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by rakeshd347 » Wed Oct 02, 2013 7:16 pm
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

Is it GMAT like question?
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by theCodeToGMAT » Wed Oct 02, 2013 8:37 pm
20! to 24! ==> Each will give power of 4
25! to 29! ==> Each will give power of 6
30! to 33! ==> Each will give power of 7

So, 6(4*5 + 6*5 + 7*4) = 468

Answer [spoiler]{A}[/spoiler]
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by ganeshrkamath » Wed Oct 02, 2013 11:31 pm
rakeshd347 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

Is it GMAT like question?
The power of 3! = 6 indicates that the number of zeros = number of zeros in (20!*21!*...*33!) * 6
So search for an option that is a multiple of 6.
Choose A

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