G7.1 There are $a$ zeros at the end of the product $1 \times 2 \times 3 \times \ldots \times 100$. Find $a$.
When each factor of 5 is multiplied by 2 , a trailing zero will appear in the product. The number of factors of 2 is clearly more than the number of factors of 5 in 100! It is sufficient to find the number of factors of 5 .
$5,10,15, \ldots, 100$; altogether 20 numbers, have at least one factor of 5 . $25,50,75,100$; altogether 4 numbers, have two factors of 5 .
$\therefore$ Total number of factors of 5 is $20+4=24$
There are 24 trailing zeros of 100 !
$$
\Rightarrow a=24
$$