7.4. How many zeros does the product $s(1) \cdot s(2) \cdot \ldots \cdot s(100)$ end with, where $s(n)$ denotes the sum of the digits of the natural number $n$?
Answer: .19 zeros. Solution. Consider the numbers from the first hundred for which the sum of the digits is divisible by 5. Such numbers have a digit sum of either 5, 10, or 15. There are 6 numbers with a digit sum of 5: these are $5, 14, 23, 32, 41, 50$. There are 9 numbers with a digit sum of 10: these are $19, 28, \ldots, 91$. There are 4 numbers with a digit sum of 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. The factor 2 in $P$ appears in a (much) higher degree (it is sufficient to consider, for example, numbers with a digit sum of 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 $\left.\geq 3 \cdot 9 = 27\right)$. Therefore, $P$ ends with 19 zeros.