1. (20 points) On the board, all two-digit numbers divisible by 5, where the number of tens is greater than the number of units, were written down. There turned out to be $A$ such numbers. Then, all two-digit numbers divisible by 5, where the number of tens is less than the number of units, were written down. There turned out to be $B$ such numbers. What is $100 B+A$?
Answer: 413.

Solution: We will write two-digit numbers in the form $\overline{x y}$, where $x$ is the number of tens, and $y$ is the number of units.

Let's calculate what $A$ is. We need two-digit numbers divisible by 5, i.e., numbers where $y$ is either 0 or 5. Note that if $y=0$, then $x$ can take any value from 1 to 9, giving us 9 numbers. And if $y=5$, then $x$ can take values only from 6 to 9, giving us another 4 numbers. In total, we get 13 numbers.

Let's calculate $B$. Here, there can only be the case where $y=5$, because $x$ cannot be less than 0. So, $x$ can take values from 1 to 4. Therefore, $B=4$. From this, we get that $100 B+A=400+13=413$.