Marcelo E. Magallon ([EMAIL PROTECTED]) wrote: > Regarding my "one step" solution, the secret can be computed as: > > Op(i=1..digits, C(i, digits)*digit[i]) > > where Op(a,b) := (0, 1..9, 1..9, ...)[a+b] and C(i, j) := j!/i!/(j-i)! > (or something like that, I can't think straight right now). This lead > to a horribly long solution that doesn't even pass the test program :-)
"One step" for very unusual bindings of "one" ;-) I also thought of this. However you can avoid messing around with factorials by simply extracting digits from the powers of 11. I'm just gutted that I had considered *1.1 and then rejected it as being too much hassle to remove the two outer digits. If only I'd realised I could have done that in 3 extra characters! Ah well. At least mine was the best entry did the magic op, the pop, and removal of the trailing digit all within one s///ge. Small comfort ...
