APWJ Problem 4 (The Google Test)... Ewart has cracked it!! The solution is so elegant it's got to be the right one. It's as beautiful as mine is ugly. And it perfectly fits Eugene's "crossword clue": "Alternately: something for nothing!"
Sum the alternate digits (both ways). Equate the two sums. It was suggested by divisibility by 11 (...I've learnt something!) Technical glitches prevent Ewart from posting the announcement himself, so let me do so by posting his letter to me. Please read the quoted text below plus Ewart's addendum to http://www.jsoftware.com/jwiki/Doc/Articles/Play211 This comes just in time. I was about to go to press. Ian Clark Subeditor, APWJ Edn 2. On Wed, Nov 25, 2009 at 2:39 PM, Shaw, Ewart <[email protected]> wrote: > Dear Ian, > > I've put a suggested solution to the pi digits problem > on the article page; note that 10 digit sequences beginning '0' > must be excluded (otherwise 0582097494 would have been included). > > Sorry - I have University-mail problems posting to the J Forums, > so currently have to use the JWiki and/or send personal emails. > > Regards, Ewart > > J.E.H.Shaw http://www.warwick.ac.uk/statsdept > Ewart Shaw http://www.ewartshaw.co.uk > > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
