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by Anurag@Gurome » Wed Mar 14, 2012 8:02 pm
jkelk wrote:Hi

Please help solve the attached.

Thanks,
J
4^17 - 2^28
= (2²)^17 - 2^28
= 2^34 - 2^28
= 2^28(2^6 - 1)
= 2^28(63)
= 2^28(7 * 3^2)

Therefore, the greatest prime factor of 4^17 - 2^28 is 7.

The correct answer is D.
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by PhDmessi » Tue Mar 20, 2012 4:44 am
Hi jkelk,

1. the base needs to be the same to factor out: 4 = 2^2
--> (2^2)17 - 2^28 = 2^34 - 2^28

2. Find the common factor: 2^28

3. Factor out: 2^28 (2^6-1)
= 2^28 (63)
= 2^28 (3*3*7)

--> Thus, 7 is the greates prime factor
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by [email protected] » Tue Mar 20, 2012 6:00 am
missed the expansion of 63 to get maximum prime...gud one
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by [email protected] » Wed Mar 21, 2012 10:52 pm
Very silly question, I do not know how am i really missing out on such questions...

= 4^17 - 2^28 {this is something similar to x^3 - x^2} (cannot do anything about it, except
take common out)

= 2^(2X17) - 2^28

= 2^(34) - 2^28


= 2^28 X (2^6 - 1) (Basic Algebra)


= [****6 X 63] (due to cyclicity we know that in 2^28 the last digit is going to be 6)

(we know 2^6 = 64 and 64 - 1 = 63 which is itself divisible by 7)

So the greatest prime number divisible by the number is 7.

Hope this really helped...
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