[math-fun] Fwd: 2015 with the other six digits
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
Zack Chroman (via Bill Gosper):
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.
I should point out that in my previous reply, which had 2102 as unreachable (using, in order, 10-1), I did NOT consider parentheses, which IS allowable in problem 259.
participants (2)
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Bill Gosper -
Hans Havermann