> What is the proof that if a the sum of the digits of a number's > decimal representation is a multiple of 3, then the number is a > multiple of 3?
(Note that "is divisible by" and "is a multiple of" mean the same thing.)
First of all, it's important to understand "linearity":
Suppose integers a, b, c satisfy the relation a + b = c. If two of the integers are divisible by 3, then the third integer also is divisible by 3.
(If you have two, then you must have three.)
Proof: Suppose a and c are divisible by 3. Then a = 3x and c = 3y. By substitution, 3x + b = 3y, and b = 3y - 3x = 3(x - x). By definition, b is divisible by 3. A similar argument works for the other two cases.
Proof of the Divisibility by 3 Test.
We prove the Test for three-digit numbers.
Let x be any three-digit number. Then x = 10h+10t+u, where h, t, and u are the hundredth's, tenth's, and unit's digits respectively (and h is not zero).
So, x = (9h+h)+(9t+t)+u
x = 9(h+t) + (h+t+u)
Since 9(h+t) is divisible by 3, by linearity x is divisible by 3 if and only if h+t+u is also.