13 coins of which 2 may be counterfeit?
You are presented with a set of 13 suspect coins. Of these, it is
known that the number of counterfeit coins is 0 - 2. All of
the counterfeit coins, if any, are lighter OR heavier than the norm.. You must identify the counterfeit coins, if
any, after 4 weightings or fewer.
Work the Problem into a guess pattern that can be applied to any of the three possible results of a two-pan balance scale.
To recap:
N(coins) = 13
The number of weighs is 4
There can be up to two counterfeit coins.
You’re right the 81 points would bring it over the required for 4 or less weighs.
Lets assume their heavy and the first weigh is ALWAYS balanced.
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You have 13 coins.
The possible arrangements are:
0 counterfeits -> 1 way
1 counterfeit (light) -> 13 ways
1 counterfeit (heavy) -> 13 ways
2 counterfeits (light) -> C(13,2) = 78 ways
2 counterfeits (heavy) -> C(13,2) = 78 ways
With any weighing you have 3 possible outcomes:
Left side heavier
Right side heavier
Balanced
And with 4 weighings you would have 3^4 = 81 pieces of information.
There is no way you can devise a pattern that would uniquely determine:
1) Number of counterfeits
2) Whether they are lighter or heavier
3) Where they are in the set of coins
if no. of counterfeit coins = 0
1. take 6 in each pan. balanced
2. take the remaining 13th coin and weight it against any of the 12 coins it’ll balance and hence no counterfeit.
if no. of counterfeit coins = 1
1. take 6 in each pan.
either of the pans has a counterfeit coin or the left out coin = counterfeit
If the two pans balance then remaining coin = counterfeit
else
2. let the counterfeit coin be of a lighter weight than the original coin. So, take the 6 coins in the lesser pan and make them into two pans of three coins each.
the lesser pan of the two has a counterfeit, and it has three coins in it.
3. so take any two and weigh them
a) if they are balanced the remaining one =counterfeit
b) else the pan with lesser wt has the counterfeit
Damn.. im running out of patience typing this.
If no. of counterfeit coins = 2 and both the counterfeits are of lesser wts
1. take 6 in each pan.
A. Each of the pans can have a counterfeit coin, or a pan can have two counterfeits
or
B. The left out coin = counterfeit and a pan with a lesser wt has one counterfeit
2.A…. If the two pans balance then ,
a) Each of the pans do have a counterfeit coin.
So use the steps 2,3 in the method mentioned in the (no.of counterfeits = 1)
b) or the pan having a lesser wt has both counterfeits in it.
3,B. the remaining coin = counterfeit and the lesser pan has one coin in it.
One heck of a typing situation easy to do with a mind rather than a keyboard =))