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Find the number of trailing zeros in the product of (1^1)*(5

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Find the number of trailing zeros in the product of (1^1)*(5 [#permalink] New post 25 Jan 2011, 06:42
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Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)………. *(50^50).

A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225

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 [#permalink] New post 25 Jan 2011, 07:24
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feruz77 wrote:
Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)………. *(50^50).
A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225

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


We have a trailing zero when we multiplying 2 by 5. Now, in the product: (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)*...*(50^50) there will be obviously more 5-s than 2-s so the # of 2-s will be limiting factor for the # of trailing zeros.

So we should count # of 2-s for even bases, basically we should factor out 2-s: 10^{10}*20^{20}*30^{30}*40^{40}*50^{50}=
=2^{10}*(2^2)^{20}*(2)^{30}*(2^3)^{40}*2^{50}*(something)=2^{10+40+30+120+50}*(something)=2^{250}*(something).

So there will be 250 trailing zeros in the above product as there are at least as many 5-s as 2-s.

Answer: C.

Theory on this topic: everything-about-factorials-on-the-gmat-85592.html
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Re: Trailing zeros question [#permalink] New post 29 Jan 2011, 02:12
Thanks Bunuel.

My remark:
Because of conditions of the stem, I think, this question is an exclusion from the general approach where one must count 5s a number of which in factorials are usually less or equal to a number of 2s.
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Re: Trailing zeros question [#permalink] New post 15 Feb 2011, 16:53
Can someone please submit correct answer ? I think answer is 150
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Re: Trailing zeros question [#permalink] New post 16 Feb 2011, 03:52
girlinslc wrote:
Can someone please submit correct answer ? I think answer is 150


Welcome to Gmat Club!

OA is given under the spoiler in the first post (and it's C). Solution is given in the second post. Please ask if anything needs further clarification.
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Re: Trailing zeros question [#permalink] New post 18 Feb 2011, 11:02
great. thanks for the post. +1
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Re: Find the number of trailing zeros in the product of (1^1)*(5 [#permalink] New post 21 Dec 2012, 05:39
Looking at the numbers it looks like

(1x5)^5
(2x5)^10
...
(10x5)^50

1. Determine the limiting factor. Is it 2 or is it 5? We know that all the numbers are multiple of 5 but not of 2. Thus, the limiting factor in this case is 2. Let's drop all the 5. Then, we count factors of 2 of even multiples.

2^10 = 10
4^20 = 20 + 20
6^30 = 30
8^40 = 40 + 40 + 40
10^50 = 50

250

Answer: C

More examples of Trailing Zeroes: Arithmetic: Trailing Zeroes
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Re: Trailing zeros question [#permalink] New post 09 Mar 2013, 22:50
Bunuel wrote:
feruz77 wrote:
Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)………. *(50^50).
A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225

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


We have a trailing zero when we multiplying 2 by 5. Now, in the product: (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)*...*(50^50) there will be obviously more 5-s than 2-s so the # of 2-s will be limiting factor for the # of trailing zeros.

So we should count # of 2-s for even bases, basically we should factor out 2-s: 10^{10}*20^{20}*30^{30}*40^{40}*50^{50}=
=2^{10}*(2^2)^{20}*(2)^{30}*(2^3)^{40}*2^{50}*(something)=2^{10+40+30+120+50}*(something)=2^{250}*(something).

So there will be 250 trailing zeros in the above product as there are at least as many 5-s as 2-s.

Answer: C.

Theory on this topic: everything-about-factorials-on-the-gmat-85592.html


Dear Bunnel :)
I think the number of trailing zeroes in the solution should be 250 and not 10^250. 10^250 will be a factor of the product but the stem requires us to calculate the number of trailing zeroes which would be 250 as per the solution provided by you. Please explain what am I missing here :roll:
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Re: Trailing zeros question [#permalink] New post 10 Mar 2013, 06:14
Dipankar6435 wrote:
Bunuel wrote:
feruz77 wrote:
Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)………. *(50^50).
A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225

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


We have a trailing zero when we multiplying 2 by 5. Now, in the product: (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)*...*(50^50) there will be obviously more 5-s than 2-s so the # of 2-s will be limiting factor for the # of trailing zeros.

So we should count # of 2-s for even bases, basically we should factor out 2-s: 10^{10}*20^{20}*30^{30}*40^{40}*50^{50}=
=2^{10}*(2^2)^{20}*(2)^{30}*(2^3)^{40}*2^{50}*(something)=2^{10+40+30+120+50}*(something)=2^{250}*(something).

So there will be 250 trailing zeros in the above product as there are at least as many 5-s as 2-s.

Answer: C.

Theory on this topic: everything-about-factorials-on-the-gmat-85592.html


Dear Bunnel :)
I think the number of trailing zeroes in the solution should be 250 and not 10^250. 10^250 will be a factor of the product but the stem requires us to calculate the number of trailing zeroes which would be 250 as per the solution provided by you. Please explain what am I missing here :roll:


I think it should be clear that there are typos in answer choices.
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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Re: Find the number of trailing zeros in the product of (1^1)*(5 [#permalink] New post 10 Mar 2013, 10:29
Thanx Bunnel.....U just validated my doubts :-D (Apologies if the error was too conspicuous to be obvious :wink: )
Re: Find the number of trailing zeros in the product of (1^1)*(5   [#permalink] 10 Mar 2013, 10:29
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