Determine the number of pairs of integers $(a, b)$ such that $1 \leqslant a \leqslant 30, 3 \leqslant b \leqslant 30$ and such that $a$ is divisible by $b$ and by $b-2$.
Answer: 22. For $b=3$, the number of multiples of $b(b-2)=3$ which are $\leqslant 30$ is equal to 10. For $b=4$, the condition is that 4 divides $a$, there are 7 solutions. We find similarly 2 solutions for $b=5$, 2 solutions for $b=6$ and 1 solution for $b=8$. It is easy to show that there are no other solutions.