Problem 959

If we represent an arbitrary natural number $a$ in the number system with base 1000, that is,

$$a=\sum_{k=0}^{m} a_{k} 1000^{k} \quad\left(0 \leq a_{k}<1000, m \in \mathbb{N}\right)$$

then regarding divisibility by 37, due to $1000 \equiv 1 \bmod 37$, the simple rule applies that $a$ is divisible by 37 if and only if its "digit sum" $\sum_{k=0}^{m} a_{k}$ is.

In the given case, the digits are simply all natural numbers from 111 to 999, and the order of these digits does not matter in terms of the digit sum. And yes, this digit sum, namely

$$\frac{111+999}{2}(999-111+1)$$

is indeed divisible by 37 because $111=3 \cdot 37$.