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2020-10-30, 11:25
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#1 |
"Mike"
Aug 2002
25×257 Posts |
November 2020
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2020-11-01, 13:33
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#2 | |
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Jul 2015
1B16 Posts |
Quote:
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2020-12-06, 19:08
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#3 |
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Jan 2020
22·32 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 |
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2020-12-07, 02:15
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#4 |
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Romulan Interpreter
Jun 2011
Thailand
100101100010112 Posts |
We here, didn't solve that puzzle. Sorry.
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