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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

Author Message
Intern
Joined: 27 Jan 2005
Posts: 15
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 [#permalink]

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03 Feb 2005, 02:02
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

(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 )

Manager
Joined: 11 Jan 2005
Posts: 101

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03 Feb 2005, 03:46
though I dont understand, here is my logic

(2^2-1)(2^2+1)(2^4+1)(2^8+1)
=(2^1)(2^3)(2^5)(2^9)
=2^(1+3+5+9)
= 2^(18)

or

(2^2-1)(2^2+1)(2^4+1)(2^8+1)
= (4-1)(4+1)(16+1)(256+1)
=3+5+17+257
=282
SVP
Joined: 03 Jan 2005
Posts: 2233

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03 Feb 2005, 10:16
Use this:
(a-b)(a+b)=a^2-b^2
(A)
VP
Joined: 18 Nov 2004
Posts: 1433

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03 Feb 2005, 11:50
A for me.

a2-b2 = (a+b)(a-b)

(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
Intern
Joined: 27 Jan 2005
Posts: 15

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03 Feb 2005, 12:19
A is correct.......

Manager
Joined: 01 Jan 2005
Posts: 166
Location: NJ

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03 Feb 2005, 12:53
Hi guys

Good logic. Initially i got confused when I saw the question. But it is easy.

Thanks everybody,
Vijo.
03 Feb 2005, 12:53
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