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Math Help - Number Theory

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    Post Number Theory

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    The digits of a five-digit number are MNOPQ. Prove that MNOPQ is divisible by 7 if and only of the number MNOP - 2 x Q is divisible by 7.

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    Quote Originally Posted by mathisgood View Post
    Help on this...

    The digits of a five-digit number are MNOPQ. Prove that MNOPQ is divisible by 7 if and only of the number MNOP - 2 x Q is divisible by 7.

    Thanks!
    Divisibility Tests -- from Wolfram MathWorld
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  3. #3
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    Hello, mathisgood!

    The digits of a five-digit number are ABCDE.
    Prove that ABCDE is divisible by 7 if and only of the number ABCD - 2E is divisible by 7.

    [1] Given: . ABCDE is divisible by 7.
    . . .Prove: . ABCD-2E is divisible by 7.

    We have: . 10(ABCD) + E \;=\;7p\:\text{ for some integer }p.

    Subtract 21E\!:\;\;10(ABCD) - 20E \;=\;7p - 21E \;=\;7(p - 3E)

    Divide by 10: . (ABCD) - 2E \;=\;\frac{7(p-3E)}{10}

    Since (ABCD)-2E is an integer, then p-3E is a multiple of 10: . 10k

    And we have: . (ABCD) - 2E \;=\;\frac{7(10k)}{10} \;=\;7k

    Therefore: . (ABCD)-2E is divisible by 7.



    [2] Given: . (ABCD)-2E is divisible by 7.
    . . .Prove: . ABCDE is divisible by 7.

    We have: . (ABCD) - 2E \;=\;7q\,\text{ for some integer }q

    . . . Then: . (ABCD) \;=\;2E + 7q

    Multiply by 10: . 10(ABCD) \;=\;20E + 7q

    . . Add E\!:\;\;10(ABCD) + E \;=\;21E + 7q

    And we have: . ABCDE \;=\;7(3E + q)

    Therefore: . ABCDE is divisible by 7.

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