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## GMATPrep Exponent Problem

This topic has 3 member replies
mvshah0101 Newbie | Next Rank: 10 Posts
Joined
30 Jul 2006
Posted:
3 messages

#### GMATPrep Exponent Problem

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?

thankont Senior | Next Rank: 100 Posts
Joined
15 Dec 2006
Posted:
41 messages
Followed by:
1 members
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)
--d--

mvshah0101 Newbie | Next Rank: 10 Posts
Joined
30 Jul 2006
Posted:
3 messages
Thu Dec 14, 2006 9:59 am
Thanks again!!! That makes complete sense!

tenpercenter76 Newbie | Next Rank: 10 Posts
Joined
01 Oct 2006
Posted:
6 messages
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

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