2008
Problem - 2389
Let both $A$ and $B$ be two-digit numbers, and their difference is $14$. If the last two digits of $A^2$ and $B^2$ are the same, what are all the possible values of $A$ and $B$.
Answer
6
Without loss of generality, let's assume $A-B=14$. The last two digits of $A^2$ and $B^2$ are the same means that the $(A^2-B^2)$ is a multiple of $100$. Meanwhile, we have $$A^2 - B^2 = (A+B)(A-B) = 14(A+B)$$
In order to make this number a multiple of $100$, $(A+B)$ must be the multiple of $50$.
If $A+B = 50$, then $A=32$, $B=18$.
If $A+B = 100$, then $A=57$, $B=43$.
If $A+B = 150$, then $A=82$, $B=68$.
If $A+B = 200$, then $A$ is no longer a $2$-digit number.
Considering $A$ and $B$ can be switched, we find the final answer is $\boxed{6}$.