20090911, 03:04  #1 
43×229 Posts 
Counterfeit Coin2
You have 120 coins, one of which differs in weight from the other 119. Yo have the typical two pan balance for identifying the odd weight coin in just five (5) weighings.

20090911, 12:09  #2 
Dec 2007
Cleves, Germany
2×5×53 Posts 
This is obviously possible since 120 < (3⁵1)/2. As an added bonus, you can even say whether the counterfeit coin weighs more or less than the genuine ones. I'll leave the actual methodology as an exercise to the interested readers who never designed a Hamming code before.

20090911, 19:15  #3 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Trade the scale at the flea market for a package of athletic socks, duct tape and a chair.
Find the counterfeiter and duct tape him to the chair. Fill a sock with the 120 coins and beat the counterfeiter with it until he tells you which one is counterfeit. 0 weighings. 
20090911, 21:00  #4  
6809 > 6502
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Aug 2003
101×103 Posts
2×11×491 Posts 
Quote:
http://www.mersenneforum.org/showthr...160#post180160 

20091019, 06:09  #5 
Feb 2007
2^{4}×3^{3} Posts 

20091019, 06:55  #6  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
1100111110000_{2} Posts 
Quote:
i.e. Weigh: A vs B C vs D E vs F G vs H I vs J then ... Results given: Left down, even, left, right, even. then ... Answer is: 

20091019, 13:35  #7 
Cranksta Rap Ayatollah
Jul 2003
1201_{8} Posts 

20091023, 02:17  #8  
Random Account
Aug 2009
Not U. + S.A.
2,383 Posts 
Quote:
I remember a scale used in Chemistry class in highschool. It was in a wooden and glass case. It was all mechanical and very sensitive. I remember the instructor asking one of the guys to write his name on a little slip of paper, but leave the t's uncrossed in his name and weigh it. Then cross the t's and weigh it again. There was a difference. I don't know what type of scale it was. It had to be protected from dust, moisture, and so on. 

20091023, 05:40  #9  
6809 > 6502
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Aug 2003
101×103 Posts
2×11×491 Posts 
Quote:
You find scales in a bathroom and at the checkout counter, but balances in the lab. 

20091023, 09:21  #10 
Oct 2009
5_{8} Posts 
Counterfeit Coin
You are given 120 coins, one of which differs in weight (and in weight only) from the other 119. You are given a pan balance to identify the odd coin in just 5 trials. To distinguish from among 120 equally likely objects requires log2 120 (= 6.907 bits) of information. You will receive log2 3 (= 1.585 bits) of information on each trial so long as each of the 3 possible outcomes (left pan up, down, or balanced) is equally likely. Since 5 log2 3 (= 7.925) is greater than log2 120, a solution is feasible in 5 trials. The need to specify whether the counterfeit coin is heavy or light requires one additional bit of information. Since log2 120 + 1 (= 7.907 bits) is less than 5 log2 3 (= 7.925 bits), a solution in 5 trials is feasible. Notice that this does not guarantee a solution unless the requirement for equally likely outcomes on each trial is met. With equally likely outcomes, the maximum number of coins that may be distinguished in 5 trials is 2 raised to the power 5 log2 3 (= 243). If the counterfeit is heavy or light, the maximum number that may be distinguished will be less than 2 raised to the power 5 log2 3  1 (e.g. less than 121.5). Note that it is meaningful to carry out the math and say less than 121.5 (as opposed to less than 122). The reason for this is that, for some fraction of the time, it is possible to get the answer for 122 coins. This will happen whenever the pans do not balance in the first trial. (If they are unequal this reduces to the solution for 121 coins.) If the pans do balance in the first trial, you will be able to discover which coin is counterfeit, or whether it is light or heavy, but not both. 
20091024, 17:43  #11  
Random Account
Aug 2009
Not U. + S.A.
2,383 Posts 
Quote:
The case this device rested in was maybe five inches tall and six inches front to back, but well over a foot in length. Possibly as much as 18 inches. 
