The number of 3-digit numbers of the form $$ab5$$ (where $$a, b$$ are digits) which are divisible by $$9$$ is ------.

The number $$100a+10b+5$$ is divisible by $$9$$ exactly when its digit sum $$a+b+5$$ is a multiple of $$9$$, i.e. $$a+b=4$$ or $$a+b=13$$ (with $$a\geq 1$$). For $$a+b=4$$ there are $$4$$ valid pairs ($$a=1,2,3,4$$), and for $$a+b=13$$ there are $$6$$ valid pairs ($$a=4$$ to $$9$$), giving $$4+6=10$$ numbers in total.