[Number Theory]
Problem: Let $A$ and $B$ be digits from the set $\{0,1,2, \ldots, 9\}$. Let $r$ be the two-digit integer $\underline{A} \underline{B}$ and let $s$ be the two-digit integer $\underline{B} \underline{A}$, so that $r$ and $s$ are members of the set $\{00,01, \ldots, 99\}$. Compute the number of ordered pairs $(A, B)$ such that $|r-s|=k^{2}$ for some integer $k$.

Solution: Because $|(10 A+B)-(10 B+A)|=9|A-B|=k^{2}$, it follows that $|A-B|$ is a perfect square. $|A-B|=0$ yields 10 pairs of integers: $(A, B)=(0,0),(1,1), \ldots,(9,9)$.

$|A-B|=1$ yields 18 pairs: the nine $(A, B)=(0,1),(1,2), \ldots,(8,9)$, and their reverses.

$|A-B|=4$ yields 12 pairs: the six $(A, B)=(0,4),(1,5), \ldots,(5,9)$, and their reverses.

$|A-B|=9$ yields 2 pairs: $(A, B)=(0,9)$ and its reverse.

Thus the total number of possible ordered pairs $(A, B)$ is $10+18+12+2=\mathbf{4 2}$.

Final answer: 42