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| − | Question: You are given 12 coins. One of them is heavier or lighter than the rest. Identify
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| − | this coin in just three weighings
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| − | Solution:
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| − | Number the coins 1 through 12 and divide them coins into 4 sets of 3...
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| − | There are multiple comparison sets possible. This is an acceptable template to find a few of them.
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| − | (This template is NOT definitive, there are other solutions that don't follow this template)
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| − | Compare (set 1 & 1st coin from set 4) against (set 2 + 2nd coin from set 4)
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| − | Compare (set 1 & 2nd coin from set 4) against (set 3 + 1st coin from set 4)
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| − | Compare (1st coin from each set) against (3rd coin from each set)
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| − | A more concise example:
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| − | Compare 1 2 3 10 against 4 5 6 11
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| − | Compare 1 2 3 11 against 7 8 9 10
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| − | Compare 1 4 7 10 against 3 6 9 12
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| − | Each weighing can have 3 possible outcomes: Left Heavy, Right Heavy, or Balanced (L,R or B)
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| − | Build a truth table to interpret outcomes...many outcomes are not possible.
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| − | Note: THE TABLE VALUES ARE DERIVED FROM THE CHOSEN COMPARISON SETS!
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| − | outcome: fake coin:
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| − | l l l 1 is heavy
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| − | r r r 1 is light
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| − | l l b 2 is heavy
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| − | r r b 2 is light
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| − | l l r 3 is heavy
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| − | r r l 3 is light
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| − | r b l 4 is heavy
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| − | l b r 4 is light
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| − | r b b 5 is heavy
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| − | l b b 5 is light
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| − | r b r 6 is heavy
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| − | l b l 6 is light
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| − | b r l 7 is heavy
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| − | b l r 7 is light
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| − | b r b 8 is heavy
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| − | b l b 8 is light
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| − | b r r 9 is heavy
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| − | b l l 9 is light
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| − | r l r 10 is heavy
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| − | l r l 10 is light
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| − | l r b 11 is heavy
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| − | r l b 11 is light
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| − | b b r 12 is heavy
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| − | b b l 12 is light
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| − |
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| − | There are multiple comparison set possibilities, each with their own comparison table solution.
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| − | ----
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| − | A simpler solution #2:
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| − | * put 6 coins on each side of the scale, one side will be heavier.
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| − | * use the heavier side from the first weighing and put 3 coins on each side of the scale.
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| − | * using the heavier side from the 2nd weighing, pick 2 coins and put 1 on each side of the scale.
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| − | If the scale is balanced then the coin you didn't weigh is the heavier one. Otherwise, the scale will show which one of the other 2 is the heavy coin.
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| − | ----
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| − | Since we do not know if the faulty coin is heavier or lighter , the soluition #2 is not correct
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| − | The only solution is solution nr 1, for which we can also use a binary tree
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| − | ----
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| − | I found the explanation at mathforum.org/library/drmath/view/55618.html to be helpful. The key is to build a truth table, but how that table was built is a little tricky because you can't just write out all possible combinations of results and use that. I tried writing a truth table and found that I had to swap some values around to make sure that each measurement step had exactly 4 on each side.
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