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

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== See also ==
 
== See also ==
 
[[Divisibility rules | Back to divisibility rules]]
 
[[Divisibility rules | Back to divisibility rules]]
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[[Category:Divisibility Rules]]

Revision as of 19:25, 6 March 2014

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 follows.

See also

Back to divisibility rules

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