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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)

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VP
Joined: 06 Feb 2007
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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) [#permalink]  28 Jul 2007, 13:38
(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!!!
Intern
Joined: 20 Jul 2005
Posts: 28
Location: Vietnam
Followers: 0

Kudos [?]: 2 [0], given: 0

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)
_________________

Regards,
Sonfbm

Senior Manager
Joined: 28 Jun 2007
Posts: 463
Followers: 2

Kudos [?]: 5 [0], given: 0

Ans : A

Multiple use of the formula

a-b * a+b = a^2 - b^2
VP
Joined: 06 Feb 2007
Posts: 1024
Followers: 17

Kudos [?]: 103 [0], given: 0

@!%\$^@\$
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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b and c................ 1 26 Apr 2009, 13:10
2^2-1)(2^2+1)(2^4+1)(2^8+1)= 2^16-1 2^16+1 2^32-1 1 10 Oct 2007, 19:52
2^2)-1))((2^2)+1))((2^4)+1))((2^8)+1))= A) (2^16)-1 B) 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 5 03 Feb 2005, 01:02
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