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.
Algebraic Proof Watch
- Thread Starter
- 28-01-2010 23:20
- 28-01-2010 23:37
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
- 28-01-2010 23:51
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.