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What is the greatest prime factor

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Source: — Problem Solving |

by mj78ind » Sun Jun 20, 2010 7:46 am
upon factorization we get 2^28*(2^6 - 1) = 2^28*63, hence the greatest prime factor of that is 7.

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by indiantiger » Sun Jun 20, 2010 8:12 am
given to us 4^17 - 2^28 and need to find its Greatest prime factor

when you see this question the first thing is to make the bases same

=(2^2)^17 - 2^28
=2^34 - 2^28
now we have same bases so we can take out 2^28
=2^28(2^6 - 1)
=2^28(64 -1)
=2^28*63
=2^28 * 7 * 3 * 3
from here you can see the greatest prime factor is 7 (answer option D)
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by ssuarezo » Sun Jun 20, 2010 9:01 am
Hi guys:

Can u believe that I wrote the -1, but I didn't use it??
When i try to do things against time, I do foolish errors like this.
Thanks for your time.

Silvia
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by Patrick_GMATFix » Sun Jun 20, 2010 9:43 am
Silvia, I can definitely believe it. I make this type of error too often.

In general, when solving questions about factors, divisibility or primes, it's a good idea to transform your numbers into their prime factors. In this case, 4^17 - 2^28 becomes 2^34 - 2^28.

To find the factors of an expression, factor it. in this case, you can factor out 2^28 from both sides to get 2^28 (2^6 - 1). The only prime factor of the portion outside the parenthesis is 2. The portion inside the parentheses simplifies to 63. It's greatest prime factor is 7. Thus the greatest prime factor of the initial expression is 7. The correct answer is D.

For more detailed explanation and a video solution, have a look at GMATPrep question 1121. If you struggle with similar questions, set topic to 'Exponents & Roots' and difficulty to '600-700 AND 700+' when you use the drill generator.

Good luck,
-Patrick
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by aagar2003 » Sun Jun 20, 2010 5:57 pm
Patrick_GMATFix wrote:Silvia, I can definitely believe it. I make this type of error too often.


For more detailed explanation and a video solution, have a look at GMATPrep question 1121.

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by Patrick_GMATFix » Sun Jun 20, 2010 6:38 pm
Send me a PM and I'll help you get free access. Or just request the free license that's offered as an alternative
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