Let \(b_k 10^{3(k-1)}\) represent a block of three numbers. For example:
$$ 4896927 = 927 + 896000 + 4000000 $$
Here \(b_1 = 927\), \(b_2 = 896\) and \(b_3 = 4\). We can represent any integer \(n\) like this
$$ n = b_1 + b_2 10^3 + b_3 10^6 + b_4 10^9 + b_5 10^{12} + b_6 10^{15} + \ldots $$
We can manually check \(10^3 ≡ - 1 \bmod 7\). This means:
$$ (10^3)^n ≡ (-1)^n \bmod 7 $$
This also means \([(10^3)^n ≡ 1 \bmod 7]\) when \(n\) is even and \([(10^3)^n ≡ -1 \bmod 7]\) when \(n\) is odd:
$$ \begin{align} 7 &| 10^{3n} + 1 \quad \text{ if } n \text{ is odd} \\ 7 &| 10^{3n} - 1 \quad \text{ if } n \text{ is even} \end{align} $$
Going back to our integer \(n\). We can group all the blocks with even \(k\)'s together and all the blocks with odd \(k\)'s together:
$$\begin{gather} n = (b_2 10^3 + b_4 10^9 + b_6 10^{15} + \ldots) + (b_1 + b_3 10^6 + b_5 10^{12} + \ldots) \\ n = (b_1 + b_3 10^{3(2)} + b_5 10^{3(4)} + b_7 10^{3(6)} + \ldots) + (b_2 10^{3(1)} + b_4 10^{3(3)} + b_6 10^{3(5)} + \ldots) \end{gather}$$
Let's divide \(n\) by 7:
$$\frac{n}{7} = \frac{b_1}{7} + \frac{b_3 10^{3(2)} + b_5 10^{3(4)} + b_7 10^{3(6)} + \ldots}{7} + \frac{b_2 10^{3(1)} + b_4 10^{3(3)} + b_6 10^{3(5)} + \ldots}{7} $$
If \(7 | 10^{3n} - 1\) when \(n\) is even, then there must be an integer \(h\) such that:
$$ \begin{align} \frac{n}{7} &= \frac{b_1 + b_3 + b_5 + \ldots}{7} + \frac{b_3 (10^{3(2)} - 1) + b_5 (10^{3(4)} - 1) + b_7 (10^{3(6)} - 1) + \ldots}{7} + \frac{b_2 10^{3(1)} + b_4 10^{3(3)} + b_6 10^{3(5)} + \ldots}{7} \\ \frac{n}{7} &= \frac{b_1 + b_3 + b_5 + \ldots}{7} + h + \frac{b_2 10^{3(1)} + b_4 10^{3(3)} + b_6 10^{3(5)} + \ldots}{7} \end{align}$$
If \(7 | 10^{3n} + 1\) when \(n\) is odd, then there must be an integer \(j\) such that:
$$ \begin{align} \frac{n}{7} &= \frac{b_1 + b_3 + b_5 + \ldots}{7} + h + \frac{b_2 (10^{3(1)} +1) + b_4 (10^{3(3)}+1) + b_6 (10^{3(5)} + 1 ) + \ldots}{7} - \frac{b_2 + b_4 + b_6 + \ldots}{7} \\ \frac{n}{7} &= \frac{b_1 + b_3 + b_5 + \ldots}{7} + h + j - \frac{b_2 + b_4 + b_6 + \ldots}{7} \end{align}$$
Rearranging:
$$ \begin{align} \frac{n}{7} &= h + j + \frac{b_1 - b_2 + b_3 -b_4 + b_5 - b_6 + \ldots}{7} \end{align}$$
This means \(n/7\) is an integer if and only if \([(b_1 - b_2 + b_3 - b_4 + \ldots)/7]\) is an integer.