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 , 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.