## Task 6 - 050936

A mother places a bowl of cherries on the table for her three children, Jürgen, Renate, and Christine, with the note that each should take a third of the cherries upon their return.

Jürgen, who comes home first, takes one cherry first because the number of cherries is not divisible by three, and then takes a third of the remaining cherries.

When Renate comes home, she thinks she is the first. She takes two cherries first because the number of cherries is not divisible by three, and then takes a third of the remaining cherries.

Christine also believes she is the first when she returns, and takes a third of the cherries in the bowl.

The mother then notices that a total of 42 cherries have been eaten.

How many cherries were in the bowl initially?
If the original number of cherries is denoted by $x$, the following can be set up: Jürgen takes

$$
1+\frac{x-1}{3}=\frac{x+2}{3}
$$

leaving

$$
x-\frac{x+2}{3}=\frac{2 x-2}{3}
$$

Renate takes

$$
2+\left(\frac{2 x-2}{3}-2\right) \cdot \frac{1}{3}=\frac{2 x+10}{9}
$$

leaving

$$
\frac{2 x-2}{3}-\frac{2 x+10}{9}=\frac{4 x-16}{9}
$$

Christine takes

$$
\frac{4 x-16}{9} \cdot \frac{1}{3}=\frac{4 x-16}{27}
$$

leaving

$$
\frac{4 x-16}{9}-\frac{4 x-16}{27}=\frac{8 x-32}{27}
$$

According to the problem,

$$
x-\frac{8 x-32}{27}=42
$$

and thus $x=58$. Initially, there were 58 cherries in the bowl.

Taken from $[5]$

### 6.8 VI. Olympiad 1966

### 6.8.1 First Round 1966, Grade 9