B Cube Math: How Many Smaller Cubes in a Larger Cube?

[bob012345]

TL;DR Summary

Find how many smaller cubes can be drawn in a larger cube of dimension nxnxn.

Here is a mild diversion. Given a cube of dimensions nxnxn, how many smaller cubes can be drawn within it. All cubes are integer dimensions and are drawn on the grid lines if you imagine the larger cube as a 3D graph paper. For example, in a 2x2x2 cube one can draw one 2x2x2, eight 1x1x1's. Try working it out for a 4x4x4 cube. Try and derive the general formula and use it for a 10x10x10 cube. Or a 100x100x100 cube. Have fun.

 

[mathman]

Do the smaller cubes fill the original cube? If so, you need to get a list of divisors. For n=10, use 1,2,5 to get $10^3+5^3+2^3$, while for n=100, use 1,2,4,5,10,20,25,50.to get $100^3+50^3+25^3+20^3+10^3+5^3+4^3+2^3$.

 

[bob012345]

Do the smaller cubes fill the original cube? If so, you need to get a list of divisors. For n=10, use 1,2,5 to get $10^3+5^3+2^3$, while for n=100, use 1,2,4,5,10,20,25,50.to get $100^3+50^3+25^3+20^3+10^3+5^3+4^3+2^3$.

All the cubes can be drawn within the bounds of the larger cube. I did not use the word fill because I don't want people to think this is about how many smaller cubes fill the volume. It's not that. In the simple example I gave of the 2x2x2 case, you have 5 possibilities. The 1x1x1's certainly fill the volume. So does the 2x2x2. For larger cubes, you will see smaller cubes overlapping each other but for each size, the total volume will always be spanned by the smaller cubes. There will be no empty spaces. If it helps, think about the problem of how many squares can be drawn within a larger square on graph paper. Some smaller squares can overlap. I hope this helps.

 

[mfb]

You can draw all the cubes, and if you draw with perspective they are all distinguishable. That gives you n³ cubes of size 1, (n-1)³ cubes of size 2 and so on. If you mean something else you'll need a clearer description.

 

[bob012345]

You can draw all the cubes, and if you draw with perspective they are all distinguishable. That gives you n³ cubes of size 1, (n-1)³ cubes of size 2 and so on. If you mean something else you'll need a clearer description.

That's correct. The answer for how many cubes can be drawn is $\sum_{j=1}^n j^3$. But this can be put in a more compact form of $(\frac {n(n+1)}{2})^2$. So, in the 4x4x4 case we have a total of 100 and 10x10x10 case we have 3025. The 100x100x100 case gives 25,502,500 possible cubes that can be drawn.

So, the next question is how many cuboids of size {$i ,j ,k = 1,n$} can be drawn in a cube of $n$x$n$x$n$ ?