62nd Ukrainian National Mathematical Olympiad

Answer: $k\in 1,2,4$

Suppose that $k\ge 5$. It is clear that among any $k$ consecutive numbers, there is one that is divisible by $5$, and one that is divisible by $2$, so their product ends in $0$, hence $k$ is divisible by $10$. It is clear that then $k\ge 10$, so in the product of $k$ consecutive numbers there are at least two numbers divisible by $5$ and at least two numbers divisible by $2$, and one of them is divisible by $k$. Hence, the product is divisible by $10k$, so there will be more zeros at the end than zeros at the end of $k$, which is where we get the contradiction. Thus, $k\le 4$.

If $k=3$, then the product will be even, but it cannot end in the digit $3$.

For $k=1,2,4$ it is enough to consider the following examples: $1$, $1\cdot 2=2$, $1\cdot 2\cdot 3\cdot 4=24$.