mersenneforum.org November 2020
 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

2610 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 111112 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 3×47×67 Posts We here, didn't solve that puzzle. Sorry.

 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 05:20.

Fri May 14 05:20:15 UTC 2021 up 36 days, 1 min, 0 users, load averages: 2.07, 1.84, 1.75