I believe Knuth sketched an algorithm for this in his latest Christmas Tree Lecture. ---------- Forwarded message ---------- From: Zack Chroman <zacchro@gmail.com> Date: Wed, Dec 31, 2014 at 6:10 AM Subject: Re: [math-fun] 2015 with the other six digits To: Bill Gosper <billgosper@gmail.com> Using 1,2,3,4,5,6,7,8,9 in that order (with concatenated digits), it's possible up to 6964. I have some Mathematica code for this because of this problem <https://projecteuler.net/problem=259>, but it's being annoying when I try to get it to go 10-1 instead of 1-9. -Zack On Wed, Dec 31, 2014 at 4:37 AM, Bill Gosper <billgosper@gmail.com> wrote:
Hans Havermann>[...]
Don't forget to count down on New Year's Eve:
10*9*8*7/6/5*4*3-2+1 --------
When's the soonest year this becomes impossible? --rwg