mersenneforum.org November 2020
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

 2020-10-30, 11:25 #1 Xyzzy     "Mike" Aug 2002 11111111001012 Posts November 2020
2020-11-01, 13:33   #2
tgan

Jul 2015

2·13 Posts

Quote:
 Originally Posted by Xyzzy https://www.research.ibm.com/haifa/p...ember2020.html
I think it is related to Perfect powers for very large numbers

 2020-12-06, 19:08 #3 0scar   Jan 2020 1F16 Posts The published four-step base solution is flawed. At the very first step, (2^2)^31 is equal to 2^(2*31), not to 2^(2^31). We can get a working four-step solution by modifying the starting list to 4,2^4,2^29: 4 < 2^4 < 50 and 2^29 < 10^9, so it still fits the constraints; moreover, (2^4)^(2^29) = 2^(4*(2^29)) = 2^(2^31). Then the remaining three steps are correct. The starting entry "4" can be replaced by any number between 2 and 50. Did someone use less than four steps for base solution? How long is your fastest bonus solution? Last fiddled with by 0scar on 2020-12-06 at 19:10
 2020-12-07, 02:15 #4 LaurV Romulan Interpreter     Jun 2011 Thailand 24E716 Posts We here, didn't solve that puzzle. Sorry.

 Thread Tools

 Similar Threads Thread Thread Starter Forum Replies Last Post Xyzzy Puzzles 89 2019-12-04 19:18 Batalov Puzzles 3 2017-12-08 14:55 Xyzzy Puzzles 1 2016-12-06 16:41 R. Gerbicz Puzzles 3 2015-12-01 17:48 Xyzzy Puzzles 1 2014-12-02 17:40

All times are UTC. The time now is 13:47.

Fri May 14 13:47:54 UTC 2021 up 36 days, 8:28, 0 users, load averages: 2.13, 1.92, 1.88

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.