Difference between revisions of "Divisibility rules/Rule 2 for 7 proof"

 
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The divisibility rule would be <math>2n_0-k</math>, where <math>k=d_110^0+d_210^1+d_310^2+...</math>, where <math>d_{n-1}</math> is the nth digit from the right (NOT the left) and we have <math>k-2n_0\equiv 2n_0+6k</math> and since 2 is relatively prime to 7, <math>2n_0+6k\equiv n_0+3k\pmod{7}</math>.  Then yet again <math>n_0+3k\equiv n_0+10k\pmod{7}</math>, and this is equivalent to our original number.
 
The divisibility rule would be <math>2n_0-k</math>, where <math>k=d_110^0+d_210^1+d_310^2+...</math>, where <math>d_{n-1}</math> is the nth digit from the right (NOT the left) and we have <math>k-2n_0\equiv 2n_0+6k</math> and since 2 is relatively prime to 7, <math>2n_0+6k\equiv n_0+3k\pmod{7}</math>.  Then yet again <math>n_0+3k\equiv n_0+10k\pmod{7}</math>, and this is equivalent to our original number.
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Revision as of 13:19, 22 August 2006

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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