## Task 5 - 110835

Gisela presents the following task at a pioneer afternoon:

"If I take nuts from this container and give the first of you half and a half nut, and then to the second, third, etc., in turn, half of the remaining nuts and a half nut, then I will have used them all up.

How large is the number of nuts that the container contained?

How many nuts would each of the five pioneers receive?"
Let the number of nuts in the container be $x$. Then the first pioneer would receive $\frac{x}{2}+\frac{1}{2}$, leaving a remainder of

$$
\frac{x}{2}-\frac{1}{2}=\frac{1}{2}(x-1)
$$

From this, the second pioneer would receive $\frac{1}{4}(x-1)+\frac{1}{2}$, leaving a remainder of $\frac{1}{4}(x-1)-\frac{1}{2}=\frac{1}{4}(x-3)$.

From this, the third pioneer would receive $\frac{1}{8}(x-3)+\frac{1}{2}$, leaving a remainder of $\frac{1}{8}(x-3)-\frac{1}{2}=\frac{1}{8}(x-7)$.

From this, the fourth pioneer would receive $\frac{1}{16}(x-7)+\frac{1}{2}$, leaving a remainder of

$$
\frac{1}{16}(x-7)-\frac{1}{2}=\frac{1}{16}(x-15)
$$

From this, the fifth pioneer would receive $\frac{1}{32}(x-15)+\frac{1}{2}$, leaving a remainder of

$$
\frac{1}{32}(x-15)-\frac{1}{2}=\frac{1}{32}(x-31)
$$

According to the problem, this remainder is zero. Therefore, $x=31$. Thus, the container contained exactly 31 nuts. Of these, the following would be received:

The 1st pioneer 16 nuts, the 2nd pioneer 8 nuts, the 3rd pioneer 4 nuts, the 4th pioneer 2 nuts, and the 5th pioneer 1 nut.