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Upgrade to 13 (Posted on 2011-02-16) Difficulty: 4 of 5
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.


No Solution Yet Submitted by Ady TZIDON    
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Solution Three Solutions for Three Arm Ratios Comment 4 of 4 |
Solution 1: 2/1 Scale

In this solution the arms are set to a 2:1 ratio.  Weighings are 6v3 or 2v1.

Divide the coins into four groups of three coins, plus the 13th coin.  Call the groups A, B, C, and D.  Make the first two weighings (A+B) vs C and (B+C) vs D.

Case 1: Both weighings are equal.
The 13th coin is fake.  For the third weighing, weigh two good coins against the fake to determine if it is lighter or heavier.

Case 2: One weighing is equal and one is unequal.
The fake is in D if the first weighing is equal or it is in A if the second weighing is equal.  The direction of the weighing determines if the fake is light or heavy.

Case 3: Both weighings are unequal.
If the inequalities are in the same direction then the fake is in B.  If they are in opposing directions then the fake is in C.  Either way, the direction of the weighing determines if the fake is light or heavy.

In either Case 2 or 3, make the third weighing a good coin plus a suspect coin against another suspect coin.  That result will indicate which of the three is fake.
For example if the fake is in D and is light, then Weigh D1+good vs D2.  If equal then D3 is fake, if D2 side is heavy then D1 is fake, and if the D2 side is light then D2 is fake.

Solution 2: 5/4 Scale

In this solution the arms are set to a 5:4 ratio.  Weighings are 5v4.  The strategy stays mostly the same as the 2:1 solution, with some changes.

The coins are divided into the same groups.  But group B is split into two subgroups B1+B2 and B3.  The weighings made are (A+B1+B2) vs (C+B3) and (B1+B2+C) vs (D+B3).

If the fake is in A, B, C, or the 13th coin then the third weighing is the same, except for some additional good coins added to make the weighing 5v4.

If the fake is in B then the possible fakes are either {B1 is heavy or B2 is heavy or B3 is light} or {B1 is light or B2 is light or B3 is heavy}.  B1 and B2 are used in the third weighing in this case.

Solution 3: 7/2 Scale

In this solution the arms are set to a 7:2 ratio.  Weighings are 7v2.  The same strategy still applies, but with more adjustments.

The coins are divided into the same groups.  But group A is split into two subgroups A1+A2 and A3, and group C is split into subgroups C1+C2 and C3.   The weighings made are (A1+A2+B+C1+C2) vs (A3+C3) and (B+C3+D) vs (C1+C2).

Depending on which group has the fake, the third weighing is made in one of the ways described in the previous two cases.

  Posted by Brian Smith on 2016-07-04 20:43:35
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