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Math Help - Modular arithmetic

  1. #1
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    Modular arithmetic

    Can someone explain why this works?

    5^101 mod 124 = 5^(101 mod 3) mod 124 = 5^2 mod 124 = 25

    How did they know to choose 3 in the problem above? If given a problem like this, I don't see how I would deduce that I need to take 101 mod 3 to make the problem easier.

    Thanks

    Edit: Is there a general formula for this? Can I use this in any problem with a large exponent?
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  2. #2
    MHF Contributor

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    Re: Modular arithmetic

    note that 53 = 125 = 1 (mod 124).

    so if n = 3k + t (that is: n = t (mod 3))

    5n = (53k)(5t) = (53)k(5t) = (1k)(5t) = 5t (mod 124).
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