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by sunilrawat » Mon Oct 03, 2011 7:38 am
If 60! is written out as an integer, with how many consecutive 0's will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

I calculated 7 0's from the product of 10, 20...60 and six from 2*5, 12*15, 22*25, 32*35, 42*45, 52*55
I am missing one 0...where?
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by rohit_gmat » Mon Oct 03, 2011 7:47 am
sunilrawat wrote:If 60! is written out as an integer, with how many consecutive 0's will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

I calculated 7 0's from the product of 10, 20...60 and six from 2*5, 12*15, 22*25, 32*35, 42*45, 52*55
I am missing one 0...where?
u shud count the number of 5s (& if needed 2s) to make a 10...
25 has 2 5s - so it actually 'contains' 2 zeros... (i.e. 25 x 4 x anyth = 100 x anything )[/b]
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by sl750 » Mon Oct 03, 2011 7:54 am
60/5 + 60/25

12+2 = 14 zeros. This is just a shortcut instead of having to count the number of 5's in 60!
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by sunilrawat » Mon Oct 03, 2011 7:56 am
I see from your example how u get the zeros from 5's.
But I don't get your logic. Can you please explain?
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by sunilrawat » Mon Oct 03, 2011 8:02 am
rohit_gmat wrote:
sunilrawat wrote:If 60! is written out as an integer, with how many consecutive 0's will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

I calculated 7 0's from the product of 10, 20...60 and six from 2*5, 12*15, 22*25, 32*35, 42*45, 52*55
I am missing one 0...where?
u shud count the number of 5s (& if needed 2s) to make a 10...
25 has 2 5s - so it actually 'contains' 2 zeros... (i.e. 25 x 4 x anyth = 100 x anything )[/b]
got it, thanks.
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