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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, multiply the last digit by 4 and add it to the rest of the number.  If this new number is divisible by 7, 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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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.
  
 
== Proof ==
 
== Proof ==

Revision as of 22:41, 23 February 2008

To test if a number $N$ 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 $n$. This process can be repeated for large numbers, as with the second divisibility rule for 7.

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 follows.

See also

Back to divisibility rules

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