Divisibility by 37

Theorem

Let $n$ be an integer which has at least $3$ digits when expressed in decimal notation.

Let the digits of $n$ be divided into groups of $3$, counting from the right, and those groups added.

Then the result is equal to a multiple of $37$ if and only if $n$ is divisible by $37$.

Write $n = \ds \sum_{i \mathop = 0}^k a_i 10^{3 i}$, where $0 \le a_i < 1000$.

This divides the digits of $n$ into groups of $3$.

Then the statement is equivalent to:

$37 \divides n \iff 37 \divides \ds \sum_{i \mathop = 0}^k a_i$

Note that $1000 = 37 \times 27 + 1 \equiv 1 \pmod {37}$.

Hence:

$\ds n$

$=$

$\ds \sum_{i \mathop = 0}^k a_i 10^{3 i}$

$\ds $

$=$

$\ds \sum_{i \mathop = 0}^k a_i 1000^i$

$\ds $

$\equiv$

$\ds \sum_{i \mathop = 0}^k a_i 1^i$

$\ds \pmod {37}$

$\ds $

$\equiv$

$\ds \sum_{i \mathop = 0}^k a_i$

$\ds \pmod {37}$

which proves our statement.

$\blacksquare$