1

Let us assume that we have a box of coins which contains 10 normal coins and 11 counterfeit coins. We know that each counterfeit coin is 1 gram less than each of the normal. One coin was taken from the box.

If we have a regular scale and we can only use it once (for one single measurement), how can we find out what was the coin which was taken out of the box (figure out whether it was a counterfeit or normal one)?

We can assume that the coins' weight can be any real number.

What I have considered so far:

  • I know that if we measure the coin alone, which was taken out of the box, we have no way of figuring out whether it is a counterfeit or a normal one without measuring the whole box - and we cannot measure it since we will be breaking our rule of having to use the scale just once.
  • I do not know how will measuring just the box would help either. Correct me if I am wrong but since we do not know how much is the weight of an individual coin, the total weight wouldn't be of any use to us, right? So I am stuck.
Giovanni
  • 113
  • What are your thoughts? What have your tried so far? (Hint: what is left in the box after your remove the coin?) – Brian61354270 Jan 26 '20 at 16:48
  • 3
    is the weight of a coin an integer ? – Laassila souhayl Jan 26 '20 at 16:51
  • @IamOptimus Please add your long comment to the question. The question needs to be updated every time you provide fresh information or useful background. – almagest Jan 26 '20 at 16:59
  • @almagest Apologies, I have updated the main post. – Giovanni Jan 26 '20 at 17:04
  • The only idea I have is to exploit integrality, which you say we can not assume. Under that assumption, the combined weight of the remaining coins is either $20w-10$ or $20w-11$ and exactly one of those is divisible by $10$. Clearly no use if we can't take $w\in \mathbb N$. – lulu Jan 26 '20 at 17:07

1 Answers1

2

For the purposes of this answer, I am going to assume that the weight of a coin is an integer. For convenience, I'll denote the weight of a normal coin by $N$.

Once you remove one coin from the box, one of two events will occur:

  1. You removed a normal count. In which case, there are now $9$ normal coins and $11$ counterfeit coins in the box. The total weight of these coins is given by $$9N + 11(N-1)=20N-11\tag 1$$
  2. You removed a counterfeit coin. In this case, there are $10$ normal coins and $10$ counterfeit coins left in the box. The total weight is there given by $$10N + 10(N-1) = 20N - 10. \tag 2$$

It should not be hard to see that regardless of the value of $N$, case (1) will give an odd number for the total weight, while case (2) will given an even number. Therefore, you can determine whether the coin you selected was normal or counterfeit simply by measuring the total weight of all of the remaining coins (one measurement) and checking whether the resulting weight is even or odd. An even weight corresponds to a counterfeit coin, while an odd weight will be indicative of a normal coin.

Brian61354270
  • 2,170
  • I am not sure about the rules but is this an answer for the given question ? the answer expected is a proof that the problem cannot be solved . – Laassila souhayl Jan 26 '20 at 17:11
  • @lessili The OP expressed some ambiguity about the context of the problem, so I figured it was quite likely that their source was poorly worded and expected a solution of this sort. As you noted, this question is unsolved if one takes the weight of a coin to be a real number, which would makes this a rather silly riddle indeed. I see now that you simultaneously give the same solution in a comment above. If you wish to convert it into an answer, I would happy to delete mine. – Brian61354270 Jan 26 '20 at 17:20