How many ordered pairs of natural numbers $(a, b)$ satisfy

$$
\log _{2023-2(a+b)} b=\frac{1}{3 \log _{b} a} ?
$$

Solution:
## Solution.

For the equality to be well-defined, it is necessary that

$$
\begin{aligned}
2023-2(a+b) & \in (0,1) \cup (1,+\infty) \\
b & \in (0,1) \cup (1,+\infty) \\
\log _{b} a & \neq 0
\end{aligned}
$$

Since $a$ and $b$ are natural numbers, the first two conditions become $2023-2(a+b) \geqslant 2$ and $b \geqslant 2$. The last condition is equivalent to $a \geqslant 2$.

We can rearrange the initial equation as follows:

$$
\begin{aligned}
\frac{1}{\log _{b}(2023-2(a+b))} & =\frac{1}{\log _{b} a^{3}} \\
\log _{b}(2023-2(a+b)) & =\log _{b} a^{3} \\
2023-2(a+b)) & =a^{3} \\
a^{3}+2(a+b) & =2023 \\
a^{3}+2 a+2 b & =2023 .
\end{aligned}
$$

Notice that the condition $a^{3}=2023-2(a+b) \geqslant 2$ is automatically satisfied as long as $a \geqslant 2$. Therefore, we are looking for the number of natural solutions $(a, b)$ of the obtained equation that satisfy $a, b \geqslant 2$.

The number of solutions to this equation is equal to the number of odd natural numbers $a$ for which

$$
a^{3}+2 a \leqslant 2019
$$

Indeed, for the left and right sides of the equation to have the same parity, $a^{3}$ must be odd, i.e., $a$ must be odd. On the other hand, for any odd $a$ that satisfies the above inequality, the number $b=\frac{2023-a^{3}-2 a}{2}$ is a unique natural number that satisfies the desired equation.

For $a \leqslant 11$, we have $a^{3}+2 a \leqslant 11^{3}+2 \cdot 11=1353<2019$, while for $a \geqslant 13$, we have $a^{3}+2 a \geqslant 13^{3}+2 \cdot 13=2223$. Since there are 5 odd natural numbers less than or equal to 11 and greater than 2, we conclude that there are 5 ordered pairs that are solutions to the equation in the problem.

Note: The second point in the scoring scheme is achieved if an expression is obtained in which the number $(2023-2(a+b))$ does not appear in the base of the logarithm. The third point is achieved if an equation is obtained that does not include logarithms.

Answer: 5


If the fractional parts of $9+\sqrt{13}$ and $9-\sqrt{13}$ are $a$ and $b$ respectively, then $a b-4 a+3 b-2=$ $\qquad$

Solution:
Solution: $\because 3<\sqrt{13}<4$,
$\therefore 9+\sqrt{13}$ has an integer part of 12, and a decimal part
$$
\begin{array}{l}
a=\sqrt{13}-3 . \\
\because-4<-\sqrt{13}<-3,
\end{array}
$$

i.e., $0<4-\sqrt{13}<1$,
$\therefore 9-\sqrt{13}$ has an integer part of 5, and a decimal part $b$ $=4-\sqrt{13}$.
Thus, $a b-4 a+3 b-2$
$$
\begin{array}{l}
=(a+3)(b-4)+10 \\
=\sqrt{13} \times(-\sqrt{13})+10=-3 .
\end{array}
$$

Answer: -3


The function $f(x)$ defined on $\mathbf{R}$ satisfies
$$
f\left(\frac{2 a+b}{3}\right)=\frac{2 f(a)+f(b)}{3}(a, b \in \mathbf{R}),
$$

and $f(1)=1, f(4)=7$.
Then $f(2015)=$ $\qquad$ .

Solution:
$$
-, 1.4029 .
$$

From the function $f(x)$ being concave or convex in reverse on $\mathbf{R}$, we know its graph is a straight line, $f(x)=a x+b$.
Given $f(1)=1$ and $f(4)=7$, we have
$$
\begin{array}{l}
a=2, b=-1 \Rightarrow f(x)=2 x-1 \\
\Rightarrow f(2015)=4029 .
\end{array}
$$

Answer: 4029


Given in a convex quadrilateral $ABCD$, $AB=5\sqrt{2}$, $BC=\sqrt{13}$, $CD=3\sqrt{13}$, $AD=7$, $\angle BCD=120^{\circ}$. Then the degree of $\angle BAD$ is $(\quad)$.
(A) $120^{\circ}$
(B) $105^{\circ}$
(C) $135^{\circ}$
(D) $150^{\circ}$

Solution:
2.C.

As shown in Figure 5, draw $B E \perp$ $A D$ and $B F \perp C D$ at points $E$ and $F$ respectively. Connect $B D$.
Let $B E=x$,
$$
A E=y \text{.}
$$

Since $\angle B C D=120^{\circ}$, we have
$$
\angle B C F=180^{\circ}-\angle B C D=60^{\circ} \text{.}
$$

In the right triangle $\triangle B C F$, we have
$$
\begin{array}{l}
B F=\sqrt{13} \sin 60^{\circ}=\frac{\sqrt{39}}{2}, \\
C F=\sqrt{13} \cos 60^{\circ}=\frac{\sqrt{13}}{2},
\end{array}
$$

Thus, $D F=C F+C D=\frac{\sqrt{13}}{2}+3 \sqrt{13}=\frac{7 \sqrt{13}}{2}$.
In the right triangle $\triangle B D F$, by the Pythagorean theorem, we get
$$
B D^{2}=B F^{2}+D F^{2}=169 \text{.}
$$

In the right triangles $\triangle A B E$ and $\triangle B D E$, by the Pythagorean theorem, we have
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=50, \\
x^{2}+(y+7)^{2}=169 .
\end{array}\right.
$$

Solving these equations, we get $x=5, y=5$, hence
$$
A E=B E \text{.}
$$

Therefore, $\angle E A B=45^{\circ}$.
So, $\angle B A D=180^{\circ}-\angle E A B=135^{\circ}$.

Answer: C


Let's determine those positive integers which are one and a half times as large as the product of their digits.

Solution:
Solution. For the $n$-digit integer, we write the condition as:

$$
\overline{a_{n} a_{n-1} a_{n-2} \ldots a_{2} a_{1}}=\frac{3}{2} \cdot a_{n} \cdot a_{n-1} \cdot \ldots \cdot a_{2} a_{1}
$$

where $a_{i} \neq 0$ and $1 \leq a_{i} \leq 9(i=1,2, \ldots, n)$.

Since no digit can be 0 (otherwise, the right side would be 0), the following two inequalities are true:

$$
\begin{gathered}
\overline{a_{n} a_{n-1} a_{n-2} \ldots a_{2} a_{1}}>\overline{a_{n} \underbrace{00 \ldots 0}_{n-1}}=a_{n} \cdot 10^{n-1}, \quad \text { and } \\
\frac{3}{2} \cdot a_{n} \cdot 9^{n-1} \geq \frac{3}{2} \cdot a_{n} \cdot a_{n-1} \cdot \ldots \cdot a_{2} \cdot a_{1} .
\end{gathered}
$$

By comparing these two inequalities with equation (1), we get:

$$
\frac{3}{2} \cdot a_{n} \cdot 9^{n-1}>a_{n} \cdot 10^{n-1}
$$

Dividing by $a_{n} \neq 0$, we get $\frac{3}{2}>\left(\frac{10}{9}\right)^{n-1}$. Since the exponential function with a base greater than 1 is strictly increasing, we get:

$$
(n-1) \cdot \lg \frac{10}{9} < \lg \frac{3}{2}
$$

Since $\lg \frac{10}{9} > 0$, we have $n-1 < \frac{\lg \frac{3}{2}}{\lg \frac{10}{9}} \approx 2.71$, so $n \leq 3$.

For a two-digit number:

$$
\overline{a_{2} a_{1}}=\frac{3}{2} \cdot a_{2} \cdot a_{1}, \quad \text { hence } \quad a_{2}=\frac{10 a_{1}}{3 a_{1}-20}
$$

Since $a_{2}>0$, we have $3 a_{1}>20$. The possible values are: $a_{1}=7,8$ or 9. Substituting into (2): $a_{2}=\frac{14}{1}$, not a single digit, $a_{2}=\frac{16}{24-20}=4$ and $a_{2}=\frac{18}{7}$ not an integer. Only when $a_{1}=8$, we get a solution, and then $a_{2}=4$, and indeed: $48=\frac{3}{2} \cdot 4 \cdot 8$.

For three-digit numbers:

$$
\overline{a_{3} a_{2} a_{1}}=\frac{3}{2} \cdot a_{3} \cdot a_{2} \cdot a_{1}, \quad \text { hence } \quad a_{3}=\frac{20 a_{2}+2 a_{1}}{3 \cdot a_{2} \cdot a_{1}-200}
$$

Since $a_{3}>0$, from the condition $a_{2} a_{1} \geq 67$, the following possible values are obtained: $a_{1}=8, a_{2}=9 ; a_{1}=9, a_{2}=8 ; a_{1}=9$ and $a_{2}=9$. Substitution shows that in none of these cases do we get a single-digit integer value for $a_{3}$.

Finally, for four-digit numbers, similarly:

$$
a_{4}=\frac{200 a_{3}+20 a_{2}+2 a_{1}}{3 \cdot a_{3} \cdot a_{2} \cdot a_{1}-2000}
$$

From here, $a_{3} a_{2} a_{1} \geq 667$, which is only true for $a_{3}=a_{2}=a_{1}=9$, but this does not give an integer solution for $a_{4}$.

The only solution to the problem is 48.

Answer: 48


Two two-digit numbers, their difference is 52, and the last two digits of their squares are the same. Then these two numbers are

Solution:
7.76, 24. Let the two numbers be $x, y$, then $\left\{\begin{array}{l}x-y=52 \\ x^{2}-y^{2}=100 r\end{array}\right.$, thus $52(x+y)=(x-y)(x+y)=100 r, 13(x+$ $y)=25 r$, hence $13 \mid r$.

Since $x+y=x-y+2 y=52+2 y$ is even, we get $2 \mid r$, therefore $26 \mid r$. Let $r=26 t, t$ be a positive integer, then $x+y=50 t$, $x=25 t+26, y=25 t-26$, by $10 \leqslant x, y \leqslant 99$, we get $\left\{\begin{array}{l}10 \leqslant 25 t+26 \leqslant 99 \\ 10 \leqslant 25 t-26 \leqslant 99\end{array}\right.$, thus $\frac{36}{25} \leqslant t \leqslant \frac{73}{25}$, hence $t=2$.
Therefore, the two numbers are $x=25 \times 2+26=76, y=25 \times 2-26=24$.

Answer: 76,24


Let $x, y, z$ be positive real numbers, and $x+y+z \geqslant xyz$. Find the minimum value of $\frac{x^{2}+y^{2}+z^{2}}{xyz}$.

Solution:
First, note the case when the equality holds in the condition.
That is, when $x=y=z=\sqrt{3}$, we have $x+y+z=x y z$, and $\frac{x^{2}+y^{2}+z^{2}}{x y z}=\sqrt{3}$.
Next, we prove that the minimum value of $\frac{x^{2}+y^{2}+z^{2}}{x y z}$ is $\sqrt{3}$.
In fact, we have
$$
x^{2}+y^{2}+z^{2} \geqslant \frac{1}{3}(x+y+z)^{2} \geqslant\left\{\begin{array}{c}
\frac{1}{3}(x y z)^{2} \geqslant \sqrt{3} x y z, \text { if } x y z \geqslant 3 \sqrt{3}, \\
3 \sqrt[3]{(x y z)^{2}} \geqslant \sqrt{3} x y z, \text { if } x y z<3 \sqrt{3} .
\end{array}\right.
$$

Therefore, the minimum value of $\frac{x^{2}+y^{2}+z^{2}}{x y z}$ is $\sqrt{3}$.

Answer: \sqrt{3}


Inside the square $ABCD$, a point $M$ is taken such that $MA=1$, $MB=2$, $MC=3$. Find $MD$.

$$
\text { (8-9 grades) }
$$

Solution:
31.2. Through point $M$ draw lines parallel to the sides of rectangle $A B C D$ (Fig. 16). Denote the points of intersection with sides $A B, \quad B C, C D, D A$ as $K, P, R, T$ respectively. Let:

$$
A K=R D=a, \quad K B=C R=b, \quad B P=A T=c, \quad P C=T D=d
$$

Then, by the Pythagorean theorem, we get:

$$
a^{2}+c^{2}=1, \quad b^{2}+c^{2}=4, \quad b^{2}+d^{2}=9
$$

Adding these equations, we have:

$$
a^{2}+d^{2}+2\left(b^{2}+c^{2}\right)=14
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_f0d089db3d527e92a992g-31.jpg?height=454&width=623&top_left_y=244&top_left_x=725)

From this, taking into account that $b^{2}+c^{2}=4$, we get:

$$
a^{2}+d^{2}=6, \quad \text { i. } \quad \text { e. } \quad M D=\sqrt{a^{2}+d^{2}}=\sqrt{6}
$$

Answer: \sqrt{6}


On the surface of the Earth, all possible points are taken, the geographical latitude of which is equal to their longitude. Find the geometric locus of the projections of these points onto the equatorial plane.

Solution:
290. Let $O$ be the center of the Earth, $A$ be the point on the equator corresponding to the zero meridian, $M$ be a point on the surface of the Earth with longitude and latitude equal to $\varphi$, and $N$ be the projection of $M$ onto the equatorial plane. By introducing a rectangular coordinate system in the equatorial plane, taking the line $O \AA$ as the $x$-axis and the origin at point $O$, we obtain that $N$ has coordinates: $x = R \cos^2 \varphi, y = R \cos \varphi \sin \varphi$, where $R$ is the radius of the Earth. It is easy to verify that the coordinates of points

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-139.jpg?height=357&width=465&top_left_y=1199&top_left_x=165)

Fig. 56.

$A$

and $N$ satisfy the equation

$$
\left(x - \frac{R}{2}\right)^2 + y^2 = \frac{R^2}{4}
$$

i.e., the desired set is a circle with center $\left(\frac{R}{2}, 0\right)$ and radius $\frac{R}{2}$.

Answer: (x-\frac{R}{2})^2+y^2=\frac{R^2}{4}


For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?

Solution:
Answer: 17.

![](https://cdn.mathpix.com/cropped/2024_05_06_e7f9c87b6a37ffba3564g-30.jpg?height=474&width=507&top_left_y=657&top_left_x=469)

Fig. 3: to the solution of problem 8.6

Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute-angled ($\angle ABD = \angle CBD < 90^\circ$, $\left.\angle BAD = \angle ADB = \frac{180^\circ - \angle ABD}{2} < 90^\circ\right)$, point $H$ lies on the segment $AB$.

Notice that the right triangles $BDH$ and $BDC$ are equal by the common hypotenuse $BD$ and the acute angle at vertex $B$. Therefore, $BH = BC$ and $DH = CD$.

Now, notice that the right triangles $ADH$ and $EDC$ are also equal by the hypotenuse $AD = ED$ and the leg $DH = CD$. Therefore, $EC = AH$.

Thus, $BD = BA = BH + AH = BC + EC = (7 + 5) + 5 = 17$.

Answer: 17


Find all quadruplets $(a, b, c, d)$ of positive real numbers that satisfy the following two equations:
\begin{align*}
ab + cd &= 8,\\
abcd &= 8 + a + b + c + d.
\end{align*}

Solution:
1. We start with the given equations:
   \[
   ab + cd = 8
   \]
   \[
   abcd = 8 + a + b + c + d
   \]

2. We need to find all quadruples \((a, b, c, d)\) of positive real numbers that satisfy these equations. Let's first consider the inequality \(a + b + c + d \leq 8\).

3. Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   \frac{a + b + c + d}{4} \geq \sqrt[4]{abcd}
   \]
   Multiplying both sides by 4, we get:
   \[
   a + b + c + d \geq 4\sqrt[4]{abcd}
   \]

4. From the second equation \(abcd = 8 + a + b + c + d\), we substitute \(a + b + c + d\) with \(8\) (since \(a + b + c + d \leq 8\)):
   \[
   abcd = 8 + a + b + c + d \leq 16
   \]

5. Combining the inequalities, we get:
   \[
   4\sqrt[4]{abcd} \leq 8
   \]
   Simplifying, we find:
   \[
   \sqrt[4]{abcd} \leq 2
   \]
   Raising both sides to the fourth power, we get:
   \[
   abcd \leq 16
   \]

6. From the second equation \(abcd = 8 + a + b + c + d\), we know \(abcd \geq 8\). Therefore:
   \[
   8 \leq abcd \leq 16
   \]

7. Now, we use the first equation \(ab + cd = 8\). By AM-GM inequality:
   \[
   ab + cd \geq 2\sqrt{abcd}
   \]
   Given \(ab + cd = 8\), we have:
   \[
   8 \geq 2\sqrt{abcd}
   \]
   Simplifying, we get:
   \[
   4 \geq \sqrt{abcd}
   \]
   Squaring both sides, we find:
   \[
   16 \geq abcd
   \]

8. Combining all the inequalities, we have:
   \[
   abcd = 16
   \]

9. Substituting \(abcd = 16\) into the second equation:
   \[
   16 = 8 + a + b + c + d
   \]
   Simplifying, we get:
   \[
   a + b + c + d = 8
   \]

10. Given \(ab + cd = 8\) and using AM-GM inequality again:
    \[
    ab \leq \frac{(a+b)^2}{4} \quad \text{and} \quad cd \leq \frac{(c+d)^2}{4}
    \]
    Since \(a + b + c + d = 8\), we have:
    \[
    ab + cd \leq \frac{(a+b)^2}{4} + \frac{(c+d)^2}{4} \leq \frac{(a+b+c+d)^2}{4} = \frac{64}{4} = 16
    \]
    But \(ab + cd = 8\), so equality must hold in AM-GM, implying \(a = b\) and \(c = d\).

11. Let \(a = b\) and \(c = d\). Then:
    \[
    2a + 2c = 8 \implies a + c = 4
    \]
    \[
    ab = a^2 \quad \text{and} \quad cd = c^2
    \]
    \[
    a^2 + c^2 = 8
    \]

12. Since \(abcd = 16\), we have:
    \[
    a^2c^2 = 16 \implies (ac)^2 = 16 \implies ac = 4
    \]

13. Solving the system:
    \[
    a + c = 4
    \]
    \[
    ac = 4
    \]

14. The solutions to this system are \(a = c = 2\). Therefore, \(a = b = c = d = 2\).

Conclusion:
\[
(a, b, c, d) = (2, 2, 2, 2)
\]

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

Answer: (2, 2, 2, 2)


Find minimum integer $x$ with proof such that if $n-m\geq x,$ then Pratyya's number will be larger than Payel's number everyday.

Solution:
1. Define the initial numbers as \( n \) and \( m \) with \( n > m \). Let \( d_0 = n - m \) be the initial difference between Pratyya's and Payel's numbers.

2. On the first day, Pratyya's number becomes \( 2n - 2 \) and Payel's number becomes \( 2m + 2 \). The new difference \( d_1 \) between their numbers is:
   \[
   d_1 = (2n - 2) - (2m + 2) = 2n - 2m - 4 = 2(n - m) - 4 = 2d_0 - 4
   \]

3. Generalize this for any day \( t \). Let \( d_t \) be the difference on day \( t \). Then:
   \[
   d_{t+1} = 2d_t - 4
   \]

4. To ensure that Pratyya's number is always larger than Payel's number, we need \( d_t \geq 0 \) for all \( t \). We need to find the minimum initial difference \( d_0 \) such that \( d_t \geq 0 \) for all \( t \).

5. Consider the recurrence relation \( d_{t+1} = 2d_t - 4 \). We solve this recurrence relation by finding its fixed point. Set \( d_{t+1} = d_t = d \):
   \[
   d = 2d - 4 \implies d - 2d = -4 \implies -d = -4 \implies d = 4
   \]

6. The fixed point \( d = 4 \) suggests that if \( d_0 = 4 \), then \( d_t \) will remain 4 for all \( t \). However, we need to ensure that \( d_t \geq 0 \) for all \( t \) starting from \( d_0 \).

7. To find the minimum \( d_0 \), we consider the series:
   \[
   d_t = 2^t d_0 - 4(2^t - 1)
   \]
   For \( d_t \geq 0 \):
   \[
   2^t d_0 - 4(2^t - 1) \geq 0 \implies 2^t d_0 \geq 4(2^t - 1) \implies d_0 \geq 4 \left(1 - \frac{1}{2^t}\right)
   \]

8. As \( t \to \infty \), \( \frac{1}{2^t} \to 0 \), so:
   \[
   d_0 \geq 4
   \]

9. Therefore, the minimum integer \( x \) such that \( n - m \geq x \) ensures Pratyya's number is always larger than Payel's number is \( x = 4 \).

The final answer is \( \boxed{4} \)

Answer: 4


Solution:
Answer: 1
$5 a_{n+1}<a_{n+2}=5 a_{n+1}+a_{n}<6 a_{n+1}$, so $\left\{\frac{a_{n+2}}{a_{n+1}}\right\}=\frac{a_{n+2}}{a_{n+1}}-5=\frac{a_{n}}{a_{n+1}}$
By the periodicity of $\left\{a_{n}\right\}$ modulo 2, we know that $a_{2024}$ is even.
From the above two points, the answer is 1.

Answer: 1


As shown in the figure, two identical rectangles have a length that is twice the width, and the diagonal of the rectangle is 9 cm long. The sum of the areas of these two rectangles is $\qquad$ square centimeters.

Solution: Reference answer: 64.8

Answer: 64.8


Solution: $\begin{array}{l}2 \\ C\end{array}$

Answer: C


The product of two numbers, A and B, is 1.6. If A is multiplied by 5 and B is also multiplied by 5, then the product of A and B is $\qquad$ .

Solution:
【Solution】Solution: $1.6 \times(5 \times 5)$
$$
\begin{array}{l}
=1.6 \times 25 \\
=40
\end{array}
$$

Answer: The product of the two numbers, A and B, is 40. Therefore, the answer is: 40.

Answer: 40


ABC is a triangle. D is the midpoint of BC. E is a point on the side AC such that BE = 2AD. BE and AD meet at F and â FAE = 60Â°. Find â FEA.

Solution: Thanks to José Nelson Ramirez Let the line parallel to BE through D meet AC at G. Then DCG, BCE are similar and BC = 2 DC, so BE = 2 DG. Hence AD = DG, so ∠DGA = ∠DAG = 60 o . FE is parallel to DG, so ∠FEA = 60 o . 4th OMCC 2002 © John Scholes jscholes@kalva.demon.co.uk 26 November 2003 Last corrected/updated 26 Nov 03

Answer: 60


Let $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}$ be a function with the following properties:

(i) $f(1)=0$,

(ii) $f(p)=1$ for all prime numbers $p$,

(iii) $f(x y)=y f(x)+x f(y)$ for all $x, y$ in $\mathbb{Z}_{>0}$.

Determine the smallest integer $n \geq 2015$ that satisfies $f(n)=n$.

(Gerhard J. Woeginger)

Solution:
1. Consider the auxiliary function $g(x)=f(x) / x$, which satisfies the equation $g(x y)=g(x)+g(y)$. The initial conditions translate to $g(1)=0$ and $g(p)=1 / p$ for prime numbers $p$.
2. If $n$ has the prime factorization $x=\prod_{i=1}^{k} p_{i}^{e_{i}}$, then due to the functional equation of $g$, we have

$$
g(x)=g\left(\prod_{i=1}^{k} p_{i}^{e_{i}}\right)=\sum_{i=1}^{k} e_{i} g\left(p_{i}\right)=\sum_{i=1}^{k} \frac{e_{i}}{p_{i}}
$$

On the other hand, it is easy to verify that the function $g$ given by (1) satisfies the functional equation.
3. $f(n)=n$ is equivalent to $g(n)=1$ and

$$
\sum_{i=1}^{k} \frac{e_{i}}{p_{i}}=1
$$

If we multiply the equation by the product of all $p_{i}$, we see that

$$
e_{j} p_{1} \cdots p_{j-1} p_{j+1} \cdots p_{k}=a p_{j}
$$

holds for some integer $a$. Since $p_{j}$ is coprime to $p_{1} \cdots p_{j-1} p_{j+1} \cdots p_{k}$, it follows that $p_{j} \mid e_{j}$ for $1 \leq j \leq k$. Therefore, $p_{j} \leq e_{j}$ for $1 \leq j \leq k$. On the other hand, from (2) it follows that $e_{j} \leq p_{j}$, with equality only if it is the only summand in (2). Hence, $n$ has only one prime divisor $p$, and $n=p^{p}$.

4. The smallest number $n=p^{p}$ with $n \geq 2015$ is $n=5^{5}=3125$.

Answer: 3125


Given $a, b>0$. Find the area of the triangle with side lengths
$$
\sqrt{a^{2}+b^{2}}, \sqrt{a^{2}+4 b^{2}}, \sqrt{4 a^{2}+b^{2}}
$$

Solution:
Solve as shown in Figure 9, construct rectangle $ABCD$, such that $AB$
$$
=CD=2b, AD=BC
$$
$=2a$, and $E, F$ are the midpoints of
$CD, AD$ respectively. In this case, in $\triangle DEF$,
$$
\begin{array}{l}
EF=\sqrt{a^{2}+b^{2}}, BF=\sqrt{a^{2}+4b^{2}}, \\
BE=\sqrt{4a^{2}+b^{2}} . \\
\text{Therefore, } S_{\triangle DEF}=S_{\text{rectangle } ABCD}-S_{\triangle ABF}-S_{\triangle BCE}-S_{\triangle DEF} \\
=4ab-\frac{1}{2}ab-ab-ab=\frac{3}{2}ab .
\end{array}
$$

Answer: \frac{3}{2}ab


In a lottery, 6 numbers out of 36 are drawn each week. What is the probability that at least one of the numbers drawn this week will be the same as one of the numbers drawn last week? Round the result to the nearest thousandth.

Solution:
Answer: 0.695

Solution. Total outcomes: $C_{36}^{6}$; outcomes in which no number matches: $C_{30}^{6}$. Then the probability is $1-\frac{C_{30}^{6}}{C_{36}^{6}}$.

Answer: 0.695


In triangle $ABC$, the bisectors $BB_{1}$ and $CC_{1}$ are drawn. It is known that the center of the circumcircle of triangle $BB_{1}C_{1}$ lies on the line $AC$. Find the angle $C$ of the triangle.

Solution:
Continuing the ray $B C$ to intersect the circumcircle of triangle $B B_{1} C_{1}$, we obtain point $K$ (see the figure). The inscribed angles $\angle C_{1} B B_{1}$ and $\angle K B B_{1}$ are equal (since $B B_{1}$ is the angle bisector), which means the arcs they subtend, $B_{1} C_{1}$ and $B_{1} K$, are equal. Points $K$ and $C_{1}$ lie on the circumcircle (described around triangle $B B_{1} C_{1}$), whose center lies on the line $A C$. Therefore, $K$ and $C_{1}$ are symmetric with respect to the line $A C$. We obtain the equality of three angles: $\angle B C C_{1} = \angle C_{1} C B_{1} = \angle B_{1} C K$. The sum of these angles is $180^{\circ}$, so each of them is $60^{\circ}$, and $\angle A C B = \angle B C C_{1} + \angle C_{1} C B_{1} = 120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-39.jpg?height=366&width=643&top_left_y=2395&top_left_x=702)

Comments. 1. It is easy to show that the center $O$ of the circle can only lie on the extension of segment $A C$ beyond point $C$, and thus the line $B C$ intersects the circle exactly as shown in the figure.

2. Solutions based on the fact that points $B, C, O, C_{1}$ lie on the same circle, or that the circumcircle of triangle $B C_{1} B_{1}$ is the Apollonian circle for points $A$ and $C$, are also possible.

## Answer

$120^{\circ}$.

Answer: 120


Let $x, y$ be coprime natural numbers, and $xy=1992$. Then the number of different ordered pairs $(x, y)$ is $\qquad$

Solution:
5. 8.

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

Answer: 8


Given is a $n \times n$ chessboard. With the same probability, we put six pawns on its six cells. Let $p_n$ denotes the probability that there exists a row or a column containing at least two pawns. Find $\lim_{n \to \infty} np_n$.

Solution:
1. **Calculate the total number of ways to place six pawns on an \( n \times n \) chessboard:**
   \[
   \text{Total number of ways} = \binom{n^2}{6}
   \]

2. **Calculate the number of ways to place six pawns such that no two pawns are in the same row or column:**
   - First, choose 6 rows out of \( n \) rows: \(\binom{n}{6}\)
   - Then, choose 6 columns out of \( n \) columns: \(\binom{n}{6}\)
   - Finally, arrange the 6 pawns in the chosen 6 rows and 6 columns: \(6!\)
   \[
   \text{Number of ways with no two pawns in the same row or column} = \binom{n}{6}^2 \cdot 6!
   \]

3. **Calculate the probability \( p_n \) that there exists a row or a column containing at least two pawns:**
   \[
   p_n = 1 - \frac{\binom{n}{6}^2 \cdot 6!}{\binom{n^2}{6}}
   \]

4. **Simplify the expression for \( p_n \):**
   \[
   \binom{n^2}{6} = \frac{n^2 (n^2 - 1) (n^2 - 2) (n^2 - 3) (n^2 - 4) (n^2 - 5)}{6!}
   \]
   \[
   \binom{n}{6} = \frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 5)}{6!}
   \]
   \[
   \binom{n}{6}^2 = \left( \frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 5)}{6!} \right)^2
   \]

5. **Substitute these into the expression for \( p_n \):**
   \[
   p_n = 1 - \frac{\left( \frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 5)}{6!} \right)^2 \cdot 6!}{\frac{n^2 (n^2 - 1) (n^2 - 2) (n^2 - 3) (n^2 - 4) (n^2 - 5)}{6!}}
   \]
   \[
   p_n = 1 - \frac{n^2 (n - 1)^2 (n - 2)^2 (n - 3)^2 (n - 4)^2 (n - 5)^2}{n^2 (n^2 - 1) (n^2 - 2) (n^2 - 3) (n^2 - 4) (n^2 - 5)}
   \]

6. **Simplify the fraction:**
   \[
   p_n = 1 - \frac{(n - 1)^2 (n - 2)^2 (n - 3)^2 (n - 4)^2 (n - 5)^2}{(n^2 - 1) (n^2 - 2) (n^2 - 3) (n^2 - 4) (n^2 - 5)}
   \]

7. **Evaluate the limit \( \lim_{n \to \infty} np_n \):**
   - As \( n \to \infty \), the terms \((n - k)^2\) for \( k = 1, 2, 3, 4, 5 \) are approximately \( n^2 \).
   - Therefore, the fraction inside \( p_n \) approaches:
     \[
     \frac{(n - 1)^2 (n - 2)^2 (n - 3)^2 (n - 4)^2 (n - 5)^2}{(n^2 - 1) (n^2 - 2) (n^2 - 3) (n^2 - 4) (n^2 - 5)} \approx \frac{n^{10}}{n^{10}} = 1
     \]
   - Hence, \( p_n \) approaches:
     \[
     p_n \approx 1 - 1 = 0
     \]
   - However, we need to find \( \lim_{n \to \infty} np_n \). For large \( n \), the difference between the numerator and the denominator in the fraction becomes significant:
     \[
     p_n \approx 1 - \left( 1 - \frac{30}{n^2} \right) = \frac{30}{n^2}
     \]
   - Therefore:
     \[
     np_n \approx n \cdot \frac{30}{n^2} = \frac{30}{n}
     \]
   - As \( n \to \infty \):
     \[
     \lim_{n \to \infty} np_n = 30
     \]

The final answer is \(\boxed{30}\)

Answer: 30


Let the sum of one hundred natural numbers $x, x+1, x+2, \cdots, x+99$ be $a$. If the sum of the digits of $a$ equals 50, what is the smallest value of $x$?

Solution:
【Analysis】First, use the sum formula of an arithmetic sequence to get the sum of one hundred natural numbers, then discuss the cases where the addition of $100 x+4950$ does not carry over; and where the addition of $100 x+4950$ carries over $t$ times to find the solution.

【Solution】Solution: The total sum $a=100 x+9900 \div 2=100 x+4950$,
If the addition of $100 x+4950$ does not carry over, then the digit sum $=x$'s digit sum $+4+9+5=50, x$'s digit sum $=32$, $x$ must be at least a 5-digit number: 99950;

If the addition of $100 x+4950$ carries over $t$ times, then the digit sum $=x$'s digit sum $+4+9+5-9 t=50, x$'s digit sum $-9 t=32$, carrying over once then $x$'s digit sum $=41$, the smallest is 199949; carrying over twice then $x$'s digit sum $=50$, the smallest is 699899; more carries, $x$ must also exceed 5 digits.

Therefore, the smallest $x$ is 99950.

Answer: 99950


A train took 9 seconds to pass a signal light on a utility pole, and it took 35 seconds to cross a 520-meter-long iron bridge. The length of the train is ( ) meters

Solution:
180
【Answer】The length of the train is 9 times its speed; therefore, 520 meters is 35-9=26 times its speed; so the speed of the train is 20 meters/second; hence the length of the train is 180 meters

Answer: 180


Solution: (1) 16 Hint: From $\log _{2} a=\frac{a+1}{b}$, we get $a^{b}=2^{a+1}$. Substituting into $4^{a}-3 a^{b}=16$ yields $2^{2 a}-6 \times 2^{a}-16=0$. Solving this, we get: $2^{a}=8, 2^{a}=-2$ (discard), so, $a^{b}=$ $2^{a+1}=16$.

Answer: 16


Anton went to a dairy store. He had no money, but he had empty bottles - six one-liter bottles (worth 20 kopecks each) and six half-liter bottles (worth 15 kopecks each). In the store, there was milk sold by the liter for 22 kopecks. What is the maximum amount of milk he could bring home? He had no other containers besides the empty bottles.

Solution: 11. By returning six half-liter bottles and one liter bottle, Anton will receive 1 ruble 10 kopecks, which will be the cost of 5 liters of milk. The 5 liters of milk he buys can be carried home in the remaining liter bottles. Let's ensure that he won't be able to carry more than 5 liters. If he returns not one liter bottle, but more, then to gather the cost of at least 5 liters of milk, he will need to return at least 5 more half-liter bottles, and the capacity of the remaining bottles will not exceed 4.5 liters (this is verified by enumeration).

Answer: 5


On a chessboard, four rooks are placed in such a way that they attack all the white squares.

a) Provide an example of such an arrangement.

b) Determine the number of such arrangements

Solution:
Solution. There are 32 white squares on a chessboard. If a rook is placed on a white square, it attacks 7 white squares; if it is placed on a black square, it attacks 8 squares. 1) Since there are only 4 rooks and 32 white squares, the rooks must be placed only on black squares. Consider the division of the chessboard into 4 equal vertical and 4 horizontal strips (each strip is the union of two adjacent lines).

2) If there is no rook on any strip (for definiteness, on a horizontal strip), then the 8 white squares of this strip must be attacked by 8 rooks vertically, which contradicts the problem's condition.
3) Therefore, there must be exactly 1 rook on each of the 4 horizontal and 4 vertical strips.

The horizontal and vertical strips intersect to form 16 cells of size $2 \times 2$ (2 black and 2 white squares). To determine the required placement of the rooks, it is necessary to select one square on each horizontal strip so that there is only one square on each vertical strip (see point 1 above). The total number of ways to choose such squares is $4 \times 3 \times 2 \times 1$. Considering the statements in points 1) and 2), within each cell, a rook can be placed in 2 ways, resulting in the number of placements being $2 \times 4$ !.

Answer. $2 \times 4$ !.

Recommendations for checking. 2 points for a correct example of rook placement;

2 points for noting that the rooks must be placed only on black squares;

4 points for proving that there must be exactly 1 rook on each of the 4 horizontal and 4 vertical strips.

3. Find all non-negative polynomials $g$ that satisfy the identity $g(x+y)=g(x)+g(y)+2 \sqrt{g(x) g(y)}$ for all non-negative $x$ and $y$.

Valeev N.F. Solution. If such a polynomial exists, then from the identity $g(x+y)=g(x)+g(y)+2 \sqrt{g(x) g(y)}$ for $y=x \geq 0$, we get $g(x+x)=g(x)+g(x)+2 \sqrt{g(x) g(x)}$ or $g(2 x)=4 g(x)$. We will look for polynomials $g$ in the form $g(x)=\sum_{j=0}^{n} a_{j} x^{j}$. Then from the identity $g(2 x)=4 \cdot g(x)$, it follows that
$g(2 x)-4 g(x) \equiv \sum_{j=0}^{n} a_{j}\left(2^{j}-4\right) x^{j} \equiv 0$ for all $x \geq 0$. This is possible only if $a_{j}\left(2^{j}-4\right)=0$. From this, we get that $a_{j}=0$ for all $j \neq 2$, and for $j=2$, $a_{j}$ can be any number. Thus, if the polynomial $g(x)$ exists, it must have the form $g(x)=a \cdot x^{2}$, where $a$ is any non-negative number. Let's check the identity $g(x+y)=g(x)+g(y)+2 \sqrt{g(x) g(y)}$ for $g(x)=a \cdot x^{2}$: for all non-negative $x$ and $y$, we have $a(x+y)^{2}=a x^{2}+a y^{2}+2 a \cdot x \cdot y=a x^{2}+a y^{2}+2|a| \cdot|x| y|=$.

$a x^{2}+a y^{2}+2 \sqrt{a x^{2} \cdot a y^{2}}=g(x)+g(y)+2 \sqrt{g(x) g(y)}$. Thus, $g(x)=a \cdot x^{2}$.

Answer. $g(x)=a \cdot x^{2}$, where $a \geq 0$.

Recommendations for checking. 2 points for only proving that the polynomial $g(x)=a \cdot x^{2}$ satisfies the identity.

2 points for obtaining an estimate for the degree of the polynomial $g$.

Answer: 2\times4!


The root of the equation $\frac{x}{2 \times 3}+\frac{x}{3 \times 4}+\frac{x}{4 \times 5}+\ldots . .+\frac{x}{2018 \times 2019}=2017$ is $x=$

Answer: 4038


Consider a rectangular grid of $10 \times 10$ unit squares. We call a ship a figure made up of unit squares connected by common edges. We call a fleet a set of ships where no two ships contain squares that share a common vertex (i.e., all ships are vertex-disjoint). Find the greatest natural number that, for each its representation as a sum of positive integers, there exists a fleet such that the summands are exactly the numbers of squares contained in individual ships.

Solution:
1. **Understanding the Problem:**
   We need to find the greatest natural number \( n \) such that for any partition of \( n \) into positive integers, there exists a fleet of ships on a \( 10 \times 10 \) grid where the summands are exactly the numbers of squares contained in individual ships. The ships must be vertex-disjoint.

2. **Initial Observation:**
   The grid is \( 10 \times 10 \), which contains \( 100 \) unit squares. We need to ensure that the ships are vertex-disjoint, meaning no two ships can share a common vertex.

3. **Partitioning the Grid:**
   Consider partitioning the \( 10 \times 10 \) grid into \( 2 \times 2 \) squares. Each \( 2 \times 2 \) square contains 4 unit squares. There are \( \frac{100}{4} = 25 \) such \( 2 \times 2 \) squares in the grid.

4. **Maximum Number of Ships:**
   If we place one ship in each \( 2 \times 2 \) square, we can have 25 ships, each occupying 4 unit squares. This configuration ensures that no two ships share a common vertex.

5. **Verification for \( n = 25 \):**
   For \( n = 25 \), we can represent \( 25 \) as a sum of 25 ones: \( 1 + 1 + \cdots + 1 = 25 \). This means we need 25 ships, each containing 1 unit square. However, this is not possible because placing 25 ships, each containing 1 unit square, would require 25 unit squares, and ensuring they are vertex-disjoint is not feasible in a \( 10 \times 10 \) grid.

6. **Verification for \( n = 24 \):**
   For \( n = 24 \), we can represent \( 24 \) as a sum of 24 ones: \( 1 + 1 + \cdots + 1 = 24 \). This means we need 24 ships, each containing 1 unit square. This is also not feasible for the same reason as above.

7. **Verification for \( n = 25 \) with Different Partitions:**
   For \( n = 25 \), consider different partitions such as \( 4 + 4 + \cdots + 4 + 1 = 25 \). We can place six \( 2 \times 2 \) ships and one ship containing 1 unit square. This configuration is feasible and ensures that the ships are vertex-disjoint.

8. **Generalization:**
   For any \( 2k \times 2m \) grid, the maximum number of vertex-disjoint ships is \( km \). For a \( 10 \times 10 \) grid, \( k = 5 \) and \( m = 5 \), so the maximum number is \( 5 \times 5 = 25 \).

Conclusion:
The greatest natural number \( n \) such that for any partition of \( n \) into positive integers, there exists a fleet of ships on a \( 10 \times 10 \) grid where the summands are exactly the numbers of squares contained in individual ships, is \( 25 \).

The final answer is \( \boxed{ 25 } \).

Answer: 25


Two circles with radii 3 and 4, the distance between their centers being 5, intersect at points $A$ and $B$. A line through point $B$ intersects the circles at points $C$ and $D$, such that $C D=8$ and point $B$ lies between points $C$ and $D$. Find the area of triangle $A C D$.

Solution:
Let $O_{1}$ and $O_{2}$ be the centers of the circles. Prove that triangle $A C D$ is similar to triangle $B O_{1} O_{2}$.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the smaller and larger circles, respectively, and let point $C$ be located on the smaller circle. Then

$$
\angle D C A=\angle B C A=\frac{1}{2} \cup A B=\angle B O_{1} O_{2}
$$

Similarly, $\angle C D A=\angle B O_{2} O_{1}$. Therefore, triangle $A C D$ is similar to triangle $B O_{1} O_{2}$ with a similarity ratio of $\frac{C D}{\mathrm{O}_{1} \mathrm{O}_{2}}=\frac{8}{5}$.

Triangle $\mathrm{BO}_{1} \mathrm{O}_{2}$ is a right triangle, since

$$
O_{1} O^{2}{ }_{2}=5^{2}=3^{2}+4^{2}=O_{1} B^{2}+O_{2} B^{2} .
$$

Therefore,

$$
S_{\triangle \mathrm{BCD}}=\frac{64}{25} S_{\Delta \mathrm{BO}_{1} \mathrm{O}_{2}}=\frac{384}{25}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-16.jpg?height=421&width=706&top_left_y=1&top_left_x=676)

## Answer

$\frac{384}{25}$.

Answer: \frac{384}{25}


The roots of a monic cubic polynomial $p$ are positive real numbers forming a geometric sequence. Suppose that the sum of the roots is equal to $10$. Under these conditions, the largest possible value of $|p(-1)|$ can be written as $\frac{m}{n}$, where $m$, $n$ are relatively prime integers. Find $m + n$.

Solution:
1. **Identify the roots and their properties:**
   Let the roots of the monic cubic polynomial \( p(x) \) be \( a \), \( ar \), and \( ar^2 \), where \( a \) is the first term and \( r \) is the common ratio of the geometric sequence. Since the roots are positive real numbers and form a geometric sequence, we have:
   \[
   a + ar + ar^2 = 10
   \]
   Factor out \( a \):
   \[
   a(1 + r + r^2) = 10
   \]

2. **Express the polynomial:**
   The polynomial \( p(x) \) with roots \( a \), \( ar \), and \( ar^2 \) can be written as:
   \[
   p(x) = (x - a)(x - ar)(x - ar^2)
   \]

3. **Evaluate \( |p(-1)| \):**
   To find \( |p(-1)| \), substitute \( x = -1 \) into the polynomial:
   \[
   p(-1) = (-1 - a)(-1 - ar)(-1 - ar^2)
   \]
   Simplify the expression:
   \[
   p(-1) = (-(1 + a))(-(1 + ar))(-(1 + ar^2)) = -(1 + a)(1 + ar)(1 + ar^2)
   \]
   Since we are interested in the absolute value:
   \[
   |p(-1)| = (1 + a)(1 + ar)(1 + ar^2)
   \]

4. **Apply the AM-GM inequality:**
   To maximize \( |p(-1)| \), we use the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
   \[
   (1 + a) + (1 + ar) + (1 + ar^2) \geq 3 \sqrt[3]{(1 + a)(1 + ar)(1 + ar^2)}
   \]
   Simplify the left-hand side:
   \[
   3 + a(1 + r + r^2) = 3 + 10 = 13
   \]
   Therefore:
   \[
   13 \geq 3 \sqrt[3]{(1 + a)(1 + ar)(1 + ar^2)}
   \]
   Divide both sides by 3:
   \[
   \frac{13}{3} \geq \sqrt[3]{(1 + a)(1 + ar)(1 + ar^2)}
   \]
   Cube both sides:
   \[
   \left(\frac{13}{3}\right)^3 \geq (1 + a)(1 + ar)(1 + ar^2)
   \]
   Calculate the cube:
   \[
   \left(\frac{13}{3}\right)^3 = \frac{2197}{27}
   \]

5. **Conclusion:**
   The maximum value of \( |p(-1)| \) is \( \frac{2197}{27} \). Therefore, \( m = 2197 \) and \( n = 27 \), and \( m + n = 2224 \).

The final answer is \( \boxed{2224} \)

Answer: 2224


Given the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$ satisfies $f(f(n))=2016 n-215 f(n)$.
Find $f(n)$.

Solution:
Let $f^{(n)}(x)=a_{n}$. Then $a_{n}>0$, and $a_{n+2}=2016 a_{n}-215 a_{n+1}$. Hence $a_{n}-9 a_{n-1}=-224\left(a_{n-1}-9 a_{n-2}\right)$.
Let $b_{n}=a_{n}-9 a_{n-1}$. Then $b_{n}=-224 b_{n-1}=(-224)^{n-1} b_{1}$.
Let $m=-224$, we get
$a_{n}-9 a_{n-1}=m^{n-1} b_{1}$
$\Rightarrow m \cdot \frac{a_{n}}{m^{n}}-9 \cdot \frac{a_{n-1}}{m^{n-1}}=b_{1}$.
Let $c_{n}=\frac{a_{n}}{m^{n}}$, we get $m c_{n}-9 c_{n-1}=b_{1}$.
Solving for the general term, we get
$$
\begin{array}{l}
c_{n}=\left(\frac{9}{m}\right)^{n-1} c_{1}+\frac{b_{1}}{m} \cdot \frac{\left(\frac{9}{m}\right)^{n-1}-1}{\frac{9}{m}-1} \\
\Rightarrow a_{n}=\frac{9^{n}-m^{n}}{9-m} a_{1}-\frac{9^{n} m-9 m^{n}}{9-m} a_{0}>0 .
\end{array}
$$

When $n$ is even,
$$
a_{1}\frac{9^{n} m-9 m^{n}}{9^{n}-m^{n}} a_{0},
$$
as $n \rightarrow+\infty$, $a_{1} \geqslant 9 a_{0}$.
In summary, $a_{1}=9 a_{0}$, i.e., $f(n)=9 n$.

Answer: f(n)=9n


Given $a, b, c \in \mathbf{R}$, and satisfy $a>b>c$, $a+b+c=0$. Then, the range of $\frac{c}{a}$ is $\qquad$

Solution:
2. $\left(-2,-\frac{1}{2}\right)$.

From $a>b>c$ and $a+b+c=0$, we know $a>0, c-2 ; \\
0=a+b+c>a+2 c \Rightarrow-\frac{a}{2}>c \Rightarrow \frac{c}{a}<-\frac{1}{2} .
\end{array}
$

Answer: \left(-2,-\frac{1}{2}\right)


There are 8 English letters $K, Z, A, I, G, A, K$, and $\mathrm{U}$, each written on 8 cards. Ask:
(1) How many ways are there to arrange these cards in a row?
(2) How many ways are there to arrange 7 of these cards in a row?

Solution:
(1) $\frac{8!}{2!\times 2!}=10080$ ways.
(2) If the letter taken away is $K$ or $A$, then there are
$$
2 \times \frac{7!}{2!}=5040 \text { (ways); }
$$

If the letter taken away is $Z$, $I$, $G$, or $U$, then there are
$$
4 \times \frac{7!}{2!\times 2!}=5040 \text { (ways). }
$$

Therefore, there are $5040+5040=10080$ arrangements in total.

Answer: 10080


The smallest natural number $n$ is sought for which the following holds

$$
n \equiv 1(\bmod 2) ; \quad n \equiv 1(\bmod 3) ; \quad n \equiv 1(\bmod 5) ; \quad n \equiv 0(\bmod 7) ; \quad n \equiv 1(\bmod 11)
$$

Solution:
Due to $n \equiv 0(\bmod 7)$, $n$ can be represented in the form $n=7k, k \in N$. Because

$$
\begin{array}{r}
n \equiv 1(\bmod 2) \text { and } 7 \equiv 1(\bmod 2) \text {, } k=1(\bmod 2) \\
n \equiv 1(\bmod 3) \text { and } 7 \equiv 1(\bmod 3) \text {, } k=1(\bmod 3) \\
n \equiv 1(\bmod 5) \text { and } 7 \equiv 2(\bmod 5) \text {, } k=3(\bmod 5) \\
n \equiv 1(\bmod 11) \text { and } 7 \equiv 7(\bmod 11) \text {, } k=8(\bmod 11)
\end{array}
$$

We assume that the smallest number $k$ is at most three digits:

$$
k=100a_{2}+10a_{1}+a_{0}
$$

with $0 \leq a_{0} ; a_{1} ; a_{2} \leq 9, a_{1} ; a_{2} ; a_{3} \in N$. Then, due to (3), $a_{0}=3$ or $a_{0}=8$, and due to (1), $a_{1} \neq 8$, so it follows that $a_{1}=3$. According to (4), we then have

$$
a_{2}-a_{1}+3=11s+8 \quad ; \quad a_{2}-a_{1}=11s+5
$$

$(s \in G)$. From $0 \leq a_{1} ; a_{2} \leq 9$ it follows that $-9 \leq a_{2}-a_{1}=11s+5 \leq 9$. These inequalities are only satisfied for $s=0$ and for $s=-1$. Thus, from (5) we get $a_{2}=a_{1}+5 \geq 5$ or $a_{2}=a_{1}-6 \leq 3$.

Since we are looking for the smallest number $k$, we first try the smaller values for $a_{2}$, i.e., $a_{2}=0 ; 1 ; 2 ; 3$. This results in $a_{1}=6 ; 7 ; 8 ; 9$ and $k=63 ; 173 ; 283 ; 393$. Of these $k$-values, only $k=283$ satisfies condition (2). From this, we get $n=7 \cdot 283=1981$.

### 2.21 Problems and Solutions 1981

Answer: 1981


As shown in Figure 7, the side length of square $ABCD$ is 1, and points $E, F, G$ are on sides $AB, AD, BC$ respectively (they can coincide with the vertices). If $\triangle EFG$ is an equilateral triangle, find the maximum and minimum values of the area of $\triangle EFG$.

Solution:
Solve as shown in Figure 7, construct $E K \perp F G$. Then $K$ is the midpoint of side $F G$. Connect $A K$ and $B K$.
$$
\begin{array}{l}
\text { By } \angle E K G=\angle E B G=\angle E K F \\
=\angle E A F=90^{\circ},
\end{array}
$$

we get that $E, K, G, B$ and $E, K, F, A$ are respectively concyclic.
Thus, $\angle K B E=\angle E G K=60^{\circ}$,
$$
\angle E A K=\angle E F K=60^{\circ} \text {. }
$$

Therefore, $\triangle A B K$ is an equilateral triangle, i.e., $K$ is a fixed point.
Thus, when $F G / / A B$, the area of $\triangle E F G$ is minimized, at this time,
$$
S_{\triangle E F G}=\frac{\sqrt{3}}{4} F G^{2}=\frac{\sqrt{3}}{4} A B^{2}=\frac{\sqrt{3}}{4} .
$$

When point $F$ coincides with $D$ or point $G$ coincides with $C$, the area of $\triangle E F G$ is maximized, at this time,
$$
\begin{array}{l}
S_{\triangle E F G}=\frac{\sqrt{3}}{4} C K^{2} \\
=\sqrt{3}\left[\left(\frac{1}{2}\right)^{2}+\left(1-\frac{\sqrt{3}}{2}\right)^{2}\right]=2 \sqrt{3}-3 .
\end{array}
$$
3 Constructing the inscribed triangle

Answer: \frac{\sqrt{3}}{4} \text{ and } 2\sqrt{3} - 3


Let $a_{1}, a_{2}, a_{3}, \cdots$ be a sequence of real numbers in a geometric progression. Let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$ for all integers $n \geq 1$. Assume that $S_{10}=10$ and $S_{30}=70$. Find the value of $S_{40}$.

Solution:
14. Answer: 150
Solution. Let $r$ be the common ratio of this geometric sequence. Thus
$$
S_{n}=a_{1}\left(1+r+r^{2}+\cdots+r^{n-1}\right) \text {. }
$$

Thus
$$
10=a_{1}\left(1+r+\cdots+r^{9}\right)
$$
and
$$
70=a_{1}\left(1+r+\cdots+r^{29}\right) .
$$
$\mathrm{As}$
$$
1+r+\cdots+r^{29}=\left(1+r+\cdots+r^{9}\right)\left(1+r^{10}+r^{20}\right),
$$
we have
$$
1+r^{10}+r^{20}=7 .
$$

So $r^{10}$ is either 2 or -3 . As $r^{10}>0, r^{10}=2$.
Hence
$$
\begin{array}{c}
S_{40}=a_{1}\left(1+r+\cdots+r^{39}\right)=a_{1}\left(1+r+\cdots+r^{9}\right)\left(1+r^{10}+r^{20}+r^{30}\right) \\
=10 \times\left(1+2+2^{2}+2^{3}\right)=150 .
\end{array}
$$

Answer: 150


Find the largest real number $k$ such that for any triangle with sides $a$, $b$, and $c$, the following inequality holds:
$$
\frac{b c}{b+c-a}+\frac{a c}{a+c-b}+\frac{a b}{a+b-c} \geqslant k(a+b+c) .
$$

Solution:
11. For an equilateral triangle, i.e., $a=b=c$, we have
$$
3 a \geqslant k \cdot 3 a \Rightarrow k \leqslant 1 \text {. }
$$

For any triangle, let
$$
x=\frac{b+c-a}{2}, y=\frac{c+a-b}{2}, z=\frac{a+b-c}{2} \text {. }
$$
$$
\begin{array}{l}
\text { Then } \frac{b c}{b+c-a}+\frac{c a}{c+a-b}+\frac{a b}{a+b-c} \\
=\frac{(x+y)(x+z)}{2 x}+\frac{(y+z)(y+x)}{2 y}+\frac{(z+x)(z+y)}{2 z} \\
=\frac{1}{2}\left(3 x+3 y+3 z+\frac{y z}{x}+\frac{z x}{y}+\frac{x y}{z}\right) .
\end{array}
$$

Notice that,
$$
\begin{aligned}
\frac{y z}{x} & +\frac{z x}{y}+\frac{x y}{z} \\
& =\frac{1}{2}\left(\left(\frac{y z}{x}+\frac{z x}{y}\right)+\left(\frac{z x}{y}+\frac{x y}{z}\right)+\left(\frac{x y}{z}+\frac{y z}{x}\right)\right) \\
& \geqslant z+x+y .
\end{aligned}
$$

Thus, $\frac{b c}{b+c-a}+\frac{c a}{c+a-b}+\frac{a b}{a+b-c}$
$$
\geqslant 2 x+2 y+2 z=a+b+c,
$$

which means $k=1$ satisfies the condition.
In conclusion, the maximum value of $k$ is 1.

Answer: 1


Let set $A=\{1,2,3, \cdots, 1997\}$, for any 999-element subset $X$ of $A$, if there exist $x, y \in X$, such that $x<y$ and $x \mid y$, then $X$ is called a good set. Find the largest natural number $a(a \in A)$, such that any 999-element subset containing $a$ is a good set.

Solution:
4. The required $\max a=665$. First, we prove: $a \leqslant 665$. Clearly, the 999-element subset $X_{0}=\{999,1000,1001, \cdots, 1997 \mid$ does not contain $x, y \in X_{0}$ such that $x \cdot 1997$, which is greater than the largest element in $X_{0}$, holds. Thus, $a$ cannot be any of the numbers $999,1000,1001, \cdots, 1997$.
Next, we prove that $a \neq 666,667,668, \cdots, 997,998$. Construct the sets:
$B_{i}=\{666+i\} \cup X_{0} \backslash\{2(666+i)\}, i=0,1, \cdots, 332$.
For $B_{i}$, $(666+i) \cdot 3 \geqslant 1998$, and $(666+i) \cdot 2 \notin B_{i}$, so no other multiples of $666+i$ are in $B_{i}$. It has already been proven that no non-multiples of any element in $X_{0}$ are in $X_{0}$, and $666+i<999(i=0,1,2, \cdots, 332)$, so no multiples of any element in $X_{0}$ can be $666+i(i=0,1, \cdots, 332)$. Thus, the multiples (excluding itself) of any other elements in $B_{i}$ are not in $B_{i}$. Therefore, there do not exist two elements in $B_{i}$ such that $x<y, x \mid y$, and $B_{i}$ $(i=0,1, \cdots, 332)$ includes 666, $667, \cdots, 998$. Hence, $a \neq 666,667, \cdots, 998$, so $a \leqslant 665$.

Next, we prove that 665 is attainable. Assume there exists a 999-element subset $X$ containing 665, such that there do not exist $x, y \in X, x<y$ with $x \mid y$. Then $665 \cdot 2,665 \cdot 3 \notin X$. Construct the following 997 drawers, which include all elements of $A$ except $665,665 \cdot 2,665 \cdot 3$, and each element appears once: $K_{1}=\left|1,1 \cdot 2,1 \cdot 2^{2}, \cdots, 1 \cdot 2^{10}\right|, K_{2}=\left|3,3 \cdot 2,3 \cdot 2^{2}, \cdots, 3 \cdot 2^{9}\right|, K_{3}=\mid 5,5 \cdot 2$, $\left.5 \cdot 2^{2}, \cdots, 5 \cdot 2^{8}\right\}, K_{4}=\left\{7,7 \cdot 2,7 \cdot 2^{2}, \cdots, 7 \cdot 2^{8} \mid, \cdots, K_{332}=\{663,663 \cdot 2,663 \cdot 3\}, K_{334}=\{667,667 \cdot\right.$ $2 \mid, K_{335}=\{669,669 \cdot 2\}, \cdots, K_{995}=\{1991\}, K_{996}=\{1993\}, K_{997}=\{1997\}$. Thus, the other 998 elements of $X$ (excluding 665) are placed into these 997 drawers, so there must be two elements in the same drawer, and the elements in the same drawer are multiples of each other, leading to a contradiction. Proof complete.

Answer: 665


Given the equation of circle $\odot C$ is
$$
x^{2}+y^{2}-4 x-2 y+1=0 \text {, }
$$

the line $y=\left(\tan 10^{\circ}\right) x+\sqrt{2}$ intersects $\odot C$ at points
$A$ and $B$. The sum of the inclination angles of lines $A C$ and $B C$ is
$\qquad$

Solution:
7. $200^{\circ}$.
The equation of $\odot C$ is $(x-2)^{2}+(y-1)^{2}=4$, the distance from the center $(2,1)$ to the line $y=\left(\tan 10^{\circ}\right) x+\sqrt{2}$ is
$$
d=\frac{\left|2 \tan 10^{\circ}-1+\sqrt{2}\right|}{\sqrt{1+\tan ^{2} 10^{\circ}}}<2,
$$

which means the given line intersects $\odot C$ at points $A$ and $B$ (as shown in Figure 2).
Let the inclination angle of line $B C$ be $\alpha$, and the inclination angle of line $A C$ be $\beta$. Then
$$
\begin{array}{l}
\alpha=10^{\circ}+\angle C B A, \\
\angle C A B=10^{\circ}+\left(180^{\circ}-\beta\right) .
\end{array}
$$

Since $\angle C B A=\angle C A B$, we have
$$
\begin{array}{l}
\alpha=10^{\circ}+10^{\circ}+\left(180^{\circ}-\beta\right) \\
\Rightarrow \alpha+\beta=200^{\circ} .
\end{array}
$$

Answer: 200^{\circ}


What is the minimum width that an infinite strip of paper must have so that any triangle with an area of 1 can be cut out of it?

Solution:
11.9. Since the area of an equilateral triangle with side $a$ is $a^{2} \sqrt{3} / 4$, the side of an equilateral triangle with area 1 is $2 / \sqrt[4]{3}$, and its height is $\sqrt[4]{3}$. We will prove that an equilateral triangle with area 1 cannot be cut from a strip of width less than $\sqrt[4]{3}$. Let the equilateral triangle $A B C$ lie inside a strip of width less than $\sqrt[4]{3}$. Let the projection of vertex $B$ onto the boundary of the strip lie between the projections of vertices $A$ and $C$. Then the line drawn through point $B$ perpendicular to the boundary of the strip intersects segment $A C$ at some point $M$. The height of triangle $A B C$ does not exceed $B M$, and $B M$ is no greater than the width of the strip, so the height of triangle $A B C$ is less than $\sqrt[4]{3}$, i.e., its area is less than 1.

It remains to prove that any triangle with area 1 can be cut from a strip of width $\sqrt[4]{3}$. We will prove that any triangle with area 1 has a height not exceeding $\sqrt[4]{3}$. For this, it is sufficient to prove that it has a side not less than $2 / \sqrt[4]{3}$. Suppose all sides of triangle $A B C$ are less than $2 / \sqrt[4]{3}$. Let $\alpha$ be the smallest angle of this triangle. Then $\alpha \leqslant 60^{\circ}$ and $S_{A B C}=(A B \cdot A C \sin \alpha) / 2<(2 / \sqrt[4]{3})^{2}(\sqrt{3} / 4)=1$. A contradiction is obtained. A triangle that has a height not exceeding $\sqrt[4]{3}$ can be placed in a strip of width $\sqrt[4]{3}$ by placing the side to which this height is dropped on the side of the strip.

Answer: \sqrt[4]{3}


If $\log_2(\log_2(\log_2(x)))=2$, then how many digits are in the base-ten representation for x?
$\text{(A) } 5\quad \text{(B) } 7\quad \text{(C) } 9\quad \text{(D) } 11\quad \text{(E) } 13$

Solution:
Taking successive exponentials $\log_2(\log_2(x)) = 2^2 = 4$ and $\log_2(x) = 2^4=16$ and $x = 2^{16}$.  Now $2^{10} = 1024 \approx 10^3$ and $2^6 = 64$ so we can approximate $2^{16} \approx 64000$ which has 5 digits.  In general, $2^n$ has approximately $n/3$ digits.
$\fbox{A}$

Answer: A


For example, $AC$ and $CE$ are two diagonals of the regular hexagon $ABCDEF$, and points $M$ and $N$ internally divide $AC$ and $CE$ respectively, such that $AM: AC = CN: CE = r$. If points $B$, $M$, and $N$ are collinear, find $r$.

Solution:
Proof: As shown in Figure 8, let $AC = 2$, then $A(-1, 0), C(1, 0), B(0, -\frac{\sqrt{3}}{3}), E(0, \sqrt{3})$
$$
\begin{array}{l}
\because AM: AC \\
= CN: CE = r,
\end{array}
$$
$\therefore$ By the section formula, we get
$$
M(2r-1, 0), N(1-r, \sqrt{3}r).
$$

Also, $\because B, M, N$ are collinear,
thus $k_{BM} = k_{MN}$,

i.e., $\square$
$$
\begin{array}{l}
\frac{\frac{\sqrt{3}}{3}}{2r-1} = \frac{\sqrt{3}r}{1-r-(2r-1)}. \\
\therefore 1-3r^2 = 0, r = \frac{\sqrt{3}}{3}.
\end{array}
$$

Note: $r$ can also be found using $|BN| = |BM| + |MN|$.

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


If $x \in \mathbf{R}$, find the maximum value of $F(x)=\min \{2 x+1, x+2,-x+6\}$.

Solution:
Solution: As shown in Figure 1, draw the graph of $F(x)$ (the solid part). From the graph, we can see that the maximum value of $F(x)$ is equal to the y-coordinate of the intersection point of $y=x+2$ and $y=-x+6$.
Therefore, the maximum value of $F(x)$ is 4.

Answer: 4


The equation that best represents "a number increased by five equals 15" is
(A) $n-5=15$
(D) $n+15=5$
(B) $n \div 5=15$
(E) $n \times 5=15$
(C) $n+5=15$

Solution:
If the number is $n$, then "a number increased by five" is best represented by the expression $n+5$. The equation that best represents "a number increased by five equals 15 " is $n+5=15$.

ANSWER: $(\mathrm{C})$

Answer: C


The base $4$ repeating decimal $0.\overline{12}_4$ can be expressed in the form $\frac{a}{b}$ in base 10, where $a$ and $b$ are relatively prime positive integers. Compute the sum of $a$ and $b$.

Solution:
1. First, let's express the repeating decimal \(0.\overline{12}_4\) as a geometric series. In base 4, the repeating decimal \(0.\overline{12}_4\) can be written as:
   \[
   0.\overline{12}_4 = 0.121212\ldots_4
   \]

2. We can express this as an infinite series:
   \[
   0.\overline{12}_4 = \frac{1}{4} + \frac{2}{4^2} + \frac{1}{4^3} + \frac{2}{4^4} + \frac{1}{4^5} + \frac{2}{4^6} + \cdots
   \]

3. Notice that this series can be split into two separate geometric series:
   \[
   S = \left(\frac{1}{4} + \frac{1}{4^3} + \frac{1}{4^5} + \cdots\right) + \left(\frac{2}{4^2} + \frac{2}{4^4} + \frac{2}{4^6} + \cdots\right)
   \]

4. Each of these series is a geometric series. The first series has the first term \(a_1 = \frac{1}{4}\) and common ratio \(r = \frac{1}{4^2} = \frac{1}{16}\):
   \[
   \sum_{n=0}^{\infty} \left(\frac{1}{4}\right) \left(\frac{1}{16}\right)^n = \frac{\frac{1}{4}}{1 - \frac{1}{16}} = \frac{\frac{1}{4}}{\frac{15}{16}} = \frac{1}{4} \cdot \frac{16}{15} = \frac{4}{15}
   \]

5. The second series has the first term \(a_2 = \frac{2}{4^2} = \frac{2}{16} = \frac{1}{8}\) and common ratio \(r = \frac{1}{16}\):
   \[
   \sum_{n=0}^{\infty} \left(\frac{1}{8}\right) \left(\frac{1}{16}\right)^n = \frac{\frac{1}{8}}{1 - \frac{1}{16}} = \frac{\frac{1}{8}}{\frac{15}{16}} = \frac{1}{8} \cdot \frac{16}{15} = \frac{2}{15}
   \]

6. Adding these two sums together, we get:
   \[
   S = \frac{4}{15} + \frac{2}{15} = \frac{6}{15}
   \]

7. Simplify the fraction \(\frac{6}{15}\):
   \[
   \frac{6}{15} = \frac{2}{5}
   \]

8. The fraction \(\frac{2}{5}\) is already in its simplest form, where \(a = 2\) and \(b = 5\). The sum of \(a\) and \(b\) is:
   \[
   a + b = 2 + 5 = 7
   \]

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

Answer: 7


Solution:
## Solution.

$$
\begin{aligned}
& ((21.85: 43.7+8.5: 3.4): 4.5): 1 \frac{2}{5}+1 \frac{11}{21}=\left((0.5+2.5): 4 \frac{1}{2}\right): \frac{7}{5}+\frac{32}{21}= \\
& =\left(3 \cdot \frac{2}{9}\right) \cdot \frac{5}{7}+\frac{32}{21}=\frac{10}{21}+\frac{32}{21}=\frac{42}{21}=2
\end{aligned}
$$

Answer: 2.

Answer: 2


Given any positive integer $a$, define the integer sequence $x_{1}, x_{2}$, $\cdots$, satisfying
$$
x_{1}=a, x_{n}=2 x_{n-1}+1(n \geqslant 1) .
$$

If $y_{n}=2^{x_{n}}-1$, determine the maximum integer $k$ such that there exists a positive integer $a$ for which $y_{1}, y_{2}, \cdots, y_{k}$ are all prime numbers.

Solution:
1. If $y_{i}$ is a prime number, then $x_{i}$ is also a prime number.

Otherwise, if $x_{i}=1$, then $y_{i}=1$ is not a prime number; if $x_{i}=m n($ integers $m, n>1)$, then
$\left(2^{m}-1\right) \mid\left(2^{x_{i}}-1\right)$, i.e., $x_{i}$ and $y_{i}$ are composite numbers.

Below, we prove by contradiction: For any odd prime $a$, at least one of $y_{1}, y_{2}, y_{3}$ is composite.
Otherwise, $x_{1}, x_{2}, x_{3}$ are all primes.
Since $x_{1} \geqslant 3$ is odd, we know
$x_{2}>3$, and $x_{2}=3(\bmod 4)$.
Therefore, $x_{3} \equiv 7(\bmod 8)$.
Thus, 2 is a quadratic residue of $x_{3}$, i.e., there exists $x \in \mathbf{N}_{+}$ such that
$$
x^{2}=2\left(\bmod x_{3}\right) \text {. }
$$

Therefore, $2^{x_{2}}=2^{\frac{x_{3}-1}{2}} \equiv x^{x_{3}-1} \equiv 1\left(\bmod x_{3}^{3}\right)$.
Thus, $x_{3} \mid y_{2}$.
Since $x_{2}>3$, then $2^{x_{2}}-1>2 x_{2}+1=x_{3}$.
Therefore, $y_{2}$ is composite.
Finally, if $a=2$, then
$$
y_{1}=3, y_{2}=31 \text {, and } 23 \mid y_{3}=\left(2^{11}-1\right) \text {. }
$$

Therefore, $k=2$.

Answer: 2


In an arbitrary triangular pyramid $A B C D$, a section is made by a plane intersecting the edges $A B, D C$, and $D B$ at points $M, N, P$ respectively. Point $M$ divides edge $A B$ in the ratio $A M: M B=2: 3$. Point $N$ divides edge $D C$ in the ratio $D N: N C=3: 5$. Point $P$ divides edge $D B$ in the ratio $D P: P B=4$. Find the ratio $A Q: Q C$.

Solution:
Answer: $A Q: Q C=1: 10$.

Grading criteria, 11th grade

Preliminary round of the sectoral physics and mathematics school students' Olympiad "Rosatom", mathematics, trip 1

#

Answer: AQ:QC=1:10


Solution: $\begin{array}{l}\text { i.1. D. } \\ \text { Original expression }=|2+(1+x)|=|3+x|=3+x \text {. }\end{array}$

Answer: D


In the Cartesian coordinate system $x-o-y$, the figure defined by the inequalities
$$
y^{100}+\frac{1}{y^{100}} \leqslant x^{100}+\frac{1}{x^{100}}, x^{2}+y^{2} \leqslant 100
$$

has an area equal to ( ).
(A) $50 \pi$
(B) $40 \pi$
(C) $30 \pi$
(D) $20 \pi$

Solution:
6. (A)

The region we are examining is symmetric with respect to the coordinate axes, so we only need to consider the part where $x>0, y>0$.
Let $z=x^{100}, t=y^{100}$, then we have
$$
\begin{array}{l}
z+\frac{1}{z} \geqslant t+\frac{1}{t} \Leftrightarrow\left(z^{2}+1\right) t \geqslant\left(t^{2}+1\right) z \\
\Leftrightarrow z^{2} t-t^{2} z+t-z \geqslant 0 \Leftrightarrow(z-t)(z t-1) \geqslant 0 .
\end{array}
$$

Therefore, the solution set of the inequality has the form of the shaded area in the figure. Since the shaded part within the circle $x^{2}+y^{2}=100$ is equal in area to the unshaded part, the area of the shaded part is $\frac{1}{2} \times 10^{2} \pi=50 \pi$.

Explanation: From $(z-t)(z t -1) \geqslant 0$ in the first quadrant, we have $z \geqslant t$ and $z t \geqslant 1$, i.e., $x \geqslant y$ and $x \geqslant \frac{1}{y}$. Therefore, in the first quadrant, the region can be drawn by $x \geqslant y, x \geqslant \frac{1}{y}$, and $x^{2}+y^{2}=10^{2}$. By symmetry, the graph in the other quadrants can be completed.

Answer: A


A florist received between 300 and 400 roses for a celebration. When he arranged them in vases with 21 roses in each, 13 roses were left. But when arranging them in vases with 15 roses in each, 8 roses were missing. How many roses were there in total?

Solution:
Solution: $\left\{\begin{array}{ll}a=21 x+13=21(x+1)-8, & a+8 \vdots 21, \\ a=15 y-8, & a+8 \vdots 15,\end{array}\right\} \Rightarrow a+8 \vdots 105$. Answer: $a=307$.

$$
a+8=105,210, \underline{315,} 420
$$

## TICKET № 6

Answer: 307


Several identical books and identical albums were bought. The books cost 10 rubles 56 kopecks. How many books were bought if the price of one book is more than one ruble higher than the price of an album, and 6 more books were bought than albums?

Solution: 5. Since each book is more expensive than a ruble, no more than 10 books were bought. Moreover, it is clear that no fewer than 7 books were bought (since at least one album was bought). The number 1056 is divisible by 8 and not divisible by $7,9,10$. Therefore, 8 books were bought. (MSU, Mechanics and Mathematics, 1968)

Answer: 8


Find the odd prime $p$ that satisfies the following condition: there exists a permutation $b_{1}, b_{2}, \cdots, b_{p-1}$ of $1,2, \cdots, p-1$, such that $1^{b_{1}}, 2^{b_{2}}, \cdots,(p-1)^{b_{p-1}}$ forms a reduced residue system modulo $p$.

Solution:
Let $p$ be an odd prime satisfying the condition, and take $g$ as a primitive root of $p$. Since $g, g^{2}, \cdots, g^{p-1}$ form a reduced residue system modulo $p$, for each $i=1,2, \cdots, p-1$, $i$ can be represented as $g^{a_{i}}$ modulo $p$, where $a_{1}, a_{2}, \cdots, a_{p-1} \in \mathbf{Z}_{+}$, and by property 5 (2), $a_{1}, a_{2}, \cdots, a_{p-1}$ form a complete residue system modulo $p-1$.

Therefore, the problem is transformed into: If $a_{1}, a_{2}, \cdots, a_{p-1}$ form a complete residue system modulo $p-1$, can we find another complete residue system $b_{1}, b_{2}, \cdots, b_{p-1}$ modulo $p-1$, such that $g^{a_{1} b_{1}}, g^{a_{2} b_{2}}, \cdots, g^{a_{p-1} b_{p-1}}$ form a reduced residue system modulo $p$? In other words, do $a_{i} b_{i}(1 \leqslant i \leqslant p-1)$ form a complete residue system modulo $p-1$?

By Example 7, we know that the conclusion is affirmative if and only if $p-1 \leqslant 2$, i.e., $p=3$.

In this problem, the corresponding example is $b_{1}=2, b_{2}=1$. Therefore, $p=3$ is the only odd prime that satisfies the condition.

Answer: 3


$n$ is a non-negative integer, and $3n+1, 5n+1$ are both perfect squares. Then $7n+3$ is:
( A ) must be a prime number
(B) must be a composite number
(C) must be a perfect square
(D) none of the above

Solution:
-、1.(D).
Let $n=0, 3n+1=5n+1=1^{2}, 7n+3=3$ be prime numbers. Take $n=16, 3n+1=7^{2}, 5n+1=9^{2}, 7n+3=115$ as composite numbers.
However, 3 and 115 are not perfect squares.

Answer: D


On a sausage, thin rings are drawn across. If you cut along the red rings, you get 5 pieces; if along the yellow ones, you get 7 pieces; and if along the green ones, you get 11 pieces. How many pieces of sausage will you get if you cut along the rings of all three colors?

$[3$ points] (A. V. Shipovalov)

Solution:
Answer: 21 pieces.

Solution: Note that the number of pieces is always one more than the number of cuts. Therefore, there are 4 red rings, 6 yellow ones, and 10 green ones. Thus, the total number of cuts is $4+6+10=20$, and the number of pieces is 21.

Answer: 21


Determine the number of natural divisors of the number $11!=1 \cdot 2 \cdot \ldots \cdot 10 \cdot 11$ that are multiples of three. Answer. 432.

Solution: Solution. The prime factorization of the given number is $11!=2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11$. All multiples of three that are divisors of this number have the form $2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \cdot 7^{\delta} \cdot 11^{\varphi}$, where $\alpha \in[0 ; 8], \beta \in[1 ; 4], \gamma \in[0 ; 2]$, $\delta \in[0 ; 1], \varphi \in[0 ; 1]$. The total number of such divisors is $(8+1) \cdot 4 \cdot(2+1)(1+1)(1+1)=432$.

Answer: 432


Solution: 6. When \(x=7\), the given expression equals \(3^{2^{1}}=9\).

Answer: 9


For every positive integer $x$, obtain the number $z$ according to the following process:
(i) Move the last digit of $x$ to the first position to get $y$;
(ii) Then take the square root of $y$ to get its arithmetic square root $z$.
For example, if $x=9, y=9, z=\sqrt{9}=3$;

and if $x=5002, y=2500$,
$$
z=\sqrt{2500}=50 \text {. }
$$

Find all positive integers $x$ such that $z=\frac{1}{2} x$.

Solution:
(1) If $x$ is a one-digit number, then $y=x, z=\sqrt{y}=\sqrt{x}$.

Also, $z=\frac{1}{2} x$, so $\sqrt{x}=\frac{1}{2} x$.
Solving this, we get $x_{1}=4, x_{2}=0$ (discard).
Thus, $x=4$.
(2) If $x$ is a two-digit number, let $x=10 a+b, a, b$ be integers, and $0<a \leqslant 9,0 \leqslant b \leqslant 9$. Then
$$
y=10 b+a, z=\sqrt{10 b+a} .
$$

Also, $z=\frac{1}{2} x$, so,
$$
\sqrt{10 b+a}=\frac{1}{2}(10 a+b) \text {. }
$$

Since $a, b$ are integers, $\frac{1}{2}(10 a+b)$ is a rational number, thus, $10 b+a$ must be a perfect square, and $\sqrt{10 b+a} \leqslant 9$. Therefore, $\frac{1}{2}(10 a+b)$ is also an integer no greater than 9. Hence, $a=1$. Thus, we have
$$
\sqrt{10 b+1}=5+\frac{1}{2} b .
$$

Solving this, we get $b_{1}=8, b_{2}=12$ (discard).
Thus, $x=18$.
(3) If $x \geqslant 100$, then $y<10 x$,
$$
\begin{array}{l}
z=\sqrt{y}<\sqrt{10 x}=\frac{\sqrt{10}}{\sqrt{x}} \cdot x \\
\leqslant \frac{\sqrt{10}}{\sqrt{100}} x=\frac{1}{\sqrt{10}} x<\frac{1}{2} x .
\end{array}
$$

Thus, $x \geqslant 100$ does not meet the condition.
Therefore, the positive integer $x=4$ or 18.

Answer: 4 \text{ or } 18


Larry can swim from Harvard to MIT (with the current of the Charles River) in $40$ minutes, or back (against the current) in $45$ minutes. How long does it take him to row from Harvard to MIT, if he rows the return trip in $15$ minutes? (Assume that the speed of the current and Larryâs swimming and rowing speeds relative to the current are all constant.) Express your answer in the format mm:ss.

Solution:
1. **Define Variables:**
   - Let \( v \) be the speed of the current.
   - Let \( l \) be Larry's swimming speed in still water.
   - Let \( d \) be the distance between Harvard and MIT.
   - Let \( r \) be Larry's rowing speed in still water.

2. **Set Up Equations for Swimming:**
   - When swimming with the current, Larry's effective speed is \( l + v \). The time taken is 40 minutes.
     \[
     l + v = \frac{d}{40}
     \]
   - When swimming against the current, Larry's effective speed is \( l - v \). The time taken is 45 minutes.
     \[
     l - v = \frac{d}{45}
     \]

3. **Solve for \( l \) and \( v \):**
   - Add the two equations:
     \[
     (l + v) + (l - v) = \frac{d}{40} + \frac{d}{45}
     \]
     \[
     2l = \frac{d}{40} + \frac{d}{45}
     \]
     \[
     2l = \frac{45d + 40d}{1800}
     \]
     \[
     2l = \frac{85d}{1800}
     \]
     \[
     l = \frac{85d}{3600}
     \]
     \[
     l = \frac{17d}{720}
     \]
   - Subtract the two equations:
     \[
     (l + v) - (l - v) = \frac{d}{40} - \frac{d}{45}
     \]
     \[
     2v = \frac{45d - 40d}{1800}
     \]
     \[
     2v = \frac{5d}{1800}
     \]
     \[
     v = \frac{5d}{3600}
     \]
     \[
     v = \frac{d}{720}
     \]

4. **Set Up Equations for Rowing:**
   - When rowing against the current, Larry's effective speed is \( r - v \). The time taken is 15 minutes.
     \[
     r - v = \frac{d}{15}
     \]
     Substitute \( v = \frac{d}{720} \):
     \[
     r - \frac{d}{720} = \frac{d}{15}
     \]
     \[
     r = \frac{d}{15} + \frac{d}{720}
     \]
     \[
     r = \frac{48d}{720} + \frac{d}{720}
     \]
     \[
     r = \frac{49d}{720}
     \]

5. **Calculate Time for Rowing with the Current:**
   - When rowing with the current, Larry's effective speed is \( r + v \).
     \[
     r + v = \frac{49d}{720} + \frac{d}{720}
     \]
     \[
     r + v = \frac{50d}{720}
     \]
     \[
     r + v = \frac{d}{14.4}
     \]
   - The time taken to row from Harvard to MIT is:
     \[
     \text{Time} = \frac{d}{r + v} = \frac{d}{\frac{d}{14.4}} = 14.4 \text{ minutes}
     \]
   - Convert 14.4 minutes to minutes and seconds:
     \[
     14.4 \text{ minutes} = 14 \text{ minutes and } 0.4 \times 60 \text{ seconds} = 14 \text{ minutes and } 24 \text{ seconds}
     \]

The final answer is \( \boxed{14:24} \).

Answer: 14:24


Martin is trying to fill each cell of a rectangular grid with 8 rows and $\mathrm{n}$ columns with one of the four letters $\mathrm{P}, \mathrm{O}, \mathrm{F}$, and $\mathrm{M}$ such that for any pair of distinct rows, there is at most one column where the intersections of the two rows are cells with the same letter. What is the largest integer $n$ for which this is possible?

Solution:
Solution to Exercise 5 In this problem, we are looking for the largest integer satisfying a certain property. Suppose we want to show that the largest integer sought is the integer c. To show that c is indeed the largest integer, we will on the one hand show that if an integer n satisfies the property, then n ≤ c, and on the other hand we will show that we can find a table with exactly c columns.

The statement provides an assumption about the rows and columns of a table.

We start by looking at the information we have about the rows of this table. There are 8 rows, so there are \(\binom{8}{2} = 28\) pairs of rows.

We then look at the information we have about the columns of the table. Since the statement's hypothesis mentions identical letters within the same column, we will count how many pairs of identical letters can belong to the same column. We fix a column of the table and see how many pairs of identical letters we can form at a minimum. After several trials on a column of size 8, we conjecture that there are always at least 4 pairs of identical letters within the same column. Indeed, in a column there are 8 letters, so a letter appears at least 2 times.

- If each letter appears at most 2 times, then each letter appears exactly 2 times since there are 8 cells in the column and 4 possible letters. We thus have 4 pairs of identical letters in the column.
- If a letter appears exactly 3 times, say the letter P, then we can form 3 pairs of letters P. Among the 5 remaining letters, at least one of the letters O, F, and M appears twice according to the pigeonhole principle, which gives us a fourth pair of identical letters.
- If a letter appears at least 4 times in the column, say the letter P, we can form 4 pairs of letters P.

Thus, in a column, we can always find 4 pairs of identical letters. Therefore, for each column, there are at least 4 pairs of rows such that the intersections with the column have the same letter. But according to the statement's hypothesis, a pair of rows can be associated with at most one column such that its intersections with the two rows are two cells containing the same letter. We deduce that there are at most \(4 \times n\) pairs of rows. Therefore, \(4n \leq 28\) and \(n \leq 7\).

Conversely, there exists a configuration with the following 7 columns:

| P | P | P | P | P | P | P |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| O | O | O | O | O | P | O |
| F | F | F | F | O | O | P |
| M | M | M | M | P | O | O |
| P | M | F | O | F | F | F |
| O | F | M | P | M | F | M |
| F | P | P | M | M | M | F |
| M | O | O | F | F | M | M |

The largest integer \(\mathfrak{n}\) of columns is therefore \(\mathfrak{n} = 7\).

Graders' Comments

The exercise is generally well done; students have understood that the constraint is that each column adds 4 pairs of identical letters, but few have provided a fully rigorous proof of this fact. Some have given a set of 7 columns and justified that 7 is optimal because it is not possible to add a column to what they have done. However, this justification is not sufficient. Indeed, just because the construction cannot be enlarged does not mean that its size is maximal: one can imagine that there exists a larger and very different construction from the one given.

Answer: 7


The diagonals of convex quadrilateral $BSCT$ meet at the midpoint $M$ of $\overline{ST}$.  Lines $BT$ and $SC$ meet at $A$, and $AB = 91$, $BC = 98$, $CA = 105$.  Given that $\overline{AM} \perp \overline{BC}$, find the positive difference between the areas of $\triangle SMC$ and $\triangle BMT$.

Solution:
1. **Identify the given information and setup the problem:**
   - The diagonals of convex quadrilateral \( BSCT \) meet at the midpoint \( M \) of \( \overline{ST} \).
   - Lines \( BT \) and \( SC \) meet at \( A \).
   - Given lengths: \( AB = 91 \), \( BC = 98 \), \( CA = 105 \).
   - \( \overline{AM} \perp \overline{BC} \).

2. **Determine the lengths \( BM \) and \( MC \):**
   - Since \( M \) is the midpoint of \( \overline{ST} \) and \( \overline{AM} \perp \overline{BC} \), we can use the fact that \( \triangle ABM \) and \( \triangle AMC \) are right triangles.
   - Using the Pythagorean theorem in \( \triangle ABM \):
     \[
     AB^2 = AM^2 + BM^2 \implies 91^2 = AM^2 + BM^2
     \]
   - Using the Pythagorean theorem in \( \triangle AMC \):
     \[
     AC^2 = AM^2 + MC^2 \implies 105^2 = AM^2 + MC^2
     \]
   - Subtract the first equation from the second:
     \[
     105^2 - 91^2 = MC^2 - BM^2
     \]
   - Simplify:
     \[
     (105 - 91)(105 + 91) = MC^2 - BM^2 \implies 14 \times 196 = MC^2 - BM^2 \implies 2744 = MC^2 - BM^2
     \]
   - Let \( BM = x \) and \( MC = y \). Then:
     \[
     y^2 - x^2 = 2744
     \]
   - Also, from the Pythagorean theorem:
     \[
     91^2 = AM^2 + x^2 \quad \text{and} \quad 105^2 = AM^2 + y^2
     \]
   - Subtracting these:
     \[
     105^2 - 91^2 = y^2 - x^2 \implies 2744 = y^2 - x^2
     \]
   - Solving for \( x \) and \( y \):
     \[
     x = 35 \quad \text{and} \quad y = 63
     \]

3. **Construct parallelogram \( BSXT \):**
   - Since \( BX \) bisects \( ST \), \( X \) is on \( BC \) and \( MX = 35 \), implying \( XC = 28 \).

4. **Calculate the areas of \( \triangle SMC \) and \( \triangle BMT \):**
   - Notice that \( [SMC] - [BMT] = [SXC] + [SMX] - [BMT] = [SXC] \), because \( BSXT \) is a parallelogram.
   - Since \( BT \parallel AB \) and \( BT \parallel SX \), \( AB \parallel SX \) and \( \triangle ABC \sim \triangle SXC \).
   - Therefore, \( SX = 26 \) and \( SC = 30 \).

5. **Calculate the area of \( \triangle SXC \):**
   - Using the similarity ratio:
     \[
     \frac{SX}{AB} = \frac{SC}{AC} \implies \frac{26}{91} = \frac{30}{105}
     \]
   - The area of \( \triangle ABC \) is:
     \[
     \text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC)
     \]
   - Since \( \triangle ABC \sim \triangle SXC \), the area ratio is:
     \[
     \left(\frac{26}{91}\right)^2 = \frac{[SXC]}{[ABC]}
     \]
   - Calculate the area of \( \triangle ABC \):
     \[
     [ABC] = \frac{1}{2} \times 91 \times 98 \times \sin(\angle ABC)
     \]
   - Using the similarity ratio:
     \[
     [SXC] = \left(\frac{26}{91}\right)^2 \times [ABC] = \left(\frac{26}{91}\right)^2 \times \frac{1}{2} \times 91 \times 98 \times \sin(\angle ABC)
     \]
   - Simplify:
     \[
     [SXC] = \left(\frac{26}{91}\right)^2 \times 91 \times 98 \times \sin(\angle ABC) = 336
     \]

The final answer is \( \boxed{336} \).

Answer: 336


Indicate the integer closest to the larger root of the equation

$$
\operatorname{arcctg}\left(\left(\frac{7 x}{10}-\frac{5}{14 x}\right)^{2}\right)-\operatorname{arcctg}\left(\left(\frac{7 x}{10}+\frac{5}{14 x}\right)^{2}\right)=\frac{\pi}{4}
$$

Solution:
Solution. The equation is equivalent to $\frac{7 x}{10}=\frac{5}{14 x} \Longleftrightarrow |x|=\frac{5}{7}$.

Answer: 1.

Answer: 1


Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{2}} \frac{\cos x \, dx}{1+\sin x+\cos x}
$$

Solution:
## Solution

We will use the universal substitution:

$$
t=\operatorname{tg} \frac{x}{2}
$$

From which:

$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left(\frac{\pi}{2}\right)}{2}=\operatorname{tg} \frac{\pi}{4}=1
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{2}} \frac{\cos x d x}{1+\sin x+\cos x}=\int_{0}^{1} \frac{\frac{1-t^{2}}{1+t^{2}} \cdot \frac{2 d t}{1+t^{2}}}{1+\frac{2 t}{1+t^{2}}+\frac{1-t^{2}}{1+t^{2}}}=\int_{0}^{1} \frac{\frac{1-t^{2}}{1+t^{2}} \cdot \frac{2 d t}{1+t^{2}}}{\frac{1+t^{2}+2 t+1-t^{2}}{1+t^{2}}}= \\
& =\int_{0}^{1} \frac{2\left(1-t^{2}\right) d t}{\left(1+t^{2}\right)(2+2 t)}=\int_{0}^{1} \frac{2(1-t)(1+t) d t}{\left(1+t^{2}\right) 2(t+1)}= \\
& =\int_{0}^{1} \frac{(1-t) d t}{1+t^{2}}=\int_{0}^{1} \frac{d t}{1+t^{2}}-\frac{1}{2} \cdot \int_{0}^{1} \frac{d\left(t^{2}+1\right)}{1+t^{2}}= \\
& =\left.\operatorname{arctg} t\right|_{0} ^{1}-\left.\frac{1}{2} \cdot \ln \left(1+t^{2}\right)\right|_{0} ^{1}= \\
& =\operatorname{arctg} 1-\operatorname{arctg} 0-\frac{1}{2} \cdot \ln \left(1+1^{2}\right)+\frac{1}{2} \cdot \ln \left(1+0^{2}\right)= \\
& =\frac{\pi}{4}-0-\frac{1}{2} \cdot \ln 2+0=\frac{\pi}{4}-\frac{1}{2} \cdot \ln 2
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D1} \% 8 \mathrm{~B}+8-15 »

Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals

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- Last edited: 10:52, May 1, 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 8-16

## Material from PlusPi

Answer: \frac{\pi}{4}-\frac{1}{2}\cdot\ln2


To bake 100 buns, it takes Mother 30 minutes, and Maya 40 minutes. Marko eats 100 buns in one hour. Mother and Maya bake buns continuously, while Marko eats continuously. After how much time from the start of baking will there be exactly 100 buns on the table?

Solution:
Solution. First method. If the mother bakes 100 rolls in 30 minutes, then in 1 minute she will bake $\frac{100}{30}=\frac{10}{3}$ rolls. If Maya bakes 100 rolls in 40 minutes, then in 1 minute she will bake $\frac{100}{40}=\frac{10}{4}$ rolls. Since the two of them bake together, then in 1 minute they will bake $\frac{10}{3}+\frac{10}{4}=\frac{35}{6}$ rolls. Marko eats 100 rolls in one hour, so in 1 minute he will eat $\frac{100}{60}=\frac{10}{6}=\frac{5}{3}$ rolls. If the time after which 100 rolls will remain on the table is denoted by $x$, then we have $\left(\frac{35}{6}-\frac{5}{3}\right) x=100$, i.e., $x=24$ minutes.

Second method. If the mother bakes 100 rolls in 30 minutes, then in 2 hours she will bake 400 rolls. If Maya bakes 100 rolls in 40 minutes, then in 2 hours she will bake 300 rolls. Marko eats 100 rolls in one hour, so in two hours he will eat 200 rolls. Then in two hours we have $400+300-200=500$ rolls. We need to have 100 rolls left, so $500: 5=100$ and it follows that the required time is $120: 5=24$ minutes.

Answer: 24


Let $a \in \mathbf{R}^{+}$,
$A=\left\{(x, y) \left\lvert\,(x-1)^{2}+(y-2)^{2} \leqslant \frac{4}{5}\right.\right\}$
and $B=\{(x, y)|| x-1|+2| y-2 \mid \leqslant a\}$

are point sets in the plane $x O y$. Then $A \subseteq B$ if
(A) $a \geqslant 2$
(B) $a \geqslant \sqrt{5}$
(C) $a \geqslant \sqrt{6}$
(D) $a \geqslant 3$

Solution:
6. (A).

Set $A$ is a circular region with center at $(1,2)$ and radius $\frac{2}{\sqrt{5}}$. Set $B$ is a rhombus with the diagonal intersection at $(1,2)$, and the two diagonals are parallel to the coordinate axes, with the diagonal parallel to the $x$-axis being twice the length of the other diagonal.
$A \subseteq B$, if and only if the circle is inside the rhombus and tangent to it, which is the minimum case for the value of $a$ (i.e., the length of $EA$). At this time,
$$
\begin{aligned}
a & \cdot \frac{a}{2}=r \cdot \sqrt{a^{2}+\left(\frac{a}{2}\right)^{2}} \\
& =\frac{2}{\sqrt{5}} \cdot a \cdot \sqrt{\frac{5}{4}} . \\
\therefore a_{\text {min }} & =2, a \geqslant 2 .
\end{aligned}
$$

Answer: A


Given positive real numbers $a, b$ satisfy: for real number $x$, when $1 \leqslant x \leqslant 4$, we have $a x+b-3 \leqslant 0$. Then the minimum value of $\frac{1}{a}-b$ is $\qquad$

Solution:
10. 1 .

Let $f(x)=a x+b-3$. Then $f(x)$ is a linear function. From the conditions, we have
$$
\left\{\begin{array}{l}
a+b-3 \leqslant 0, \\
4 a+b-3 \leqslant 0
\end{array} \Rightarrow a \leqslant \frac{1}{4}(3-b)\right. \text {. }
$$

Also, since $a>0$, we know $0<b<3$.
Thus, $\frac{1}{a}-b \geqslant \frac{4}{3-b}-b=\frac{4}{3-b}+(3-b)-3$
$$
\geqslant 2 \sqrt{\frac{4}{3-b}(3-b)}-3=1 \text {. }
$$

When $b=1, a=\frac{1}{2}$, the equality holds.
Therefore, the minimum value of $\frac{1}{a}-b$ is 1.

Answer: 1


How many sets of positive integer solutions does the equation $3 x+5 y=501$ have?

Solution:
Solve: From $3 x+5 y=501$, we get
$$
x=\frac{501-5 y}{3}=167-2 y+\frac{y}{3}.
$$

Since $x, y$ are positive integers, $\frac{y}{3}$ must also be a positive integer. Let $\frac{y}{3}=t, t$ be a positive integer. Then
$$
y=3 t, x=167-6 t+t=167-5 t.
$$

Therefore,
$$
x=167-5 t, y=3 t \text{ ( } x, y, t \text{ are all positive integers). }
$$

Let $t=1$ to get $(x, y)=(162,3)$;
Let $t=2$ to get $(x, y)=(157,6)$,
......
Let $t=33$ to get $(x, y)=(2,99)$.
That is, when $t=1,2, \cdots, 33$, we get the positive integer solutions $(x, y)$, and there are no other positive integer solutions.
So the number of positive integer solutions is 33.
In fact, we can give a general formula to find the number of solutions for a general linear Diophantine equation.
Theorem If $(a, b)=1, a>0, b>0$, then
$$
\begin{array}{c}
a x+b y=N \quad(x \geqslant 0, y \geqslant 0) \\
\frac{N-(b l+a m)}{a b}+1.
\end{array}
$$

The number of solutions is
where $l$ is the smallest non-negative number satisfying $b l=N(\bmod a)$, and $m$ is the smallest non-negative number satisfying $a m=N$ $(\bmod b)$.

Proof: Let $x_{0}, y_{0}$ be a solution satisfying $x \geqslant 0, y \geqslant 0$ and $x_{0}$ is the smallest, then
$$
x=x_{0}+b t, y=y_{0}-a t.
$$

gives all integer solutions of $a x+b y=N$, where $t$ is an integer. Let $y_{0}$ be the largest one, then from $y=y_{0}-a t$ we get
$$
0 \leqslant t=\frac{y_{0}}{a}-\frac{y}{a} \leqslant\left[\frac{y_{0}}{a}\right].
$$

Also, from
$$
b l=N(\bmod a), a m=N(\bmod b),
$$

we have
$$
N=a r+b l, N=b T+a m,
$$

From the meaning of $l, m$ we know
$$
y_{0}=T=\frac{N-a m}{b}.
$$

Clearly, $l<a$, otherwise, from
$$
\begin{array}{c}
a r+b l=N, \\
a x+b y=N
\end{array}
$$

we know $r, l$ is a set of non-negative solutions, and $l$ is the smallest. And
$$
a x+b y=N
$$

all solutions can be given by
$$
x=r+s b, y=l-s a.
$$

Taking $s=1, y=1-a \geqslant 0$ is still a solution, which contradicts the minimality of $l$. On the other hand, from the conditions we have
$$
\begin{array}{c}
N \equiv b l \equiv b l+a m(\bmod a), \\
N \equiv a m \equiv a m+b l(\bmod b).
\end{array}
$$

Given the condition $(a, b)=1$, we get
$$
\begin{array}{c}
N \equiv a m+b l(\bmod a b), \\
a b \mid N-(a m+b l).
\end{array}
$$

This means $\frac{N-(a m+b l)}{a b}$ is an integer. Also, since
$$
\frac{y_{0}}{a}=\frac{T}{a}=\frac{N-a m}{a b}=\frac{N-(a m+b l)}{a b}+\frac{1}{a},
$$

and $l<a,\left[\frac{1}{a}\right]=0$, we have
$$
\begin{aligned}
{\left[\frac{y_{0}}{a}\right] } & =\left[\frac{N-(a m+b l)}{a b}+\frac{l}{a}\right] \\
& =\frac{N-(a m+b l)}{a b}+\left[\frac{l}{a}\right]=\frac{N-(a m+b l)}{a b},
\end{aligned}
$$

So $0 \leqslant t \leqslant\left[\frac{y_{0}}{a}\right]=\frac{N-(a m+b l)}{a b}$.
Hence $t$ can take $\frac{N-(a m+b l)}{a b}+1$ values, which is the number of solutions satisfying the conditions:
$$
\frac{N-(a m+b l)}{a b}+1.
$$

Answer: 33


Solution:
## Solution.

Domain of definition: $\frac{3 x-2 y}{2 x}>0$.

Transform the first equation of the system

$$
\sqrt{\frac{3 x-2 y}{2 x}}+\frac{1}{\sqrt{\frac{3 x-2 y}{2 x}}}-2=0 \Leftrightarrow\left(\sqrt{\frac{3 x-2 y}{2 x}}\right)^{2}-2 \sqrt{\frac{3 x-2 y}{2 x}}+1=0
$$

where $\sqrt{\frac{3 x-2 y}{2 x}} \neq 0,\left(\sqrt{\frac{3 x-2 y}{2 x}}-1\right)^{2}=0 \Rightarrow \sqrt{\frac{3 x-2 y}{2 x}}=1, \frac{3 x-2 y}{2 x}=1, x=2 y$.

From the second equation of the system we have $(2 y)^{2}-18=2 y(4 y-9) \Leftrightarrow$ $\Leftrightarrow 2 y^{2}-9 y+9=0$, from which $y_{1}=\frac{3}{2}, y_{2}=3 ; x_{1}=3, x_{2}=6$.

Answer: $\left(3 ; \frac{3}{2}\right),(6 ; 3)$.

Answer: (3;\frac{3}{2}),(6;3)


Two circles are constructed on a plane such that each passes through the center of the other. $P$ and $Q$ are the two points of their intersection. A line through point $P$ intersects the first circle at point $A$ and the second circle at point $B$, with $P$ lying between $A$ and $B$. A line through point $Q$, parallel to $AB$, intersects the first circle at point $D$ and the second circle at point $C$. Find the area of quadrilateral $ABCD$, given that $AP=2, PB=1$.

Solution:
Ex. 105. Answer: $\frac{9 \sqrt{3}}{2}$. Solution. Let the center of the first circle be $O_{1}$, the center of the second circle be $O_{2}$, the angle $\angle A O_{1} P=\alpha, \angle D O_{1} Q=\beta$. Note that the arc $\mathrm{PO}_{2} Q$ is equal to the arc $A D$ and is equal to $120^{\circ}$. Therefore, it is true that $\alpha+\beta=\frac{2 \pi}{3}$. Draw the line $P Q$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_91ff4e46083e03d62f1eg-38.jpg?height=443&width=554&top_left_y=755&top_left_x=751)

The area of the parallelogram $A B C D$ consists of two areas of identical trapezoids $A P Q D$ and $P B C Q$. The area of trapezoid $A P Q D$, in turn, consists of four areas of triangles with vertices at point $O_{1}$. Thus, $S_{A B C D}=$

$=2\left(S_{A P O_{1}}+S_{P Q O_{1}}+S_{Q D O_{1}}+S_{D A O_{1}}\right)=2 \cdot \frac{1}{2} r^{2}\left(\sin \alpha+\sin 120^{\circ}+\sin \beta+\sin 120^{\circ}\right)=$ $=r^{2}\left(\sqrt{3}+2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}\right)=r^{2} \sqrt{3}\left(1+\cos \frac{\alpha-\beta}{2}\right)=r^{2} \sqrt{3}\left(1+\cos \left(\alpha-\frac{\pi}{3}\right)\right)$. From the conditions of the problem, it follows that $2=2 r \sin \frac{\alpha}{2}$ and $1=2 r \sin \frac{\beta}{2}$, from which we get that $2 \sin \frac{\beta}{2}=\sin \frac{\alpha}{2}$. Considering that $\frac{\beta}{2}=\frac{\pi}{3}-\frac{\alpha}{2}$, rewrite the equation as $2 \sin \left(\frac{\pi}{3}-\frac{\alpha}{2}\right)=\sin \frac{\alpha}{2} \Rightarrow \sqrt{3} \cos \frac{\alpha}{2}=2 \sin \frac{\alpha}{2} \Rightarrow \operatorname{tg} \frac{\alpha}{2}=\frac{\sqrt{3}}{2} \Rightarrow$ $\sin \frac{\alpha}{2}=\frac{\sqrt{3}}{\sqrt{7}}, \cos \frac{\alpha}{2}=\frac{2}{\sqrt{7}}, r=\frac{\sqrt{7}}{\sqrt{3}}$. From this, it follows that $\sin \alpha=\frac{4 \sqrt{3}}{7}, \cos \alpha=$ $\frac{1}{7}, \cos \left(\alpha-\frac{\pi}{3}\right)=\frac{13}{14}$. Thus, $\quad S_{A B C D}=\frac{7}{3} \sqrt{3}\left(1+\frac{13}{14}\right)=\frac{9 \sqrt{3}}{2}$.

Answer: \frac{9\sqrt{3}}{2}


In a class, there are boys and girls, such that the number of boys divides the number of girls. Florin distributes 1000 candies equally among the students in the class. Knowing that there are at least 10 boys in the class, and if 10 more students were to join, Florin could, again, distribute the 1000 candies equally among the students, find out how many candies each student in the class receives.

Solution:
## Solution.

Let $b$ be the number of boys, $f$ be the number of girls, and $n=b+f$ be the number of students in the class. According to the problem, $n \mid 1000$. (1p)

From $b \mid f$ and $b \geq 10$, it follows that $n=b+f \geq 2b \geq 20$. (1p)

Therefore, $n \in \{20, 25, 40, 50, 100, 125, 200, 250, 500, 1000\}$. (1p)

From $(n+10) \mid 1000$, it follows that $n=40$. (2p)

Then each student receives $1000 \div 40 = 25$ candies. (1p)

It remains to show that the situation indicated by the problem is possible. Thus, from $b \mid f$ and $b+f=n=40$, we deduce $b \mid 40$. Since $b \geq 10$, we obtain two possible variants: a) $b=10$ and $f=30$; b) $b=20$ and $f=20$. (1p)

Answer: 25


Arrange the digits from 1 to 6 (each must be used exactly once) so that the sum of the three numbers located on each of the 7 lines is equal to 15. In your answer, indicate which digits should be placed at positions $A-F$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6da73bfd3e09e8b55e3fg-09.jpg?height=363&width=456&top_left_y=612&top_left_x=499)

(a) Instead of the letter $A$

(b) Instead of the letter $B$

(c) Instead of the letter $C$

(d) Instead of the letter $D$

(e) Instead of the letter $E$

(f) Instead of the letter $F$
(1) the digit 1 should be placed

(2) the digit 2 should be placed

(3) the digit 3 should be placed

(4) the digit 4 should be placed

(5) the digit 5 should be placed

(6) the digit 6 should be placed

Solution:
Answer: $A=4, B=1, C=2, D=5, E=6, F=3$.

Solution. According to the condition, $A, D, E$ are different digits, not exceeding 6, the sum of which is 15. If these digits are taken to be the maximum possible, then their sum is $4+5+6=15$. Therefore, $A, D, E$ are $4,5,6$ in some order (if at least one of the digits is no more than 3, then the sum of all three digits is no more than $3+5+6=14$) and $D \neq 6$ (otherwise $B+D+9>15$). Therefore, $E=6$.

Since $7+C+E=15$ and $E=6$, we get $C=2$.

Since $9+C+A=15$ and $C=2$, we get $A=4$.

Since $A+8+F=15$ and $A=4$, we get $F=3$.

Since $7+D+F=15$ and $F=3$, we get $D=5$.

Since $9+D+B=15$ and $D=5$, we get $B=1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6da73bfd3e09e8b55e3fg-10.jpg?height=361&width=456&top_left_y=90&top_left_x=499)

It is easy to verify that the obtained arrangement satisfies all the conditions.

Answer: A=4,B=1,C=2,D=5,E=6,F=3


Solution:
Given that there are three independent equations (equations) and exactly three letters, it should be possible to solve for $x, y, z$, but in practice, it is quite difficult. Inspired by the "chain equality" type problem from the previous question, we can introduce the letter $k$ - try.

Assume $1995 x^{3}=1996 y^{3}=1997 z^{3}=k$, then
$$
x=\sqrt[3]{\frac{k}{1995}}, y=\sqrt[3]{\frac{k}{1996}}, z=\sqrt[3]{\frac{k}{1997}} .
$$

From the given conditions, $x, y, z, k$ have the same sign, and since $x y z>0$, we know $x, y, z, k>0$. Substituting into the last equation given, we get
$$
\sqrt[3]{\frac{k}{x}+\frac{k}{y}+\frac{k}{z}}=\sqrt[3]{\frac{k}{x^{3}}}+\sqrt[3]{\frac{k}{y^{3}}}+\sqrt[3]{\frac{k}{z^{3}}} .
$$

Rearranging, we get
$$
\sqrt[3]{k}\left(\sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right)=0 \text{. }
$$

Since $k>0$, we know $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is a root of $\sqrt[3]{u}-u=0$.

Therefore, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. (The other two roots $0, -1$ do not meet the conditions)

Answer: 1


Sixteen counters, which are black on one side and white on the other, are arranged in a 4 by 4 square. Initially all the counters are facing black side up. In one 'move', you must choose a 2 by 2 square within the square and turn all four counters over once.
Describe a sequence of 'moves' of minimum length that finishes with the colours of the counters of the 4 by 4 square alternating (as shown in the diagram).

Solution:
Solution
Define a set of coordinates from the bottom left corner, such that $(1,1)$ relates to the counter in the bottom left corner, $(1,4)$ relates to the counter in the top left corner, $(4,1)$ relates to the counter in the bottom right corner, etc.
Consider the counter at $(1,1)$. This must be flipped and can only be flipped by flipping the bottom left 2 by 2 square.

Now both the counter at $(1,2)$ and $(2,1)$ need to be flipped and it is not possible to flip both together without also changing $(1,1)$. Therefore flip the counters in the 2 by 2 square with lower left corner at $(2,1)$ and in the 2 by 2 square with lower left corner at $(1,2)$. Since the counter at $(2,2)$ will have then been flipped three times, it will be of the opposite colour to its original colour as required. Therefore we have completed the square in the bottom left corner with three 'moves'.

A similar argument can be applied to the counter at $(4,4)$ and the top right 2 by 2 square which has not been affected by any of the moves so far. Three more 'moves' are required to turn all four counters in this 2 by 2 square to the required colour.

In fact, this completes the pattern of the 4 by 4 square as a whole. Therefore a minimum of six 'moves' is required.

Answer: 6


If numbers $a_{1}, a_{2}, a_{3}$ are taken in ascending order from the set $1,2, \cdots, 14$, such that both $a_{2}-a_{1} \geqslant 3$ and $a_{3}-a_{2} \geqslant 3$ are satisfied. Then, the number of all different ways to select the numbers is $\qquad$ kinds.

Solution:
Solution: Given $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$, $14-a_{3} \geqslant 0$, let $a_{1}=x_{1}, a_{2}-a_{1}=x_{2}, a_{3}-a_{2}=x_{3}, 14-a_{3}=x_{4}$, then
$$
x_{1}+x_{2}+x_{3}+x_{4}=14 .
$$

Thus, the problem is transformed into finding the number of integer solutions to the equation under the conditions
$$
x_{1} \geqslant 1, x_{2} \geqslant 3, x_{3} \geqslant 3, x_{4} \geqslant 0
$$

Transform the equation into
$$
\begin{array}{l}
x_{1}+\left(x_{2}-2\right)+\left(x_{3}-2\right)+\left(x_{4}+1\right)=11 . \\
\text { Let } x_{1}=y_{1}, x_{2}-2=y_{2}, x_{3}-2=y_{3}, \\
x_{4}+1=y_{4} .
\end{array}
$$

The number of positive integer solutions to $y_{1}+y_{2}+y_{3}+y_{4}=11$ is $\mathrm{C}_{10}^{3}$, so the number of different ways to choose $a_{1}, a_{2}, a_{3}$ that meet the conditions is $\mathrm{C}_{10}^{3}=120$.

Answer: 120


What is the minimal $k$ for which the spiders are sure to catch the fly on a $2019 \times 2019$ grid, given that on their turn, the fly can move 1 square and the $k$ spiders can each move 1 square?

Solution: We show that 2 spiders suffice: when the fly moves to a coordinate, the spider on that coordinate replicates the movement, the other approaches.

Answer: 2


$\frac{\sin 2 \alpha+\cos 2 \alpha-\cos 6 \alpha-\sin 6 \alpha}{\sin 4 \alpha+2 \sin ^{2} 2 \alpha-1}$

Solution:
Solution.

$$
\begin{aligned}
& \frac{\sin 2 \alpha+\cos 2 \alpha-\cos 6 \alpha-\sin 6 \alpha}{\sin 4 \alpha+2 \sin ^{2} 2 \alpha-1}= \\
& =\frac{(\sin 2 \alpha-\sin 6 \alpha)+(\cos 2 \alpha-\cos 6 \alpha)}{\sin 4 \alpha-\left(1-2 \sin ^{2} 2 \alpha\right)}= \\
& =\left[\sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right. \\
& \left.\cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} ; 1-2 \sin ^{2} x=\cos 2 x\right]= \\
& =\frac{2 \cos 4 \alpha(-\sin 2 \alpha)+(-2 \sin 4 \alpha(-\sin 2 \alpha))}{\sin 4 \alpha-\cos 4 \alpha}= \\
& =\frac{-2 \cos 4 \alpha \sin 2 \alpha+2 \sin 4 \alpha \sin 2 \alpha}{\sin 4 \alpha-\cos 4 \alpha}=\frac{2 \sin 2 \alpha(\sin 4 \alpha-\cos 4 \alpha)}{\sin 4 \alpha-\cos 4 \alpha}=2 \sin 2 \alpha .
\end{aligned}
$$

Answer: $2 \sin 2 \alpha$.

Answer: 2\sin2\alpha


If you were to calculate the result of

$$
\underbrace{999 \ldots 99}_{2014 \text { nines }} \times \underbrace{444 \ldots 44}_{2014 \text { fours }}
$$

and then add up all the digits of the result, what would the outcome be?

Solution:
B3. 18126 A good strategy is to first look at small examples. This way, we find successively

$$
\begin{array}{rccc}
9 \times 4 & = & 40-4 & =36, \\
99 \times 44 & = & 4400-44 & =4356, \\
999 \times 444 & = & 444000-444 & =443556, \\
9999 \times 4444 & = & 44440000-4444 & =44435556 .
\end{array}
$$

The pattern will be clear. To answer the question, we note that $999 \ldots 99 = 1000 \ldots 00 - 1$, where the first number has 2014 zeros. The product is therefore equal to

$$
\underbrace{444 \ldots 44}_{2014 \text { fours }} \underbrace{000 \ldots 00}_{2014 \text { zeros }} - \underbrace{444 \ldots 44}_{2014 \text { fours }} = \underbrace{444 \ldots 44}_{2013 \text { fours }} 3 \underbrace{555 \ldots 55}_{2013 \text { fives }} 6 .
$$

If we add the digits of this number, we get $2013 \cdot 4 + 3 + 2013 \cdot 5 + 6 = 2013 \cdot 9 + 9 = 18126$.

Answer: 18126


Let $a, b$ be positive integers. When $a^{2}+b^{2}$ is divided by $a+b$, the quotient is $q$ and the remainder is $r$. Find all pairs $(a, b)$ such that $q^{2}+r=1977$.

Solution:
[Solution] Since $a^{2}+b^{2}=q(a+b)+r$, where $0 \leqslant r < q$, and if $r > 0$ it leads to a contradiction, so $q \leqslant 44$, and $\frac{a^{2}+b^{2}}{a+b} > 128$. Without loss of generality, assume $a \geqslant b$, then $a \geqslant \frac{1}{2}(a+b)$.

When $a \leqslant \frac{2}{3}(a+b)$, $b \geqslant \frac{1}{3}(a+b)$,
$$\frac{a^{2}+b^{2}}{a+b} \geqslant \frac{\left[\frac{1}{2}(a+b)\right]^{2}+\left[\frac{1}{3}(a+b)\right]^{2}}{a+b}=\frac{13}{36}(a+b).$$

Since $(a+b) > 128$,
so $\frac{a^{2}+b^{2}}{a+b} > 46$, (this contradicts $\frac{a^{2}+b^{2}}{a+b} < 46$).

When $a > \frac{2}{3}(a+b)$,
$$\frac{a^{2}+b^{2}}{a+b} > \frac{\left[\frac{2}{3}(a+b)\right]^{2}}{a+b}=\frac{4}{9}(a+b) > 56$$

This contradicts $\frac{a^{2}+b^{2}}{a+b} < 56$.
Therefore, $46 < \frac{a^{2}+b^{2}}{a+b} < 56$.
Since $q \leqslant 44$, hence $q=44$ and $r=41$, so we have
$$a^{2}+b^{2}=44(a+b)+41$$

That is, $(a-22)^{2}+(b-22)^{2}=1009$.
Solving this, we get $\left\{\begin{array}{l}a_{1}=50, \\ b_{1}=37 ;\end{array} \quad\left\{\begin{array}{l}a_{2}=50, \\ b_{2}=7 ;\end{array}\right.\right.$
$$\left\{\begin{array} { l }
{ a _ { 3 } = 3 7 , } \\
{ b _ { 3 } = 5 0 ; }
\end{array} \quad \left\{\begin{array}{l}
a_{4}=7 \\
b_{4}=50
\end{array}\right.\right.$$

Thus, the pairs $(a, b)$ are:
$$(50,37),(50,7),(37,50),(7,50)$$

Answer: (50,37),(50,7),(37,50),(7,50)


Suppose two functions $ f(x)=x^4-x,\ g(x)=ax^3+bx^2+cx+d$ satisfy $ f(1)=g(1),\ f(-1)=g(-1)$.

Find the values of $ a,\ b,\ c,\ d$ such that $ \int_{-1}^1 (f(x)-g(x))^2dx$ is minimal.

Solution:
1. Given the functions \( f(x) = x^4 - x \) and \( g(x) = ax^3 + bx^2 + cx + d \), we need to find the values of \( a, b, c, \) and \( d \) such that \( \int_{-1}^1 (f(x) - g(x))^2 \, dx \) is minimized, subject to the conditions \( f(1) = g(1) \) and \( f(-1) = g(-1) \).

2. First, we calculate \( f(1) \) and \( f(-1) \):
   \[
   f(1) = 1^4 - 1 = 0
   \]
   \[
   f(-1) = (-1)^4 - (-1) = 1 + 1 = 2
   \]

3. Using the conditions \( f(1) = g(1) \) and \( f(-1) = g(-1) \), we get:
   \[
   g(1) = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d = 0
   \]
   \[
   g(-1) = a(-1)^3 + b(-1)^2 + c(-1) + d = -a + b - c + d = 2
   \]

4. Solving these two equations:
   \[
   a + b + c + d = 0
   \]
   \[
   -a + b - c + d = 2
   \]

5. Adding and subtracting these equations:
   \[
   (a + b + c + d) + (-a + b - c + d) = 0 + 2 \implies 2b + 2d = 2 \implies b + d = 1
   \]
   \[
   (a + b + c + d) - (-a + b - c + d) = 0 - 2 \implies 2a + 2c = -2 \implies a + c = -1
   \]

6. Therefore, we have:
   \[
   d = 1 - b
   \]
   \[
   c = -1 - a
   \]

7. Substituting these into \( h(x) = f(x) - g(x) \):
   \[
   h(x) = x^4 - x - (ax^3 + bx^2 + cx + d)
   \]
   \[
   h(x) = x^4 - ax^3 - bx^2 - (c+1)x - d
   \]
   \[
   h(x) = x^4 - ax^3 - bx^2 - (-1 - a + 1)x - (1 - b)
   \]
   \[
   h(x) = x^4 - ax^3 - bx^2 + ax - b + 1
   \]

8. To minimize \( \int_{-1}^1 (h(x))^2 \, dx \), we compute:
   \[
   \int_{-1}^1 (x^4 - ax^3 - bx^2 + ax - b + 1)^2 \, dx
   \]

9. By symmetry and properties of even and odd functions, we can simplify the integral:
   \[
   \int_{-1}^1 (h(x))^2 \, dx = 2 \int_0^1 (h(x))^2 \, dx
   \]

10. Expanding \( (h(x))^2 \):
    \[
    (h(x))^2 = (x^4 - ax^3 - bx^2 + ax - b + 1)^2
    \]

11. Integrating term by term and simplifying:
    \[
    \int_0^1 (x^4 - ax^3 - bx^2 + ax - b + 1)^2 \, dx
    \]

12. After detailed calculations (omitted for brevity), we find:
    \[
    \int_{-1}^1 (h(x))^2 \, dx = \frac{16}{105}a^2 + \frac{16}{15}\left(b - \frac{8}{7}\right)^2 + \frac{64}{2205}
    \]

13. To minimize this expression, set \( a = 0 \) and \( b = \frac{8}{7} \):
    \[
    c = -1 - a = -1
    \]
    \[
    d = 1 - b = 1 - \frac{8}{7} = -\frac{1}{7}
    \]

The final answer is \( \boxed{ a = 0, b = \frac{8}{7}, c = -1, d = -\frac{1}{7} } \)

Answer: a = 0, b = \frac{8}{7}, c = -1, d = -\frac{1}{7}


Amir is 8 kg heavier than Ilnur, and Daniyar is 4 kg heavier than Bulat. The sum of the weights of the heaviest and lightest boys is 2 kg less than the sum of the weights of the other two. The total weight of all four is 250 kg. How many kilograms does Amir weigh?

Solution:
Answer: 66 kg.

Solution. Let the weight of Bulat in kilograms be denoted by $x$, and the weight of Ilnur by $y$. Then Amir weighs $y+8$, and Daniyar weighs $-x+4$. Note that the heaviest is either Amir or Daniyar, and the lightest is either Ilnur or Bulat.

If the heaviest is Daniyar, then $x+4>y+8$, so $x>y$ and the lightest is Ilnur. In this case, the combined weight of the heaviest and the lightest is $x+4+y$, which is 4 kg less than the combined weight of the other two. This is a contradiction. Therefore, the heaviest is Amir.

Suppose Bulat is the lightest. Then together with Amir, they weigh $x+y+8$, which is 4 kg more than the other two boys. This is a contradiction.

The only remaining case is when the lightest is Ilnur. In this case, Ilnur and Amir together weigh $2y+8$. From the problem statement, it follows that Bulat and Daniyar together weigh $2y+10$. Therefore, the total weight of all four boys is $4y+18=250$, from which $y=58$. Therefore, Amir weighs 66 kg.

Criteria. Only the correct answer with an incorrect or missing solution -3 points. Missing cases during enumeration - no more than 10 points depending on the number and severity of the missed cases.

In the case of a solution like the jury's, missing any of the first two cases (with the rest of the solution correct) - 10 points; missing both of the first two cases with the third case correctly analyzed - 6 points.

Answer: 66


Point $K$ is the midpoint of edge $A A_{1}$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, point $L$ lies on edge $B C$. Segment $K L$ is tangent to the sphere inscribed in the cube. In what ratio does the point of tangency divide segment $K L$?

Solution: 57. $\frac{4}{5}$, starting from point $K$.

Answer: \frac{4}{5}