Show that a natural number is divisible by 3 if and only if the sum of the digits is divisible by 3.
A natural number $n$ is of the form
$$ n = (a_0 \times 10^0) + (a_1 \times 10^1) + (a_2 \times 10^2) + \ldots + (a_m \times 10^m) $$
where $m$ is an integer, and $a_i$ are the digits of the number in base 10.
Propositions 3.6 and 3.8 with $10 \equiv 1 \pmod 3$ allow us to reduce this as follows
$$ \begin{align} n & = (a_0 \times 10^0) + (a_1 \times 10^1) + (a_2 \times 10^2) + \ldots + (a_m \times 10^m) \\ \\ & \equiv (a_0 \times 1^0) + (a_1 \times 1^1) + (a_2 \times 1^2) + \ldots + (a_m \times 1^m) \pmod 3 \\ \\ n & \equiv a_0 + a_1 + a_2 + \ldots + a_m \pmod 3 \end{align} $$
($\implies$) If $n$ is divisible by 3, then $n \equiv 0 \pmod 3$. This means the sum of the digits must also be congruent to 0 modulo 0, that is, divisible by 3.
($\impliedby$) If the sum of digits is divisible by 3, then that sum is congruent to 0 modulo 3. That means $n \equiv 0 \pmod 3$, that is, $n$ is divisible by 3.
And so a natural number is divisible by 3 if and only if the sum of the digits is divisible by 3.