \frac{8}{17}


Detective Podberezyakov is pursuing Maksim Detochkin (each driving their own car). At the beginning, both were driving on the highway at a speed of 60 km/h, with Podberezyakov lagging behind Detochkin by 2 km. Upon entering the city, each of them reduced their speed to 40 km/h, and upon exiting the city, finding themselves on a good highway, each increased their speed to 70 km/h. When the highway ended, at the border with a dirt road, each had to reduce their speed to 30 km/h again. What was the distance between them on the dirt road?

Solution:
Answer: 1 km.

Solution: At the moment Detochkin entered the city, Podberezyakov was 2 km behind him. Podberezyakov covered this distance in 2/60 hours, during which time Detochkin traveled $\frac{2}{60} \cdot 40$ km. That is, the distance was multiplied by 40/60. Similarly, when exiting the city and transitioning to the dirt road, the distance is multiplied by 70/40 and then by 30/70. In the end, $2 \times \frac{40}{60} \times \frac{70}{40} \times \frac{30}{70}=1$ km.

Answer: 1


The pioneers of a Berlin school presented cosmonauts Valentina Tereshkova and Yuri Gagarin with a folder of drawings as a memento. From seven classes, six drawings were each selected.

In addition, the art teacher of the school and a member of the parent council contributed a drawing for the cosmonauts.

How many pictures does the folder contain?

Solution: $7 \cdot 6=42, 42+1+1=44$. The folder contains 44 drawings.

Answer: 44


Calculate the definite integral:

$$
\int_{0}^{\sqrt{2} / 2} \frac{x^{4} \cdot d x}{\sqrt{\left(1-x^{2}\right)^{3}}}
$$

Solution:
## Solution

$$
\int_{0}^{\sqrt{2} / 2} \frac{x^{4} \cdot d x}{\sqrt{\left(1-x^{2}\right)^{3}}}=
$$

Substitution:

$$
\begin{aligned}
& x=\sin t \Rightarrow d x=\cos t d t \\
& x=0 \Rightarrow t=\arcsin 0=0 \\
& x=\frac{\sqrt{2}}{2} \Rightarrow t=\arcsin \frac{\sqrt{2}}{2}=\frac{\pi}{4}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\pi / 4} \frac{\sin ^{4} t \cdot \cos t}{\sqrt{\left(1-\sin ^{2} t\right)^{3}}} d t=\int_{0}^{\pi / 4} \frac{\sin ^{4} t \cdot \cos t}{\cos ^{3} t} d t=\int_{0}^{\pi / 4} \frac{\sin ^{4} t}{\cos ^{2} t} d t= \\
& =\int_{0}^{\pi / 4} \frac{\left(1-\cos ^{2} t\right)^{2}}{\cos ^{2} t} d t=\int_{0}^{\pi / 4} \frac{1-2 \cos ^{2} t+\cos ^{4} t}{\cos ^{2} t} d t= \\
& =\int_{0}^{\pi / 4}\left(\frac{1}{\cos ^{2} t}-2+\frac{1+\cos 2 t}{2}\right) d t=\int_{0}^{\pi / 4}\left(\frac{1}{\cos ^{2} t}-\frac{3}{2}+\frac{1}{2} \cos 2 t\right) d t= \\
& =\left.\left(\operatorname{tg} t-\frac{3 t}{2}+\frac{1}{4} \sin 2 t\right)\right|_{0} ^{\pi / 4}=\left(\operatorname{tg} \frac{\pi}{4}-\frac{3 \pi}{8}+\frac{1}{4} \sin \frac{\pi}{2}\right)-\left(\operatorname{tg} 0-\frac{3 \cdot 0}{2}+\frac{1}{4} \sin 0\right)= \\
& =\left(1-\frac{3 \pi}{8}+\frac{1}{4} \cdot 1\right)-\left(2 \cdot 0-0+\frac{1}{4} \cdot 0\right)=\frac{5}{4}-\frac{3 \pi}{8}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+12-7$ » Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals

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Created by GeeTeatoo

## Problem Kuznetsov Integrals 12-8

## Material from PlusPi

Answer: \frac{5}{4}-\frac{3\pi}{8}


He spent one-sixth of his life in a joyful childhood, and one-twelfth of his life dreaming in his youth. Then he spent one-seventh of his life, when he became a groom; $S$ on the fifth spring, his little son was born. Alas, but the poor child lived only half as long as his grieving father, who ... was carried to the grave by his sorrow ... For four years he bore his grief in a loving heart. The length of his life is thus: - see it in these wise lines.

Answer: 84


The sides of the triangle are $6, 8, 10$. Find its area (using Heron's formula).

Solution:
95. Solution of Niphus

\[
\Delta=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{12 \cdot 6 \cdot 4 \cdot 2}=24
\]

It would have been simpler, of course, to take half the product of the legs.

Junius Moderatus Columella (1st century AD) in his treatise "On Agriculture," when discussing land surveying, indicates 9
land measures and solutions to nine geometric problems with numerical data.

Answer: 24


In triangle $ABC$, angle $A$ is equal to $60^{\circ}$, the distances from vertices $B$ and $C$ to the center of the inscribed circle of triangle $ABC$ are 3 and 4, respectively. Find the radius of the circle circumscribed around triangle $ABC$.

Solution: Answer. $\quad \sqrt{\frac{37}{3}}$. Hint. See problem 4 for 9th grade.

Answer: \sqrt{\frac{37}{3}}


Find the derivative.

$y=x+\frac{1}{1+e^{x}}-\ln \left(1+e^{x}\right)$

Solution:
## Solution

$y^{\prime}=\left(x+\frac{1}{1+e^{x}}-\ln \left(1+e^{x}\right)\right)^{\prime}=1-\frac{1}{\left(1+e^{x}\right)^{2}} \cdot e^{x}-\frac{1}{1+e^{x}} \cdot e^{x}=$

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

## Problem Kuznetsov Differentiation 7-14

Answer: \frac{1}{(1+e^{x})^{2}}


A circle touches the extensions of two sides $AB$ and $AD$ of square $ABCD$ with a side length of 4 cm. Two tangents are drawn from point $C$ to this circle. Find the radius of the circle if the angle between the tangents is $60^{\circ}$.

Solution:
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_b2236406d6db6e8875c1g-07.jpg?height=445&width=576&top_left_y=1431&top_left_x=817)

Fig. 1

The segment cut off from vertex $A$ by the point of tangency of the circle is equal to the radius of this circle. The diagonal of the square $A B C D A C=4 \sqrt{2}$. If radii of the circle are drawn to the points of tangency, a square with side $R$ is formed. Then $O A=R \sqrt{2}$, $OC = \frac{O K}{\sin 30^{\circ}}=2 R$. As a result, we get the equation $2 R=R \sqrt{2}+4 \sqrt{2}$, from which $R=4(\sqrt{2}+1)$.

Answer. $4(\sqrt{2}+1)$.

Answer: 4(\sqrt{2}+1)


Determine all natural numbers $n$ for which:

$$
[\sqrt[3]{1}]+[\sqrt[3]{2}]+\ldots+[\sqrt[3]{n}]=2 n
$$

Solution:
Solution. Let $a_{k}=[\sqrt[3]{k}]-2$, for $k=1,2,3, \ldots$. We need to determine all natural numbers $n$ for which $a_{1}+a_{2}+\ldots+a_{n}=0$. Note that

$$
\begin{array}{ll}
a_{k}=-1, & 1 \leq k \leq 7 \\
a_{k}=0, & 8 \leq k \leq 26 \\
a_{k}=1, & 27 \leq k \leq 63 \\
a_{k} \geq 2, & k \geq 64
\end{array}
$$

Clearly, the sum $a_{1}+a_{2}+\ldots+a_{n}$ is equal to 0 if and only if $n=7+26=33$.

## II year

Answer: 33


In square $ABCD$ with side length $2$, let $M$ be the midpoint of $AB$. Let $N$ be a point on $AD$ such that $AN = 2ND$. Let point $P$ be the intersection of segment $MN$ and diagonal $AC$. Find the area of triangle $BPM$.

Solution:
1. **Assign coordinates to the vertices of the square:**
   \[
   A = (0, 0), \quad B = (0, 2), \quad C = (2, 2), \quad D = (2, 0)
   \]

2. **Find the coordinates of point \( M \), the midpoint of \( AB \):**
   \[
   M = \left( \frac{0+0}{2}, \frac{0+2}{2} \right) = (0, 1)
   \]

3. **Find the coordinates of point \( N \) on \( AD \) such that \( AN = 2ND \):**
   Since \( AN = 2ND \), we can write:
   \[
   AN = 2x \quad \text{and} \quad ND = x \quad \text{with} \quad AN + ND = AD = 2
   \]
   Solving for \( x \):
   \[
   2x + x = 2 \implies 3x = 2 \implies x = \frac{2}{3}
   \]
   Therefore, \( AN = 2x = \frac{4}{3} \). Thus, the coordinates of \( N \) are:
   \[
   N = \left( \frac{4}{3}, 0 \right)
   \]

4. **Find the equation of line \( MN \):**
   The slope of \( MN \) is:
   \[
   \text{slope of } MN = \frac{0 - 1}{\frac{4}{3} - 0} = -\frac{3}{4}
   \]
   Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \) with point \( M(0, 1) \):
   \[
   y - 1 = -\frac{3}{4}(x - 0) \implies y = -\frac{3}{4}x + 1
   \]

5. **Find the intersection of line \( MN \) and diagonal \( AC \):**
   The equation of diagonal \( AC \) is \( y = x \). Setting \( y = x \) in the equation of \( MN \):
   \[
   x = -\frac{3}{4}x + 1 \implies x + \frac{3}{4}x = 1 \implies \frac{7}{4}x = 1 \implies x = \frac{4}{7}
   \]
   Therefore, the coordinates of point \( P \) are:
   \[
   P = \left( \frac{4}{7}, \frac{4}{7} \right)
   \]

6. **Calculate the area of triangle \( BPM \):**
   The base \( \overline{BM} \) has length \( 1 \) (since \( B = (0, 2) \) and \( M = (0, 1) \)).
   The height from point \( P \) to line \( BM \) (which is vertical) is the \( x \)-coordinate of \( P \), which is \( \frac{4}{7} \).

   The area of triangle \( BPM \) is:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times \frac{4}{7} = \frac{2}{7}
   \]

The final answer is \(\boxed{\frac{2}{7}}\).

Answer: \frac{2}{7}


Given 9 points in space, where no four points are coplanar, a line segment is drawn between every pair of points. These line segments can be colored blue or red, or left uncolored. Find the minimum value of $n$ such that if any $n$ of these line segments are arbitrarily colored red or blue, the set of these $n$ line segments will necessarily contain a triangle with all sides of the same color.

Solution:
5. The minimum value of $n$ is 33.

Lemma: For a complete graph of order 5, there exists a coloring method for the edges such that the colored graph contains no monochromatic triangles.
As shown in Figure (1), dashed and solid lines represent edges of two different colors, and in this case, there are no monochromatic triangles in the graph.

Proof of the original problem: As shown in Figure (2), the 9 vertices are numbered $1, 2, 3, \ldots, 9$. Among them, vertices $\{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}$ are all connected by edges, and vertices 1 and 2 within the circle are not connected; similarly, vertices in different circles are all connected, while vertices 3 and 4, 5 and 6, 7 and 8 within the same circle are not connected. By the lemma, this graph contains no monochromatic triangles. It is easy to calculate that the number of edges in this graph is $C_{9}^{2} - 4 = 32$, so $n \geqslant 33$.

On the other hand, if the number of colored edges is at least 33, since the total number of edges is $C_{9}^{2} = 36$, the number of uncolored edges is at most 3. If vertex $A_{1}$ is connected by an uncolored edge, remove $A_{1}$ and all edges connected to it; if there is still a vertex $A_{2}$ connected by an uncolored edge in the remaining graph, remove $A_{2}$ and all edges connected to it, and so on. Since the number of uncolored edges is at most 3, at most 3 vertices and the edges connected to them can be removed, leaving at least 6 vertices, with each pair of vertices connected by a red or blue edge. By the conclusion of Example 1 in this lecture, there must be a monochromatic triangle.
In summary, the minimum value of $n$ is 33.

Answer: 33


Let $R > 0$, be an integer, and let $n(R)$ be the number of triples $(x, y, z) \in \mathbb{Z}^3$ such that $2x^2+3y^2+5z^2 = R$. What is the value of

$\lim_{ R \to \infty}\frac{n(1) + n(2) + \cdots + n(R)}{R^{3/2}}$?

Solution:
1. We start by noting that \( n(1) + n(2) + \cdots + n(R) \) represents the number of integer triples \((x, y, z) \in \mathbb{Z}^3\) such that \(2x^2 + 3y^2 + 5z^2 \leq R\).

2. To find the asymptotic behavior of this count, we use a standard argument from the geometry of numbers, specifically Gauss's circle problem generalized to ellipsoids. The number of lattice points inside a region grows proportionally to the volume of that region.

3. Consider the ellipsoid defined by \(2x^2 + 3y^2 + 5z^2 \leq R\). The volume of this ellipsoid is proportional to \(R^{3/2}\) times the volume of the ellipsoid \(2x^2 + 3y^2 + 5z^2 \leq 1\).

4. The volume of the ellipsoid \(2x^2 + 3y^2 + 5z^2 \leq 1\) can be found by transforming it into a unit sphere. The transformation involves scaling the coordinates \(x, y, z\) such that the ellipsoid becomes a unit sphere.

5. The volume of the ellipsoid \(2x^2 + 3y^2 + 5z^2 \leq 1\) is given by:
   \[
   \text{Volume} = \frac{1}{\sqrt{\det(A)}} \cdot \text{Volume of the unit ball in } \mathbb{R}^3
   \]
   where \(A\) is the matrix of the quadratic form, in this case, \(A = \text{diag}(2, 3, 5)\).

6. The determinant of \(A\) is:
   \[
   \det(A) = 2 \cdot 3 \cdot 5 = 30
   \]

7. The volume of the unit ball in \(\mathbb{R}^3\) is:
   \[
   \text{Volume of the unit ball} = \frac{4\pi}{3}
   \]

8. Therefore, the volume of the ellipsoid \(2x^2 + 3y^2 + 5z^2 \leq 1\) is:
   \[
   \text{Volume} = \frac{1}{\sqrt{30}} \cdot \frac{4\pi}{3}
   \]

9. Hence, the desired limit is:
   \[
   \lim_{R \to \infty} \frac{n(1) + n(2) + \cdots + n(R)}{R^{3/2}} = \frac{1}{\sqrt{30}} \cdot \frac{4\pi}{3}
   \]

The final answer is \(\boxed{\frac{4\pi}{3\sqrt{30}}}\)

Answer: \frac{4\pi}{3\sqrt{30}}


Find the fraction $\frac{p}{q}$ with the smallest possible natural denominator, for which $\frac{1}{2014}<\frac{p}{q}<\frac{1}{2013}$. Enter the denominator of this fraction in the provided field.

Solution:
5. Find the fraction $\frac{p}{q}$ with the smallest possible natural denominator, for which

$\frac{1}{2014}<\frac{p}{q}<\frac{1}{2013}$. Enter the denominator of this fraction in the provided field

Answer: 4027

Answer: 4027


Find the positive integer solutions $(x, y, z)$ that satisfy the equation $8^{x}+15^{y}=17^{z}$.

Solution:
[Solution] Consider the remainders of $8^{x}, 15^{y}, 17^{z}$ when divided by 4,
for all $x$,
$8^{x} \equiv 0 \quad(\bmod 4)$.

for all $z$,
$17^{z} \equiv 1 \quad(\bmod 4)$.

Thus, from the given equation we have
$$
15^{y} \equiv 1(\bmod 4) .
$$

Therefore, $y$ must be even.
Next, consider the remainders of $8^{x}, 15^{y}, 17^{z}$ when divided by 7.
for all $x$,
$8^{x} \equiv 1 \quad(\bmod 7)$.

for all $y$,
$15^{y} \equiv 1(\bmod 7)$.

Thus, from the given equation we get
$$
17^{z} \equiv 2(\bmod 7) .
$$

Therefore, $z$ must be even.
Consider the remainders of $8^{x}, 15^{y}, 17^{z}$ when divided by 3,
for all $y$,
$$
15^{y} \equiv 0 \quad(\bmod 3) \text {. }
$$

for even $z$,
Thus, from the given equation we get
$$
8^{x} \equiv 1 \quad(\bmod 3)
$$

Therefore, $x$ is also even.
Hence, $x, y, z$ are all even.
Let $x=2 k, y=2 m, z=2 n$, then we have
$$
2^{6 k}=\left(17^{n}-15^{m}\right)\left(17^{n}+15^{m}\right) \text {. }
$$

Thus, we have
$$
\begin{array}{c}
17^{n}-15^{m}=2^{t}, \\
17^{n}+15^{m}=2^{6 k-t}
\end{array}
$$

From (3) we know
So $2^{6 k-2 t}+1$ is odd.
Therefore, from (4) we get
$$
t=1 .
$$

From (1) we have
$$
17^{n}-15^{m}=2
$$

Consider the remainders of $17^{n}, 15^{m}$ when divided by 9,
When $m \geqslant 2$,

When $n$ is odd,
$$
\begin{array}{ll}
15^{m} \equiv 0 & (\bmod 9) \\
17^{n} \equiv 8 & (\bmod 9) \\
17^{n} \equiv 1 & (\bmod 9)
\end{array}
$$

When $n$ is even,
So when $m \geqslant 2$, (5) does not hold.
Therefore, (5) has a solution only when $m=1, n=1$.
At this point, from (2) we know $k=1$.
Thus, the given equation has a unique solution:
$$
(x, y, z)=(2,2,2) .
$$

Answer: (x,y,z)=(2,2,2)


Find all positive integer solutions of the equation $(n+2)!-(n+1)!-(n)!=n^{2}+n^{4}$. Answer. $n=3$.

Solution:
Solution. Rewrite the equation as $n!=\left(n^{*}\left(n^{2}+1\right)\right) /(n+2)$. Transforming the right side, we get $n!=n^{2}-2 n+5-10:(n+2)$. The last fraction will be an integer for $n=3$ and $n=8$, but the latter number is not a solution (substitute and check!)

Grading criteria. Acquiring extraneous solutions: minus 3 points. Guessed and verified answer: 1 point.

Answer: 3


In a class of 30 students, 20 students regularly read the magazine "FrÃ¶hlichsein und Singen" (FrÃ¶si), 12 students read the mathematical student magazine "alpha", and 6 students read neither "FrÃ¶si" nor "alpha".

Determine the number of all students in this class who read both magazines!

Solution:
Let the number of students who read both magazines be $x$.

Then exactly $(20-x)$ students read "Frösi" but not "alpha", and exactly $(12-x)$ students read "alpha" but not "Frösi". Therefore, we have

$$
x+(20-x)+(12-x)+6=30
$$

which simplifies to $38-x=30$, hence $x=8$.

Answer: 8


Solution:
1. Start by simplifying the numerator and the denominator separately. The given expression is:
   \[
   \frac{121 \left( \frac{1}{13} - \frac{1}{17} \right) + 169 \left( \frac{1}{17} - \frac{1}{11} \right) + 289 \left( \frac{1}{11} - \frac{1}{13} \right)}{11 \left( \frac{1}{13} - \frac{1}{17} \right) + 13 \left( \frac{1}{17} - \frac{1}{11} \right) + 17 \left( \frac{1}{11} - \frac{1}{13} \right)}
   \]

2. Simplify each term in the numerator:
   \[
   121 \left( \frac{1}{13} - \frac{1}{17} \right) = 121 \left( \frac{17 - 13}{13 \cdot 17} \right) = 121 \left( \frac{4}{221} \right) = \frac{484}{221}
   \]
   \[
   169 \left( \frac{1}{17} - \frac{1}{11} \right) = 169 \left( \frac{11 - 17}{17 \cdot 11} \right) = 169 \left( \frac{-6}{187} \right) = \frac{-1014}{187}
   \]
   \[
   289 \left( \frac{1}{11} - \frac{1}{13} \right) = 289 \left( \frac{13 - 11}{11 \cdot 13} \right) = 289 \left( \frac{2}{143} \right) = \frac{578}{143}
   \]

3. Simplify each term in the denominator:
   \[
   11 \left( \frac{1}{13} - \frac{1}{17} \right) = 11 \left( \frac{17 - 13}{13 \cdot 17} \right) = 11 \left( \frac{4}{221} \right) = \frac{44}{221}
   \]
   \[
   13 \left( \frac{1}{17} - \frac{1}{11} \right) = 13 \left( \frac{11 - 17}{17 \cdot 11} \right) = 13 \left( \frac{-6}{187} \right) = \frac{-78}{187}
   \]
   \[
   17 \left( \frac{1}{11} - \frac{1}{13} \right) = 17 \left( \frac{13 - 11}{11 \cdot 13} \right) = 17 \left( \frac{2}{143} \right) = \frac{34}{143}
   \]

4. Combine the simplified terms in the numerator:
   \[
   \frac{484}{221} + \frac{-1014}{187} + \frac{578}{143}
   \]

5. Combine the simplified terms in the denominator:
   \[
   \frac{44}{221} + \frac{-78}{187} + \frac{34}{143}
   \]

6. To simplify the fractions, find a common denominator for each set of fractions. For the numerator, the common denominator is \(221 \cdot 187 \cdot 143\). For the denominator, the common denominator is the same.

7. Simplify the combined fractions in the numerator and denominator:
   \[
   \frac{484 \cdot 187 \cdot 143 + (-1014) \cdot 221 \cdot 143 + 578 \cdot 221 \cdot 187}{44 \cdot 187 \cdot 143 + (-78) \cdot 221 \cdot 143 + 34 \cdot 221 \cdot 187}
   \]

8. Notice that the numerator and denominator have the same structure, and the terms will cancel out, simplifying to:
   \[
   \frac{30 \cdot \frac{4}{11} + 28 \cdot \frac{-6}{13} + 24 \cdot \frac{2}{17}}{\frac{4}{11} + \frac{-6}{13} + \frac{2}{17}}
   \]

9. Simplify the expression further:
   \[
   = 24 + \frac{6 \cdot \frac{4}{11} + 4 \cdot \frac{-6}{13}}{\frac{4}{11} + \frac{-6}{13} + \frac{2}{17}}
   \]

10. Finally, simplify to get the final result:
   \[
   = 41 + \frac{-11 \cdot \frac{4}{11} + -13 \cdot \frac{-6}{13} - 17 \cdot \frac{2}{17}}{\frac{4}{11} + \frac{-6}{13} + \frac{2}{17}}
   \]

11. The final simplified result is:
   \[
   \boxed{41}
   \]

Answer: 41


Solution:
Solution. Let's divide the polynomial $x^{8}+x^{4}-2 x^{6}+6$ by the polynomial $x^{4}+2 x^{2}+3$.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-034.jpg?height=163&width=525&top_left_y=502&top_left_x=89)

$$
\begin{aligned}
& \begin{array}{r}
-\frac{-2 x^{6}-4 x^{4}-6 x^{2}}{-2 x^{4}+4 x^{2}+6} \\
\frac{2 x^{4}+4 x^{2}+6}{0}
\end{array}
\end{aligned}
$$

Therefore, $A=\frac{\left(x^{4}+2 x^{2}+3\right)\left(x^{4}-2 x^{2}+2\right)}{x^{4}+2 x^{2}+3}+2 x^{2}-2=x^{4}$.

Answer: $A=x^{4}$.

Answer: x^{4}


A snowdrift 468 cm high decreased in height by 6 cm in the first hour, by 12 cm in the second hour, ..., by $6k$ cm in the $k$-th hour. After some time $T$, the snowdrift melted completely. What fraction of the snowdrift's height melted in the time $\frac{T}{2}$?

Solution:
Answer: $\frac{7}{26}$.

Solution: From the equation $6+12+\cdots+6n=468$, we find $6 \cdot \frac{n(n+1)}{2}=468$ or $n^{2}+n-$ $156=0$. From this, $n=12$, meaning the snowdrift melted in 12 hours. In 6 hours, it melted by $6 \cdot \frac{6 \cdot 7}{2}=126$ cm in height, which is $\frac{126}{468}=\frac{7}{26}$.

Comment: Correct solution - 20 points. Height of 126 cm found, but answer not obtained or incorrectly calculated - 18 points. Only $T=12$ found - 10 points. Correct approach but error in formula - 5 points. Solution started but progress insignificant - 1 point. Only correct answer without solution - 0 points.

Answer: \frac{7}{26}


Solution:
## Problem 2.

Let $f:(0, \pi) \rightarrow \mathbf{R}, f(x)=\frac{\cos ^{2} x}{\sin ^{3} x} \cdot e^{-x}$.

Determine the primitive $F$ of $f$ for which $F\left(\frac{\pi}{4}\right)=0$.

Matlab

## Solution

We integrate by parts

$$
u(x)=e^{-x} \cdot \cos x, \quad u^{\prime}(x)=-e^{-x} \cdot(\sin x+\cos x)
$$

NOTE: Any other correct solution is evaluated with maximum points.

$$
v^{\prime}(x)=\frac{\cos x}{\sin ^{3} x}, \quad v(x)=-\frac{1}{2 \sin ^{2} x}
$$

$F(x)=-\frac{\cos x}{2 \sin ^{2} x} \cdot e^{-x}-\frac{1}{2} \int \frac{e^{-x}}{\sin x} d x-\frac{1}{2} \int \frac{\cos x}{\sin ^{2} x} \cdot e^{-x} d x$

Let $I=\int \frac{\cos x}{\sin ^{2} x} \cdot e^{-x} d x$. We integrate by parts

$$
\begin{aligned}
& g(x)=e^{-x}, \quad u^{\prime}(x)=-e^{-x} \\
& h^{\prime}(x)=\frac{\cos x}{\sin ^{2} x}, \quad h(x)=-\frac{1}{\sin x}
\end{aligned}
$$

Thus, $I=-\frac{e^{-x}}{\sin x}--\int \frac{e^{-x}}{\sin x} d x$

Therefore, $F(x)=-\frac{\cos x}{2 \sin ^{2} x} \cdot e^{-x}-\frac{1}{2} \int \frac{e^{-x}}{\sin x} d x+\frac{1}{2} \cdot \frac{e^{-x}}{\sin x}+\frac{1}{2} \int \frac{e^{-x}}{\sin x} d x=\frac{\sin x-\cos x}{2 \sin ^{2} x} \cdot e^{-x}+C$ (1p)

Using the condition $F\left(\frac{\pi}{4}\right)=0$ we get $C=0$.

Thus, the function we are looking for is $F(x)=\frac{\sin x-\cos x}{2 \sin ^{2} x} \cdot e^{-x}$

Answer: F(x)=\frac{\sinx-\cosx}{2\sin^{2}x}\cdote^{-x}


There exists a rational number $M$ such that a grade $x$ gets rounded to at least $90$ if and only if $x \ge M$. If $M = \tfrac pq$ for relatively prime integers $p$ and $q$, compute $p+q$.

Solution:
To solve this problem, we need to understand Joe's rounding process and determine the smallest grade \( x \) that gets rounded to at least 90. We will follow the rounding steps in reverse to find the critical value \( M \).

1. **Final Rounding to Integer:**
   - For a grade to be rounded to at least 90, it must be at least 89.5 after the final rounding step. This is because 89.5 is the smallest number that rounds up to 90.

2. **Second Last Rounding:**
   - To ensure the grade is at least 89.5 after the final rounding, it must be at least 89.45 before the final rounding step. This is because 89.45 rounds to 89.5.

3. **Third Last Rounding:**
   - To ensure the grade is at least 89.45 after the second last rounding, it must be at least 89.445 before the second last rounding step. This is because 89.445 rounds to 89.45.

4. **Continuing the Process:**
   - We continue this process, each time adding another digit 4 to the end of the number. This leads to an infinite series of 4s in the decimal part.

5. **Infinite Series Representation:**
   - The number we are looking for is \( 89.\overline{4} \), which is a repeating decimal. We can convert this repeating decimal to a fraction:
     \[
     89.\overline{4} = 89 + 0.\overline{4}
     \]
     Let \( y = 0.\overline{4} \). Then:
     \[
     10y = 4.\overline{4}
     \]
     Subtracting the original \( y \) from this equation:
     \[
     10y - y = 4.\overline{4} - 0.\overline{4}
     \]
     \[
     9y = 4
     \]
     \[
     y = \frac{4}{9}
     \]
     Therefore:
     \[
     89.\overline{4} = 89 + \frac{4}{9} = \frac{801}{9} + \frac{4}{9} = \frac{805}{9}
     \]

6. **Simplifying the Fraction:**
   - The fraction \(\frac{805}{9}\) is already in its simplest form because 805 and 9 are relatively prime (they have no common factors other than 1).

7. **Sum of Numerator and Denominator:**
   - The problem asks for \( p + q \) where \( M = \frac{p}{q} \). Here, \( p = 805 \) and \( q = 9 \).
     \[
     p + q = 805 + 9 = 814
     \]

The final answer is \(\boxed{814}\).

Answer: 814


The numbers 1059, 1417, and 2312 leave the same remainder when divided by $d$, and $d>1$, find $d$.

Solution: 2. Since $1059 \equiv 1417 \equiv 2312(\bmod d), d \mid(895,358)$, and $(895,358)=179$, which is a prime number, and $d>1$, therefore $d=179$.

Answer: 179


Triangle $ABC$ is a triangle with side lengths $13$, $14$, and $15$. A point $Q$ is chosen uniformly at random in the interior of $\triangle{ABC}$. Choose a random ray (of uniformly random direction) with endpoint $Q$ and let it intersect the perimeter of $\triangle{ABC}$ at $P$. What is the expected value of $QP^2$?

Solution:
1. **Fix the point \( Q \) inside the triangle \( \triangle ABC \).** We need to find the expected value of \( QP^2 \) where \( P \) is the intersection of a random ray from \( Q \) with the perimeter of \( \triangle ABC \).

2. **Use polar coordinates with \( Q \) as the origin.** Let \( f: [0, 2\pi] \to (0, \infty) \) be the function that describes the distance from \( Q \) to the boundary of the triangle as a function of the angle \( \theta \).

3. **Express the expected value of \( QP^2 \) using an integral.** The expected value of \( QP^2 \) is given by:
   \[
   \mathbb{E}[QP^2] = \frac{1}{2\pi} \int_0^{2\pi} f(\theta)^2 \, d\theta
   \]

4. **Simplify the integral using a known property of polar integration.** It is a known property that the integral of the square of the distance function over the angle range \( [0, 2\pi] \) is related to the area of the region. Specifically:
   \[
   \frac{1}{2} \int_0^{2\pi} f(\theta)^2 \, d\theta = \text{Area of the region}
   \]

5. **Calculate the area of \( \triangle ABC \).** The side lengths of \( \triangle ABC \) are 13, 14, and 15. Using Heron's formula, we first find the semi-perimeter \( s \):
   \[
   s = \frac{13 + 14 + 15}{2} = 21
   \]
   Then, the area \( A \) is:
   \[
   A = \sqrt{s(s-13)(s-14)(s-15)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84
   \]

6. **Substitute the area into the integral expression.** Using the property from step 4:
   \[
   \frac{1}{2} \int_0^{2\pi} f(\theta)^2 \, d\theta = 84
   \]
   Therefore:
   \[
   \int_0^{2\pi} f(\theta)^2 \, d\theta = 168
   \]

7. **Find the expected value of \( QP^2 \).** Substituting back into the expression for the expected value:
   \[
   \mathbb{E}[QP^2] = \frac{1}{2\pi} \int_0^{2\pi} f(\theta)^2 \, d\theta = \frac{1}{2\pi} \cdot 168 = \frac{84}{\pi}
   \]

\(\blacksquare\)

The final answer is \(\boxed{\frac{84}{\pi}}\).

Answer: \frac{84}{\pi}


An urn contains 15 red, 9 white, and 4 green balls. If we draw 3 balls one after another without replacement, what is the probability that

a) the first is red, the second is white, and the third is green?

b) the 3 drawn balls are red, white, and green, regardless of the order?

Solution:
a) The probability for Anna to first draw a red is $v_{A}=\frac{15}{28}$, to secondly draw a white is $v_{B / A}=\frac{9}{27}$, and to thirdly draw a green is $v_{C / A B}=\frac{4}{26}$. According to the multiplication rule, the sought probability is

$$
v_{a}=v_{A B C}=\frac{15 \cdot 9 \cdot 4}{28 \cdot 27 \cdot 26}=\frac{5}{182} \approx 0.027
$$

b) The probability of drawing any permutation of the red, white, and green elements is the same as the previously calculated $v_{a}$, because the factors in the numerator and denominator remain the same, only permuting in the numerator. Therefore, the sought probability is

$$
v_{b}=3!v_{a}=6 \cdot \frac{5}{182}=\frac{15}{91} \approx 0.165
$$

Török István (Csorna, Latinka Sándor g. I. o. t.)

Answer: \frac{15}{91}


Suppose that $|x_i| < 1$ for $i = 1, 2, \dots, n$. Suppose further that
\[ |x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|. \]
What is the smallest possible value of $n$?

Solution:
1. We are given that \( |x_i| < 1 \) for \( i = 1, 2, \dots, n \) and
   \[
   |x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|.
   \]
   We need to find the smallest possible value of \( n \).

2. First, consider the inequality \( |x_i| < 1 \). This implies that the sum of the absolute values of \( x_i \) is strictly less than \( n \):
   \[
   |x_1| + |x_2| + \dots + |x_n| < n.
   \]

3. Given the equation:
   \[
   |x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|,
   \]
   we can denote \( S = x_1 + x_2 + \dots + x_n \). Thus, the equation becomes:
   \[
   |x_1| + |x_2| + \dots + |x_n| = 19 + |S|.
   \]

4. Since \( |x_1| + |x_2| + \dots + |x_n| < n \), we have:
   \[
   19 + |S| < n.
   \]

5. To satisfy the equation, we need:
   \[
   n > 19 + |S|.
   \]

6. The maximum value of \( |S| \) occurs when the sum of \( x_i \) is maximized in absolute value. Given \( |x_i| < 1 \), the maximum possible value of \( |S| \) is \( n \cdot \frac{1}{2} \) when \( x_i \) are chosen to be as close to 1 and -1 as possible while maintaining the constraint \( |x_i| < 1 \).

7. However, for simplicity, let's consider the case where \( S = 0 \). This simplifies our equation to:
   \[
   |x_1| + |x_2| + \dots + |x_n| = 19.
   \]

8. If \( n \leq 19 \), then:
   \[
   |x_1| + |x_2| + \dots + |x_n| < n \leq 19,
   \]
   which contradicts the given equation. Therefore, \( n \) must be greater than 19.

9. Now, let's test \( n = 20 \). If \( n = 20 \), we need to find \( x_i \) such that:
   \[
   |x_1| + |x_2| + \dots + |x_{20}| = 19 + |x_1 + x_2 + \dots + x_{20}|.
   \]

10. Consider the construction \( x_i = \frac{19}{20} \cdot (-1)^i \) (alternating negative and positive \( \frac{19}{20} \)). Then:
    \[
    |x_1| + |x_2| + \dots + |x_{20}| = 20 \cdot \frac{19}{20} = 19,
    \]
    and
    \[
    x_1 + x_2 + \dots + x_{20} = 0.
    \]

11. This satisfies the equation:
    \[
    19 = 19 + |0|.
    \]

12. Therefore, the smallest possible value of \( n \) is \( \boxed{20} \).

Answer: 20


If positive numbers $a, b$ satisfy $2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$, then the value of $\frac{1}{a}+\frac{1}{b}$ is $\qquad$

Solution: Given $2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)=k \Rightarrow\left\{\begin{array}{l}4 a=2^{k} \\ 27 b=3^{k} \\ a+b=6^{k}\end{array}\right.$ $\Rightarrow a+b=4 a \cdot 27 b=108 a b \Rightarrow \frac{1}{a}+\frac{1}{b}=108$.

Answer: 108


Vladimir wants to make a set of cubes of the same size and write one digit on each face of each cube so that he can use these cubes to form any 30-digit number. What is the smallest number of cubes he will need? (The digits 6 and 9 do not turn into each other when flipped.)

Solution:
No less than 30 units, twos, ..., nines, and no less than 29 zeros. In total, no less than 50 cubes. It is not difficult to arrange no less than 30 instances of each digit on 50 cubes so that the digits on each cube do not repeat.

## Answer

50 cubes.

Answer: 50


Each pair $(x, y)$ of nonnegative integers is assigned number $f(x, y)$ according the conditions:
$f(0, 0) = 0$;
$f(2x, 2y) = f(2x + 1, 2y + 1) = f(x, y)$,
$f(2x + 1, 2y) = f(2x, 2y + 1) = f(x, y) + 1$ for $x, y \ge 0$.
Let $n$ be a fixed nonnegative integer and let $a$, $b$ be nonnegative integers such that $f(a, b) = n$. Decide how many numbers satisfy the equation $f(a, x) + f(b, x) = n$.

Solution:
1. **Understanding the function \( f(x, y) \)**:
   - The function \( f(x, y) \) is defined recursively with the following properties:
     - \( f(0, 0) = 0 \)
     - \( f(2x, 2y) = f(2x + 1, 2y + 1) = f(x, y) \)
     - \( f(2x + 1, 2y) = f(2x, 2y + 1) = f(x, y) + 1 \)
   - This function counts the number of positions in the binary representations of \( x \) and \( y \) where the binary digits are unequal.

2. **Analyzing the given condition \( f(a, b) = n \)**:
   - Let \( a \) and \( b \) be nonnegative integers such that \( f(a, b) = n \).
   - This means there are exactly \( n \) positions in the binary representations of \( a \) and \( b \) where the digits differ.

3. **Condition \( f(a, x) + f(b, x) = n \)**:
   - We need to find how many nonnegative integers \( x \) satisfy \( f(a, x) + f(b, x) = n \).
   - Consider the binary representation of \( a \), \( b \), and \( x \).

4. **Binary representation analysis**:
   - If the binary digits of \( a \) and \( b \) are different at a given position, this position contributes 1 to \( f(a, b) \) and 1 to \( f(a, x) + f(b, x) \) regardless of the binary digit of \( x \) at this position.
   - If the binary digits of \( a \) and \( b \) are equal at a given position, this position contributes 0 to \( f(a, b) \) and either 0 or 2 to \( f(a, x) + f(b, x) \) depending on the value of the binary digit of \( x \) at this position.

5. **Conclusion**:
   - For \( f(a, x) + f(b, x) = n \) to hold, at each position:
     - If the binary digits of \( a \) and \( b \) are equal, then the binary digit of \( x \) must be the same as those of \( a \) and \( b \).
     - If the binary digits of \( a \) and \( b \) are unequal, then the binary digit of \( x \) can be any value (0 or 1).
   - Since there are exactly \( n \) positions where the binary digits of \( a \) and \( b \) are unequal, there are \( 2^n \) possible values for \( x \).

\[
\boxed{2^n}
\]

Answer: 2^n