GMATPrep Exponent Problem

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GMATPrep Exponent Problem

by mvshah0101 » Thu Dec 14, 2006 7:17 am
Hello all:

I ran into this problem on the GMATPrep test.

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

A. 2
B. 3
C. 5
D. 7
E. 11

This was a tough one because had never dealt with adding and subtracting numbers w/exponents before. I converted 4^17 to 2^34 and I was stumped from there. My natural instinct had me choose A. However, the OA is D. I tried to work through the problem to see exactly why it is D.

By quickly analyizing the patterns of the powers of 2, I could see that every 4th power of 2 ends in the units digit of 6 (ie, 2^4 =16, 2^8 = 256), and every 2nd power of ends in a units digit of 4 (ie, 2^2 = 4, 2^6 = 64). Therefore I assumed that 2^34 (4^17) would end in the units digit of 4, and 2^28 would end in the units digit of 6, the difference of which would have a units digit of 8.

That is as far as I got and I could only eliminate 5 from the answers.

Could anyone help me out with this?

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by tenpercenter76 » Thu Dec 14, 2006 8:47 am
4^17 - 2^28 =

2^34 - 2^28 =

//factor out 2^28
2^28(2^6 - 1)=

2^28(63)

the greatest prime factor of 2^28 is 2, but the greatest prime factor of 63 is 7

D is the answer.

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by mvshah0101 » Thu Dec 14, 2006 9:59 am
Thanks again!!! That makes complete sense!

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by thankont » Fri Dec 15, 2006 2:11 pm
you can solve it like this:
4^17-4^14=4^14(4^3-1)=4^14*(63)=4^14(3^2*7)
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by Scott@TargetTestPrep » Wed Jun 20, 2018 4:30 pm
mvshah0101 wrote:Hello all:

I ran into this problem on the GMATPrep test.

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

A. 2
B. 3
C. 5
D. 7
E. 11
We need to determine the greatest prime factor of 4^17 - 2^28. We can start by breaking 4^17 into prime factors.

4^17 = (2^2)^17 = 2^34

Now our equation is as follows:

2^34 - 2^28

Note that the common factor in each term is 2^28; thus, the expression can be simplified as follows:

2^28(2^6 - 1)

2^28(64 - 1)

2^28(63)

2^28 x 9 x 7

2^28 x 3^2 x 7

We see that the greatest prime factor must be 7.

Answer: D

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by Brent@GMATPrepNow » Wed Jun 20, 2018 4:37 pm
mvshah0101 wrote: What is the greatest prime factor of 4^17 - 2^28?

A. 2
B. 3
C. 5
D. 7
E. 11
4^17 - 2^28 = (2²)^17 - 2^28
= 2^34 - 2^28
= 2^28(2^6 - 1)
= 2^26(64- 1)
= (2^26)(63)
= (2^26)(3)(3)(7)

So, the PRIME factors are 2, 3, and 7

Answer: D

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