9.5. Let $s(n)$ denote the sum of the digits of a natural number $n$. How many zeros does the number equal to the product $s(1) \cdot s(2) \cdot \ldots \cdot s(100)$ end with?
Answer: 19 zeros. Hint. Consider the numbers from the first hundred for which the sum of the digits is divisible by 5. Such numbers have a sum of digits of either 5, 10, or 15. There are 6 numbers with a sum of 5: these are $5,14,23,32,41,50$. There are 9 numbers with a sum of digits 10: these are $19,28, \ldots, 91$. There are 4 numbers with a sum of digits 15: these are $69,78,87,96$. Thus, in the product $P=s(1) \cdot s(2) \cdot \ldots \cdot s(100)$, the factor 5 appears 19 times. And the factor 2 in $P$ appears in a (much) higher degree (it is enough to consider, for example, numbers with a sum of digits 8; there are 9 such numbers, and each contributes a factor of $2^{3}$ to $P$, so the factor 2 appears in $P$ in a degree $\geq 3 \cdot 9=27$). Therefore, $P$ ends with 19 zeros.