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GMATPrep Exponent Problem

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

GMATPrep Exponent Problem

Post Thu Dec 14, 2006 7:17 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    tenpercenter76 Newbie | Next Rank: 10 Posts Default Avatar
    Joined
    01 Oct 2006
    Posted:
    6 messages
    Post 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.

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

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

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