2017 AIME I Problems/Problem 11
Revision as of 16:41, 8 March 2017 by Mathguy5041 (talk | contribs)
We know that if 
is a median, then will be the median of the medians.
WLOG, assume is in the upper left corner. One of the two other values in the top row needs to be below , and the other needs to be above . This can be done in 
ways.
The other 
can be arranged in 
ways.
Finally, accounting for when is in every other space, our answer is 
. But we only need the last 
digits, so 
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 
There is a 
chance that exactly one of 1, 2, 3 is in the same row.
There is a 
chance that the other two smaller numbers end up in the same row.
.
