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Math Help - Mod 15

  1. #1
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    Mod 15

    Q: If 5x=5y (mod 15), then could someone please explain to me why it's NOT true to say that x=y (mod 15) but it IS true to say x=y (mod 3)? Just a little confused as to why I can't divide both sides by 5 in the first instance to give x=y (mod 15) and yet I can for x=y (mod 3).
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  2. #2
    Member Henderson's Avatar
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    Would a counterexample be enough? Try x=4, y=7.
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  3. #3
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    Hello,

    5x=5y \bmod 15 \implies 5x-5y=15k \quad k \in \mathbb{Z}

    Simplifying : x-y=3k, that is to say x=y \bmod 3

    This is partly because if you assume that x-y=15k' (which would yield x=y mod 15), it means that k was a multiple of 5, and nothing lets you say that.
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