Divisibility rules/Rule 2 for 7 proof

Proof for Rule 2:

The divisibility rule would be $2n_0-k$ , where $k=d_110^0+d_210^1+d_310^2+...$ , where $d_{n-1}$ is the nth digit from the right (NOT the left) and we have $k-2n_0\equiv 2n_0+6k$ and since 2 is relatively prime to 7, $2n_0+6k\equiv n_0+3k\pmod{7}$ . Then yet again $n_0+3k\equiv n_0+10k\pmod{7}$ , and this is equivalent to our original number.

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Divisibility rules/Rule 2 for 7 proof

Proof for Rule 2: