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 $\mathrm{a}, \mathrm{b}$, $\mathrm{c}$ is said to form an arithmetic progression if $\mathrm{a}+\mathrm{c}=2 \mathrm{~b}$.

A correct numerical answer without justification will earn 4 points. To earn full points, a detailed reasoning is required.
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 $\mathrm{n}$ is one of the sought numbers, then $a$ and $\mathrm{c}$ are digits of the same parity, and we have $\mathrm{a} b=(\mathrm{a}+\mathrm{c}) / 2$. Thus, once the pair ( $\mathrm{a}, \mathrm{c}$ ) is determined, with $a$ and $\mathrm{c}$ of the same parity, we can associate a unique integer $\mathrm{n}$ 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 ( $\mathrm{a}, \mathrm{c}$ ) as sought, and the expected answer is therefore 45.

Graders' Comments Exercise well done. Some students were particularly efficient in their arguments, while others patiently counted all the solutions. The danger of manual search was a few omissions or additions. Be careful not to rush in reasoning and to verify that you can prove all the rules you state!