Why does multiple of 3 add up digits trick work?

You must be referring to the quick way to determine whether a given integer is divisible by 3; that is, 3 divides an integer whenever the digits of sum to a multiple of 3.

Suppose has digits. We can expand it as a sum of the numbers in any given digits place by the corresponding power of 10. For example, if , we can write



More generally, if



(where denotes the numeral in the -th's place), then we have the expansion



Notice that for any integer , we have



which is clearly divisible by 3. So from each power of 10 in the expansion of , we can add and subtract 1, then rearrange the terms of the sum:





We know is divisible by 3, which means the remainder upon dividing by 3 is just the sum of the digits of . If this remainder is divisible by 3, then so must be the original number, .

Back to our previous example: if , then we have the expansion





Dividing through by 3, we get a remainder of , which is divisible by 3, and so 2148 must also be a multiple of 3.

In case for some reason you're not convinced 15 is a multiple of 3, you can apply the same trick:




Dividing through by 3 leaves a remainder of , which is also a multiple of 3, so that 15 must be, too.