Difference between revisions of "Divisibility rules/Rule 2 for 7 proof"
m |
|||
| Line 2: | Line 2: | ||
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. | ||
| + | |||
| + | |||
| + | [[Divisibility rules | Back to Divisibility Rules]] | ||
Revision as of 13:19, 22 August 2006
Proof for Rule 2:
The divisibility rule would be 
, where 
, where 
is the nth digit from the right (NOT the left) and we have 
and since 2 is relatively prime to 7, 
. Then yet again 
, and this is equivalent to our original number.
