AIME 2017 I, Problem 11

Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$, $a_2$, and $a_3$ be the medians of the numbers in rows $1$, $2$, and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$. Let $Q$ be the number of arrangements for which $m = 5$. Find the remainder when $Q$ is divided by $1000$.

Solution: Assume that $5 \in \{a_1, a_2, a_3\}$, $m \neq 5$, and WLOG, $\max{(a_1, a_2, a_3)} = 5$. Then we know that the other two medians in $\{a_1, a_2, a_3\}$ and the smallest number of rows 1, 2, and 3 are all less than 5. But there are only 4 numbers less than 5 in $1, 2, 3, \dots, 9$, a Contradiction. Thus, if $5 \in \{a_1, a_2, a_3\}$, then $m = 5$. WLOG, assume $5$ is in the upper left corner. One of the two other values in the top row needs to be below $5$, and the other needs to be above $5$. This can be done in $4\cdot4\cdot2=32$ ways. The other $6$ can be arranged in $6!=720$ ways. Finally, accounting for when $5$ is in every other space, our answer is $32\cdot720\cdot9$, which is $207360$. But we only need the last $3$ digits, so $\fbox{360}$ is our answer. ~Solution by SuperSaiyanOver9000, mathics42, edited by zhaohm

Answer: 360