690. Find the divisibility rule for 2 in a number system with any odd base.
Let's denote the base of the numeral system as $d$. Take any number in this system

$$
x=\overline{a_{1} a_{2} a_{3} \ldots a_{n-1} a_{n d}}=a_{1} \cdot d^{n-1}+a_{2} \cdot d^{n-2}+a_{3} \cdot d^{n-3}+\ldots+a_{n-1} \cdot d+a_{n}
$$

Add and subtract the sum $a_{1}+a_{2}+a_{3}+\ldots+a_{n}$ (the "plus-minus" trick).

We get:

$$
\begin{gathered}
x=a \cdot\left(d^{n-1}-1\right)+a_{2} \cdot\left(d^{n-2}-1\right)+a_{3} \cdot\left(d^{n-3}-1\right)+\ldots \\
\ldots+a_{n-1} \cdot(d-1)+\left(a_{1}+a_{2}+a_{3}+\ldots+a_{n}\right)
\end{gathered}
$$

Since $d$ is odd, each of the differences

$$
d^{n-1}-1, \quad d^{n-2}-1, \ldots, \quad d-1
$$

is even. Therefore, the number $x$ is divisible by 2 if and only if the sum $a_{1}+a_{2}+a_{3}+\ldots+a_{n}$ is divisible by 2.

Answer: A number is divisible by 2 if and only if the sum of its digits is divisible by 2.