🚀 go-pugleaf

freitag@elrond.toppoint.de (Claus Schoenleber) writes:
>erick@fraser.sfu.ca (Erick Bryce Wong) writes:
>
>I don't think so.  Here you have to write (11 - i) if i=1 is the leftmost
>position, because the first digit has to be multiplied with 10, the following
>with decrement 1, etc. Then the sum has to be subtracted from the next greater
>number dividable by 11. Your formula stops at the same point as my first
>(and wrong) one.

But, don't the two operations of subtracting i from 11 and of subtracting the
total from a multiple of 11 cancel out?  (Two wrongs make a right? :-)

>There are several ways to write it down, of course. It was meant as an

Quite true.  I just figured that simplifying it would make it look, well,
um, er, simpler... ;)

-- 
 -- Erick, the perfect square  :-)