Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

http://blog.tanyakhovanova.com/?p=279

This is an extension of the problem posted here.

You have 14 coins. You know that 7 are real, and 7 are counterfeit. The real coins are 1000g, and the fake coins are 999kg.

You have a balance accurate enough to do the measurement. However, this balance is a peculiar balance. Assume that if you weigh multiple coins, you do not place them incrementally. That is, if you weigh 3 coins against another 3 coins, you would place all the coins in the balance scale at the same time.

1)If the weight one one side is heavier, the balance incinerates one coin selected at random from the heavier side. (Recall that the real coins are heaver than the fake ones). 2)If the weight is equal, you take back all the coins.

Find a methodology in which one can necessarily identify a single real coin that has not been incinerated. Either find the strategy, or prove that such a strategy does not exist.

share|improve this question
7  
Welcome to MathSE. I see that this is your first question. So I wanted to let you know a few things about MathSE. Posting questions in the imperative (i.e. Compute all such, Prove that...), is considered rude by some of the members, so it would be nice of you to change that wording; perhaps by adding what are your thoughts or what you have tried in trying to answer the problem. These sort of pleasantries usually result in more and better answers. Thank you. –  Arturo Magidin Oct 7 '11 at 20:00
4  
That's some heavy coinage. –  Henning Makholm Oct 7 '11 at 20:44
    
See #4 here usamts.org/Tests/Problems_23_1.pdf from an ongoing contest. nice one. –  user17294 Oct 8 '11 at 3:17
    
My goodness. This was given to me by a student of mine (I am a teacher). I am sorry for any inconveniences... –  Stewart Oct 8 '11 at 12:27
    
I will adhere to this in the future. Thank you. –  Stewart Oct 8 '11 at 12:43
add comment

1 Answer 1

Here is how you do it.

Weigh coins 1,2 against 3,4; 5,6 against 7,8; 9, 10 against 11, 12.

If any of the pans are not in balance, weigh the two coins in the lighter pan against each other. If they do not balance (so 1 is light and 1 is heavy), the remaining coin from the heavier pan in the first weighing is heavy and real and you are done. If they do balance, you know they are both fake -- the remaining coin from the heavy pan in the first weighing can be either real or fake.

If one of these 2-vs-2 weighings is in balance (suppose e.g. it is 1,2 vs 3,4), either all coins are real, all are fake, or 1 from each pan is real and one is fake. Weigh 1,3 against 2,4. If this is not in balance, the remaining coin from the heavier pan is real and you are done. If it is in balance, weigh 1,4 against 2,3. If this is not in balance, the remaining coin from the heavier pan is real and you are done. If it is in balance, either they are all real or all fake.

Our remaining cases have 3 sets of coins, which are either 4 coins of the same type or 3 coins of which 2 have been identified as fakes, and they have two as-yet-unweighed coins (13 and 14). If the case is 3,3,3, the remaining non-identified coins from each set have at most one fake among them: weigh one of these against another of these; an imbalance means the third is real and you're done, a balance means they're both real and you're done. If the case is 4,3,3, each of the 4 coins must be real and you're done. If the case is 4,4,3, one of the two sets of 4 is all-fake and one is all-real. Weigh them against each other. You are done. You cannot have 4,4,4.

You never wind up weighing coins 13 and 14.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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