Divisibility by $8$ and $9$. How many ordered pairs $(a,b)$ exist such that the four-digit number, $a04b$, is divisible by both $8$ and $9$?
How should I approach this? (without modular arithmetic?)

We can write:
$$a04b=1000\cdot a+100\cdot 0+10\cdot 4+b$$
Once $1000$ and $40$ are divisible by $8$ then $b$ must be divisible by $8$ and then $b=8$.
For division by $9$ we can write:
$$a048= 999\cdot a + a+40+8$$
Once $999$ is divisible by $9$ then $a+48$ has to be divisible by $9$. Once $0\le a\le 9$ the only value that fits is $a=6$.
