A popular question and many solutions were submitted. Well done to Chen of The Chinese High School Singapore, Sana, Jenny, Chris, Rosion of Madras College, St. Andrews, Andrei of School 205 Bucharest, Mary of Birchwood Community High School. The following is an amalgamation of their solutions.

The sum of the areas of all the squares, 182 + 152 + 142 + 102 + 92 + 82 + 72 + 42 + 12 = 1056.

1056 into 2 factors one of which must be larger than or equal to 32, we have 1056 = 32*33.

If all these squares can be fitted together to form a rectangle, the shorter side of the rectangle must be at least 32cm (18+14) as 18+15 or 18+14 will definitely form part of a side.

From this, we can arrange the squares cleverly, i.e. the 14 and 4 square are most likely to be placed next to the 18 square, etc and through a little bit of trial and error, it can be concluded that such a rectangle is possible to construct and is shown in the attached picture, where the unnamed square in the centre is the square with side 1.

Where

Red – 18 cm
Yellow - 15 cm
Blue – 14 cm
Purple – 10 cm
Light green – 9 cm
Magenta – 8 cm
Dark green – 7 cm
White – 4 cm
Black – 1 cm