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If we consider an n dimensional shape, that has s sub-shapes. If we then use it to generate an (n+1)-dimensional shape, we see that if the extra dimension is of size 1 then there are also s sub-shapes of the the (n+1)-dimensional shape. If the extra dimension is of size 2 then there are s sub-shapes with size 2 in the new dimension and 2s sub-shapes with size 1 in the new dimension, giving 3s sub-shapes overall. Following this reasoning we see that if we add a new dimension to the original shape with size k, then we end up with T(k)s sub-shapes of the (n+1) dimensional shape. (If maybe easier to see this if you consider going from 1-dimension to 2-dimensions, or 2-dimensions to 3-dimensions).

So in general the number of sub-shapes of an n-dimensional shape is the product of the triangular numbers that measure each dimension, and in particular for 3-dimensional blocks with dimensions of

x,yandz, the number of sub-blocks is:And this fits with the examples given:

This Python program finds the

x,y,zdimensions of a block consisting of 105105 sub-blocks in 33msSolution:The dimensions of the block are 6 × 10 × 13.To solve the problem manually we see that we want:

Re-writing the product as the product of thee adjacent pairs of integers:

hence:

x=6,y=10,z=13.