1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Assistance Required

  1. Nov 5, 2007 #1
    What are the last three digits of 2007^2007^2007^2007?

    So far, I have tried doing it one by one to see if there was a pattern in the units, but haven't had much luck.

    Any help?
     
  2. jcsd
  3. Nov 5, 2007 #2
    Perhaps think of it in a different manner: what is the remainder of 2007^n when divided by 1000 ?
     
  4. Nov 5, 2007 #3

    Avodyne

    User Avatar
    Science Advisor

    This is a pretty tricky problem in modular arithemetic. Here's a simpler problem of the same sort: what is the remainder when 2^1000 is divided by 13 ?

    Solution: notice that 64 = -1 mod 13, since 5*13 = 65. Also, 64 is a power of 2: 2^6 = 64. Then, 1000 = 6*166 + 4, so 2^1000 = (2^6)^166 * 2^4 = 64^166 * 2^4. Then, 2^1000 mod 13 = (64^166 mod 13)*(2^4 mod 13) = (64 mod 13)^166 * (16 mod 13) = (-1)^166 * 3 = 3. So the answer is 3.

    Your problem is considerably more difficult.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Assistance Required
  1. Please assist! (Replies: 1)

  2. Paid Assistance? (Replies: 2)

Loading...