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Evaluate \(\dfrac{(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)}{2^{32}-1}\)

 May 11, 2020
 #1
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+1

Reduce the expression, if possible, by cancelling the common factors.

 

answer = 1

 May 11, 2020
 #2
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+1

Note that  we  can write  2^32  - 1  as

 

(2^16 + 1) (2^16 - 1)  =

 

(2^16 + 1) (2^8 + 1) (2^8 - 1) =

 

(2^16 + 1) ( 2*8 + 1) ( 2^4 + 1) (2^4  - 1)  =

 

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

 

(2^2 - 1) (2^2 + 1) ( 2^4 + 1) ( 2^8 + 1) ( 216 + 1)

 

Putting all of this together  we have

 

(2 + 1) (2^2 + 1) (2^4 + 1) (2 ^16 + 1)

_______________________________________   =

(2^2 - 1) ( 2^2 + 1) ( 2^4 + 1) (2^8+1) (2^16 + 1)

 

 

(2 + 1)

_______  =

(2^2 - 1)

 

(2 + 1) 

___________  = 

(2 + 1) (2 - 1)

 

   3

_____  =

3 * 1

 

 

1

 

 

cool cool cool

 May 11, 2020

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