'd like to take issue with the answer to the puzzler of 3-2-03, and it may be with the wording of the puzzle. I believe the question was - what is the minimum number of weighs necessary to find the counterfeit coin, while you answer of 4 is the maximum number of necessary weighs.

50 coins, divided to two groups of 2 - and weigh. One group weighs more. You’re now down to 25 coins.

25 coins, divided to two groups of 12, with one left out - and weigh. The two groups weight the same.
The coin left out is bogus - two weighs - you’re done. (Although, I do think dividing by 3 is quite clever. I didn’t think of that.)

What do you think?

Jim Blaney

The question says the “minimum number of weighs NECESSARY”, not POSSIBLE. NECESSARY has to include all choices of coins, not just a lucky choice.
That’s what I think.
David

There’s also another answer to this Puzzler:

Divide the 50 coins into three piles of 21, 21 and 8. Weigh the two piles of 21 [WEIGHING 1]. If they are the same weight, set aside the two piles of 21 coins then take the pile of 8 coins and divide it in half. Weigh each pile of 4 coins [WEIGHING 2]. Split the heavier pile of 4 coins into two piles of two coins and weigh them [WEIGHING 3]. Take the heavier pile of two coins and split it into separate coins and weigh them [WEIGHING 4]. The heavier coin is the one we’re looking for.

If one of the piles of 21 coins is heavier than the other, discard the lighter one and split the heavier one into three piles of 7 coins each. Weigh two of the piles [WEIGHING 2]. If the piles have the same weight, discard them and work on the third pile. If one of the piles of 7 coins is heavier than the other, use the heavier one. No matter what, now you’re working with a pile of 7 coins.

Divide the 7 coins into groups of 3, 3 and 1. Weigh the two groups of 3 coins [WEIGHING 3]. If these two groups of coins weigh the same, the coin we’re looking for is the lone coin left out. If one of the groups of 3 coins is heavier than the other, take the heavier one and divide it into individual coins. Weigh two of the coins [WEIGHING 4]. If the coins weigh the same, the lone coin left out is the one we’re looking for. If one of the coins is heavier than the other, the heavier coin is the one we’re looking for.

There are many ways to organize the coins for the 4 weighings. This is because it works for up to 81 coins.

As for the first comment, by that logic I guess the minimum number of weighings is zero - just pick a coin, if you’re lucky it’s the one…

It’s possible to find the bogus coin in one weighing, not two. Just pick two coins at random. If one weighs more, that’s it.

But that doesn’t satisfy the requirement that the method you propose must guarantee to find the coin, not only that it might find the coin. In the OP’er proposed method, there is a good chance the method wouldn’t identify which coin it was in two weighing, if the two groups of 12 didn’t weigh the same. Which would be the more likely case.

The only question remaining: Can the bogus coin among the 12 remaining still be identified in two weighings? If so, the method would still satify the puzzler requirements.

Try it: Divide the 12 remaining coins into 3 piles of 4 for the 3rd weighing. That would narrow it to 4 coins. For the 4th and final weighing then, you’d still have 4 coins left, which is one too many coins to guarantee you’d determine the bogus one, so you couldn’t guarantee to find the bogus coin using the proposed method in 4 weighings. It could take as many as 5 weighings using the OP’ers method.