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# GRE Math Challenge #95-If (5^4 -1)/n is an integer

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GRE Math Challenge #95-If (5^4 -1)/n is an integer [#permalink]  09 May 2015, 11:39
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Question Stats:

100% (01:41) correct 0% (00:00) wrong based on 5 sessions
If $$\frac{(5^4 - 1)}{n}$$ is an integer an n is an integer, then n could be each of the following EXCEPT

(A) 4
(B) 6
(C) 13
(D) 25
(E) 26
[Reveal] Spoiler: OA

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Sandy
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Re: GRE Math Challenge #95-If (5^4 +1)/n is an integer [#permalink]  19 Dec 2017, 11:09
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sandy wrote:
If $$\frac{(5^4 - 1)}{n}$$ is an integer an n is an integer, then n could be each of the following EXCEPT

(A) 4
(B) 6
(C) 13
(D) 25
(E) 26

Here,

$$\frac{(5^4 - 1)}{n}$$ = $$\frac{(625 - 1)}{n}$$ = $$\frac{624}{n}$$

Since n is an integer and the fraction $$\frac{624}{n}$$ is an integer so the value of n could be 4,6,13 or 26.

Hence n cannot be equal to 25.
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Re: GRE Math Challenge #95-If (5^4 -1)/n is an integer [#permalink]  19 Dec 2017, 12:42
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$$\frac{(5^4 - 1)}{n}$$

=$$\frac{624}{n}$$

=$$\frac{(2^4*3*13)}{n}$$

n could not be 25
Director
Joined: 20 Apr 2016
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WE: Engineering (Energy and Utilities)
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Kudos [?]: 603 [0], given: 124

Re: GRE Math Challenge #95-If (5^4 -1)/n is an integer [#permalink]  19 Dec 2017, 21:35
shawon wrote:
$$\frac{(5^4 - 1)}{n}$$

=$$\frac{624}{n}$$

=$$\frac{(2^4*3*13)}{n}$$

n could not be 25

I have edited to the correct mathematical expression

Here there is also an explanation how to use correct mathematical expression on forum. Please read before posting
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Re: GRE Math Challenge #95-If (5^4 -1)/n is an integer   [#permalink] 19 Dec 2017, 21:35
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