What happened? 317 people die...

[TxHockeyGuy]

If the 4 vs 4 weighing contains an inbalance, note which way the imbalance is. Remove 2 coins from one of the sides (let's pick the right side), and take one coin from the left and put it on the right. Now, if the scale balances, you know that one of the two coins that you removed from the right side is counterfeit, and you can use your third weighing to figure that out. If the imbalance stays the same, you know it is one of the two coins on the right side, and you use your third weighing to figure that out. If the imbalance changes, you know that the coin that you moved from the left is the culprit.

You are assuming the coin is heavy, it could be light. This solution won't work.

 

[TxHockeyGuy]

Ok split into 3 groups.
#1 Weight 4 vs 4. If they are the same weight you know the coin isn't in one of those.So you do 3 of the coins from the original weighing as your "controls" and weigh 3 of the remaining coins. If those are the same you know the coin you didn't weigh is the counterfit, weigh it against any one of the original coins to see if it is heavier or lighter. If the 3 vs 3 results in an imbalance (if it is heavier or lighter, you now know which direction the coin goes). Of the group that is off you weigh 1 vs 1. If they balance # 3 is the counterfeit. If they don't balance which ever one is off in the proper direction is the counterfit.

#2 if in your 4 vs 4 you have an imbalance. You know that the 4 coins that were set asside does not contain a fake. From here I am stumped because you don't know which side of the original weighing has a fake. So I need an extra weighing in order to figure it out.

The first part works.

As for the second part, the other poster was on the right track with swapping some coins around. That's your second hint.

 

[Trinigordo]

See packhorses' explaination.

This one was a mind tester if you were thinking math, which is easy to do.

Of course that really screws with you since the last number one I posted was really quacky and required you to change the base of the number to figure it out.


These are fun to mess with my fellow 'rocket scientists' at work with. Some will really rack their brains trying to do too much analysis of it... some will sit and work them in their heads in minutes..... We used to get them and put them up on the whiteboard in the lab for all to work. heh

I used to think I was kind of smart, but thank you for putting me back in my place

 

[fairybasslet]

Sorry to hear that your head is about to explode.

Do you think he's laugh if you told him that since he used him own money for the CPR card, here's a dollar to use with the $10 overpayment from the CPR book for the stethoscope?

:lol: :lol: Good one.
BTW, I'm still trying to find the answer to the "What do you make more of and get less," but I even forgot the question by now.

 

[Trinigordo]

Fairy, speaking from experience any parent without exact change is looking for trouble :evil:

 

[bob1dp]

The Conterfeit Coin

WEIGHING ONE
Weigh three coins against three coins with six standing by not being weighed. If the weighing produces uneven results you know that the six coins not weighed are non counterfeit.

WEIGHING TWO
You then take the three coins that weighed heavy and weigh them against three non counterfeit coins from the six not weighed. If the previously weighed three coins still weigh heavy you know the counterfeit coin is in that group and is heavy. If the three coins weigh the same as the non counterfeit coins then the counterfeit is one of the three coins previously weighed is counterfeit and is light.

WEIGHING THREE
Depending on your results you will either weigh the three that weighed heavy or the three that weighed light. For this example lets say you determined the counterfeit is light. You take two of the three coins and weigh them against each other. If they weigh un-evenly the coin that weighs light is the counterfeit. If they are equal in weight the un-weighed coin is the counterfeit.

If on WEIGHING ONE the scale balances out you know that none of the weighed coins contain the counterfeit. You then simple move on with WEIGHING TWO. Except now you take three of the un-weighed coins and weigh them against three of the already weighed but non counterfeit coins.

 

[TxHockeyGuy]

The Conterfeit Coin

WEIGHING ONE
Weigh three coins against three coins with six standing by not being weighed. If the weighing produces uneven results you know that the six coins not weighed are non counterfeit.

WEIGHING TWO
You then take the three coins that weighed heavy and weigh them against three non counterfeit coins from the six not weighed. If the previously weighed three coins still weigh heavy you know the counterfeit coin is in that group and is heavy. If the three coins weigh the same as the non counterfeit coins then the counterfeit is one of the three coins previously weighed is counterfeit and is light.

WEIGHING THREE
Depending on your results you will either weigh the three that weighed heavy or the three that weighed light. For this example lets say you determined the counterfeit is light. You take two of the three coins and weigh them against each other. If they weigh un-evenly the coin that weighs light is the counterfeit. If they are equal in weight the un-weighed coin is the counterfeit.

If on WEIGHING ONE the scale balances out you know that none of the weighed coins contain the counterfeit. You then simple move on with WEIGHING TWO. Except now you take three of the un-weighed coins and weigh them against three of the already weighed but non counterfeit coins.

Oh so very close. However, your solution does not solve this particular instance.

WEIGHING ONE
Weigh three coins against three coins, say they balance.

WEIGHING TWO
Weigh three unkown coins vs 3 known coins, say they balance.

WEIGHING THREE
Now you have 3 unknown coins and only can only weigh once. There's no way to determine which coin is counterfeit for sure and if it is heavy or light. If you're lucky enough to pick the 2 that are not counterfeit, you know the other is but not if it is heavy or light. If you happen to test the counterfeit coin since you don't know if it is heavy or light you won't know which one is counterfeit.

You are very close in your methodology. Remember my previous hint though, first weighing has to be 4v4 for this to work.

 

[scubafool]

TxHockeyGuy, do you realize how much of my valuable SB time you have consumed with this? I mean, really, do you appreciate how destructive this obsession has become?

 

[TxHockeyGuy]

TxHockeyGuy, do you realize how much of my valuable SB time you have consumed with this? I mean, really, do you appreciate how destructive this obsession has become?

Yes, I spent a good 6 hours myself when I originally was given this puzzle and no one game me any hints. You could always ask for the answer...

 

[bob1dp]

Sorry about the strange formatting but the board doesn’t do an outline very well. Man was this a tough one. He is my take on it.

Take three groups of four: A, B, and C. Weigh group A against group B

1. If group A = group B the counterfeit is in group C. Take 3xC and weigh it against 3xA. Set aside one C.

A. If group 3xA=group 3xC then the set aside C that wasn't weighed contains the counterfeit. Weigh one A against un-weighed set aside C to get weight of counterfeit C.

B. If group 3xA>group 3xC then one of 3xC is the counterfeit and light. Weigh C.1 against C.2 set aside C.3. If C.1 = C.2 then C.3 is the counterfeit and is light. If C.1 > C.2 then C.2 is the counterfeit and is light. If C.1 < C.2 then C.1 is the counterfeit and is light.

2. If group A > group B, then group C contains no counterfeit coins and that either group A has a heavy counterfeit or group B has a light counterfeit. Set aside 1xA and 2xB. Move 2xA to group B, Move 1xB to group A. Add 1xC to group A. You are now weighing 1A*+1B+1C against 2A+1B* and setting aside 1xA and 2xB.

A. If 1A*+1B+1C >2A+1B* which was the original uneven weighing. Then 1A* or 1B* are the counterfeits because they are the only element that didn't change. Weigh 1B* against 1C and if 1B*=1C then 1A* is the counterfeit and heavier. If 1B*<1C then it is B* and it is lighter

B. If 1A+1B+1C =2A+1B Then set aside 1xA or 2xB Are the counterfeit. Weigh 2xC against (1xA + 1xB) and set aside B. If (1xA + 1xB) weighs heavy 1xA is the counterfeit and is heavy, if it weighs equal the set aside B is the counterfeit and is light. If it weighs is light then the 1xB being weighed is the counterfeit and light.

C. If 1A+1B+1C <2A+1B which is the opposite of the original mismatch. The 1B in group A and 2A in group B are the counterfeits. Weigh 2xC against 1A + 1B and set aside A. If it weighs heavy A is the counterfeit and is heavy, if it weighs equal the set aside A is the counterfeit and is heavy. If it weighs light B is the counterfeit and light.