2013 AMC 10B Problem 22

Let $S$ be defined as: $$\begin{aligned} S &= A+J+E \\ &=B+J+F \\ &=C+J+G \\ &=D+J+H \end{aligned}$$ $$\begin{aligned} 4S &= A+B+C+D+E \\ &\quad+F+G+H+4J \end{aligned}$$ $$4S = 45+3J$$ $$45+3J \equiv 0 \mod 4$$ $$3J \equiv 3 \mod 4$$ $$J \equiv 1 \mod 4$$ This means that $J=1,5,9.$ From here, let’s assume $J=1.$ We will see that the other cases are similar enough to omit. If $J=1,$ then we know that the pairs of numbers that satisfy the equality above are: $$2+9 = 3+8 = 4+7 = 5+6$$ There are $4!$ ways to distribute the pairs over the four groups, and then $2^4$ ways for these groups to swap elements (i.e. $2+9\iff 9+2$). Now, if we look at the $J=5$ and $J=9$ cases, we see a similar pattern in the number of groupings and swaps. As such, we have: $$3\cdot 4! \cdot 2^4 = 1152$$ possibilities. Thus, the correct answer is C .