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Math Help - Divisibility

  1. #1
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    Divisibility

    Could someone help me prove the following using congruences?

    Prove that a number n is divisible by 7 if and only if n' − 2n0 is divisible by
    7 where n0 is the last digit and n' is the number obtained from n by deleting
    the last digit.

    Thanks.
    Jimmy
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  2. #2
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    Hello, Jimmy!

    Could someone help me prove the following using congruences?

    Prove that a number n is divisible by 7 if and only if n' − 2n_o is divisible by 7
    where n_o is the last digit and n' is the number obtained from n by deleting the last digit.

    Let: n .= .10a + b, then: a = n' and b = n_o

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    [1] If n = 10a + b is divisible by 7, then a - 2b is divisible by 7.

    We are given: . 10a + b .= .0 (mod 7)

    Then: . . . . . . . . . .10a .= .-b (mod 7)

    Or: . . . . . . . . . . . . 3a .= .6b (mod 7)

    Multiply by 5: . . . . 15a .= .30b (mod 7)

    Or: . . . . . . . . . . . . . a .= .2b (mod 7)

    Hence: . . . . . . . a - 2b .= .7k for some integer k.

    Therefore: .a - 2b is divisible by 7.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    [2] If a - 2b is divisible by 7, then n = 10a + b is divisible by 7.

    We are given: . a - 2b .= .0 (mod 7)

    Then: . . . . . . . . . . a .= .2b (mod 7)

    Multiply by 10: . . 10a .= .20b (mod 7)

    Or: . . . . . . . . . . 10a .= .6b (mod 7)

    Add b: . . . . . 10a + b .= .7b (mod 7)

    Then: . . . . . .10a + b .= .0 (mod 7)

    Hence: . . . . . 10a + b .= .7k for some integer k.

    Therefore: .n = 10a + b is divisible by 7.

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