Timeline for Twenty-four coins
Current License: CC BY-SA 4.0
31 events
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Feb 28, 2023 at 19:19 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 28, 2023 at 7:25 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 27, 2023 at 19:54 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 27, 2023 at 19:18 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 27, 2023 at 19:10 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 24, 2023 at 11:35 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 24, 2023 at 7:19 | comment | added | Kris Van Bael | Of course, thank you. | |
Feb 22, 2023 at 8:48 | comment | added | Nautilus | ++++ means the fakes average out to a normal coin, and both are in the same split group of 3. It should make $3!*8=48$, not $96$. | |
Feb 21, 2023 at 22:59 | comment | added | Kris Van Bael | Do all these tracks actually end within 8 measurements? Eg. I believe that the ++++ branch leaves you with 96 combinations, which is not possible in 4 weighting’s. | |
Feb 19, 2023 at 16:00 | comment | added | Nautilus | I just hope now that I didn't overlook anything. | |
Feb 19, 2023 at 15:59 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 19, 2023 at 9:02 | comment | added | Pumbaa | Could the steps after ++++ be described in more detail, please? The omitted following steps might be not that obvious/self-evident to some readers. :D | |
Feb 18, 2023 at 22:27 | comment | added | Retudin | You are missing the case that if the first group is heavier than both the second and the third, the fakes can be in group 1 and 4 | |
Feb 18, 2023 at 18:16 | comment | added | Nautilus | Edited, sorry about that. | |
Feb 18, 2023 at 18:16 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 18, 2023 at 15:40 | comment | added | Pumbaa | An error in step ++: If we also get a balanced result from that, the fakes must be in the 1st (average is normal) or 4th. - No, the fakes might also be in the 2nd (average is normal) or 3rd (average is normal). | |
Feb 18, 2023 at 13:16 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 18, 2023 at 13:08 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 17, 2023 at 11:06 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 17, 2023 at 11:04 | comment | added | Nautilus | We just need to find the two fakes in no particular order, but even so you can't identify them without differentiating them from each other when their average weight equals a normal coin's. | |
Feb 17, 2023 at 10:02 | comment | added | Bernardo Recamán Santos | No way of reducing answer to 7 using both devices? | |
Feb 16, 2023 at 16:11 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 16, 2023 at 16:10 | comment | added | Trenin | NVM - I see now what you have done. I am still not sure I agree, but I see now how it has been accounted for. | |
Feb 16, 2023 at 16:05 | comment | added | Trenin | I think you need to differentiate the case where "both are lighter than normal and the same weight" from "both are lighter than normal, but one is heavier than the other". | |
Feb 16, 2023 at 16:03 | comment | added | Nautilus | I already took them all into account: "both could be lighter than normal, both could be heavier, or when only one is lighter, the sum could be heavier/lighter/equal than/to two normal coins." | |
Feb 16, 2023 at 15:57 | comment | added | Trenin | I think there are even more possibilities. Both lighter and the same, Both lighter and different, Both heavier and the same, both heavier and different, the average is heavier, the average is lighter, the average is genuine. So 7 total. | |
Feb 16, 2023 at 15:33 | history | undeleted | Nautilus | ||
Feb 16, 2023 at 15:32 | history | edited | Nautilus | CC BY-SA 4.0 |
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Feb 16, 2023 at 15:23 | history | deleted | Nautilus | via Vote | |
Feb 16, 2023 at 15:22 | comment | added | Trenin | Do you need to differentiate the following possibilities as well: the two coins together are lighter than two normal coins, and the two coins together are heavier than two normal coins. | |
Feb 16, 2023 at 15:08 | history | answered | Nautilus | CC BY-SA 4.0 |