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Prime Factorization

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by harsh.champ » Sun Feb 07, 2010 2:34 pm
skorolkova wrote:What is the greatest prime factor of 4^(17) - 2^28?

I thought it would be 2 but apparently not
Now,(2^28) can be written as (4^14).
So, 4^(17) - 2^28 becomes 4^(17) - 4^(14)
= (4^14)[(4^3)-1]
= (4^14)[63]
=(4^14) [7x(3^2)]
So, the greatest prime factor is 7.
Last edited by harsh.champ on Sun Feb 07, 2010 2:41 pm, edited 1 time in total.
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by skorolkova » Sun Feb 07, 2010 2:36 pm
harsh.champ wrote:
skorolkova wrote:What is the greatest prime factor of 4^(17) - 2^28?

I thought it would be 2 but apparently not
I am also getting 2 only

The answer says 7... I'm wondering if that's related to the 28... but then why not 17?
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by harsh.champ » Sun Feb 07, 2010 2:46 pm
skorolkova wrote:
harsh.champ wrote:
skorolkova wrote:What is the greatest prime factor of 4^(17) - 2^28?

I thought it would be 2 but apparently not
I am also getting 2 only

The answer says 7... I'm wondering if that's related to the 28... but then why not 17?
Hey skorolkova,
I found out my mistake.The answer indeed is 7 only.
I was not considering the negative sign and just answered the question taking both the terms as separate entities.
I think you also made the same mistake.

Just check my solution post and tell whether it is doubtful .



Ofcourse,it is not 28 or 17. The power is never related to the divisor in any way.
The method of prime factorization itself goes like N = (a^x)(b^y)(c^z) where a,b,c are the prime divisors and the powers don't have any relation with the prime factors.:)
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker :P
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