> ...Then I did it a way that actually makes sense, but naturally my connection decided to quit for the day. Maybe this is better.
>
> N-4R=0
> N-4[(N+4)/5]=0
> N-4N/5-16/5=0
> N/5=16/5
> N=16
>
> Adding 4 to N in the second step is to make it a multiple of 5. Adding 1, 2, and 3 is also possible, but they produce lower values of N.
>
> gab"Or maybe not."by

Well, that is the right answer, but I'm not sure I understand the steps (why would N have to be a multiple of 4 in the first step?). And in general, if the maximum number of columns is C, then the most items you can have without using the last column is (C-1)^2. I will try to post the proof that I wrote.

--gremlinn