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I used a variety of tools to solve this puzzle.

Firstly I labelled the 13 squares in the diagram from (apparently) smallest to (apparently) largest,

a, b, c, d, e, f, g, h, i, j, k, l, m.Then by considering vertical slices through the diagram we get 7 equations linking subsets of these variables to the height of the rectangle,

x:By considering horizontal slices through the diagram we get 6 equations linking subsets of these variables to the width of the rectangle,

y:These, along with the area constraint:

Give us 14 equations in 15 variables, so we can rewrite the system to give us the values in the term of a single variable.

I used the SymPy symbolic maths library to do this:

This runs in 3.9s and gives me the following results:

For positive integers

n:We require the values to be all different (which they are), and for two of the values to be consecutive, which means

n=1 (the consecutive values arefandg, having values of 19 and 20 respectively).So we now know that George packed a 3×3, 5×5, 9×9, 11×11, 14×14, 19×19, 20×20, 24×24, 31×31, 33×33, 36×36, 39×39 and a 42×42 square into a 75×112 rectangle. Thus:

The 7 squares in the perimeter of George’s packing are: 19, 24, 31, 33, 36, 39, 42.

George’s Mum also packed the squares into a 75×112 rectangle, but in a (non-trivially) different way

I wrote my own rectangle packer for

Enigma 1251(based on my code forEnigma 17), but here I’ve solved the packing problem with a MiniZinc model:(Note that the orientation constraint could be simplified here, as the rectangles we are packing are squares).

This program runs in 718ms (using:

mzn-g12lazy -a enigma1241.mzn), and produces two separate packings, George’s shown above, and the packing shown below (along with their rotations and reflections).This second packing is the one that George’s Mum made.

The 8 squares in the perimeter of this packing are: 19, 20, 24, 31, 33, 36, 39, 42.

Solution:The extra edge piece in George’s Mum’s packing is the 20×20 square.