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If p is a natural number and p! ends with y trailing zeros
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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.
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Re: trailing zeros question (logical approach needed)
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25 Jan 2011, 07:10
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: everythingaboutfactorialsonthegmat85592.html) > now, # of trailing zeros for \((5p)!\) will be \(\frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+...=p+(\frac{p}{5}+\frac{p}{5^2}+...)=p+y\). Answer: A.
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Re: trailing zeros question (logical approach needed)
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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: trailing zeros question (logical approach needed)
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03 May 2011, 10:44
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)
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03 May 2011, 20:34
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: If p is a natural number and p! ends with y trailing zeros
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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
which is our answer...opt A



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Re: If p is a natural number and p! ends with y trailing zeros
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26 Sep 2013, 00:20
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: If p is a natural number and p! ends with y trailing zeros
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25 Dec 2013, 14:46
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
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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
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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
Answer (A).



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