Divisibility rules/Rule 1 for 13 proof

To test if a number $N$ 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 $n$ . 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.

Page

Toolbox

Search

Divisibility rules/Rule 1 for 13 proof

Proof

See also