Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.
Choices
A) 468	B) 469	C) 470	D) 467	E) 471
Let me give you one quick way to solve such questions.   3! = 6 =&gt; the answer must be of the form 106n106n   Only A and C have even powers. Hence B C E are out   Out of A and C only A's power is divisible by 3, hence C is out.   Hence 468 is the answer
<result>A</result>
Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.