The problem is stated here. Join the Discord if you have something to add.


You can place a square touching more than 2 squares. If you do so, you have to add up all the squares it touches.

Mr Pickle (Gord Hamilton), advised me against my original strategy of keeping everything as condensed as possible. He also told me that you can prove the optimal solution is finite, but if you allow the squares to be arbitrarily large, then you can place any number of squares.


If you can place squares which are arbitrarily large, why does that allow you to place any number of squares?

Known results:

There is only 1 way to make 2: 1 + 1. For all other numbers n, the number of ways in which it can be made, is given by \(\lfloor (n - 1)/2 \rfloor.\) For example, 6 can be made in \(\lfloor (6 - 1)/2 \rfloor = 2\) ways: 1 + 5, and 2 + 4. 7 can be made in \(\lfloor (7 - 1)/2 \rfloor = 3\) ways: 1 + 6, 2 + 5, and 3 + 4.


How many squares can you place if each square labelled \(n\) must be \(n \times n?\)

Progress so far:

I managed to place 8 squares. If you manage to place more than 8, please let me know.


Make a program for drawing the squares, and writing numbers inside them. Maybe at first there are just 3 tools: draw square, select square (backspace after to delete), alter text (maybe by clicking inside the square).