Divisibility rules/Rule 1 for 13 proof

Proof

Let $n = d_0\cdot10^0 + d_1\cdot 10^1 +d_2\cdot 10^2 + \ldots$ be a positive integer with units digit $d_0$ , tens digit $d_1$ and so on. Then $k=d_110^0+d_210^1+d_310^2+...$ 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.

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Divisibility rules/Rule 1 for 13 proof

Proof