You have 12 coins, of which one is counterfeit and has a different weight (you don’t know whether it’s lighter or heavier). You must determine which one is counterfeit by using a balance scale 3 times.
(Warning: It IS complicated. If you don’t have the head for it, don’t bother.)
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Divide into 4 parts of three
Weigh a and b
Then weigh c and d and see which is the odd one out
Then from those 3 weigh 2 if they are even u know it’s the one not on the scale, otherwise it’s the one that’s either heavier or lighter depending on what u know from the first 2 weigh ins if he odd one is heavier or lighter
OK. Same riddle as the chessboard, but there are only 2 fields.
evil king hides key to your freedom under one of two fields. On the fields there are coins he can arrange randomly, heads or tails, he shows you the field where the key is hidden, you must turn one coin, your friend has to know with one guess where the key is, otherwise you will be hanged...
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Make up with your friend that if he finds one heads one tails, the heads one is correct. If both heads it’s on the left. If both tails it’s on the right.
So if king leaves both heads or both tails, you flip one to make the heads be the correct one. If king leaves one of each, you see if you need to point right or left and flip one accordingly.
I don’t see at all how one would answer with 64 coins though. But then I didn’t check out that answer in the link yet.
Divide into 4 parts of three
Weigh a and b
Then weigh c and d and see which is the oddone out
Then from those 3 weigh 2 if they are even u know it’s the one not on the scale, otherwise it’s the one that’s either heavier or lighter depending on what u know from the first 2 weigh ins if he odd one is heavier or lighter
When they don’t balance, you can’t tell if there’s a heavier coin on side A, or a lighter coin on side B, so you’re still left with 6 possible counterfeit coins......nice try though!
You have 12 coins, of which one is counterfeit and has a different weight (you don’t know whether it’s lighter or heavier). You must determine which one is counterfeit by using a balance scale 3 times.
(Warning: It IS complicated. If you don’t have the head for it, don’t bother.)
Ok I think I got this.
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#1 you weigh 4&4. I’ll go with the hardest scenario and say that the counterfeit one is here and we now don’t know whether it’s one of the 4 weighing more or 4 weighing less. (If the scale was even then it’s easy to finish off. I’ll only explain the hardest scenario)
#2 on side A- you weigh 3 of the previous light ones plus 2 heavy ones. On side B- you weigh 1 light one plus the 4 good ones from #1. (If it’s even you are left with only 2 heavy ones to weigh at the 3rd weighing and see which is heavier)
*In the event side A is lighter you do for #3 2 of the light ones and see whether one is lighter, or the third is the counterfeit.
*in the event side A is heavier, you stay with a shaila between the 2 heavy ones. You weigh those on round three. If one is heavier then that’s the counterfeit, if they’re even then the light from side B is the counterfeit.
Ok, I’m dizzy but I’m glad I got it!
Last edited by ExtraCredit on Tue, Oct 06 2020, 8:13 pm; edited 1 time in total
Divide into 4 parts of three
Weigh a and b
Then weigh c and d and see which is the odd one out
Then from those 3 weigh 2 if they are even u know it’s the one not on the scale, otherwise it’s the one that’s either heavier or lighter depending on what u know from the first 2 weigh ins if he odd one is heavier or lighter
This is incorrect because
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Since you don’t know whether the counterfeit is heavier or lighter, how do you know between C and D ‘which is the odd one out?’
When they don’t balance, you can’t tell if there’s a heavier coin on side A, or a lighter coin on side B, so you’re still left with 6 possible counterfeit coins......nice try though!
Oh, I only saw your reply after I commented on her answer....
#1 you weigh 4&4. I’ll go with the hardest scenario and say that the counterfeit one is here and we now don’t know whether it’s one of the 4 weighing more or 4 weighing less. (If the scale was even then it’s easy to finish off. I’ll only explain the hardest scenario)
#2 on side A- you weigh 3 of the previous light ones plus 2 heavy ones. On side B- you weigh 1 light one plus the 4 good ones from #1. (If it’s even you are left with only 2 heavy ones to weigh at the 3rd weighing and see which is heavier)
*In the event side A is lighter you do for #3 2 of the light ones and see whether one is lighter, or the third is the counterfeit.
*in the event side A is heavier, you stay with a shaila between the 2 heavy ones. You weigh those on round three. If one is heavier then that’s the counterfeit, if they’re even then the light from side B is the counterfeit.
Ok, I’m dizzy but I’m glad I got it!
Not sure if I’d be able to follow along w/o knowing the answer.....but you got it! My solution is slightly simpler, but same reasoning.
Not sure if I’d be able to follow along w/o knowing the answer.....but you got it! My solution is slightly simpler, but same reasoning.
I was gonna write that this megilla is strictly for Tamari because it will just confuse others. I’d love to see your simpler version.
Btw what I love about this riddle is no calculator or spread sheet will help you. Plain old fashioned cheshbonos.
Make up with your friend that if he finds one heads one tails, the heads one is correct. If both heads it’s on the left. If both tails it’s on the right.
So if king leaves both heads or both tails, you flip one to make the heads be the correct one. If king leaves one of each, you see if you need to point right or left and flip one accordingly.
I don’t see at all how one would answer with 64 coins though. But then I didn’t check out that answer in the link yet.
ETA yay my spoiler tags worked
Your answer would work. It's somewhat similar to what I would propose, just does not follow the same mathematical paradigm.
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You could consider binary numbers: the first place is zero, nothing happens when you flip this coin, the second place tells you where the key is: 0 Heads) for field 0 and 1 (tails) for field 1 (I.e. 0 (head) for left, 1 (tails) for right)
So if the right coin is in the right position, you flip the left coin, which has no significance... If not you flip the right coin...
This is a method you could expand to bigger fields with an astute arrangement of how you count your figures...
I was gonna write that this megilla is strictly for Tamari because it will just confuse others. I’d love to see your simpler version.
Btw what I love about this riddle is no calculator or spread sheet will help you. Plain old fashioned cheshbonos.
A megilla indeed.....Hopefully it's clear enough for all to follow...Here goes...
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Let's label the coins 1-12 for explanation purposes. Divide them into 3.
Weigh #1: Side A- #1,2,3,4 and Side B- #5,6,7,8. We'll take the simplest scenario first. If these two balance, we know it's among the last four. Then;
Weigh #2 Side A- #1,2 and Side B- #9,10. This will tell you whether it's one of 9,10 and also if it's heavier or lighter. So if Side B is heavier here, obviously then weigh 9 against 10 to see which one's heavier and vice versa. If the second weighing is balanced, then proceed by weighing #1 against #11, which will finalize whether it's 11 or 12.
On to the more complicated scenario, where the first weighing doesn't balance. Here you have to take into account which side weighed down. So suppose side A (1234) weighed down. That means, either one of 1234 might be a heavy counterfeit coin OR one of 5678 is a lighter coin. So,
Weigh #2: Side A- 1,2,3,5 and Side B- 4,9,10,11. This weighing will leave you with 3 possible outcomes. Option One: If Side A is heavier, then it must be that one of 1,2,3 is heavier since we didn't put any of the possible lightweight coins on side B this time. Option Two: If Side B weighs down, then it must be either 5 is a lightweight coin or that 4 is a heavy coin, or Option Three: If they balance, then we're left with one of 6,7,8 being a lightweight coin. Then proceed, based on these 3 outcomes, respectively:
Weigh #3 #1 against #2, see which one's heavier, or if it's even, then it's #3.
OR, #4 against #1 to see whether it's 4, if not it's 5.
OR, #6 against #7 to see which one's lighter, or if even, then it's #8.
A megilla indeed.....Hopefully it's clear enough for all to follow...Here goes...
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Let's label the coins 1-12 for explanation purposes. Divide them into 3.
Weigh #1: Side A- #1,2,3,4 and Side B- #5,6,7,8. We'll take the simplest scenario first. If these two balance, we know it's among the last four. Then;
Weigh #2 Side A- #1,2 and Side B- #9,10. This will tell you whether it's one of 9,10 and also if it's heavier or lighter. So if Side B is heavier here, obviously then weigh 9 against 10 to see which one's heavier and vice versa. If the second weighing is balanced, then proceed by weighing #1 against #11, which will finalize whether it's 11 or 12.
On to the more complicated scenario, where the first weighing doesn't balance. Here you have to take into account which side weighed down. So suppose side A (1234) weighed down. That means, either one of 1234 might be a heavy counterfeit coin OR one of 5678 is a lighter coin. So,
Weigh #2: Side A- 1,2,3,5 and Side B- 4,9,10,11. This weighing will leave you with 3 possible outcomes. Option One: If Side A is heavier, then it must be that one of 1,2,3 is heavier since we didn't put any of the possible lightweight coins on side B this time. Option Two: If Side B weighs down, then it must be either 5 is a lightweight coin or that 4 is a heavy coin, or Option Three: If they balance, then we're left with one of 6,7,8 being a lightweight coin. Then proceed, based on these 3 outcomes, respectively:
Weigh #3 #1 against #2, see which one's heavier, or if it's even, then it's #3.
OR, #4 against #1 to see whether it's 4, if not it's 5.
OR, #6 against #7 to see which one's lighter, or if even, then it's #8.
Lol. I think this megilla beats mine
I think that here too, if someone didn’t figure it out themselves they’ll have a hard time following.
Your answer would work. It's somewhat similar to what I would propose, just does not follow the same mathematical paradigm.
Hidden:
You could consider binary numbers: the first place is zero, nothing happens when you flip this coin, the second place tells you where the key is: 0 Heads) for field 0 and 1 (tails) for field 1 (I.e. 0 (head) for left, 1 (tails) for right)
So if the right coin is in the right position, you flip the left coin, which has no significance... If not you flip the right coin...
This is a method you could expand to bigger fields with an astute arrangement of how you count your figures...
You found in your great-grandmother’s attic five short chains each made of four gold links. You'd love to combine all of them to have a nice necklace of 20 links. So you bring it to a jeweler, who tells you the cost of making the necklace will be $10 for each gold link that she has to break and then reseal. How much will it cost you?
You found in your great-grandmother’s attic five short chains each made of four gold links. You'd love to combine all of them to have a nice necklace of 20 links. So you bring it to a jeweler, who tells you the cost of making the necklace will be $10 for each gold link that she has to break and then reseal. How much will it cost you?
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40$. I open each link from one chain and use them to join the remaining four short chains together.
although, honestly, I would invest a few extra dollars for a nice clip to open and close the chain...
40$. I open each link from one chain and use them to join the remaining four short chains together.
although, honestly, I would invest a few extra dollars for a nice clip to open and close the chain...
Yup!!
It's antique, though, considering where you found it. Not sure you'd find a clip that goes well with it
You found in your great-grandmother’s attic five short chains each made of four gold links. You'd love to combine all of them to have a nice necklace of 20 links. So you bring it to a jeweler, who tells you the cost of making the necklace will be $10 for each gold link that she has to break and then reseal. How much will it cost you?
40$. I open each link from one chain and use them to join the remaining four short chains together.
although, honestly, I would invest a few extra dollars for a nice clip to open and close the chain...
Ohhhhh brilliant!
But (always a but if I couldn’t figure it out ) it should’ve been worded: “What would be the cheapest way to do it”. Because technically I answered the question correctly too.
