Exercise 5. How many numbers between 100 and 999 (inclusive) have digits that form an arithmetic progression when read from left to right?

A sequence of three numbers \(a, b, c\) is said to form an arithmetic progression if \(a + c = 2b\).

A correct numerical answer without justification will earn 4 points. To earn full points, a detailed reasoning is expected.
Solution to Exercise 5: Let $n$ be a number between 100 and 999, and let $a, b, c$ be its three digits, read from left to right. If $n$ is one of the sought numbers, then $a$ and $c$ are digits of the same parity, and $b = (a + c) / 2$. Thus, once the pair $(a, c)$ is determined, with $a$ and $c$ of the same parity, we can associate a unique integer $n$ among those sought.

Therefore, it is a matter of determining how many pairs of integers $(a, c)$ of the same parity exist such that $1 \leqslant a \leqslant 9$ and $0 \leqslant c \leqslant 9$. For each possible value of $a$, there are exactly 5 suitable values of $c$. Therefore, there are $9 \times 5 = 45$ pairs $(a, c)$ as sought, and the expected answer is therefore 45.