A giant cube is made of $n^3$ small cubes so that its dimensions are $nxnxn$. How many cubes of any size can be made of small cubes in the giant cube?

Question

I am familiar with a similar problem that asks the number of squares that can be made in an $n x n$ chessboard. The answer was $\sum_{k=1}^{n} k^2$. I "feel" that the answer to the cube problem is $\sum_{k=1}^{n} k^3$, however, am struggling to find a way to prove this. Perhaps it has something to do with choosing the number of edges of the cubes? Does anybody have any thoughts on how to prove this (that is, if it's correct)? Thank you!

Answer 1

Let $\space n=1\space$ as the size of a unit cube. We know there must be a multiple $\space (m) \space $ of cubes on an edge to make a larger cube.

$\quad m=1\longrightarrow 1^3=1\space $ unit cubes in the giant one.

$\quad m=2\longrightarrow 2^3=8\space $ unit cubes in the giant one.

$\quad m=3\longrightarrow 3^3=27\space $ unit cubes in the giant one.

$\quad m=6\longrightarrow 6^3=216\space $ unit cubes in the giant one.

$\quad m=10\longrightarrow 10^3=1000\space $ unit cubes in the giant one.

If there are $\space m \space $ unit cubes on an edge, there are $\space (m)^3 \space$ unit cubes in the giant cube.$\quad m=2\longrightarrow 2^3=8\space $ unit cubes in the giant one.

If, however, you want to include one set of all possible cube multiples $\space (n) \space$ of $\space 1,\space$ it involves the floor values of "cubes" dividing $\space m.\quad$ For instance, if $\space m=6,\space$ the sum of cube sizes $\space 1,2,3,4,5,6\space$ is $\space 6^3+3^2+2^3+4^0+5^0+6^0=236.\quad$ If $\space m=5,\space$ the sum of cube sizes $\space 1,2,3,4,5\space$ is $\space 5^2 +2^2 +3^0+4^0+5^0=32.\space$

If you want to include all possible positions of non-integer divisions within the cube, you must use $\space \bigg(n\pmod m + \bigg\lfloor \dfrac{m}{n}\bigg\rfloor\bigg)^3\space$

for each non-integer cube division in the sum. For instance, if $\space m=6 \space$ and the "unit" size is $\space 4,\space$ we know that the $\space 4$-cube can fit $\space 3\space$ spaces up/down, left/right, forward/backward and the count for "that" cube size is $\space \bigg(6\pmod 4 + \bigg\lfloor \dfrac{6}{4} \bigg\rfloor\bigg)^3 = (2+1)^3=27.\quad$

For $\space m=6, n=2,\space$ we know that $\space 3\space $ cubes on each side fit in $\space 0\space $ different positions on that side.

$\space \bigg(6\pmod 2 + \bigg\lfloor \dfrac{6}{2} \bigg\rfloor\bigg)^3 = (0+3)^3=27.\quad$

For $\space m=5, n=2,\space$ we know that $\space 2\space $ cubes on each side fit in $\space 2\space $ spaces on that side.

$\space \bigg(5\pmod 2 + \bigg\lfloor \dfrac{5}{2} \bigg\rfloor\bigg)^3 = (2+2)^3=64.\quad$

Calculate the number of unit cubes of each size from $\space 1\text{-to-}m\space$ add them up for your total.