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trailing zeros

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by sanju09 » Tue Feb 24, 2009 6:03 am
How many trailing zeros will be there after the rightmost non-zero digit in the value of 25!?

A. 25
B. 8
C. 6
D. 5
E. 2
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by DanaJ » Tue Feb 24, 2009 6:24 am
You've got plenty of 2's in there, so you're looking for the 5's (since 2*5 = 10).
You've got:
5 in 5
5 in 10
5 in 15
5 in 20
2 5's in 25
This makes 6 fives or 6 trailing zeros.
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by dendude » Tue Feb 24, 2009 10:19 am
DanaJ wrote:You've got plenty of 2's in there, so you're looking for the 5's (since 2*5 = 10).
You've got:
5 in 5
5 in 10
5 in 15
5 in 20
2 5's in 25
This makes 6 fives or 6 trailing zeros.
But are'nt we forgetting the 0's themselves in 10 and 20.
So there should be two more trailing zeroes.
Ans should be B. 8
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by odod » Tue Feb 24, 2009 10:24 am
I don't understanding anything from the post above..can you elaborate for me?
ODOD
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by DanaJ » Tue Feb 24, 2009 10:38 am
dendude: but if you take the 5's out of 10 and 20 you are left with "humble" 2 and 4 respectively. This is how I see it, but I may be wrong.

odod: What we are looking for is the number of trailing zeros, or the number of times you have the number 10 in 25! (since multiplying a number by 10 just adds a zero at the end of the number). Now, 10 = 2*5. There are plenty of 2's in 25!, since every even number has 2 in it. This means that the important stuff is actually 5, so you're looking for 5's in 1*2*3*...*24*25 = 25!. You have 5's in:
5 - one 5
10 - one 5
15 - one 5
20 - one 5
25 - two 5's (since 25 = 5^2).
This means that you have exactly 6 fives in 25! and, since there are plenty of 2's to go around, you have 10 times 6 in 25!.

Hope you understand now...
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by deepoe » Tue Feb 24, 2009 10:49 am
https://www.purplemath.com/modules/factzero.htm <- explains this:$ Now I understand this exercise too :$
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by Mr2Bits » Tue Feb 24, 2009 11:54 am
deepoe wrote:https://www.purplemath.com/modules/factzero.htm <- explains this:$ Now I understand this exercise too :$
Thanks for the link, learn something new everyday.
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by dendude » Tue Feb 24, 2009 12:20 pm
DanaJ wrote:dendude: but if you take the 5's out of 10 and 20 you are left with "humble" 2 and 4 respectively. This is how I see it, but I may be wrong.
DanaJ: Yes you're right! I see my mistake now. There would be only 6 trailing 0's
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by welcome » Tue Feb 24, 2009 12:41 pm
Good Link, great learning.

There is another fast was of doing this.

Devide by 25 once , and devide by 5 once, add both. (Take only quetioned.) (for less than 125.)

Eg. 25! ---(so devided 25 by 25 =1 )+ (25! so devided 25 by 5= 5) = 6.

101! ---(101/25) + (101/5) = 4+20 = 24.
Shubham.
590 >> 630 >> 640 >> 610 >> 600 >> 640 >> 590 >> 640 >> 590 >> 590
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by Vidushi Jain » Wed Aug 22, 2012 12:55 am
But when we divide 200/25+200/5= 48 where as 200! has 49 trailing zeros is there a seprate rule for no.s after 101!??
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by vk_vinayak » Wed Aug 22, 2012 2:14 am
Vidushi Jain wrote:But when we divide 200/25+200/5= 48 where as 200! has 49 trailing zeros is there a seprate rule for no.s after 101!??
You forgot 125, which has 3 multiples of 5.

200/125 + 200/25 + 200/5 = 49
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