We had a few solutions sent in for this challenge. I think that maybe there were other pupils who got so far, on their own or with the teacher, and decided it was not worth sending in. We would like to reassure you that all journeys in the solving of challenges - however far you get - we welcome!

Here is the excellent result that we received from class $5$A at St. Christopher's School in Penang, Malaysia.

Well Done !

From Jenthghe and Olivia we received.

For this problem we used a trial and error method!

For solid cube number $1$ we did, $3$x$3$x$3$=$27$

For frame cube number $1$ we did $3$x$3$x$3$=$27-7=20$ we took away $7$ because there was seven holes in the cube.

We did the same for solid and frame cube number $2$ but this time we found that we had to do $4$x$4$x$4$=$64$ (for the solid cube), then find what the frame cube had which was $4$x$4$x$4$=$64-24=40$ (for the frame) we took $24$ away because that was how many holes were in the cube.

For solid and frame number $3$ we did $5$x$5$x$5$=$125$ (for the solid cube) and for the frame we did $5$x$5$x$5$=$125-65=60$. we took away $65$ because there were $65$ cubes missing.

This was a good start to this challenge but a closer look at the frame cubes would be necessary. Then you might see that the $24$ you take away from $64$ cubes for the second cube only takes account of the $4$ cubes on each face that were removed. There was also some cubes removed from the centre of the solid cube and they would form a $2$ by $2$ by $2$ cube. This would mean a total of $32$ cubes altogether.