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 Post subject: prime factorPosted: Mon May 25, 2009 10:08 am

Joined: Thu May 21, 2009 2:51 pm
Posts: 5
what is the greatest prime factor of 4^17-2^28?

I know it's 7, but why

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 Post subject: Re: prime factorPosted: Mon May 25, 2009 5:38 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
4^17-2^28

(2^2)^17 - 2^28

2^34 - 2^28

(2^28)(2^6 - 1)

(2^28)(7*3*3), so greatest prime factor is 7

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 Post subject: Re: prime factorPosted: Fri Jun 12, 2009 8:33 am

Joined: Fri Jun 12, 2009 8:21 am
Posts: 1
Question on this...how do you know to combine the 17 and 2?

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 Post subject: Re: prime factorPosted: Fri Jun 12, 2009 10:25 am

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
4^17-2^28
(2^2)^17 - 2^28

One of our first steps is to take all the base numbers down to prime factors. Since 4 is really 2^2, we replace 4 with 2^2 and then raise that to the 17th. This becomes 2^34.

Veritas Help

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 Post subject: Re: prime factorPosted: Wed Sep 08, 2010 7:49 pm

Joined: Tue Aug 31, 2010 9:07 am
Posts: 2
Is there an easier way to factor the term (2^6-1) other than calculating it out?

Is this the intended calculation, or is there a shortcut?
2*2=4, 4*2=8...32*2=64, 64-1=63

Then 63=7*9=7*3*3

If, instead, the term was (2^15-1), calculating it out would seem tedious. How would you factor this?

Thanks,

Cooper

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 Post subject: Re: prime factorPosted: Sun Sep 12, 2010 6:26 am

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
You should probably know the powers of 2 up through 2^6 or 2^7. They won't usually ask you to calculate 2^15, BUT, if they did, you could use 2^5 * 2^5 * 2^5 or something comparable.
(On 2^6 you could also calculate 2^3 * 2^3 and arrive at the right point and it would be faster than the other route...)

Veritas Help

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