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    A fairly well known result is that a number is a multiple of 3 if and only if the sum of its digits is a multiple of 3.

    a) Prove that for a 3 digit number, written 'abc', then the number itself i.e 100a + 10b + c is a multiple of 3 if and only if a + b + c is also a multiple of 3.
    b) Explain how you can generate your proof from a 3-digit number, to a number with any number of digits.

    a) Use the modulo as a way of answering the question. I'm not using LaTex cause I cba but here goes...

    3 | 100a+10b+1c <=>

    100a+10b+1c = 0 mod 3

    99 = 0 mod 3 as does 9

    therefore 100 = 1 mod 3, 10 = 1 mod 3 and 1 = mod 3 therefore 1a+1b+1c = 0 mod 3

    that'll work out I think, since a,b,c are irrelevant as they are multiples of 100, 10 and 1 respectively which we've seen to be congruent to 0 mod 3.

    that's a conjecture. i'd wait for further assistance

    good luck

    As above really, hes got all the details right. You might want to prove that you can do arithmetic with congruences, but apart from that it's fairly standard.
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Updated: January 28, 2010

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