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# If p is a natural number and p! ends with y trailing zeros

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

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25 Jan 2011, 06:55
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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.
[Reveal] Spoiler: OA

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Re: trailing zeros question (logical approach needed) [#permalink]

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25 Jan 2011, 07:10
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feruz77 wrote:
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 $$\frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+...=p+(\frac{p}{5}+\frac{p}{5^2}+...)=p+y$$.

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Re: trailing zeros question (logical approach needed) [#permalink]

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03 May 2011, 10:44
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feruz77 wrote:
It is very logical and simple approach.
Excellent, Bunuel!

It seems to me there is no alternative solution method. If there is one I would appreciate your contributions. Thanks.

I solved it this way:

Since the p! has trailing zeros, Let's assume p=10

10! will have 2 trailing 0s (by the method provided by Bunuel)
p = 10 y = 2
5p! i.e 50! will have 12 trailing 0s = 10 + 2 = p + y

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Re: trailing zeros question (logical approach needed) [#permalink]

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03 May 2011, 20:34
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for p = 10 trailing 0's =
10/5 = 2

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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Re: trailing zeros question (logical approach needed) [#permalink]

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04 Apr 2013, 19:14
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Bunuel wrote:
feruz77 wrote:
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 $$\frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+...=p+(\frac{5p}{5}+\frac{5p}{5^2}+...)=p+y$$.

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

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Re: trailing zeros question (logical approach needed) [#permalink]

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05 Apr 2013, 04:21
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Expert's post
Amateur wrote:
Bunuel wrote:
feruz77 wrote:
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 $$\frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+...=p+(\frac{5p}{5}+\frac{5p}{5^2}+...)=p+y$$.

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

Thank you. Edited. +1.
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Re: If p is a natural number and p! ends with y trailing zeros [#permalink]

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26 Sep 2013, 00:20
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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.

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Re: trailing zeros question (logical approach needed) [#permalink]

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25 Jan 2011, 07:16
It is very logical and simple approach.
Excellent, Bunuel!

It seems to me there is no alternative solution method. If there is one I would appreciate your contributions. Thanks.

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

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16 Nov 2012, 18:05
suppose p=6
I am taking 6 because it will have 5 & 2 both to become trailing zeroes.

now 6! has 1 trailing zero.

=6/5 = 1 =y

so 5p = 5*6 = 30

new number is 30!.
Trailing zeroes in (5p)! = 30/5 + 30/25 = 6+1 = 7
now here we get 7=6+1=p+y

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

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25 Dec 2013, 14:46
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feruz77 wrote:
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.

...Or use smart numbers

p= 5! we have 1 trailing zero

5*5! = 25! we have 6 trailing zeroes

Since p = 5 and 'y' = 1

The only answer choice that will satisfy is A

Hope it helps
Cheers!
J

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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...

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

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11 Feb 2014, 11:11
Let p = 1. p! = 1! = 1, which means y= 0

(5p)! = 5! = 120, trailing zeros = 1

1 = 1 + 0 = p + y

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

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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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Re: If p is a natural number and p! ends with y trailing zeros   [#permalink] 20 Apr 2014, 09:05
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