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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.
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    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
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    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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