Little math challenge (no complete automatism allowed)

[Valentin Albillo]

(04-26-2018 04:11 PM)pier4r Wrote:  having
\[ 1^{2}, 2^{2}, 3^{2}, ... , 27^{2} \]
divide them in 3 groups (no duplicated) with the same sum.

By the usual formula, the sum of the squares of integers from 1 to 27 is 6930 so each group's sum must be 6930/3 = 2310.

Knowing this, the rest is just a simple matter of trying by hand a few cyclic permutations of 9-element sets and, in order not to spoil the fun for others, I'll give here just the first, middle and last element for each group:

1^2 + ... + 15^2 + ... + 26^2 = 2310
2^2 + ... + 13^2 + ... + 27^2 = 2310
3^2 + ... + 14^2 + ... + 25^2 = 2310

I've only considered the case of equal-size groups (9 elements each) but perhaps unequally-sized groups having the same sum are possible as well.

Regards.
V.
.