If 73! has 16 zeroes at the end, how many zeroes will 80! have at the end? Choose the correct choice from the following answers:
A) 16
B) 17
C) 18
D) 19
E) 20
Prime Factoring: you need a 2 and a 5 to make a 10 (a "zero"), and there are TONS of 2's so let's skip these and focus on the (rarer) 5s:   80! = 1*2*3*4*5*6...*78*79*80   Since there are 80 consecutive numbers, there are 16 multiples of 5 in there, but if we're prime factoring, we need to remember that SOME multiples of 5 actually contain more than just one 5. Which? 25 comes to mind -- it's got two of them! So all the multiples of 25 actually contain two 5's (ie: 50 and 75)   So, to recap, we have 16 of them, plus 3 more (the additional 5's in 25, 50, and 75), so that makes 19, and since we have more than enough 2's, we know our number will have exactly 19 zeros at the end.   ANSWER:19
<result>D</result>
If 73! has 16 zeroes at the end, how many zeroes will 80! have at the end?