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

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''An understanding of [[Introduction to modular arithmetic | basic modular arithmetic]] is necessary for this proof.'' | ''An understanding of [[Introduction to modular arithmetic | basic modular arithmetic]] is necessary for this proof.'' | ||

− | The divisibility rule would be <math> | + | The divisibility rule would be <math>2d_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-2d_0\equiv 2d_0+6k</math> and since 2 is relatively prime to 7, <math>2d_0+6k\equiv d_0+3k\pmod{7}</math>. Then yet again <math>d_0+3k\equiv d_0+10k\pmod{7}</math>, and this is equivalent to our original number. |

==See also== | ==See also== | ||

[[Divisibility rules | Back to Divisibility Rules]] | [[Divisibility rules | Back to Divisibility Rules]] | ||

[[Category:Divisibility Rules]] | [[Category:Divisibility Rules]] |

## Latest revision as of 23:52, 21 September 2020

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

*An understanding of basic modular arithmetic is necessary for this 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.