Factor Problem

kmittal82: Master | Next Rank: 500 Posts; Posts: 324; Joined: Mon Jul 05, 2010 6:44 am; Location: London; Thanked: 70 times; Followed by:3 members

by kmittal82 » Fri Jul 09, 2010 7:57 am

4^17 = 2^34

4^17 - 2^28 = 2^34 - 2^28 = 2^28(2^6 - 1)

Now, 2^6 - 1 = 63, and the greatest prime factor for that is 7 (63 = 3x3x7)

Hence, answer should be (D)

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amising6: Master | Next Rank: 500 Posts; Posts: 294; Joined: Wed May 05, 2010 4:01 am; Location: india; Thanked: 57 times

by amising6 » Fri Jul 09, 2010 7:59 am

monteleone82 wrote:Please help with the question below!

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

A. 2
B. 3
C. 5
D. 7
E. 11

4^17 - 2^28
(2^2)^17 - 2^28
2^34-2^28
2^28(2^6-1)
2^28(64-1)
2^28(63)
2^28*3*3*7
so gretaest prime factor=7

Ideation without execution is delusion

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monteleone82: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Thu Sep 03, 2009 11:38 am

by monteleone82 » Fri Jul 09, 2010 8:10 am

How do you get (2^6-1)? I am confused with that part

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kmittal82: Master | Next Rank: 500 Posts; Posts: 324; Joined: Mon Jul 05, 2010 6:44 am; Location: London; Thanked: 70 times; Followed by:3 members

by kmittal82 » Fri Jul 09, 2010 8:11 am

monteleone82 wrote:How do you get (2^6-1)? I am confused with that part

2^34 - 2^28

Taking 2^28 as common:

2^28(2^6 - 1)

2^28 x 2^6 = 2^34 [a^b x a^ c = a^(b+c) ]
2^28 x 1 = 2^28

Any clearer ?

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monteleone82: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Thu Sep 03, 2009 11:38 am

by monteleone82 » Fri Jul 09, 2010 8:15 am

Got it. thanks.

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