When solving sum placement puzzles, it helps to look at extremes.

Keep reading to see a sample of this kind of thinking, or jump ahead straight to today's challenge.

On the puzzle above, each tile gets used once, and the numbers on the sides indicate what the sums are of their corresponding row or column.

$8$ is fairly large compared to the other tiles: where could it go? It can't be in the first or third column because they would become too large, so it gets forced in the middle.

The $6$ is too large to go into the right column, so it has to go in the left. If it was on top, the sum $6 + 8$ is too large to make $11 ,$ so it has to be on bottom.

The other number in the column with the $6$ needs to be a $1$ to make a sum of $7 .$ Try placing the last values yourself to solve the puzzle, and then go on and tackle our daily challenge.