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Math Help - Need help on an easy proof

  1. #1
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    Question Need help on an easy proof

    need to prove that:
    if m, d, and k are nonnegative integers and d does not equal 0, then (m + dk) mod d = m mod d.

    I'm pretty sure you need to use the quotient remainder theorem.

    thanks in advance.
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  2. #2
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    m + dk \mod d \equiv m \mod d + dk \mod d \equiv m \mod d + 0 \mod d \equiv m \mod d. Is there any reason you can't use this argument?
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  3. #3
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    how do you know (m + dk) mod d = m mod d + dk mod d?

    is that a rule that works for all integers?

    also

    wouldn't m mod + 0 mod d = m mod d + d and not m mod d
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  4. #4
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    Quote Originally Posted by Mathwizard4 View Post
    how do you know (m + dk) mod d = m mod d + dk mod d?

    is that a rule that works for all integers?
    Yes, it does. The remainder when a + b is divided by d is congruent to the sum of the remainder of a when divided by d and the remainder of b when divided by d.
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  5. #5
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    ok thanks but what about this

    wouldn't m mod + 0 mod d = m mod d + d and not m mod d ?
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  6. #6
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    Quote Originally Posted by Mathwizard4 View Post
    wouldn't m mod + 0 mod d = m mod d + d and not m mod d
    Not sure what you're asking here. Do you mean m \mod (d + d) or m \mod d + d \mod d?
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  7. #7
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    how do you know....

    m \mod d + dk \mod d \equiv m \mod d + 0 \mod d
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  8. #8
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    can anyone else help?
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  9. #9
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    answer?

    sorry to hop on your thread with a question, but did you ever figure out how to prove that? I have the same exact problem and cannot figure it out at all, driving me insane and its due in about 6 hours
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  10. #10
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    Quote Originally Posted by discreteDilema View Post
    sorry to hop on your thread with a question, but did you ever figure out how to prove that? I have the same exact problem and cannot figure it out at all, driving me insane and its due in about 6 hours
    not yet I gave up. but I don't need to have it for anything. sorry senor
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