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trailing numbers

Expert replies
Source: — Problem Solving |

by Rich@VeritasPrep » Fri Jun 18, 2010 6:41 am
What they're really asking is:

How many factors of 10 are there in 25!. A factor of 10 corresponds to an extra zero.

10 is equal to 2 times 5, so we need both a factor of 2 and a factor of 5 to create a factor of 10.

You're multiplying 25*24*23* ... *3*2*1.

Of the numbers, only 25, 20, 15, 10, and 5 have a factor of 5, and there are two factors of 5 in 25. That's a total of 6 factors of 5.

There are obviously many more factors of 2, but you can only get 6 factors of 10.

There will be 6 trailing zeros.
Last edited by Rich@VeritasPrep on Fri Jun 18, 2010 7:18 am, edited 2 times in total.
Rich Zwelling
GMAT Instructor, Veritas Prep
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by francoisph » Fri Jun 18, 2010 6:47 am
you missed 25=5^2

answer will be 6 I think
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by jube » Fri Jun 18, 2010 6:55 am
No. of 0s depend on the no. of 10s in 25!
No. of 10s depend on no. of 2s & 5s.

So 25! has:
No. of 2s - 17
No. of 5s- 6

Therefore no. of 10s will be 6 (since 5 is just 6). Therefore, 25! will have 6 zeroes after the rightmost non-zero digit
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by mj78ind » Fri Jun 18, 2010 6:57 am
francoisph wrote:How many trailing zeros will be there after the rightmost non-zero digit in the value of 25!?
Any combination of 5 and 2 AND 10 will give one trailing zero:

5 X even # (from the 5 in 25!)
10
3 X 5 X even # (from the 15 in 25!)
20
5 X even # AND 5 X even # (Note: 25 will give 2 trailing zeros and not one, one for each 5 in 25)

Hence a total of 6 trailing zeros.
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by Rich@VeritasPrep » Fri Jun 18, 2010 7:03 am
Indeed I did miss 5^2 = 25. Nice catch!

Correcting my previous post now :)
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by amising6 » Fri Jun 18, 2010 7:33 am
How many trailing zeros will be there after the rightmost non-zero digit in the value of 25!?

for zero you need to find number of 5 and 0

now to find number of 5 in 25!



here goes the formula 25/5 you get 5 quiotent
then you divide quiotent 5 by 5 you get 1
which is not furthur divided so number of 5 will be 5+1=6

similairly no of 2 in 25 !
25/2 you get 12 ,12 divided by 2=6 when 6 divided by 2 you get 3 when 3 divided by 2 you get 1
so number of 2 's (add all the quotient dont worry about remainder) 12+6+3+1=22 two's

now there are 6 5's along with 6 2's will give 6 trailing zeroes
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