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Old 2020-10-30, 11:25   #1
Xyzzy
 
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"Mike"
Aug 2002

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Default November 2020

https://www.research.ibm.com/haifa/p...ember2020.html
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Old 2020-11-01, 13:33   #2
tgan
 
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Quote:
Originally Posted by Xyzzy View Post
I think it is related to Perfect powers for very large numbers
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Old 2020-12-06, 19:08   #3
0scar
 
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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
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Old 2020-12-07, 02:15   #4
LaurV
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We here, didn't solve that puzzle. Sorry.
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