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# (2^2-1)(2^2+1)(2^4+1)(2^8+1)=

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Manager
Joined: 16 Feb 2012
Posts: 220

Kudos [?]: 408 [1], given: 121

Concentration: Finance, Economics

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05 Jul 2012, 02:26
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Difficulty:

15% (low)

Question Stats:

79% (00:51) correct 21% (01:52) wrong based on 149 sessions

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(2^2-1)(2^2+1)(2^4+1)(2^8+1)=

A. 2^16 - 1
B. 2^16 + 1
C. 2^32 - 1
D. 2^128 - 1
E. 2^16(2^16-1)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 05 Jul 2012, 02:53, edited 1 time in total.
Edited the question.

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05 Jul 2012, 02:55
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Stiv wrote:
(2^2-1)(2^2+1)(2^4+1)(2^8+1)=

A. 2^16 - 1
B. 2^16 + 1
C. 2^32 - 1
D. 2^128 - 1
E. 2^16(2^16-1)

Apply $$(a-b)(a+b)=a^2-b^2$$:

$$(2^2-1)(2^2+1)(2^4+1)(2^8+1)=(2^4-1)(2^4+1)(2^8+1)=(2^8-1)(2^8+1)=2^{16}-1$$.

Hope it's clear.
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25 Aug 2014, 09:24
Hello from the GMAT Club BumpBot!

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03 Sep 2014, 02:45
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$$(2^2-1)(2^2+1)(2^4+1)(2^8+1)= 2^{(2+2+4+8)} - 1 = 2^{16} - 1$$

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28 Nov 2015, 15:56
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: (2^2-1)(2^2+1)(2^4+1)(2^8+1)=   [#permalink] 28 Nov 2015, 15:56
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