74. How many zeros does the product of all integers from 1 to 100 inclusive end with?

In the future, we will use the following notation:

$$
1 \cdot 2 \cdot 3 \cdot 4 \ldots(n-1) \cdot n=n!
$$

(read as: $n$ factorial). Thus, the previous problem can be more succinctly stated as: how many zeros does the number 100! end with?
74. The number of bullets at the end of a number indicates how many times 10 is a factor in that number. The number 10 is equal to the product of $2 \cdot 5$; in the product of all integers from 1 to 100, the factor 2 appears in a higher power than the factor 5. Therefore, the product $1 \cdot 2 \cdot 3 \ldots 100$ is divisible by such a power of 10 (i.e., ends with such a number of zeros) as this product contains factors of 5. But up to 100, there are 20 numbers that are multiples of five, and four of them (25, 50, 75, and 100) are also multiples of 25, i.e., contain two factors of 5 each. Therefore, in the product $1 \cdot 2 \cdot 3 \ldots 100$, the number 5 appears as a factor 24 times; hence, there will be 24 zeros at the end of this product.