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If p is a natural number and p! ends with y trailing zeros, then the number of zeros that (5p)! ends with will be

a) (p+y) trailing zeros b) (5p+y) trailing zeros c) (5p+5y) trailing zeros d) (p+5y) trailing zeros e) none of them above

Can someone help me how to solve this question? I think, there must be more than one solution method.

Given: \(p!\) has \(y\) trailing zeros: \(\frac{p}{5}+\frac{p}{5^2}+\frac{p}{5^3}+...=y\) (check for theory on this topic: everything-about-factorials-on-the-gmat-85592.html) --> now, # of trailing zeros for \((5p)!\) will be \(\frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+...=p+(\frac{p}{5}+\frac{p}{5^2}+...)=p+y\).

for 5p, 50/5 = 10, 50/25 = 2 hence total is 12 Similarly, for p = 20 trailing 0's = 20/5 = 4 for 5p, trailing 0's = 50/5 = 10, 50/25 = 2 hence total is 10 + 2 = 12 thus we observe, number of 0's = p + y example p = 20, y = 4 giving 24.

Hence A.
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If p is a natural number and p! ends with y trailing zeros, then the number of zeros that (5p)! ends with will be

a) (p+y) trailing zeros b) (5p+y) trailing zeros c) (5p+5y) trailing zeros d) (p+5y) trailing zeros e) none of them above

Can someone help me how to solve this question? I think, there must be more than one solution method.

Given: \(p!\) has \(y\) trailing zeros: \(\frac{p}{5}+\frac{p}{5^2}+\frac{p}{5^3}+...=y\) (check for theory on this topic: everything-about-factorials-on-the-gmat-85592.html) --> now, # of trailing zeros for \((5p)!\) will be \(\frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+...=p+(\frac{5p}{5}+\frac{5p}{5^2}+...)=p+y\).

Answer: A.

Hello Bunuel.... There is a typo in the equation just before putting down p+y... others may get confused....

If p is a natural number and p! ends with y trailing zeros, then the number of zeros that (5p)! ends with will be

a) (p+y) trailing zeros b) (5p+y) trailing zeros c) (5p+5y) trailing zeros d) (p+5y) trailing zeros e) none of them above

Can someone help me how to solve this question? I think, there must be more than one solution method.

Given: \(p!\) has \(y\) trailing zeros: \(\frac{p}{5}+\frac{p}{5^2}+\frac{p}{5^3}+...=y\) (check for theory on this topic: everything-about-factorials-on-the-gmat-85592.html) --> now, # of trailing zeros for \((5p)!\) will be \(\frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+...=p+(\frac{5p}{5}+\frac{5p}{5^2}+...)=p+y\).

Answer: A.

Hello Bunuel.... There is a typo in the equation just before putting down p+y... others may get confused....

Re: If p is a natural number and p! ends with y trailing zeros [#permalink]

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11 Feb 2014, 09:48

Chandni170 wrote:

Hi Bunuel,

I'm trying to understand the explanation here and am unable to understand how we got the last step:

now, # of trailing zeros for (5p)! will be \frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+...=p+(\frac{p}{5}+\frac{p}{5^2}+...)=p+y.

I understand that the # of trailing zeros for (5p)! will be \frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+...

But how is this equal to p+(\frac{p}{5}+\frac{p}{5^2}+...)=p+y?

Infact when I solve \frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+... I factor out 5 and get 5(\frac{p}{5}+\frac{p}{5^2}+\frac{p}{5^3}+...)= 5y

This might be a silly question and I'm definitely missing something out here... but can't figure out where I'm going wrong.

Kindly help me out.

Thanks.

P.S: This is the first time I'm posting on this forum... Not sure If I've done it right. Please let me know if anything needs to be changed.

Chandni170 . you have a problem with your last denominator.... - assume the last denominator for the P! division is 5^n . - then the last denominator of (5P)! is not 5^n but 5^(n+1) . now if you factor out 5 you will end up with 5 *(y+P/5^(n+1)) and not 5*y . Hope it helps...

Re: If p is a natural number and p! ends with y trailing zeros [#permalink]

Show Tags

20 Apr 2014, 09:05

p=5 p!= 120 has one zero

So, p=5 then y=1

Then

5p! = 25!

Counting zeros in 25!

5*2 =10 10 15*6 = 90 20 24*5=120 25*4=100

So 25!= 5p! has six zeros

Number of zeros in 5p! = 6 = 5 + 1 = p +y
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