20. [11] There exist unique nonnegative integers $A, B$ between 0 and 9 , inclusive, such that
$$
(1001 \cdot A+110 \cdot B)^{2}=57,108,249 \text {. }
$$

Find $10 \cdot A+B$.
Answer: 75 We only need to bound for $A B 00$; in other words, $A B^{2} \leq 5710$ but $(A B+1)^{2} \geq 5710$. A quick check gives $A B=75$. (Lots of ways to get this...)