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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 $\displaystyle ABCDE.$
    Prove that $\displaystyle ABCDE$ is divisible by 7 if and only of the number $\displaystyle ABCD - 2E$ is divisible by 7.

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

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

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

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

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

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

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



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

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

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

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

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

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

    Therefore: .$\displaystyle ABCDE$ is divisible by 7.

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