If x = a!, which of these following values of a is the least possible value (in this list) for which the last 9 digits of the integer x will all be zero?  Choose the correct answer:
A) 26 B) 32 C) 37 D) 44 E) 49
2*5 will give one 0 at the end. We need 2^9*5^9 to get 9 zeroes at the end.   There are many 2's, so we need to find the first 9 appearances of 5.   These multiples of 5 are: 5, 10, 15, 20, 5*5, 30, 35, 40...which gives us 9.   To have at least 9 zeroes at the end, a =&gt; 40.   The answer is 44.
<result>D</result>
If x = a!, which of these following values of a is the least possible value (in this list) for which the last 9 digits of the integer x will all be zero?