If x = a!, which of these following values of a is the least possible value (in this list) for which the last 8 digits of the integer x will all be zero?  Pick
A) 21  B) 29  C) 37  D) 41  E) 80
2*5 will give one 0 at the end. We need 2^8*5^8 to get 8 zeroes at the end.   There are many 2's, so we need to find the first 8 appearances of 5.   These multiples of 5 are: 5, 10, 15, 20, 5*5, 30, 35...which gives us 8.   To have at least 8 zeroes at the end, a =&gt; 35.   The answer is 37.
<result>C</result>
If x = a!, which of these following values of a is the least possible value (in this list) for which the last 8 digits of the integer x will all be zero?