Can you solve the counterfeit coin riddle? - Jennifer Lu
video description
Divide the coins into 3 groups of 4, like the video. Take 2 groups and weigh them against each other. Take the heavier group or the third group if the two you weighed were equal. Divide those 4 into 2 and 2 and weigh them. Take the heavier of those and divide into 1 and 1 and weigh them. That should give you the counterfeit.
Date: 2020-08-22
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Comments and reviews: 8
Antar
I solved this, but with an easier method that prevents the scenario where you have the 4vs4 coins with an imbalance (leading to those complicated steps afterwards. So. For the few interested: 1. You balance 3vs3. If its even, the odd coin is amongst the other 6 coins. If its uneven, one of those measured 6 is the odd one. 2. You take 3 out of the 6 coins where the odd one is and measure those against 3 of the other 6 (the ones you know are standard for sure. If there is an imbalance, you know that the odd coin is amongst those 3 and you also get to know if it is either heavier or lighter than the others. Out of the last 3 coins you measure 2 and determine it as in the video. If you have the case that there is a balance when you take the 3vs3 from step 2, you know that the odd coin is amongst the other 3 you did not measure. I just realised while finishing that sentence, that there is no way of determining the odd one in this case since you still don't know if its lighter or heavier, thus making the 1vs1 measurement in the end useless and fcking up my whole approach. Anyway, took me way to long to write this so im going to post it anyway for your entertainment. Thanks for reading: )
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I solved this, but with an easier method that prevents the scenario where you have the 4vs4 coins with an imbalance (leading to those complicated steps afterwards. So. For the few interested: 1. You balance 3vs3. If its even, the odd coin is amongst the other 6 coins. If its uneven, one of those measured 6 is the odd one. 2. You take 3 out of the 6 coins where the odd one is and measure those against 3 of the other 6 (the ones you know are standard for sure. If there is an imbalance, you know that the odd coin is amongst those 3 and you also get to know if it is either heavier or lighter than the others. Out of the last 3 coins you measure 2 and determine it as in the video. If you have the case that there is a balance when you take the 3vs3 from step 2, you know that the odd coin is amongst the other 3 you did not measure. I just realised while finishing that sentence, that there is no way of determining the odd one in this case since you still don't know if its lighter or heavier, thus making the 1vs1 measurement in the end useless and fcking up my whole approach. Anyway, took me way to long to write this so im going to post it anyway for your entertainment. Thanks for reading: )
reply
Whorabl3
There's another (dare I say simpler) way of doing this without even using the marker but it relies on the semantics of what entails a use of the scale. If a use is define as complete when all of the coins used in the weighing have been removed by the scale, then take 6 coins and put them on each side (you'll notice a discrepancy of course. Then take one coin off of each side in pairs until you notice that the discrepancy is no longer present. Then you can deduce that the one of the two coins you just removed is the fake and any of the remaining ten can be used a control. Next you simply weigh each coin against the control coin you've pick and look for the discrepancy which will in this case indicate the fake. Then you use your marker autographs.
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There's another (dare I say simpler) way of doing this without even using the marker but it relies on the semantics of what entails a use of the scale. If a use is define as complete when all of the coins used in the weighing have been removed by the scale, then take 6 coins and put them on each side (you'll notice a discrepancy of course. Then take one coin off of each side in pairs until you notice that the discrepancy is no longer present. Then you can deduce that the one of the two coins you just removed is the fake and any of the remaining ten can be used a control. Next you simply weigh each coin against the control coin you've pick and look for the discrepancy which will in this case indicate the fake. Then you use your marker autographs.
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Marklee
I criticise your poor tax laws. get thrown in jail. called to get the fake coin yet you have the AUDACITY to tell me to use the coin thrice. That already goes to show that it didn't matter to him that much. So I can just choose a random coin and lie that it's the counterfeit. But that begs the question. how did he figure that there was a fake to begin with. he obviously didn't figure it out by weighing. because he wouldve known how to isolate the coin by himself
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I criticise your poor tax laws. get thrown in jail. called to get the fake coin yet you have the AUDACITY to tell me to use the coin thrice. That already goes to show that it didn't matter to him that much. So I can just choose a random coin and lie that it's the counterfeit. But that begs the question. how did he figure that there was a fake to begin with. he obviously didn't figure it out by weighing. because he wouldve known how to isolate the coin by himself
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Rina
I feel like this was easy to solve and you didn't need the marker. All you have to do is half the 12 coins. Put 6 on one side and 6 on the other. Whatever is lighter side half that. 3 on one side 3 on the other. Then which ever side is lighter take 2 coins, 1 on one side 1 on the other. Whichever is lighter is the fake. If they weigh the same then the one you left out is the fake. He doing too much, there's no need for the marker lol
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I feel like this was easy to solve and you didn't need the marker. All you have to do is half the 12 coins. Put 6 on one side and 6 on the other. Whatever is lighter side half that. 3 on one side 3 on the other. Then which ever side is lighter take 2 coins, 1 on one side 1 on the other. Whichever is lighter is the fake. If they weigh the same then the one you left out is the fake. He doing too much, there's no need for the marker lol
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Solea
you don't even need a marker or anything just do it like
1) split the coins in 3 piles of four -> weigh 2 of them, if they balance contiue with the third pile if not continue with the lighter one
2)split this pile in 2 piles of 2 -> weigh them, continue with the lighter one as for this time one of them has to be
3) weigh the 2 remaining coins against each other et voila you just identified the fake coin
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you don't even need a marker or anything just do it like
1) split the coins in 3 piles of four -> weigh 2 of them, if they balance contiue with the third pile if not continue with the lighter one
2)split this pile in 2 piles of 2 -> weigh them, continue with the lighter one as for this time one of them has to be
3) weigh the 2 remaining coins against each other et voila you just identified the fake coin
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Mattias
The solution presented is to complicated. You dont need the marker.
Step 1: Divide the coins in three stacks. Weigh two of them. Just as in the video.
Step 2: Take the four coins in the lighter stack and weigh them two by two. The side with the counterfeit coin is lighter.
Step 3: Take the two coins and weigh them against each other. The lightest coin is the counterfeit one.
Done!
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The solution presented is to complicated. You dont need the marker.
Step 1: Divide the coins in three stacks. Weigh two of them. Just as in the video.
Step 2: Take the four coins in the lighter stack and weigh them two by two. The side with the counterfeit coin is lighter.
Step 3: Take the two coins and weigh them against each other. The lightest coin is the counterfeit one.
Done!
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zi_ang
But why dont you just do: 1. Splitt into 3 equal stacks 2. Weight two stacks (if theyre equal = third stack is the one with the fake. If they arent, the lighter one ist the one with the fake) 3. Split the stack with the fake equally (2/2) then weight them (the one thats lighter is the one with the fake) 4. Split the one with the fake equally(1/1) and the one thats lighter is the fake on.
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But why dont you just do: 1. Splitt into 3 equal stacks 2. Weight two stacks (if theyre equal = third stack is the one with the fake. If they arent, the lighter one ist the one with the fake) 3. Split the stack with the fake equally (2/2) then weight them (the one thats lighter is the one with the fake) 4. Split the one with the fake equally(1/1) and the one thats lighter is the fake on.
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byedawae
Easier solution, is imperfect unfortunately - weigh 6 on 6, then 3 on 3 for the lighter side, which if the counterfeit coin is there, you can determine by weigh 2 of them against each other.
If its not in the lighter pile, you can get 50/50 odds by weighing 2 on 2. In total the odds are on your side with a 7/8 chance you identify it.
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Easier solution, is imperfect unfortunately - weigh 6 on 6, then 3 on 3 for the lighter side, which if the counterfeit coin is there, you can determine by weigh 2 of them against each other.
If its not in the lighter pile, you can get 50/50 odds by weighing 2 on 2. In total the odds are on your side with a 7/8 chance you identify it.
reply
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