# 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 =))