I propose the following solution to the 12 coins Problem:weigh 4 coins against 4:
Left Pan .. Right Pan
1) 1,2,3,4 .. 5,6,7,8
A) If they balance, then 1-8 are all "true" coins. Weigh two true coins with coins 9,10. This will tell you if the counterfeit is in 9,10 or 11,12. Weigh a true coin against a coin from the group which is not true (its either 9,10 or 11,12). If they balance it's the other coin and if they don't balance it's the coin you weighed against the "true". Let the above sequence be called "sequence A"
B) If they do not balance, then 9-12 are all "true". Put aside coins 7,8 and shuffle the remaining coins between the pans as follows.
Left Pan .. Right Pan
2) 1,4,5 .. 2,3,6
C) If they balance, the counterfeit is 7 or 8. Use the last stage of "sequence A" to determine which of them is the counterfeit.
D) If they do not balance, then 7,8 are "true". Put aside coins 5,6 and shuffle the coins between the pans as follows.
Left Pan .. Right Pan
3) 1,3 .. 2,4
E) If they balance, the counterfeit is 5 or 6. If the scale was slanted in the same direction in weighs (1) and (2) then the counterfeit is 6, if not- it is 5.
D) If they do not balance, the counterfeit is 1,2,3 or 4. If the scale was slanted three times in the same direction the counterfeit is 1. If the scale was slanted in one direction once, then twice in the other- the counterfeit is 2. If the scale was slanted in one direction on the first weighing, slanted the other way on the second and returned to the first direction on the third-- the counterfeit is 3. If it was slanted twice in the same direction and the changed- it's 4.
Sum-up:
Left Pan .. Right Pan
1) 1,2,3,4 .. 5,6,7,8
2) 1,4,5 .. 2,3,6
3) 1,3 .. 2,4
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