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by ramyaravindran » Fri Jul 25, 2008 9:12 am
Is the question correct?. From how I read the question when you simplify the equation it comes out to be zero. 4^14 - 2^28 => 2^28 -2^28 => 0.
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by Mclaughlin » Fri Jul 25, 2008 9:42 am
oh shoot i'm sorry, i typed it wrong. heres the right question...

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

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by ramyaravindran » Fri Jul 25, 2008 1:14 pm
You can simplify the above equation as follows

4^17 - 2 ^ 28

(2^2)^17 - 2 ^ 28

2^34 - 2^28

this can be written as

2^28(2^6 - 1)

2^28(64-1)

2^28 * 63 => 2^28*7*3*3

The greatest prime number for the above would be 7. So the answer should be D.
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by pepeprepa » Fri Jul 25, 2008 2:24 pm
The same: 7
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by uncbeers » Fri Jul 25, 2008 11:51 pm
ramyaravindran wrote:You can simplify the above equation as follows

4^17 - 2 ^ 28

(2^2)^17 - 2 ^ 28

2^34 - 2^28

this can be written as

2^28(2^6 - 1)

2^28(64-1)

2^28 * 63 => 2^28*7*3*3

The greatest prime number for the above would be 7. So the answer should be D.
For some reason I am missing how you went from step 3 to step 4 here. How does 2^34 - 2^28 become 2^28(2^6 - 1)?

Thank you!
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by pepeprepa » Sat Jul 26, 2008 12:32 am
uncbeers wrote: For some reason I am missing how you went from step 3 to step 4 here. How does 2^34 - 2^28 become 2^28(2^6 - 1)?

Thank you!
2^34=2^28 * 2^6
a^(b+c)= a^b * a^c
So we have
A= 2^34 - 2^28
A= 2^28 * 2^6 - 2^28
A= 2^28 ( 2^6 -1)
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by uncbeers » Sat Jul 26, 2008 12:59 am
Should've guessed...lol. Thanks-makes perfect sense now!
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by sanju09 » Wed Apr 08, 2009 4:22 am
4^17 = (2^2)^17 = 2^34

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

so 7 :)
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by anthony.j » Sat May 09, 2009 6:00 pm
Ahhh.... I finally figured it out after writing it down on paper. You turned 2^34 into two number 2^6 and 2^28. Now the two 2^28 cancel each out out leaving 2^1 * 2^6

The rest is just simplifying.

Thanks!
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