1,2,3,4,5,6,7,8,9,10,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27
Weigh 1,2,3,4,5,6,7,8,9 and 10,11,12,13,14,15,16,17,18
Balance weigh 19,20,21 and 22,23,24
Balance weigh 25 and 26
Balance 27 heavy.
Right 26 heavy.
Left 25 heavy.
Right weigh 22 and 23
Balance 24 heavy.
Right 23 heavy.
Left 22 heavy.
Left weigh 19 and 20
Balance 21 heavy.
Right 20 heavy.
Left 19 heavy.
Right weigh 10,11,12 and 13,14,15
Balance weigh 16 and 17
Balance 18 heavy.
Right 17 heavy.
Left 16 heavy.
Right weigh 13 and 14
Balance 15 heavy.
Right 14 heavy.
Left 13 heavy.
Left weigh 10 and 11
Balance 12 heavy.
Right 11 heavy.
Left 10 heavy.
Left weigh 1,2,3 and 4,5,6
Balance weigh 7 and 8
Balance 9 heavy.
Right 8 heavy.
Left 7 heavy.
Right weigh 4 and 5
Balance 6 heavy.
Right 5 heavy.
Left 4 heavy.
Left weigh 1 and 2
Balance 3 heavy.
Right 2 heavy.
Left 1 heavy.
Jim FitzSimoons
--- In Math4u@yahoogroups.com, "Michael S." <M.Suesserott@...> wrote:
>
> The answer to your below question can easily be deduced from the
> entropies and information contents involved. Let k equal the
number of
> weighings (in your case, k=3), and let n equal the number of
coins, then
> the resulting inequality is
>
> log(n) / log(3) <= k
>
> With k=3, this yields n<=27. So if you know there is one fake coin
> heavier than the others, the maximum number of coins for which the
> problem can be solved with three weighings is 27.
>
> Best regards,
>
> Michael
>
>
> sinichi7kudo wrote:
> > I just have another question: If you know that the fake coin is
> > heavier than other coins. And you are allowed to do 3
weightings. So
> > what is the maximum number of coins?
> ...
> > Thank you.
>
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