Prime Factorization

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Apl28: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Jun 28, 2008 10:38 am

by Apl28 » Tue Dec 02, 2008 8:16 pm

This is a question from a GMAT Prep test I just took ...

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

Answers:
2
3
5
7
11

7

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Source: Beat The GMAT — Problem Solving | Tue Dec 02, 2008 8:16 pm

austin: Master | Next Rank: 500 Posts; Posts: 110; Joined: Thu Oct 23, 2008 9:12 amFollowed by: 1 member; Location: India; Thanked: 9 times

by austin » Tue Dec 02, 2008 8:48 pm

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

The prime factors are 2,3 and 7.

The greatest prime factor is 7.

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CMoore: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Mon Aug 18, 2008 4:48 pm

by CMoore » Mon Dec 08, 2008 9:54 pm

quick question..why did you subtract 1 from the 64? I am trying to figure this one out and Im still kind of confused. Thanks

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Mon Dec 08, 2008 9:59 pm

quick question..why did you subtract 1 from the 64? I am trying to figure this one out and Im still kind of confused. Thanks

If u take the common factor 2^28 u will get a 1 inside braces

2^34 - 2 ^ 28

= 2^28 ( 2 ^ 6 - 1) (which = 2^28*2^6 - 2^28*1 = 2^34 - 2^28)

Hope this helps!

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vish150783: Senior | Next Rank: 100 Posts; Posts: 58; Joined: Sun Jun 17, 2007 8:24 am

by vish150783 » Fri Dec 12, 2008 7:57 am

CMoore wrote:quick question..why did you subtract 1 from the 64? I am trying to figure this one out and Im still kind of confused. Thanks

A^n - A^m = A^(n-m). hope this helps. Property of powers. (ALGEBRA)

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nytivofan: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sat Dec 13, 2008 7:52 pm

by nytivofan » Sat Dec 13, 2008 7:58 pm

vish150783 wrote:
CMoore wrote:quick question..why did you subtract 1 from the 64? I am trying to figure this one out and Im still kind of confused. Thanks

A^n - A^m = A^(n-m). hope this helps. Property of powers. (ALGEBRA)

Am I missing something here? A^n - A^m does NOT = A^(n-m). That's why you factor out. e.g. 2^4 - 2^2 doesn't = 2^2. It's 2^2 (2^2 - 1) = 4 (3) = 12.

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Mozartain: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Tue Nov 11, 2008 8:38 pm; Thanked: 3 times; GMAT Score:740

by Mozartain » Mon Dec 15, 2008 2:44 am

vish150783 wrote:A^n - A^m = A^(n-m). hope this helps. Property of powers. (ALGEBRA)

Nope!

A^n / A^m = A^(n-m)

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mowie: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Tue Nov 18, 2008 2:23 pm

by mowie » Mon Dec 15, 2008 6:45 am

I would solved it like this:

2^34-2^28
2^28 * 2^6 - 2^28
2^28 (2^6 - 1)
2^28 (63)
=> 7

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