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Exponents - Quick Question

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Exponents - Quick Question

by cgiles01 » Thu Jul 05, 2007 9:21 am
I had trouble with the following question from GMAT Prep Test 2:

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

There is probably a very simple way of solving this problem. Any assistance is greatly appreciated!

(Answer is 7)

Chris
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by givemeanid » Thu Jul 05, 2007 9:42 am
4^17 - 2^28
= (2^2)^17 - 2^28
= 2^34 - 2^28 -> From the rule (x^y)^z = x^(yz)
= 2^28(2^6 - 1) -> From the rule x^(r+s) = x^r * x^s
= 2^28 * 63
= 2^28 * 3 * 3 * 7


Answer is 7.
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by cgiles01 » Thu Jul 05, 2007 9:53 am
I can get to 2^34 - 2^28, but still a little lost how you got to 2^28(2^6 - 1)

Thanks for your help
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by givemeanid » Thu Jul 05, 2007 10:05 am
cgiles01 wrote:I can get to 2^34 - 2^28, but still a little lost how you got to 2^28(2^6 - 1)

Thanks for your help
2^34 = 2^(28+6) = 2^28 * 2^6
This is from the rule of exponents: x^(r+s) = x^r * x^s. In this case, x = 2, r = 28, s = 6
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by cgiles01 » Thu Jul 05, 2007 10:12 am
...and the -1 comes from factoring out the 2^28. Got it! Your awesome!

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another clarification pls,

by gmat_2007 » Fri Jul 20, 2007 4:47 pm
how does 2^28 * 3 * 3* 7 equal 7??
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Re: another clarification pls,

by givemeanid » Fri Jul 20, 2007 5:42 pm
gmat_2007 wrote:how does 2^28 * 3 * 3* 7 equal 7??
The question is asking for the greatest prime factor. After factoring, you can see 7 fits the bill. Hence, the answer is 7.
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by gmat_2007 » Thu Jul 26, 2007 2:02 pm
Thanks for the reply. I was in a mental funk when I asked that questions!
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