2. When adding two natural numbers, Alyosha accidentally added an extra digit, so instead of the number 23456, he got the number 55555.

Determine which digit Alyosha accidentally added.

## solution.

Let Alyosha be adding numbers $x$ and $y$. When he appended an extra digit $a$ to one of them (let's say to $x$), the number became $10x + a$. We get the system of equations: $\left\{\begin{array}{c}x + y = 23456; \\ (10x + a) + y = 55555\end{array}9x + a = 320999x = 32099 - a \\ a = 5 \\ 9x = 32094$

$x = 3566 \quad y = 19890$. Answer: 5;

## №5: Percentages.

A flask is filled with a 96% hydrochloric acid solution. From it, 12 liters of acid were poured out and the flask was topped up with water. Then, 18 liters were poured out again and the flask was topped up with water, after which the concentration of acid in the flask was 32%. Find the volume of the flask.
Answer: 36 liters

## №6: Plane Geometry.

№2. From points B and C of the acute angle BAC, perpendiculars CM and BK are drawn, intersecting at point E, such that point M lies on AB, and K lies on AC. A circle is drawn through points A, M, and C, intersecting BK at point O, and $BE=17, OK=72$. AK: $KC=1:4$. Find the tangent of angle BAC. (the ratio of BK to AK).

![](https://cdn.mathpix.com/cropped/2024_05_06_166d4fdfa0c802c9882dg-3.jpg?height=728&width=629&top_left_y=750&top_left_x=705)

## Solution.

1) Let $\mathrm{OE}=x=\succ \mathrm{BO}=\mathrm{BE}-\mathrm{OE}=17=x$ and $\mathrm{BK}=\mathrm{BO}+\mathrm{OK}=17-x+72=89-x$.
2) $\mathrm{BK} \perp \mathrm{AC}=\succ \mathrm{KO}=\mathrm{KH}=72=\succ \mathrm{BH}=\mathrm{BO}+\mathrm{OH}=17-x+144=161-x$.
3) $\triangle \mathrm{ABK} \sim \triangle \mathrm{EMB} \Rightarrow \succ \mathrm{BM} / \mathrm{BK}=\mathrm{BE} / \mathrm{AB}=\succ \mathrm{BM} \cdot \mathrm{AB}=\mathrm{BK} \cdot \mathrm{BE}$.
4) $\mathrm{BA}$ and $\mathrm{BE}-$ secants $=\succ \mathrm{BM} \cdot \mathrm{AB}=\mathrm{BO} \cdot \mathrm{BH}$.
5) From (3 and 4) $\Rightarrow \succ \mathrm{BO} \cdot \mathrm{BH}=\mathrm{BK} \cdot \mathrm{BE}=\succ \quad(17-x)(161-x)=(89-x) 17=\succ x^{2}-$ $161 x+1224=0=\succ x=8$ or $x=153=\succ \mathrm{OE}=8 ; \mathrm{BO}=9 ; \mathrm{BK}=81$.
6) $\mathrm{AK} \cdot \mathrm{KC}=K \mathrm{KO} \cdot \mathrm{H}=\succ 4 \mathrm{AK}^{2}=72^{2} \Rightarrow \succ \mathrm{AK}=36=\succ \mathrm{BK} / \mathrm{AK}=81 / 36=2.25$.

![](https://cdn.mathpix.com/cropped/2024_05_06_166d4fdfa0c802c9882dg-4.jpg?height=796&width=691&top_left_y=316&top_left_x=862)

Answer: 2.25.

## №7: Number Theory.

Find the smallest natural $\mathrm{n}$, for which $1999!$ does not divide $34^{\mathrm{n}}$. (n!=1$\cdot$2$\cdot$3$\cdot$... $\mathrm{n}$)

## Solution:

$34=2 \cdot 17$. Determine the power to which the number 17 will enter the factorization of 1999! (2 will obviously enter this factorization in a higher power). $117<\frac{1999}{17}<118 ; 6<\frac{1999}{17^{2}}<7$ from this, 17 enters $1999!$ 123 times, and to not divide, we add one more.

Answer: 124

## №8: Parameter.