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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.

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.

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.

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

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21 Dec 2012, 05:39

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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.

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

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26 Jun 2014, 07:53

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.

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?? Please explain in detail

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.

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?? Please explain in detail

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.

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.

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

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01 Dec 2015, 08:44

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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 2016, 18:56

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