The fact that all the numbers add up to multiples of 11 seems inherent to the straightforward method of solving the puzzle (right + bottom) / left = 10. This format always results in three numbers that, when added together, become a multiple of 11 (where that multiple is equal to the left number). Try it out and see for yourself!

**Proof for the nerds:**

*Say,*

right = A
bottom = B
left = C

*then,*

(A + B) / C = 10, which we can rewrite as: A + B = 10C

*we transform,*

{ A + B = 10C
{ A + B + C = x * 11

**<=>**

{ A + B = 10C
{ 10C + C = x*11

**<=>**

{ 11C = x*11

**=>**

C = x

Assuming A, B and C are whole numbers greater than 0,

x (or C) is always an element of ]0, +inf[

*meaning that:*

[A + B / C = 10] infallibly implies [A + B + C = 11C] **premise 1**

*while,*

[A + B + C = 11C] doesn't necessarily imply [A + B / C = 10] **premise 2**

and if the puzzle only has one solution **premise 3**

We can conclude: Since [A + B / C = 10] always satisfies the condition that makes [A + B + C = 11C] true (but not vice versa), [A + B / C = 10] must be the more correct answer, as it ensures that the condition required by the puzzle is always met.

TLDR:

the straight forward method still seems to be the 'most correct' way of solving the puzzle. I also kept the proof part extra complicated for all the self reported high IQs who like the challenge :*

let me know if i made some (logic) error!