Euler_Problem-006.b93

"d"10p000p>10g:00g+00p1-:#v_$010p>"c"10g::1+:*\:5:+%\5:+/6+p`#v_       v
>$.@      ^            p01<      ^                   p01+1g01 <        0
     >                                   v
|!:\<|p01-1:g01<$$_v#!:g+6/+:5\%+:5:g01 :< p01"c"+<_v#!p01:-1g01<p01:g0<
\   ^<             >` >                 #v_v
>         v            v-g+6/+:5\%+:5:g01<          >00g        ^
##########>                                       ^
##########             >010g:5:+%\5:+/6+pv
##########                                
##########     ^                         < <
##########
##########
##########
##########
##########
##########


---------------------------------------

My solution here is *far* more complex than necessary, mostly because I thought the temporary numbers would get too big for int32 to handle (especially sum(1,100)^2).
Later I found out I could just calculate the sum of the squares and the square of the sum and subtract them ...

But I like my solution so I'll leave it here.
While I calculate the the second value (using the distributive property), I subtract in every step as much parts of the first value (the single elements of the addition) as I can (without going negative).

That way my temporary values never greatly exceed the final result.