The Fake Among Eight Coins

Ready for some brain gymnastics? Let’s start.

There are eight identical-looking coins; one of these coins is counterfeit and is known to be lighter than the genuine coins. What is the minimum number of weighings needed to identify the fake coin with a two-pan balance scale without weights?

I’ll give you some time to think now.

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Okay! Let’s solve this puzzle! Let me first give you the wrong answer, and probably the one you think about at first glance. The answer is NOT 3. As you might have already guessed by now, dividing the pile into 2 piles of 4 coins and then taking the lighter pile and diving it into two piles of 2 coins and then, again, weighting the two coins in a random pile to finally arrive at the fake coin is not the optimal way of doing this.

So now, let me give the real answer now. The answer is actually 2. How come? First, you break down the 8 coins into 4 piles that have 3, 3, 1, and 1 coins respectively. Then you weigh the two 3-coin-piles. If they weigh the same then you weigh the left coins to determine the lighter one, aka the fake one. If the two 3-coin-piles do not weigh the same, then you pick the lighter one and randomly choose two coins from that 3-coin-pile. You weigh these two coins if they are the same then the left coin in the 3-coin-pile is the fake one and if they do not weigh the same then the lighter one is the fake one.

There is one more thing before concluding our solution. And that is showing that 1 weighing cannot work at all. But it is pretty obvious that 1 weighing is not enough. (I mean you can’t point out the fake coin in a single weighing; you can at most find a group of coins that contains the fake coin.)

That was it from our puzzle corner. See you at many more funnier-to-solve puzzles!