Dimitri, let's try to explain my approach...
I started thinking to use Combination, but then I realize it can't be applied to this situation. Then, I start to observe the evolution of the amounts for each possible combination.
So, that was what you proposed:
The a) and d) answers are obvious, of course, but to get the other values I used the same approach of you by making the combinations one-by-one by hand. So, I think I have used the overlapped options too. Take a look at the pictures:
![[Image: 1432360272.png]](http://www.putimg.net/thumbnails/1432360272.png)
![[Image: 1432360333.png]](http://www.putimg.net/thumbnails/1432360333.png)
![[Image: 1432360371.png]](http://www.putimg.net/thumbnails/1432360371.png)
![[Image: 1432360399.png]](http://www.putimg.net/thumbnails/1432360399.png)
based on that I found these answers:
a) 64; b) 27; c) 8; d) 1;
and thinking of them, I noticed that 64=4³; 27=3³; 8=2³; 1 that could be 1³ (why not
).
that was the way I did the relationship between these data:
Value = (Md - size tested + 1)³
a) size tested = 1 => (4 - 1 + 1)³ => 4³
b) size tested = 2 => (4 - 2 + 1)³ => 3³
c) size tested = 3 => (4 - 3 + 1)³ => 2³
d) size tested = 4 => (4 - 4 + 1)³ => 1³
The equation I put in the previous represent the way to determine the sum of these values for any Md. That exponent ³ is related to the dimension used and thinking about your last example, its as two dimensional representation. Now, we could say that the same role would be applied to your two dimensional squares: 4²+3²+2²+1² => 30
but, the equation for calculate the sum of squares for any dimension (n) will be this one:

for n=64 (your example), Tc=(64x(64+1)x(2x64+1))/6 => 89440 !!!
Even if I'm wrong, that was a very interesting challenge Dimitri. Thanks.
I started thinking to use Combination, but then I realize it can't be applied to this situation. Then, I start to observe the evolution of the amounts for each possible combination.
So, that was what you proposed:
Quote:"Mega cube" dimension = Md = 4 smaller cubes (as you proposed)
a) how many cubes of 1 smaller cube size can be place into the Mega Cube?
b) how many cubes of 2 smaller cube size can be place into the Mega Cube?
c) how many cubes of 3 smaller cube size can be place into the Mega Cube?
d) how many cubes of 4 smaller cube size can be place into the Mega Cube?
The a) and d) answers are obvious, of course, but to get the other values I used the same approach of you by making the combinations one-by-one by hand. So, I think I have used the overlapped options too. Take a look at the pictures:
![[Image: 1432360272.png]](http://www.putimg.net/thumbnails/1432360272.png)
![[Image: 1432360333.png]](http://www.putimg.net/thumbnails/1432360333.png)
![[Image: 1432360371.png]](http://www.putimg.net/thumbnails/1432360371.png)
![[Image: 1432360399.png]](http://www.putimg.net/thumbnails/1432360399.png)
based on that I found these answers:
a) 64; b) 27; c) 8; d) 1;
and thinking of them, I noticed that 64=4³; 27=3³; 8=2³; 1 that could be 1³ (why not

that was the way I did the relationship between these data:
Value = (Md - size tested + 1)³
a) size tested = 1 => (4 - 1 + 1)³ => 4³
b) size tested = 2 => (4 - 2 + 1)³ => 3³
c) size tested = 3 => (4 - 3 + 1)³ => 2³
d) size tested = 4 => (4 - 4 + 1)³ => 1³
The equation I put in the previous represent the way to determine the sum of these values for any Md. That exponent ³ is related to the dimension used and thinking about your last example, its as two dimensional representation. Now, we could say that the same role would be applied to your two dimensional squares: 4²+3²+2²+1² => 30
but, the equation for calculate the sum of squares for any dimension (n) will be this one:

for n=64 (your example), Tc=(64x(64+1)x(2x64+1))/6 => 89440 !!!
Even if I'm wrong, that was a very interesting challenge Dimitri. Thanks.