Casting Out in Threes

Here is a quick test for divisibility by $37$, resting on the fact that $37 \times 27 = 999$. Use it to decide whether $73926$ is divisible by $37$.

Solution

Because $999$ is a multiple of $37$, the number $1000$ leaves remainder $1$ on division by $37$. So a number leaves the same remainder as the sum of its digits taken in three-digit groups from the right. For $73926$ the groups are $926$ and $073$, and $$926 + 73 = 999 = 27 \times 37,$$ a multiple of $37$ (and of $27$), so $73926$ is divisible by both. Indeed $73926 = 37 \times 1998$.

The companion fact $1001 = 7 \times 11 \times 13$ gives a matching rule: since $1000$ leaves remainder $-1$ on division by $1001$, a number leaves the same remainder as the alternating sum of its three-digit groups, on division by $7$, $11$ and $13$ all at once.