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GRE Math Challenge #39-positive integer n such that 2^n

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GRE Math Challenge #39-positive integer n such that 2^n [#permalink]  09 Sep 2014, 01:29
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Question Stats:

83% (00:21) correct 16% (00:00) wrong based on 12 sessions
What is the greatest positive integer n such that $$2^n$$ is a factor of $$12^{10}$$?

A. 10
B. 12
C. 16
D. 20
E. 60
[Reveal] Spoiler: OA

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Sandy
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Re: GRE Math Challenge #39-positive integer n such that 2^n [#permalink]  17 Jul 2018, 14:31
sandy wrote:
What is the greatest positive integer n such that $$2^n$$ is a factor of $$12^{10}$$?

A. 10
B. 12
C. 16
D. 20
E. 60

Given

$$12^{10}$$

= $$(2*2*3)^{10}$$

= $$2^{10}$$ $$2^{10}$$ $$3^{10}$$

= $$2^{10+10}$$ $$3^{10}$$

So , $$2^{20}$$ and $$3^{10}$$

n has the maximum value of 20 as a power of 2.

Re: GRE Math Challenge #39-positive integer n such that 2^n   [#permalink] 17 Jul 2018, 14:31
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