Two pirates were playing for gold coins. First, the first pirate lost half of his coins (gave them to the second), then the second pirate lost half of his, then the first pirate lost half of his again. As a result, the first pirate ended up with 15 coins, and the second with 33. How many coins did the first pirate have before the game started?
Let's try to trace "from the end" how many coins each pirate had after each game. In the last game, the first pirate lost half of his coins to the second pirate, after which he was left with 15 coins. But this is exactly the amount he just gave to the second pirate! This means that before this, the first pirate had $15 \cdot 2=30$ coins, and the second pirate had 33 - 15 = 18 coins. Similarly, in the second game, the second pirate lost exactly the amount he had left after the second game, which is 18 coins. This means that before the second game, the second pirate had $18 \cdot 2=36$ coins, and the first pirate had $30-18=12$ coins. By performing one more step of reasoning, you will find that initially, each pirate had 24 coins.

Answer

## 24

Author: Kovaldzhi A.K.

A football is made up of 32 patches: white hexagons and black pentagons. Each black patch borders only white ones, and each white patch borders three black and three white ones. How many white patches are there?

## Hint

Count the number of borders between white and black patches in two different ways.

## Solution

Let the number of white patches be denoted by $x$. Then the number of black patches is 32 - $x$. To form an equation, we will count the number of "borders" between white and black patches in two different ways.

Each white patch borders three black patches. Therefore, the number of borders is $3 \cdot x$.

On the other hand, each black patch borders five white patches. Therefore, the number of borders is 5 $\cdot$ (32 - $x$).

We get the equation $3 x=5 \cdot(32-x)$. From this, $8 x=160$ and $x=20$.

## Answer

20.