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

Question

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.

Bunuel · Answer

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 . Answer: A.

feruz77 · Answer

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.

arp · Answer

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

amit2k9 · Answer

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.

bhavinshah5685 · Answer

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

Chandni170 · Answer

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.

jlgdr · Answer

...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

elmouj · Answer

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...

Donnie84 · Answer

Let p = 1. p! = 1! = 1, which means y= 0

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

1 = 1 + 0 = p + y

Answer (A).

TGC · Answer

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

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