2017 AIME I Problems/Problem 11

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 1

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

Solution 2

(Complementary Counting with probability)

Notice that m can only equal 4, 5, or 6, and 4 and 6 are symmetric.

WLOG let $m=4$

There is a $\frac{15}{28}$ chance that exactly one of 1, 2, 3 is in the same row.

There is a $\frac{2}{5}$ chance that the other two smaller numbers end up in the same row.

$9!(1-2*\frac{15}{28}*\frac{2}{5})=362880*\frac{4}{7}=207\boxed{360}$ .

2017 AIME I (Problems • Answer Key • Resources)
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2017 AIME I Problems/Problem 11

Contents

Problem 11

Solution 1

Solution 2

See Also