Beat The GMAT

vkb16 · Tue Aug 11, 2009 5:26 am

If 60! is written out as an integer, with how many consecutive 0's will that integer end?

6
12
14
42
56

OA is C but I think is B - my method is--
from 60 to 51, there are two 10s (60, and 65x62)
from 50-41 there are two 10s (50, and 55x52)
similarly, there are in total 6x2=12 zeroes.
I didnt quite get mgmat's explanation

Thanks

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DanaJ · Tue Aug 11, 2009 6:29 am

You see this type of question a lot. You're looking for the number of trailing zeroes in 60!, which means that you need the number of 10s in 60!. Think of it this way: 7000 has three trailing zeroes, since 7000 = 10*10*10*7 (three 10s).

The number of 10s is actually provided by the number of 5s in 60!, since 10 = 2*5. There are plenty of 2s in 60! (at least one every other number, in even numbers), so 5s are more rare. This is why you're trying to count those.

There's this really short formula for finding out the number of 5s in 60! (a formula discovered on this forum):

60/5 + 60/(5^2) = 12 + 2 = 14 (we only take the quotient)

What you have to do is divide the factorial by the powers of 5 up to the greatest power smaller than 60 (5^2 = 25 - fits; 5^3 = 125 - doesn't fit).

Another example: The number of trailing zeroes in 140:

140/5 + 140/(5^2) + 140/(5^3) = 28 + 5 + 1 = 34.

LATER EDIT: the reason you got only 12 zeroes is because you didn't count the fact that 25 has two 5s and so does 50. While you did count them once, they should be counted twice!

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ket · Fri Nov 20, 2009 5:23 am

I am confused about this problem. I am ok with explanation of C it seems logical. I just wanted to check and calculated factorial of 60 in EXCEl with formula FACT(60), and the result is not 14 zeros :O ... I must be doing something wrong but the excel answer is this (23 zeros!)

8,320,987,112,741,390,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

would be nice to hear smbdy's explanation...

Cheers Smile