Biquanimous numbers

[Don Reble]

> So, let's call a number biquanimous if its digits can be split into
> two sets with the same sum.
> ...
> ... the base-2 and base-3 biquanimous
> numbers are, respectively, 2-regular and 3-regular sets. My gut
> feeling is that this generalizes to all finite bases...

    My gut feeling is that it doesn't. Determining biquanimosity
    is a special case of the subset-sum problem, and that problem
    is NP-complete. Maybe this problem is, too.

    Do tell us, if you find a clever way to solve it for really
    large bases.
--
Don Reble       djr at nk.ca