1. Determine the greatest value of the natural number $n$ such that the number $39^{n}$ is a divisor of the number 39! Explain your answer.
Solution. We have $39=3 \cdot 13$. In the number 39!, the prime number 13 appears as a divisor of 13, 26, and 39. Therefore, the greatest value of the number $n$ is 3.