Triple thunder

Note: this is a rewriting of my initial numerical solution as requested by Rajdeep Brahma.

Let's interpret the sum geometrically. We are adding over all the points $(x,y,z)\in\{1,2,\ldots,99,100\}^3$ which I'll visualize as little unit cubes forming a big $100^3$ cube. The function $\min(x,y,z)$ take values by layers on this cube as represented by the figure

• In the bottom layer $\min(x,y,z)=1$
• In the next layer $\min(x,y,z)=2$

$~~~~~\vdots$

• In the top layer $\min(x,y,z)=100$ , it's just a single unit cube corresponding to the point $(100,100,100)$

We need to figure out how many points are in each layer and add them together. Let's label them by an index $k$

With this convention, each $k$ -layer can be decomposed into easy-to-count boxes adding up to $1+3k+3k^2$ unit cubes

Then we can rewrite the triple sum as a single summation over $k$

$\sum_{x=1}^{100} \sum_{y=1}^{100} \sum_{z=1}^{100} \min(x,y,z) = \sum_{k=0}^{99} {\mbox{Value of min}\choose\mbox{on the k-layer}}{\mbox{Number of cubes}\choose\mbox{on the k-layer}} =\sum_{k=0}^{99} (100-k)(1+3k+3k^2)$ $=\sum_{k=0}^{99}100+299\sum_{k=0}^{99}k+297\sum_{k=0}^{99}k^2-3\sum_{k=0}^{99}k^3=\boxed{25502500}$

a simple python code checks the result.

1
2
3
4
5
6

sum = 0
for x in range(1,101):
    for y in range(1,101):
        for z in range(1,101):
            sum += min(x,y,z)
print(sum)

Ok so when will 5 be the minimum? In $96^3$ cases at least 5 is minimum and in $95^3$ cases 6 is atleast minimum,so 5 is the minimum in $96^3-95^3$ cases.Now by similar logic for all other numbers from 1 to 100 find out the number of occurences when it is minimum and so the required sum is $1*(100^3-99^3)+2*(99^3-98^3)+3*(98^3-97^3)....+100(1^3-0^3)$ = $1^3+2^3+3^3....+100^3$ = $25502500$ .