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1 vote

Minimum K for detecting fake pearls in one weighing

An easily found decent? upper bound
Retudin's user avatar
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7 votes

Minimum K for detecting fake pearls in one weighing

An upper bound from https://oeis.org/A096858 is attained by
RobPratt's user avatar
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2 votes

Minimum K for detecting fake pearls in one weighing

To add to what @HaganvonEitzen said in their answer, I have found a slightly smaller set that is solved in the same way. I have no clue if this is the minimum however.
not_not_Alex's user avatar
3 votes

Minimum K for detecting fake pearls in one weighing

Instead of $K=512$, I suppose that already is enough (and perhaps the best possible choice). I make use of a special property of the set With positive integers $a$ and $b$ specified below, set If $...
Hagen von Eitzen's user avatar
2 votes

Minimum K for detecting fake pearls in one weighing

I think you need at least pearls per box. You grab and put these all together on the scale to get some weight x. You have a total of pearls and now compute
quarague's user avatar
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0 votes

How many solutions to the twelve coins problem are there?

Note that there are many (I believe 10) possible solutions in which three non adaptive weighings tells you the odd coin out, but not whether it’s heavier or lighter. e.g. Note also that if heavy ...
Retudin's user avatar
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5 votes
Accepted

How many solutions to the twelve coins problem are there?

I have an answer to the non-adaptive version of the problem, that is, the version where one must declare ahead of time all three weighings one will perform, and their algorithm is not allowed to ...
Feryll's user avatar
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