1. Dima wrote down three consecutive natural numbers, where the first (the smallest) is divisible by 5, the second is divisible by 4, and the third is divisible by 3. What numbers could Dima have written down? (Provide at least one example.)
# Solution:

For example, the following sequence works: 55, 56, 57 (these are the three smallest natural numbers with the required property). There are different ways to arrive at this solution. Firstly, a logical enumeration works: the first number must be divisible by 5 and also odd (so that the next number is divisible by 4), meaning the first number ends in 5. We can try $5, 15, 25, 35, 45$—none of these work (either the next number is not divisible by 4 or the one after that is not divisible by 3), but 55 works. Secondly, we can reason as follows: according to the condition, $n$ is divisible by 5, $n+1$ is divisible by 4, and $n+2$ is divisible by 3, from which it follows that $n+5$ is divisible by 5, 4, and 3. For example, the least common multiple (LCM) of $(5, 4, 3) = 60$, so $n = 55$. Of course, there are other examples (for instance, 115, 116, 117 or 655, 656, 657).

Naturally, in the olympiad participants' solutions, this is not required—full marks are given for simply providing at least one correct example.

## Criteria:

Correct example of three such numbers - 7 points,

no correct example - 0 points (including cases where some of the three numbers meet the condition and some do not).

