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# 2^2-1)(2^2+1)(2^4+1)(2^8+1)= 2^16-1 2^16+1 2^32-1

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2^2-1)(2^2+1)(2^4+1)(2^8+1)= 2^16-1 2^16+1 2^32-1  [#permalink]

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10 Oct 2007, 20:52
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(2^2-1)(2^2+1)(2^4+1)(2^8+1)=

2^16-1
2^16+1
2^32-1
2^128-1
2^16(2^16-1)
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Re: PS simplification exponents [#permalink]

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10 Oct 2007, 20:59
young_gun wrote:
(2^2-1)(2^2+1)(2^4+1)(2^8+1)=

2^16-1
2^16+1
2^32-1
2^128-1
2^16(2^16-1)

For: (2^2-1)(2^2+1)
x^2 - y^2 = (x+y)*(x-y)
So you have:
(2^2-1)(2^2+1) = 2^4 - 1
Take the third multiple:
(2^4 - 1)*(2^4+1) = 2^8 - 1
Take the last multiple:
(2^8 - 1)*(2^8+1) = 2^16-1
Re: PS simplification exponents   [#permalink] 10 Oct 2007, 20:59
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# 2^2-1)(2^2+1)(2^4+1)(2^8+1)= 2^16-1 2^16+1 2^32-1

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