How do you prove this divisibility rule for 7 and what is the great resource except Wiki to learn to prove?

Truncate the last digit of N, double that digit, and subtract it from the rest of the number (or vice-versa). N is divisible by 7 if and only if the result is divisible by 7.

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3 Responses to “How do you prove this divisibility rule for 7 and what is the great resource except Wiki to learn to prove?”

  1. G K said:

    Dividing by 7 (2 Tests)

    Take the last digit in a number.
    Double and subtract the last digit in your number from the rest of the digits.
    Repeat the process for larger numbers.
    Example: 357 (Double the 7 to get 14. Subtract 14 from 35 to get 21 which is divisible by 7 and we can now say that 357 is divisible by 7.

    NEXT TEST
    Take the number and multiply each digit beginning on the right hand side (ones) by 1, 3, 2, 6, 4, 5. Repeat this sequence as necessary
    Add the products.
    If the sum is divisible by 7 – so is your number.
    Example: Is 2016 divisible by 7?
    6(1) + 1(3) + 0(2) + 2(6) = 21
    21 is divisible by 7 and we can now say that 2016 is also divisible by 7.

  2. Evan said:

    I am not sure I understand exactly how this rule works, but let’s give an example.

    Say the number is 154. Truncate the last digit, 4, and multiply it by 2 = 8. Subtract that from the rest of the number (15). 15 – 8 = 7, which is divisible by 7, so the original number is divisible by 7.

    If this is correct, consider the “truncated” number A, and the last digit B, then the original number is

    10 * A + B

    We want to show that

    (10 * A + B) mod 7 = 0 if and only if A – 2*B mod 7 = 0. [Think about that, A – 2*B is twice the truncated digit subtracted from the ‘rest of the number’]

    It is known that a + b mod c = a mod c + b mod c, and that a * b mod c = a mod c * b mod c

    Also, 10 = 3 mod 7, -1 mod 7 = 6 mod 7,
    -2 mod 7 = 5 mod 7 (generally x mod 7 = x + 7 mod 7)

    So that means

    if 10 * A + B mod 7 = 0 mod 7, then

    10 mod 7 * A mod 7 + B mod 7 = 0 mod 7
    or

    3 mod 7 * A mod 7 = -B mod 7 (3 = 10 mod 7)

    3 mod 7 * A mod 7 = -1 mod 7 * B mod 7
    3 mod 7 * A mod 7 = 6 mod 7 * B mod 7 (-1 = 6 mod 7)

    1 mod 7 * A mod 7 = 2 mod 7 * B mod 7
    A mod 7 = 2B mod 7

    A – 2B mod 7 = 0

    This is the other side of the rule.

  3. smci said:

    Doesn’t work. Counterexample: 49
    truncates to 9, 2*9=18, 49-18=31 which is not divisible by 7.
    Whatever the rule was, you misstated it.




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