Many people produced good results and it
was interesting to see the different ways the problem was
Did you decide that you can add another
cube onto one that you have already added to or not? So, in the
following, can the latest blue cube go in the places pictured in
the 2nd and 3rd diagrams?
Michael wrote in to say:
There were five places where a cube could be added on to each of
the five Wag Worms, and $5\times5 = 25$. So, there are $25$
different types of Wag Worm in its third year.
The Gateway School sent in two solutions
of $25$ as well.
Rowena wrote the following good
I know that a cube has $6$ faces. I looked at the Wag Worm in its
second year, you can add a cube to all of the $6$ faces except the
one with eyes, so there are $5$ possible Wag Worms.
For the Wag Worms in their third year, I looked at adding a cube to
the 'non-eye cube' and the 'eye cube'.
For the 'non-eye cube' there are $5$ visible faces (one is attached
to the 'eye cube') so I could make $5$ different Wag Worms.
For the 'eye cube' again there are $5$ visible faces, but one has
eyes, so I could have $4$ places to put the cube.
This is $5+4=9$ possible Wag Worms for each of the 5 Worms in their
second year. So in total there would be $5\times9 = 45$ different
Wag Worms in their third year.
Emma and Oscar went further and tried
the fifth year of growth. Here are some of their