2017 AIME I Problems/Problem 11

We know that if $5$ is a median, then $5$ will be the median of the medians.

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 $32\cdot720\cdot9$ . But we only need the last $3$ digits, so $\boxed{360}$ is our answer.

~Solution by SuperSaiyanOver9000, mathics42

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2017 AIME I Problems/Problem 11