1. How many five-digit numbers of the form $\overline{37 a b c}$ are there such that each of the numbers $\overline{37 a b c}, \overline{37 b c a}$, and $\overline{37 c a b}$ is divisible by 37?
The solution. The five-digit number $\overline{37 a b c}$ is divisible by 37 if and only if the number $\overline{a b c}$ is divisible by 37. Let's denote $x=\overline{a b c}, y=\overline{b c a}$, and $z=\overline{c a b}$. Then,

$$
x=100 a+10 b+c, \quad y=100 b+10 c+a, \quad z=100 c+10 a+b
$$

or

$$
10 x-y=999 a, \quad 10 y-z=999 b, \quad 10 z-x=999 c
$$

Since 999 is a multiple of 37, i.e., $999=37 \cdot 27$, it follows from the above that if any of the numbers $x, y$, or $z$ is divisible by 37, then the others are as well. Therefore, all the required numbers are those whose last three digits form a number that is a multiple of 37, i.e., $37000, 37037, 37074, 37111, \ldots, 37999$.

Since $999=37 \cdot 27$, there are exactly 28 such numbers.