|Spoiler Inside: Solution to Puzzle||SelectShow>|
Since no two counts can add to 12, both 3 and 9, both 4 and 8, and both 5 and 7 can not occur. Thus, at most, three of these value can appear in the solution. That leaves three other numbers so exactly three of them are in the solution. Three of the counts are 1, 2, and 6, and the other three are one each of 3 and 9, 4 and 8, and 5 and 7.
Since there is only one square and 1 is already known to be one of the counts, 4 and 9 are eliminated. Thus, 3 and 8 are in the solution.
There is only one sequence of three counts, and 1, 2, and 3 are in the solution. If 7 were in the solution, there would be a second such (7, 8, 9). Therefore, the last count is 5.