Suppose an integer \(n\) has \(k\) digits and let \(a_k\) be the \(k\)th digit. We can represent \(n\) like this:
$$\begin{align} n &= a_k (10^{k-1}) + a_{k-1} (10^{k-2}) + \cdots + a_3 (10^2) + a_2 (10^1) +a_1 \\ n &= a_k (99\ldots9) + a_{k-1} (99\ldots9) + \cdots + a_3 (99) + a_2 (9) + a_1 + (a_2 + a_3 + \cdots + a_{k-1} + a_k) \end{align}$$
Since 9 is divisible by 3, then \((99\ldots9)\) is also divisible by 3. So if we divide \(n\) by 3, then there exists an integer \(h\) such that:
$$\begin{align} \frac{n}{3} &= \frac{a_k (99\ldots9) + a_{k-1} (99\ldots9) + \cdots + a_3 (99) + a_2 (9)}{3} + \frac{a_1 + a_2 + a_3 + \cdots + a_{k-1} + a_k}{3} \\ \frac{n}{3} &= h + \frac{a_1 + a_2 + a_3 + \cdots + a_{k-1} + a_k}{3} \end{align}$$
This means \(\frac{n}{3}\) would be an integer if and only if \(\frac{a_1 + a_2 + a_3 + \cdots + a_{k-1} + a_k}{3}\) is an integer.