How many zeroes are at the end of $42!$ (42 factorial)?  (Reminder: The number $n!$ is the product of the integers from 1 to $n$.  For example, $5!=5\cdot 4\cdot3\cdot2\cdot 1= 120$.)
I know that the zeroes at the end of a number come from the factors of 10 in its prime factorization. To find the factors of 10, I need to count how many times 2 and 5 appear as factors in the product $42!$. I notice that there are more factors of 2 than factors of 5, so the number of factors of 5 will determine the number of zeroes. To count the factors of 5, I can use a trick: I divide 42 by 5, and get 8 with a remainder of 2. This means that there are 8 multiples of 5 between 1 and 42, namely 5, 10, 15, 20, 25, 30, 35, and 40. Each of these contributes one factor of 5, except for 25, which contributes two factors of 5, since $25=5\cdot 5$. So the total number of factors of 5 is 8+1=9. Therefore, there are 9 zeroes at the end of $42!$. 