The sum of the digits of the largest 2-digits square in any base is always equal to its square root.

Prove it.

In base k, the smallest 3 digit square = k*k = 100 (base k)

The smallest 2 digit square is therefore (k-1)*(k-1)

= k*k - 2k + 1 = (k-2)*k + 1

In base k, its digits are (k-2) and 1, and they sum to (k-1), which is its square root

*Edited on ***June 17, 2013, 11:05 am**