Have You Thought About Slicing Cubes?

I ended up with the same calculation as Otto here, but used a geometric model, thinking of the numbers to be added as elements in a $10\times 10\times 10$ grid.

Consider a set of nested cubes:

• cube 1 is the entire $10\times 10\times 10$ grid;

• cube 2 is the $9\times 9\times 9$ grid consisting of all elements with $x, y, z \geq 2$ ;

• cube 3 is the $8\times 8\times 8$ grid consisting of all elements with $x, y, z \geq 3$ ;

• etc. until cube 10, which consists only of $(10, 10, 10)$ .

It is not difficult to see that $\text{min}(x,y,z)$ is equal to the number of cubes in which $(x,y,z)$ is contained. Therefore $\sum_{x,y,z} \text{min}(x,y,z) = \sum_{x,y,z} \text{\# of cubes containing}\ (x,y,z) \\ = \sum_{n=1}^{10} \text{elements contained in cube}\ n \\ = \sum_{n=1}^{10} n^3 = \left(\sum_{n=1}^{10} n\right)^2 = 55^2 = \boxed{3025}.$

We can show that $\sum_{1\leq x,y,z\leq N}\min(x,y,z)=\sum_{k=1}^{N}k^3=\left(\frac{N(N+1)}{2}\right)^2$

For $N=10$ this gives $\boxed{3025.}$

By induction, it suffices to show that $\sum_{\max(x,y,z)=N}\min(x,y,z)=N^3$ . Indeed, paying attention to the number of $N's$ in $(x,y,z)$ , we find this sum to be $N+3\sum_{k=1}^{N-1}k+3\sum_{k=1}^{N-1}k^2=N^3$ .

(See my alternative solution here .)

Very interesting problem. I divided the sum according to the value of min (x,y,z). If min (x,y,z) is "n", then it is because only one of the three numbers is n [ (3(10-n)^2) cases], two of them are n [3(10-n) cases] or the three of them are n [1 case]. If we multiply the number of cases were min (x,y,z) is n, by the value of n, and we sum it for n=1 to n=10, we get the solution :)