July 2018 little math problem

[pier4r]

(07-26-2018 03:43 PM)Thomas Klemm Wrote:  But since \(s=a+b+c=c+d+e=e+f+g=g+h+i\) and \(a+b+c+d+e+f+g+h+i=\sum_{k=1}^{9}k=45\) we know that their sum is \(4s=45+c+e+g\) which must be divisible by 4.
This reduces the possibilities further down to 60.

This was the same point I reached, then I worked on the multiples of 4. I was not expecting such an healthy thread (n1) for a problem apparently simple.

n1: thanks to this thread I also realize that even if someone posted already a super efficient solution, having other point of views - even with less efficient solutions - gives other perspective and increase the possibility to confront/compose ideas. Thus making the thread more enjoyable.