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

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(2^2-1)(2^2+1)(2^4+1)(2^8+1)= [#permalink] New post 28 Jul 2007, 13:38
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80% (01:56) correct 20% (01:22) wrong based on 61 sessions
(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!!! :oops:

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

Last edited by Bunuel on 11 Aug 2014, 23:34, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: (2^2-1)(2^2+1)(2^4+1)(2^8+1)= [#permalink] New post 28 Jul 2007, 13:55
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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Re: (2^2-1)(2^2+1)(2^4+1)(2^8+1)= [#permalink] New post 28 Jul 2007, 15:33
Ans : A

Multiple use of the formula

a-b * a+b = a^2 - b^2
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Re: (2^2-1)(2^2+1)(2^4+1)(2^8+1)= [#permalink] New post 28 Jul 2007, 21:07
@!%$^@$
:oops: 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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Re: (2^2-1)(2^2+1)(2^4+1)(2^8+1)= [#permalink] New post 03 Aug 2014, 13:40
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Re: (2^2-1)(2^2+1)(2^4+1)(2^8+1)= [#permalink] New post 11 Aug 2014, 23:35
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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!!! :oops:


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.

Answer: A.

OPEN DISCUSSION OF THIS QUESTION IS HERE: 2-135319.html
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Re: (2^2-1)(2^2+1)(2^4+1)(2^8+1)=   [#permalink] 11 Aug 2014, 23:35
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2 Experts publish their posts in the topic (2^2-1)(2^2+1)(2^4+1)(2^8+1)= Stiv 3 05 Jul 2012, 01:26
2^2-1)(2^2+1)(2^4+1)(2^8+1)= 2^16-1 2^16+1 2^32-1 young_gun 1 10 Oct 2007, 19:52
2^2)-1))((2^2)+1))((2^4)+1))((2^8)+1))= A) (2^16)-1 B) SimaQ 1 26 Nov 2006, 04:35
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 over13 5 03 Feb 2005, 01:02
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