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.
Suppose
More generally, if
(where
Notice that for any integer
which is clearly divisible by 3. So from each power of 10 in the expansion of
We know
Back to our previous example: if
Dividing through by 3, we get a remainder of
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
