1. Can we find five different natural numbers, each of which is not divisible by 3, 4, or 5, but the sum of any three is divisible by 3, the sum of any four is divisible by 4, and the sum of all five is divisible by 5?
Answer: for example, 1, 61, 121, 181, and 241.

Solution. The five numbers mentioned can be written in the form $60k + 1$, where $k$ takes the values 0, 1, 2, 3, 4. Therefore, the sum of any three numbers

$$
\left(60k_{1} + 1\right) + \left(60k_{2} + 1\right) + \left(60k_{3} + 1\right) = 60\left(k_{1} + k_{2} + k_{3}\right) + 3
$$

is divisible by 3. Similarly, it can be proven that the sums of any four and the sum of all five numbers are divisible by 4 and 5, respectively.

Remark. It can be proven that all numbers have the same remainder when divided by 12.

Criteria. Any correct example - 7 points. Noted that the numbers have the same remainder when divided by 3 or 4, but an incorrect example is provided -2 points.