Find the smallest natural number such that its half is divisible by three, its third is divisible by four, its quarter is divisible by eleven, and its half gives a remainder of five when divided by seven.

(E. Patáková)
If half of a number is divisible by three, then the number must be divisible by both 2 and 3. For similar reasons, it must also be divisible by 3 and 4, and by 4 and 11.

The least common multiple of all these numbers is the product $3 \cdot 4 \cdot 11 = 132$; the sought number must be a multiple of 132. Half of the sought number is thus a multiple of 66, and it only remains to check the remainder when divided by seven:

| half of the sought number | remainder when divided by seven |
| :---: | :---: |
| 66 | 3 |
| 132 | 6 |
| 198 | 2 |
| 264 | 5 |

The smallest multiple of 66 that gives a remainder of 5 when divided by seven is 264. The sought number is thus $2 \cdot 264 = 528$.

Scoring. 1 point for breaking down the problem (e.g., that the first piece of information means divisibility by 2 and 3); 3 points for finding the least common multiple 132 and interpreting that the sought number is a multiple of this number, or its equivalent (the term "multiple" does not have to be explicitly stated); 2 points for finding the number 528 (the process must clearly rule out all smaller numbers).