Question

Why does multiple of 3 add up digits trick work?

Answer 1

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

Suppose $x$ has $n\ge1$ 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 $x=2148$, we can write

$2148=2\times10^3+1\times10^2+4\times10^1+8\times10^0$

More generally, if

$x=d_{n-1}d_{n-2}\ldots d_1d_0$

(where $d_i$ denotes the numeral in the $10^i$-th's place), then we have the expansion

$x=d_{n-1}10^{n-1}+d_{n-2}10^{n-2}+\cdots+d_110^1+d_0$

Notice that for any integer $k$, we have

$10^k-1=\underbrace{99\ldots99}_{k\text{ 9s}}$

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

$x=d_{n-1}10^{n-1}+\cdots+d_110^1+d_0$
$x=d_{n-1}(10^{n-1}-1+1)+\cdots+d_1(10^1-1+1)+d_0$
$x=d_{n-1}(10^{n-1}-1)+\cdots+d_1(10^1-1)+(d_{n-1}+\cdots+d_1+d_0)$

We know $10^k-1$ is divisible by 3, which means the remainder upon dividing $x$ by 3 is just the sum of the digits of $x$. If this remainder is divisible by 3, then so must be the original number, $x$.

Back to our previous example: if $x=2148$, then we have the expansion

$2148=2\times10^3+10^2+4\times10+8$
$2148=2(999+1)+(99+1)+4(9+1)+8$
$2148=2\times999+99+4\times9+(2+1+4+8)$

Dividing through by 3, we get a remainder of $2+1+4+8=15$, 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:

$15=10+5$
$15=9+1+5$

Dividing through by 3 leaves a remainder of $1+5=6$, which is also a multiple of 3, so that 15 must be, too.