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Math Help - Very interesting problem that I can't solve.

  1. #1
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    Very interesting problem that I can't solve.

    This was on my grade 10 Cayley math contest a few days ago. It is really interesting, but super hard.

    We have a circle with 10 points (think of a clock that goes from 0-9). The dial starts at 0. In each move Tom moves the dial n^n times. So on the first move Tom moves 1^1 times, and the dial is at 1. Second move Tom moves 2^2 times, and the dial is at 5, and third move Tom moves 3^3 times, and the dial is at 2. What number would the dial be pointing at after 1234 moves?

    Essentially it is asking what \sum_{n=1}^{1234} n^n \text { mod } 10 is.
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  2. #2
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    Try this:

    1. Break this into 2 problems:
    (your sum) = a mod 2.
    (your sum) = b mod 5.

    If you can find a and b, you can find the equivalent:
    (your sum) = c mod 10. (use Chinese remainder theorem) and you are done.

    Now to find b: (finding a is easier)
    Remember that the numbers
    1,2,3,4,5,6,7,8,9,10... = 1,2,3,4,0,1,2,3,4,0...mod (5).
    So your 'n' in the bottom doesn't vary all that much.

    Also, remember Fermat's little theorem => if a isn't a multiple of p, then
    a^(p-1)=1(p). In other words, things like:
    4^4=1(5), 12^12=1(5), etc.

    This should simplify your sum a lot.
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  3. #3
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    Thanks for your attempt at helping me.

    I'm not sure if I understand your explanation. It seems to be a university level solution.

    This question is from a grade 10 contest, so it should be solved with grade 10 level mathematics and a serious amount of cleverness.
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  4. #4
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    Ok, try this.

    It should be obvious that 11^11 is the same as 1^11 because all you care about is the last digit.

    Then try pairing: 1^1 and 9^9, since 9 = -1 (going 1 unit in the other direction). Also, tricks like 5^anything = 5(10), 6^anything = 6 (10), etc.
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  5. #5
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    Quote Originally Posted by qmech View Post
    Ok, try this.

    It should be obvious that 11^11 is the same as 1^11 because all you care about is the last digit.

    Then try pairing: 1^1 and 9^9, since 9 = -1 (going 1 unit in the other direction). Also, tricks like 5^anything = 5(10), 6^anything = 6 (10), etc.
    I still don't see how that can help me. I can see it being helpful for the first 10 terms...
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  6. #6
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    Note that a move of (10k)^{10k} moves the dial to the same position it was before.

    It should also be obvious that (10k + a)^{10k + a} is equivalent to a^{10k + a} = a^{10k} \cdot a^a.

    Then just look for helpful patterns. For instance:

    2^2 \equiv 4, 2^{12} \equiv 6, 2^{22} \equiv 4, 2^{32} \equiv 6

    so consecutive pairs of these terms will cancel out.
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  7. #7
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    Quote Originally Posted by icemanfan View Post
    Note that a move of (10k)^{10k} moves the dial to the same position it was before.

    It should also be obvious that (10k + a)^{10k + a} is equivalent to a^{10k + a} = a^{10k} \cdot a^a.

    Then just look for helpful patterns. For instance:

    2^2 \equiv 4, 2^{12} \equiv 6, 2^{22} \equiv 4, 2^{32} \equiv 6

    so consecutive pairs of these terms will cancel out.
    So I only need to find out what 1231^{1231}+1232^{1232}+1233^{1233}+1234^{1234} mod 10 is?

    Am I doing this right?

    1231^{1231}=[123(10)+1]^{123(10)+1}=1^{1231}=1?
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