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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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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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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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by simunda » Sat Nov 06, 2010 12:00 pm
perfect, thank you so much, got it now!
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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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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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