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'L-triominoes' printed from http://nrich.maths.org/
Neerajan and Eilidh both noticed that the
number of L-triominoes needed to tile a size $n$ L-triominoes was
equal to $n^2$.
Philip from Wilson's School explained his
To tile a size 4 triomino you would simply put 4 size 2 triominoes
together the same way you put four size 1 triominoes together to
make a size 2.
Therefore you would put 4 size 4 triominoes together in the same
way to form a size 8 triomino. Then 4 size 8 triominoes to make a
size 16, and so on.
The triominoes could therefore tile size 8, 16, 32... $2^n$. These
can therefore be made up of size 1 triominoes.
As it takes $4^1$ size 1 triominoes to tile a size 2
triomino, and $4^2$ size 1 triominoes to tile a size 4 triomino, it
would therefore take $4^3=64$ size-1 triominoes to tile a size 8
It would then take $4^4=256$ size 1 triominoes to tile a size 16
triomino and $4^5=1024$ size 1 triominoes to tile a size 32
Finally, it would take $4^x$ triominoes to tile a size $2^x$
To tile the size 3 triomino you simply use the 2x3 rectangular
blocks at the top and on the right and then the rest you use a size
2 triomino and the size 1 in the middle.
To tile a size 5 triomino you simply add another size 2 triomino in
the bottom left and for the rest you can use the 2x3 rectangular
boxes. It's the same with a size 7 triomino, add a size 2 and then
use 2x3 boxes to tile the rest. This works with all odd
Therefore all size triominoes can be tiled. If the size number is
even you use the first method and if the size number is odd you use
Isaac from Hampton School also sent us his
solution, which you can read here.