feruz77 wrote:
Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.
A. 468
B. 469
C. 470
D. 467
E. 471
Can someone help me how to solve this question? I think, there must be more than one solution method.
Do questions of such a level of difficulty appear on the actualGMAT?
It is not difficult question , rather it tests the concept of trailing zero's,
(20!) has 4
21! has 4
22! has 4
23! has 4
24! has 4
25! has(25!/5+25!/5^2\(\) ) =6
26! has 6
27! has 6
28! has 6
29! has 6
30!
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