Can someone help with this question from the free software?

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simunda: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sun May 02, 2010 12:39 pm

by simunda » Sat Nov 06, 2010 9:34 am

Hope I'm not breaking any rules here but I have come across this question a couple of times on the free official software CATs and just cannot work out how the answer is reached. Can anyone help?

What is the greatest prime factor of 4 to power 17 minus two to power 28?

Thanks.....

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Source: Beat The GMAT — Problem Solving | Sat Nov 06, 2010 9:34 am

Bharat: Senior | Next Rank: 100 Posts; Posts: 70; Joined: Sun Jul 22, 2007 11:43 pm; Thanked: 8 times

by Bharat » Sat Nov 06, 2010 9:45 am

Answer: 7

4^17 - 2^28
= 2^34 - 2^28
= 2^28 ( 2^6 - 1)
= 2^28 (64-1)
= 2^28 (7*9)
= 2^28 * 3^2 * 7 --> 7 is the largest prime here.

Regards.

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simunda: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sun May 02, 2010 12:39 pm

by simunda » Sat Nov 06, 2010 9:59 am

Thanks , that helps a lot. I kept getting the first few steps but got the difference to be 2^6.....could you explain why its 2^6-1 please?

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Bharat: Senior | Next Rank: 100 Posts; Posts: 70; Joined: Sun Jul 22, 2007 11:43 pm; Thanked: 8 times

by Bharat » Sat Nov 06, 2010 10:19 am

Bharat wrote:Answer: 7

4^17 - 2^28
= 2^34 - 2^28
= 2^28 ( 2^6 - 1)
= 2^28 (64-1)
= 2^28 (7*9)
= 2^28 * 3^2 * 7 --> 7 is the largest prime here.

Regards.

2^34 = 2^28 * 2^6 [exponents of the same base are added during multiplication]
revisit the 2nd line of the answer again:
2^34 - 2^28
= 2^28 * 2^6 - 2^28 * 1
=2^28 (2^6 - 1) [2^28 is common]
Let me know if this does not help.

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simunda: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sun May 02, 2010 12:39 pm

by simunda » Sat Nov 06, 2010 12:00 pm

perfect, thank you so much, got it now!

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zachthegnome: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Sun Nov 07, 2010 8:02 am; Thanked: 1 times

by zachthegnome » Mon Nov 08, 2010 9:11 pm

I am still missing something on the 2^6-1 step.

Please elaborate further. Thanks.

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Bharat: Senior | Next Rank: 100 Posts; Posts: 70; Joined: Sun Jul 22, 2007 11:43 pm; Thanked: 8 times

by Bharat » Mon Nov 08, 2010 9:41 pm

zachthegnome wrote:I am still missing something on the 2^6-1 step.

Please elaborate further. Thanks.

I have used the following two formulas:
1. a = a*1
2. a*b - a*1 = a (b - 1)

In these, replace "a" with "2^28" & "b" with "2^6"; & you will get the requisite step.

Let me know if there are any ambiguities. Thanks.

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