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Prime Factorization

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by austin » Tue Dec 02, 2008 8:48 pm
4^17 - 2^28
= 2^34 - 2^28
= 2^28 (2^6 - 1)
= 2^28 (64 - 1)
= 2^28 (63)
= 2^28 (7* 3^2)

The prime factors are 2,3 and 7.

The greatest prime factor is 7.
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by CMoore » Mon Dec 08, 2008 9:54 pm
quick question..why did you subtract 1 from the 64? I am trying to figure this one out and Im still kind of confused. Thanks
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by cramya » Mon Dec 08, 2008 9:59 pm
quick question..why did you subtract 1 from the 64? I am trying to figure this one out and Im still kind of confused. Thanks
If u take the common factor 2^28 u will get a 1 inside braces

2^34 - 2 ^ 28

= 2^28 ( 2 ^ 6 - 1) (which = 2^28*2^6 - 2^28*1 = 2^34 - 2^28)

Hope this helps!
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by vish150783 » Fri Dec 12, 2008 7:57 am
CMoore wrote:quick question..why did you subtract 1 from the 64? I am trying to figure this one out and Im still kind of confused. Thanks


A^n - A^m = A^(n-m). hope this helps. Property of powers. (ALGEBRA)
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by nytivofan » Sat Dec 13, 2008 7:58 pm
vish150783 wrote:
CMoore wrote:quick question..why did you subtract 1 from the 64? I am trying to figure this one out and Im still kind of confused. Thanks


A^n - A^m = A^(n-m). hope this helps. Property of powers. (ALGEBRA)
Am I missing something here? A^n - A^m does NOT = A^(n-m). That's why you factor out. e.g. 2^4 - 2^2 doesn't = 2^2. It's 2^2 (2^2 - 1) = 4 (3) = 12.
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by Mozartain » Mon Dec 15, 2008 2:44 am
vish150783 wrote:A^n - A^m = A^(n-m). hope this helps. Property of powers. (ALGEBRA)
Nope!

A^n / A^m = A^(n-m)
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by mowie » Mon Dec 15, 2008 6:45 am
I would solved it like this:

2^34-2^28
2^28 * 2^6 - 2^28
2^28 (2^6 - 1)
2^28 (63)
=> 7
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