Problem Solving

What is the greatest prime factor of 4^(17) - 2^28?

I thought it would be 2 but apparently not

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.

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?

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.