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One can see the greatest number of cubes when looking at three faces at once.

We count the cubes on each face, giving $3\times 11^2=363$ cubes, but have to subtract from this the cubes along the three edges that have been counted twice, and then add back the cube at the corner (which has three faces visible).

The final quantity is $363-(3\times 11)+1=331$ cubes.


This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.