How many 7-digit numbers divisible by 9 are there, whose second last digit is 5?
We denote the 7-digit numbers in the usual way:

$$
\overline{a_{1} a_{2} a_{3} a_{4} a_{5} 5 a_{7}}
$$

where the digits $a_{2}, \ldots, a_{7}$ can take any value from 0 to 9, and $a_{1}$ can be any number from 1 to 9.

It is known that a number is divisible by 9 if and only if the sum of its digits is divisible by 9. First, we assign arbitrary values to $a_{2}, a_{3}, a_{4}, a_{5}, a_{7}$. Then, we can always uniquely determine the value of $a_{1}$ such that the resulting 7-digit number is divisible by 9. The value of $a_{1}$ depends on the remainder when the sum of the other six digits is divided by 9. This remainder can take 9 different values, so among the numbers from 1 to 9, there is always exactly one value for $a_{1}$ that, when added to the sum, makes it divisible by 9.

The digits $a_{2}, a_{3}, a_{4}, a_{5}, a_{7}$ provide $10 \cdot 10 \cdot 10 \cdot 10 \cdot 10=10^{5}$ possible choices, meaning there are this many 7-digit numbers that meet the criteria.

Gábor Tóth (Győr, Jedlik Á. Inf. Szki. és Gimn., 12. o.t.)

Note. Consider whether the solution to the problem changes if we do not fix the second-to-last digit, or if the fixed digit is not 5.