The four digit number $AABB$ can be divided by $6$ without remainder. Determine the greatest and least value of $AABB$.
On the condition that a number can be divided by $6$, it must be divisible by both $2$ and $3$
$$AABB \equiv 0 \pmod{2}\tag{1}$$ $$AABB \equiv 0 \pmod{3}\tag{2}$$
Reducing $AABB \equiv 0 \pmod{2}$,
$$AABB \equiv 0 \pmod{2} \implies 1000A + 100A + 10B + B \implies B \equiv 0 \pmod{2}$$
Now reducing $AABB \equiv 0 \pmod{3}$,
$$AABB \equiv 0 \pmod{3} \implies 1000A + 100A +10B + B \implies A + A + B + B \equiv 0 \pmod{3} \implies 2(A+B) \equiv 0 \pmod{3}$$
I think I've gone wrong so far.
You have shown that we must have $B$ even and $A+B$ divisible by $3$. To make $AABB$ as large as possible, we'd like to have $A=9$ If this is so, can we find an even number $B$ such that $3|(9+B)?$ Similarly, to make $AABB$ as small as possible, we'd like to have $A=1$. What are the choices for $B?$