2. A natural number ends with zero, and the greatest of its divisors, not equal to itself, is a power of a prime number. Find the second-to-last digit of this number.
Answer: 1 or 5. Solution: A natural number is divisible by 2 and 5. Then its greatest proper divisor is half of the number, and the number itself has the form $2 \cdot 5^{k}$. For $k=1$, the second-to-last digit will be 1, and for $k>1$ it will be 5, since $5^{k}$ in this case ends in 25.