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# 2^2-1) (2^2+1) (2^4+4) (2^8+1)= (a)2^16-1 (b)2^16+1

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Manager
Joined: 11 Nov 2004
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2^2-1) (2^2+1) (2^4+4) (2^8+1)= (a)2^16-1 (b)2^16+1  [#permalink]  14 Jan 2005, 18:20
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0% (00:00) correct 0% (00:00) wrong based on 1 sessions
(2^2-1) (2^2+1) (2^4+4) (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)
Director
Joined: 07 Jun 2004
Posts: 614
Location: PA
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I am note sure whether i am reading the Q right My answer to this wud be E
Senior Manager
Joined: 22 Jun 2004
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Re: PS [#permalink]  15 Jan 2005, 06:40
I think it is (2^4+1) than (2^4+4) . If that is so, I choose A.

Otherwise, none of the answers match.

Janice wrote:
(2^2-1) (2^2+1) (2^4+4) (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)

_________________

Awaiting response,

Thnx & Rgds,
Chandra

Manager
Joined: 11 Nov 2004
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sorry guys the question should be like this:

(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)

...and OA is A.Please explain how you solved it.
Intern
Joined: 15 Jan 2005
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Hello Janice!
If u use the formula (a-b)(a+b)=a^2-b^2 u will easily solve this problem! So (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)
Good Luck!
Manager
Joined: 11 Nov 2004
Posts: 69
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AniAngel wrote:
Hello Janice!
If u use the formula (a-b)(a+b)=a^2-b^2 u will easily solve this problem! So (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)
Good Luck!

thanks
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