If x = a!, which of these following values of a is the least possible value (in this list) for which the last 7 digits of the integer x will all be zero? Choose the correct choice from the following options
A) 28	B) 32	C) 36	D) 40	E) 44
2*5 will give one 0 at the end. We need 2^7*5^7 to get 7 zeroes at the end.   There are many 2's, so we need to find the first 7 appearances of 5.   These multiples of 5 are: 5, 10, 15, 20, 5*5, 30...which gives us 7.   To have at least 7 zeroes at the end, a =&gt; 30.   The answer is 32.
<result>B</result>
If x = a!, which of these following values of a is the least possible value (in this list) for which the last 7 digits of the integer x will all be zero?