Difference between revisions of "Divisibility rules/Rule 2 for 7 proof"
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| − | + | Truncate the last digit of <math>N</math>, double that digit, and subtract it from the rest of the number (or vice-versa). <math>N</math> is divisible by 7 if and only if the result is divisible by 7. | |
| + | ==Proof== | ||
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. | ||
| − | + | ==See also== | |
[[Divisibility rules | Back to Divisibility Rules]] | [[Divisibility rules | Back to Divisibility Rules]] | ||
Revision as of 12:39, 17 June 2008
Truncate the last digit of 
, double that digit, and subtract it from the rest of the number (or vice-versa). is divisible by 7 if and only if the result is divisible by 7.
Proof
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.
