5 | 8 | 0 | 2 | |||||||||||||||

0 | ||||||||||||||||||

2 | 3 | |||||||||||||||||

1 | 4 | |||||||||||||||||

3 | ||||||||||||||||||

8 | ||||||||||||||||||

1 | 6 | 9 | 7 | 5 | ||||||||||||||

3 | 7 | |||||||||||||||||

4 | 2 | 0 | ||||||||||||||||

9 | ||||||||||||||||||

The blue area and the red area must contain the same set of values. This is because the blue area plus the gold area is a complete set of three rows, and the red area plus the gold area is a complete set of three boxes. There are at least five boundary areas in this jigsaw where the law of leftovers can be applied—some using rows and some using columns.

This solve method was first described by Bob Harris. He has an excellent presentation at bumblebeagle.org.