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 actual GMAT?
Asked: Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.
(20!*21!*22! ……… *33!)^3!.
Highest Power of 5 in 20! = 4
Highest Power of 5 in 21! = 4
Highest Power of 5 in 22! = 4
Highest Power of 5 in 23! = 4
Highest Power of 5 in 24! = 4
Highest Power of 5 in 25! = 5 + 1 = 6
Highest Power of 5 in 26! = 5 + 1 = 6
Highest Power of 5 in 27! = 5 + 1 = 6
Highest Power of 5 in 28! = 5 + 1 = 6
Highest Power of 5 in 29! = 5 + 1 = 6
Highest Power of 5 in 30! = 6 + 1 = 7
Highest Power of 5 in 31! = 6 + 1 = 7
Highest Power of 5 in 32! = 6 + 1 = 7
Highest Power of 5 in 33! = 6 + 1 = 7
Number of trailing zeros in (20!*21!*22! ……… *33!)^3!. = (4*5 + 6*5 + 4*7)*6 = (20 + 30 + 28)*6 = 78*6 = 468
IMO A
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