Find the number of trailing zeros in the product of (1^1)*(5 : GMAT Problem Solving (PS)
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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]

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25 Jan 2011, 05:42
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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. 150
B. 200
C. 250
D. 245
E. 225
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25 Jan 2011, 06:24
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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.

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Re: Find the number of trailing zeros in the product of (1^1)*(5 [#permalink]

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21 Dec 2012, 04:39
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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

More examples of Trailing Zeroes: Arithmetic: Trailing Zeroes
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Re: Find the number of trailing zeros in the product of (1^1)*(5 [#permalink]

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26 Nov 2014, 20:01
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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. 150
B. 200
C. 250
D. 245
E. 225

Responding to a pm:

The method discussed in my post is useful while finding the maximum power of a number in a factorial. The given product is not in factorial form and hence the method needs to be suitably modified. That said, it should not be a big problem to modify the method if you understand the basics. Zeroes are produced by multiplying a 2 and a 5. So number of 0s in this product will depend on how many matching 2s and 5s we have here. In factorials, we have more 2s than 5s because we have consecutive numbers so we usually don't bother about finding the number of 2s. Here we have handpicked numbers so we need to ensure that we have both.

From where do we get 2s? From even multiples of 5.
10^10, 20^20, 30^30, 40^40, 50^50
$$(2*5)^{10}, (2^2*5)^{20}, (2*15)^{30}, (2^3*5)^{40}, (2*25)^{50}$$
So number of 2s is 10 + 40 + 30 + 120 + 50 = 250

Now, let's see the number of 5s. Each term of the product has a 5. So the number of 5s is at least 5 + 10 + 15 + 20 + 25 + ... + 50
Then we also need to account for terms that have multiple 5s such as 25 and 50 but let's get to that later.

5 + 10 + 15 + 20 + 25 + ... + 50 = 5(1 + 2 + 3 + ..10) = 5*10*11/2 = 275

Note that number of 5s will be even more than 275 while the number of 2s is only 250. So there will be 250 trailing zeroes.
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29 Jan 2011, 01:12
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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15 Feb 2011, 15:53
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16 Feb 2011, 02:52
girlinslc wrote:

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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18 Feb 2011, 10:02
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]

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16 Jul 2013, 23:37
From 100 hardest questions
Bumping for review and further discussion.
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17 Jun 2014, 18:20
How come (2)^30 had no power inside with the 2?
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17 Jun 2014, 23:43
sagnik242 wrote:
How come (2)^30 had no power inside with the 2?

$$30^{30}=(2*15)^{30}=2^{30}*15^{30}$$
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Re: Find the number of trailing zeros in the product of (1^1)*(5 [#permalink]

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26 Jun 2014, 06:53
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.

Hi

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

Why we counted only 2? what does it mean by limiting factor and whats the importance of it??

Thanks a lot
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Posts: 35912
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Kudos [?]: 90031 [0], given: 10402

Re: Find the number of trailing zeros in the product of (1^1)*(5 [#permalink]

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26 Jun 2014, 07:11
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GGMAT730 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.

Hi

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

Why we counted only 2? what does it mean by limiting factor and whats the importance of it??

Thanks a lot

We have a trailing zero when we multiplying 2 by 5. So, each pair of 2 and 5 gives one more 0 at the end of the number. Our expression gives more 5's than 2's, so the number of 2 will determine the number of 0: for each 2 we have a 5, which when multiples will give 0.

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Re: Find the number of trailing zeros in the product of (1^1)*(5 [#permalink]

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01 Dec 2015, 07:44
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Re: Find the number of trailing zeros in the product of (1^1)*(5   [#permalink] 01 Dec 2015, 07:44
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