3. Apples were divided into two unequal piles. When half of the apples from the first pile were moved to the second pile, and then half of the apples from the second pile were moved back to the first pile, there were 18 apples in the first pile and 8 in the second. How many apples were in each pile originally?
3. The first method of solving.
1) $8 \times 2=16$ (beans) - was in the second pile after half was moved from the first pile;
2) $18-8=10$ (beans) - was in the first pile after half was moved from it;
3) $10 \times 2=20$ (beans) - was originally in the first pile;
4) $16-10=6$ (beans) - was originally in the second pile.

The second method of solving (Sofia Bekesova, student of Gymnasium No. 24).

| I pile | II pile |  |
| :---: | :---: | :--- |
| 18 | 8 | became after moving from the second pile to the first |
| 10 | 16 | became after moving half from the first pile |
| 20 | 6 | was originally |

Answer: originally, there were 20 in the first pile and 6 in the second.