Calculate $3 \times 37$ then $27 \times 37$, then give a criterion for divisibility by 37.
We easily verify that $3 \times 37=111$, and then that $27 \times 37=9 \times 3 \times 37=999$. This means that $10^{3} \equiv$ $1(\bmod 37)$. Furthermore, $10^{2} \equiv-11(\bmod 37)$. Consequently, a number $n={\overline{n_{d} n_{d-1} \ldots n_{1} n_{0}}}^{10}$ is divisible by 37 if and only if

$$
\sum_{k=0}^{d / 3}\left(n_{3 k}+10 n_{3 k+1}-11 n_{3 k+2}\right)
$$

is divisible by 37.