This question already has an answer here:

I am facing problem on understanding this problem---

You have 12 coins and a balance scale, one of which is fake. All the real coins weigh the same, but the fake coin weighs less than the rest. All the coins visually appear the same, and the difference in weight is imperceptible to your senses. In at most 3 weighings, give a strategy that detects the fake coin. (Note: the scale in this problem is a scale with two dishes, which tips toward the side that is heavier. For clarification, do an image search for “balance scale”).


marked as duplicate by Ross Millikan, hardmath, Jyrki Lahtonen Sep 13 '15 at 16:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ A start: Put $4$ coins in one pan and $4$ in the other. $\endgroup$ – André Nicolas Sep 13 '15 at 4:21
  • $\begingroup$ You do not need to be given that the fake coin is light-you can find the fake coin and tell whether it is heavy or light in three weighings. $\endgroup$ – Ross Millikan Sep 13 '15 at 4:27
  • $\begingroup$ Also see a slightly modified version for 13 coins. $\endgroup$ – hardmath Sep 13 '15 at 4:33
  • $\begingroup$ Try this more difficult problem $\endgroup$ – Count Iblis Sep 13 '15 at 4:46

This is a really old riddle. The strategy is:

Step 1: divide the coins into 2 groups (each of 6 members) and weigh them against each other, you should know then which group of 6 coins contains the fake one.

Step 2: take the group of 6 coins containing the fake one, and divide it into two groups each with 3 coins, then weigh them against each other.

Step 3: you are left with 3 coins where one of them is fake. Weigh any two of them against each other.

Step 4: Conclude.

This is the easy version. The hard version is where the fake coin is unknown whether it weighs more or less than the rest (and for this version, our strategy here doesn't work).


weighing #1 - weigh 4 against 4

if one pan is light, the fake is one of the 4 in that pan

if the pans balance, you know the fake is one of the 4 unweighed coins

This method works for anything up to $3^n$ coins where $n$ is the number of weighings.


Not the answer you're looking for? Browse other questions tagged or ask your own question.