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

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VP
Joined: 06 Feb 2007
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28 Jul 2007, 14:38
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Difficulty:

15% (low)

Question Stats:

81% (00:55) correct 19% (01:24) wrong based on 175 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)

I know that this question should be easy but I just can't figure out how to simplify!!!

OPEN DISCUSSION OF THIS QUESTION IS HERE: 2-135319.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Aug 2014, 00:34, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.

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Intern
Joined: 20 Jul 2005
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Location: Vietnam

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28 Jul 2007, 14:55
A: is correct

(2^2-1)* (2^2+1)=(2^4-1)
(2^4-1)*(2^4+1)=2^8-1
(2^8-1)*(2^8 +1)=2^16-1

You can apply to this formula: a^2- b^2=(a-b)*(a+b)
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Sonfbm

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Senior Manager
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28 Jul 2007, 16:33
Ans : A

Multiple use of the formula

a-b * a+b = a^2 - b^2

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VP
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28 Jul 2007, 22:07
@!%$^@$
I only applied it to the first part and then set there trying to figure out what to do next!
Thanks, guys!

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03 Aug 2014, 14:40
Hello from the GMAT Club BumpBot!

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12 Aug 2014, 00:35
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nervousgmat 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)

I know that this question should be easy but I just can't figure out how to simplify!!!

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

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: 2-135319.html
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Kudos [?]: 128826 [2], given: 12183

Re: (2^2-1)(2^2+1)(2^4+1)(2^8+1)=   [#permalink] 12 Aug 2014, 00:35
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