Divisibility rules/Rule 2 for 7 proof

Revision as of 13:19, 22 August 2006 by JBL (talk | contribs)

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.


Back to Divisibility Rules