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>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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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 $N$, double that digit, and subtract it from the rest of the number (or vice-versa). $N$ 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 $2d_0-k$, 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 $k-2d_0\equiv 2d_0+6k$ and since 2 is relatively prime to 7, $2d_0+6k\equiv d_0+3k\pmod{7}$. Then yet again $d_0+3k\equiv d_0+10k\pmod{7}$, and this is equivalent to our original number.

See also

Back to Divisibility Rules

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