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At even odds I'd bet it's 5 tries, although if someone gave me 10:1 odds I
might put a dollar on 4.
Calling the solution to the 12-ball-bearing problem our "Basic Solution",
please follow the following logic.
(1) If it is known that a coin is light (or heavy), then any number up to
3^n coins can be tested in n tries. Group the coins into 3 piles of 3^(n-1) and
weigh 2 against each other. If one pile is light, that is the light pile.
If neither pile is light, the third pile is the light pile. Repeat with
3^(n-1) until you get 1 coin per pile.
Note - if you have a number that is not a power of 3, pick the next larger
power of 3 for n, make the piles as even as possible, and assign any missing
coins to be in the pile not weighed.
(2) The "Basic Solution" will determine in 3 tries whether the error is
light or heavy for 12 "ball bearings", which can represent any number
of coins in an equal pile.
(3) Therefore, in 4 tries you can determine up to 3*12 = 36 coins by
settting them in piles of 3. In 5 tries you can determine up to
9*12 = 108 coins by setting them in piles of 9.
(4A) Roughly, the solution for 50 is, set the coins into 4 piles (ABCD) of
12 and 1 pile (E) of 2.
(4B) Use the basic method on piles A and B, treating each pile of 12 as if
it were 4 ball bearings (each made of 3 coins). If A and B are unequal, finish
the Basic Method to find which pile of 3 coins is light (or heavy). Then
compare 2 of the 3 coins against each other to determine the light one. This solves
the puzzle in 4.
(4C) If A and B are equal, compare A + B to C + D. If AB and CD are
unequal, then you know whether the bad coin in CD is light (or heavy).
Throw 3 regular coins into pile CD to make 27 total. Divide into 3 piles
of 9 and find the light pile. Divide into 3 piles of 3 and find the light
pile. Divide into 3 piles of 1 and find the light pile. This solves the puzzle in 5 tests.
(4D) If AB = CD, the odd coin is one of the 2 in pile E, which you can
determine in 2 tests, solving the puzzle in 4. |
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