Question

How to divide `1^2, 2^2, ..., 81^2` set into 3 subsets with equal sum and 27 numbers in each?

Answer 1

See explanation...

Explanation:

Let's try a cyclic permutation on the first `9` squares:

`1^2+5^2+9^2 = 107`

`2^2+6^2+7^2 = 89`

`3^2+4^2+8^2 = 89`

Applying the same permutation to the next `9` squares we get:

`10^2+14^2+18^2 = 620`

`11^2+15^2+16^2 = 602`

`12^2+13^2+17^2 = 602`

Applying the same permutation to the next `9` square we get:

`19^2+23^2+27^2 = 1619`

`20^2+24^2+25^2 = 1601`

`21^2+22^2+26^2 = 1601`

Notice that in each of these groups of three sums, the first sum is `18` more than the second and third.

So we can combine these sums cyclically permuted to get:

`1^2+5^2+9^2+11^2+15^2+16^2+21^2+22^2+26^2 = 2310`

`2^2+6^2+7^2+12^2+13^2+17^2+19^2+23^2+27^2 = 2310`

`3^2+4^2+8^2+10^2+14^2+18^2+20^2+24^2+25^2 = 2310`

Apply the same permutation to the second and third groups of `27` squares to get a solution for `81` squares...

`color(white)(0)1^2+color(white)(0)5^2+color(white)(0)9^2+11^2+15^2+16^2+21^2+22^2+26^2+`
`28^2+32^2+36^2+38^2+42^2+43^2+48^2+49^2+53^2+`
`55^2+59^2+63^2+65^2+69^2+70^2+75^2+76^2+80^2 = 60147`

`color(white)(0)2^2+color(white)(0)6^2+color(white)(0)7^2+12^2+13^2+17^2+19^2+23^2+27^2+`
`29^2+33^2+34^2+39^2+40^2+44^2+46^2+50^2+54^2+`
`56^2+60^2+61^2+66^2+67^2+71^2+73^2+77^2+81^2 = 60147`

`color(white)(0)3^2+color(white)(0)4^2+color(white)(0)8^2+10^2+14^2+18^2+20^2+24^2+25^2+`
`30^2+31^2+35^2+37^2+41^2+45^2+47^2+51^2+52^2+`
`57^2+58^2+62^2+64^2+68^2+72^2+74^2+78^2+79^2 = 60147`