1. (10 points) On the board, all three-digit numbers divisible by 5 were written, where the hundreds digit is greater than the tens digit, and the tens digit is greater than the units digit. There turned out to be $A$ such numbers. Then, all three-digit numbers divisible by 5 were written, where the hundreds digit is less than the tens digit, and the tens digit is less than the units digit. There turned out to be $B$ such numbers. Choose the correct statements.
a) $A>B$
б) $B>10$;
в) $A+B>100$
г) $A<10 ; \partial)$ among the listed answers, there is no correct one.
Answer: a)

Solution: Let's denote the three-digit number as $\overline{x y z}$, where $x$ is the hundreds digit, $y$ is the tens digit, and $z$ is the units digit. According to the problem, the numbers we are looking for are divisible by 5, which means $z$ is either 0 or 5.

Let's determine what $A$ is. If $z=0$, we are interested in the number of pairs $(x, y): 9 \geq x > y > 0$. There are $8+7+\ldots+1=36$ (8 - the number of numbers starting with 9; 7 - the number of numbers starting with 8; ...; 1 - the number of numbers starting with 2, i.e., the number 21). If $z=5$, the number of pairs $(x, y)$ where $9 \geq x > y > 5$ will be $3+2+1=6$. Therefore, $A=36+6=42$.

Now, let's calculate $B$. Here, only the case $z=5$ is possible, because the tens digit cannot be less than zero. Therefore, we need to find the number of pairs $(x, y)$ such that $0 < x < y < 5$. There are $3+2+1=6$ (3 - the number of numbers starting with 1; 2 - the number of numbers starting with 2; and the number 34). Thus, $B=6$.

Therefore, the correct answer is a).