Callum and Josh submitted some ways of making the largest number of brush loads and the smallest number of brush loads for 5, 6 and 7 cubes:
Lan explored the pattern for greater numbers of cubes, as did Rohaan from New Zealand. Lan describes the method used:From counting cubes, we find:
This is interesting, isn't it? If we think about the least number of BLs, we are trying to make as close to a rectangular arrangement of cubes as possible. This helps us to see why we only add one face going from an odd to an even number of cubes, but three faces going from an even to an odd.
However, is a rectangular array always the best for minimising the number of brush loads? If you had 9 cubes, how about making a square arrangement? How many brush loads of paint would you need then?
Rohaan then spotted that the largest number of strokes is always (4 x number of cubes) +1, and that the smallest number depends on whether the number of cubes is odd or even:Odd: (n-5) x2 +15
Does this work for a square arrangement too?