Divisibility by 3

Statement

Let $n$ be a natural number with decimal digits $d_k d_{k-1} \dots d_0$, so that $n = \sum_{i=0}^{k} d_i 10^i$. Then $3 \mid n$ if and only if $3 \mid d_0 + d_1 + \dots + d_k$.

Remarks

More precisely, $n \equiv d_0 + d_1 + \dots + d_k \pmod 3$. This holds because $10 \equiv 1 \pmod 3$, so $10^i \equiv 1 \pmod 3$ by compatibility with powers. For example, $3 \mid 7{,}341$, because $7 + 3 + 4 + 1 = 15$ and $3 \mid 15$. The rule for divisibility by 9 has the same proof.

Explanations

L1 human

Add the digits of the number. If the sum is a multiple of $3$, the number is a multiple of $3$, and if not, it is not. You can repeat: for $987{,}654$ the sum is $39$, and $3 + 9 = 12$ is a multiple of $3$. So $3 \mid 987{,}654$.