We consider the quantities of stickers and stamps that meet the last mentioned condition, then we check the other requirements:

One sticker and two stamps can be obtained for 17 points $(5+2 \cdot 6=17)$. If 17 points were exchanged only for stickers, two points would be unused (the remainder of $17: 5$ is 2), which does not match the problem statement.

For $s$ stickers, we have $2s$ stamps and $17s$ points. We are looking for the smallest $s$ such that $17s$ gives a remainder of one when divided by five and a remainder of three when divided by six:

| stickers | 1 | 2 | 3 |
| :---: | :---: | :---: | :---: |
| stamps | 2 | 4 | 6 |
| points | 17 | 34 | $\mathbf{51}$ |
| remainder when divided by 5 | 2 | 4 | 1 |
| remainder when divided by 6 | 5 | 4 | 3 |

Jara had at least 51 points before the exchange.

Another solution. We consider the quantities of stickers and stamps that meet the first two conditions, then we check the other requirements:

If points were exchanged only for stickers, one point would be unused; possible numbers are (multiples of five increased by one)

$$
6,11,16, \mathbf{21}, 26,31,36,41,46, \mathbf{51}, \ldots
$$

If points were exchanged only for stamps, three points would be unused; possible numbers are (multiples of six increased by three)

$$
9,15, \mathbf{21}, 27,33,39,45, \mathbf{51}, \ldots
$$

Values that satisfy both conditions simultaneously are marked in bold.

21 points can be exchanged without remainder only for three stickers and one stamp $(21=3 \cdot 5+1 \cdot 6)$. In this case, the number of stamps would not be double the number of stickers.

51 points can be exchanged without remainder either for three stickers and six stamps, or for nine stickers and one stamp $(51=3 \cdot 5+6 \cdot 6=9 \cdot 5+6)$. In the first case, the number of stamps is double the number of stickers.

Jara had at least 51 points before the exchange.

Note. The conditions discussed above for the unknown number of points are:

- the remainder when divided by five is one,
- the remainder when divided by six is three,
- the remainder when divided by seventeen is zero.

All natural numbers that satisfy these three conditions are of the form $51+5 \cdot 6 \cdot 17 \cdot k=51+510k$, where $k$ is any non-negative integer. The problem is related to the so-called Chinese Remainder Theorem.

Evaluation. 2 points for finding a system for checking possibilities (multiples of 17 in the first solution, correct remainders in the second solution); 2 points for solving the problem; 2 points for completeness and quality of the commentary.
Children at the camp earned points for completing tasks. These points could be further exchanged: five points for a sticker, six points for a badge. Jára calculated that if he wanted only stickers, one point would remain unused. If he chose only badges, three points would remain unused. In the end, he managed to use all his points, and he received twice as many badges as stickers.

How many points did Jára have before exchanging?

(E. Semerádová)