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 Upgrade to 13 (Posted on 2011-02-16)
The classical problem of eliminating a false coin out of 12 given can be solved using three steps of weighing by a simple level-balance, even without prior knowledge of the false coin being heavier or lighter than the standard coins.

Solve the same problem for 13 coins (i.e. 12 normal, one false , 3 steps to establish which is the different one and whether it is lighter or heavier), provided you may use a balance-scale with non-equal arms.

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 possible solution | Comment 3 of 4 |

Set the scales to 2a=b, assuming all coins were good.
Weigh ABCDEF against GHI.
If they balance, mark all those good coins x; then
a-weigh JKLx against xx.
b-If they balance, M is bad and the 3rd weighing
c-M against any pair of good coins will determine whether it is heavy or light.
d-if they don't balance, we will know whether the bad coin is light or heavy, and also that it is J,K, or L. Then
e-Weighing J against Kx will either balance,
f-in which case the bad coin is L,  and we will know whether L is light or heavy; or not,
g-in which case it is whichever is lighter/heavier (see d) of those 2.

If ABCDEF against GHI does not balance: we mark all 4 remaining (good) coins x, leaving either abcdef<GHI or ABCDEF>ghi using small letters for the light side.
(1) Since we have already marked light/heavy, if we can tell which coin is bad, we will at once know whether it is lighter/heavier.
(2) We have a method for determining which of 3 coins is bad in one weighing, if we know light/heavy - see d to f above.

Assume WLOG that abcdef<GHI, since this makes no difference to the method .
Take out one group of 3, say abc.
Weigh defGHI against xxx. If light, we use the last weighing to test d against f; if heavy we use the last weighing to test G against H; if they are the same then we use the last weighing to test a against b.

 Posted by broll on 2011-02-18 04:48:36

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