|
2020-10-30, 11:25
|
#1 |
"Mike"
Aug 2002
1F0816 Posts |
November 2020
|
|
|
|
2020-11-01, 13:33
|
#2 | |
|
Jul 2015
23 Posts |
Quote:
|
|
|
|
|
|
2020-12-06, 19:08
|
#3 |
|
Jan 2020
22×7 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 |
|
Romulan Interpreter
Jun 2011
Thailand
2×3×52×61 Posts |
We here, didn't solve that puzzle. Sorry.
|
|
|
|
 |
| Thread Tools | |
Similar Threads
|
||||
| Thread | Thread Starter | Forum | Replies | Last Post |
| November 2019 | Xyzzy | Puzzles | 89 | 2019-12-04 19:18 |
| November 2017 | Batalov | Puzzles | 3 | 2017-12-08 14:55 |
| November 2016 | Xyzzy | Puzzles | 1 | 2016-12-06 16:41 |
| November 2015 | R. Gerbicz | Puzzles | 3 | 2015-12-01 17:48 |
| November 2014 | Xyzzy | Puzzles | 1 | 2014-12-02 17:40 |
