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Difference between revisions of "Divisibility rules/Rule 1 for 13 proof"

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To test if a number <math>N</math> is divisible by 13, then take off the last digit, multiply it by 4 and add it to the rest of the number. If this new number is divisible by 13, then so is <math>n</math>. This process can be repeated for large numbers, as with the second divisibility rule for 7.
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Truncate the last digit, multiply it by 4 and add it to the rest of the number. The result is divisible by 13 if and only if the original number was divisble by 13. This process can be repeated for large numbers, as with the second divisibility rule for 7.
  
 
== Proof ==
 
== Proof ==
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Let <math>N = d_0\cdot10^0 + d_1\cdot 10^1 +d_2\cdot 10^2 + \cdots</math> be a positive integer with units digit <math>d_0</math>, tens digit <math>d_1</math> and so on.  Then <math>k=d_110^0+d_210^1+d_310^2+\cdots</math> is the result of truncating the last digit from <math>N</math>.  Note that <math>N = 10k + d_0 \equiv d_0 - 3k \pmod {13}</math>.  Now <math>N \equiv 0 \pmod {13}</math> if and only if <math>4N \equiv 0 \pmod {13}</math>, so <math>n \equiv 0 \pmod{13}</math> if and only if <math>4d_0 - 12k \equiv 0 \pmod{13}</math>.  But <math>-12k \equiv k \pmod{13}</math>, and the result follows.
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''An understanding of [[Introduction to modular arithmetic | basic modular arithmetic]] is necessary for this proof.''
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Let <math>N = d_0\cdot10^0 + d_1\cdot 10^1 +d_2\cdot 10^2 + \cdots</math> be a positive integer with units digit <math>d_0</math>, tens digit <math>d_1</math> and so on.  Then <math>k=d_110^0+d_210^1+d_310^2+\cdots</math> is the result of truncating the last digit from <math>N</math>.  Note that <math>N = 10k + d_0 \equiv d_0 - 3k \pmod {13}</math>.  Now <math>N \equiv 0 \pmod {13}</math> if and only if <math>4N \equiv 0 \pmod {13}</math>, so <math>n \equiv 0 \pmod{13}</math> if and only if <math>4d_0 - 12k \equiv 0 \pmod{13}</math>.  But <math>-12k \equiv k \pmod{13}</math>, and the result
  
 
== See also ==
 
== See also ==
 
[[Divisibility rules | Back to divisibility rules]]
 
[[Divisibility rules | Back to divisibility rules]]
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[[Category:Divisibility Rules]]

Latest revision as of 13:20, 12 April 2021

Truncate the last digit, multiply it by 4 and add it to the rest of the number. The result is divisible by 13 if and only if the original number was divisble by 13. This process can be repeated for large numbers, as with the second divisibility rule for 7.

Proof

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

Let $N = d_0\cdot10^0 + d_1\cdot 10^1 +d_2\cdot 10^2 + \cdots$ be a positive integer with units digit $d_0$, tens digit $d_1$ and so on. Then $k=d_110^0+d_210^1+d_310^2+\cdots$ is the result of truncating the last digit from $N$. Note that $N = 10k + d_0 \equiv d_0 - 3k \pmod {13}$. Now $N \equiv 0 \pmod {13}$ if and only if $4N \equiv 0 \pmod {13}$, so $n \equiv 0 \pmod{13}$ if and only if $4d_0 - 12k \equiv 0 \pmod{13}$. But $-12k \equiv k \pmod{13}$, and the result

See also

Back to divisibility rules

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