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M26-04

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Math Expert
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M26-04  [#permalink]

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New post 16 Sep 2014, 01:24
1
14
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

51% (01:38) correct 49% (01:53) wrong based on 139 sessions

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Math Expert
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Re M26-04  [#permalink]

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New post 16 Sep 2014, 01:24
8
6
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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Re M26-04  [#permalink]

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New post 03 Mar 2015, 05:49
1
I think this question is good and helpful.
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M26-04  [#permalink]

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New post 16 Jan 2018, 09:31
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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
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Re: M26-04  [#permalink]

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New post 23 Jul 2018, 10:45
Wow, this is a great question. I would have guessed and skipped this question if I saw it on the real exam.
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Re: M26-04  [#permalink]

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New post 25 Jul 2018, 10:21
hanamana wrote:
Wow, this is a great question. I would have guessed and skipped this question if I saw it on the real exam.


Tricky but good question!
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Re: M26-04  [#permalink]

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New post 21 Feb 2019, 13:04
beautiful question
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Re: M26-04   [#permalink] 21 Feb 2019, 13:04
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