# I have another math ?

you have 12 identical-looking coins, one of which is counterfeit. The counterfeit coin is either heavier or lighter than the rest. the only scale available is a simple balance. using the scale only 3 times. find the counterfeit coin

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It’s a little more complicated than that, because The Above answer simply assumes the coin is heavier. But we don’t know if the Coin is Heavier or Lighter, So you could be re-measuring the Heavy Stack when you should have measured the Light Stack.

Number the coins A through L. Weigh coins ABCD against coins EFGH

If the are equal, Weigh coin I vs J and then I vs K

If I and J are equal and I and K are not Equal, it’s K

If I and J are not Equal and I and K are Equal it’s J

If both I and J and I and K are equal, It’s L

If Neither are equal, it’s I

If first weighing the side with coins EFGH is heavier than the side with coins ABCD This means that either ABCD is light or EFGH is heavy.

Weigh ABE against CFI If they balance, it means that either G or H is heavy or D is light.

By weighing G and H we get the answer, because if they balance, then D has to be light. If G and H do not balance, then the heavier coin is it.

If when we weigh ABE against CFI, the CFI side is heavier, then either F is heavy or A is light or B is light.

By weighing A against B the solution is obtained the same as above.

If when we weigh ABE against CFI the CFI side is lighter, then either C is light or E is heavy. By weighing C against a Any other coin, say A, Then If C is lighter, It’s C and If they are equal it’s E

That may be a little more complicated than you were hoping, but that’s the answer.

If you knew ahead of time if the coin was lighter or heavier Then the first guy’s answer would work. But you don’t know, this method actually uses the fact it could be lighter or heavier to your advantage.

The Key is in the ABE/CFI weighing. By arranging the coins in this way You’ve split everything into groups with attached expected values of Equal, Underweight, Overweight, injecting the I (or J or K or L) coin as a control.

The rest is just a matter of finding the coin that meets either the over or underweight value.

divide 12 coins into 2 set of 6 coins, weigh them on the simple balance, and let’s just say that the fake coin is heavier, then take the which ever set of 6 is heavier; then devide that set out into 2 other set of 3 coins, and then weigh them the second time; take the heavier set of 3 coins, and take 2 coins out of that 3; weigh them again for the third time; if the 2 coin is equal, then the third coin is fake, but if one of the 2 coins that weighed heavier, then it is the fake coin; overall, you only use the simple balance 3 times.

Same concept if the fake coin is lighter; if this is the case, then always keep the set of coin which is lighter.

Is this the whole problem? Does it give any hint if the fake coin is heavier?

That will work if you know the counterfeit coin is heavier, but it may be lighter. The method is correct, but it may cost you two more steps.

Start weighting 3/3 (half of the total)

P.S.(edited): My congratulations to Crazy_Girly for this wonderful question and to Enders_Knight

who found the right solution, it was super!