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# M26-04

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Math Expert
Joined: 02 Sep 2009
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M26-04 [#permalink]

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16 Sep 2014, 00:24
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Question Stats:

35% (01:37) correct 65% (02:15) wrong based on 23 sessions

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If $$x=23^2*25^4*27^6*29^8$$ and is a multiple of $$26^n$$, where $$n$$ is a non-negative integer, then what is the value of $$n^{26}-26^n$$?

A. -26
B. -25
C. -1
D. 0
E. 1
[Reveal] Spoiler: OA

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Posts: 43831
Re M26-04 [#permalink]

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16 Sep 2014, 00:24
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Official Solution:

If $$x=23^2*25^4*27^6*29^8$$ and is a multiple of $$26^n$$, where $$n$$ is a non-negative integer, then what is the value of $$n^{26}-26^n$$?

A. -26
B. -25
C. -1
D. 0
E. 1

$$23^2*25^4*27^6*29^8=odd*odd*odd*odd=odd$$ so $$x$$ is an odd number. The only way for it to be a multiple of $$26^n$$ (even number in integer power) is when $$n=0$$, in this case $$26^n=26^0=1$$ and 1 is a factor of every integer. Thus $$n=0$$ therefore $$n^{26}-26^n=0^{26}-26^0=0-1=-1$$.

Must know for the GMAT: $$a^0=1$$, for $$a \ne 0$$ - any nonzero number to the power of 0 is 1.

Important note: the case of $$0^0$$ is not tested on the GMAT.

Answer: C
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Joined: 24 Nov 2014
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Re M26-04 [#permalink]

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03 Mar 2015, 04:49
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I think this question is good and helpful.
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Marty Murray
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m.w.murray@hotmail.com
http://infinitemindprep.com

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Posts: 11
M26-04 [#permalink]

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16 Jan 2018, 08:31
another way to solve this is by noting that the numbers in the stem can be easily prime factorised and therefore
x = (23^2) x (5^8) x (3^18) x (29^8) and 26^n = (2^n)(13^n)
since x has neither 2 as a prime factor, nor 13, n must be 0. so the answer to the question = 0 -1 = -1
M26-04   [#permalink] 16 Jan 2018, 08:31
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# M26-04

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