Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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]
11 Feb 2014, 08:48

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...