If x = a!, which of these following values of a is the least possible value (in this list) for which the last 10 digits of the integer x will all be zero? Choose the correct answer.
A) 16 B) 27 C) 38 D) 44 E) 63
2*5 will give one 0 at the end. We need 2^10*5^10 to get 10 zeroes at the end.   There are many 2's, so we need to find the first 10 appearances of 5.   These multiples of 5 are: 5, 10, 15, 20, 5*5, 30, 35, 40, 45...which gives us 10.   To have at least 10 zeroes at the end, a =&gt; 45.   The answer is 63.
<result>E</result>
If x = a!, which of these following values of a is the least possible value (in this list) for which the last 10 digits of the integer x will all be zero?