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Santa Posted 22 years ago
Jokes, Puzzles & Riddles
More difficult than the Einstiens
Here is a puzzle which will surely trouble many:
You have twelve pennies. One of them is a bad penny, and is either heavier or lighter than the others-you dont know which.You have a balancing scale - no pennies will be weighed; there are just two sides that will either balance or not;according to which pennies you put on each side.
The problem : Within just three weighings or balancings, can you find out which one is bad - and I also need to know if it is lighter or heavier than others.
You have twelve pennies. One of them is a bad penny, and is either heavier or lighter than the others-you dont know which.You have a balancing scale - no pennies will be weighed; there are just two sides that will either balance or not;according to which pennies you put on each side.
The problem : Within just three weighings or balancings, can you find out which one is bad - and I also need to know if it is lighter or heavier than others.
Top answer
It is possible to find out, but it's all down to your luck, because the most pennies that you can weigh in three turns is 6.
- It is possible to find out, but it's all down to your luck, because the most pennies that you can weigh in three turns is 6.
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30 Answers
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It is possible to find out, but it's all down to your luck, because the most pennies that you can weigh in three turns is 6.
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No luck involved here!! Just pure genius and intellect.And it is possible to solve this in more than one way!
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Well, as far as I can determine, the information to confidently guess which penny is what can be obtained, but a definite answer cannot.
If such a problem does require so much genius and intellect, and you have it, cure cancer or something damnit. Don't waste time on stuff like this. Any how, I am privileged enough to be able to waste my time here. good day.
If such a problem does require so much genius and intellect, and you have it, cure cancer or something damnit. Don't waste time on stuff like this. Any how, I am privileged enough to be able to waste my time here. good day.
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Calm down, it's only a puzzle. I'll have a go at it...
Separate the 12 pennies into 3 groups of 4. Balance 2 of the groups against each other. If they are the same weight, the bad penny isn't in either group, so you know it is in the one you didn't weigh. You can then balance 1 penny against another (from the 4 you didn't weigh). If they weigh the same, the bad penny is neither, so it m
Separate the 12 pennies into 3 groups of 4. Balance 2 of the groups against each other. If they are the same weight, the bad penny isn't in either group, so you know it is in the one you didn't weigh. You can then balance 1 penny against another (from the 4 you didn't weigh). If they weigh the same, the bad penny is neither, so it m
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You were very close indeed. But there is a way to determine the exact penny. I am sure if you think again then you should be able to crack it. If you need a hint then I will post it tomorrow.
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Here goes... Separate the 12 pennies into 3 groups of 4. Balance 2 of the groups against each other. If:
- they are the same weight: the bad penny isn't in either group, so you know it is in the one you didn't weigh. Balance 3 from one of the groups you now know doesn't contain the bad penny against 3 from the group you now know does contain the bad penny. If:
- - they are t
- they are the same weight: the bad penny isn't in either group, so you know it is in the one you didn't weigh. Balance 3 from one of the groups you now know doesn't contain the bad penny against 3 from the group you now know does contain the bad penny. If:
- - they are t
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Fleogan. You are basing it all that you are excluding the group with the penny. That is just a 33% chance. If you find you have chosen that pile you need to do one more weighing in order to establish which pile.
Wasn't it supposed to be a method that works everytime and not just if you randomly choose the correct piles/coins?
Wasn't it supposed to be a method that works everytime and not just if you randomly choose the correct piles/coins?
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Meanwhile a tried another approach on this dilemma. The result wasn't fully satisfactionary but it will give you an 83% chance of finding the wrong penny in 3 weighs. With 4 i could give 100%.
Step 1.
Divide the 12 pennys into two groups of 6 each, A and B.
Section A1
Step 2.
Take group A and divide it into two groups of 3. Weigh those groups against eachot
Step 1.
Divide the 12 pennys into two groups of 6 each, A and B.
Section A1
Step 2.
Take group A and divide it into two groups of 3. Weigh those groups against eachot
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Good try, Hoogard.But I want a foolproof result. You should find out in 3 weighings which penny is bad and if it is lighter or heavier - with a 100% result
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