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Homework Help: 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

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    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.
     
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