Task 1. The fractions $\frac{\overline{3 a 5 b}}{36}$ and $\frac{\overline{4 c 7 d}}{45}$ are natural numbers, where $a, b, c, d$ are digits. Arrange all such numbers in ascending order.
Solution. The fractions $\frac{\overline{3 a 5 b}}{36}$ and $\frac{\overline{4 c 7 d}}{45}$ are natural numbers if $\overline{3 a 5 b}$ is divisible by 36 (meaning by 4 and 9) and $\overline{4 c 7 d}$ is divisible by 45 (meaning by 5 and 9). In the first case, this is possible if the last digit is 2 or 6, so due to divisibility by 9, the possible numbers are: $\frac{3456}{36}=96$ and $\frac{3852}{36}=107$. In the second case, the last digit is 0 or 5, so due to divisibility by 9, the possible cases are: $\frac{4275}{45}=95$ and $\frac{4770}{45}=106$. Therefore, ordered by size, the numbers are

$$
\frac{4275}{45}<\frac{3456}{36}<\frac{4770}{45}<\frac{3852}{36}
$$