Breaking Balance (Part C)
$begingroup$
For a starting number of otherwise identical coins there are among them TWO IDENTICAL counterfeit coins which are either heavier or lighter than the rest. Using a three-pan balance (described in detail below), and three weighings, what is the most number of starting coins (including the counterfeits) you can have while being able to always determine the counterfeits and whether they are both heavier or both lighter?
This should not be confused with the balance detailed in Find 2 heavy coins among 27 with a 3-pan balance as it has a bunch of different rules.
So there isn't confusion, I want to explicitly state how this thing works. It has three equal-length arms supporting three pans (120deg apart), and is balancing on a point. It works like other balances in that the center-of-gravity is below the balance point and thus will tilt to counteract imbalance in the pans.
There are exactly 7 outcomes the balance will give you. The first is the "balanced" outcome, where all pans stay at the same height. The other six are variations on the following:
One "lone" pan will go up or down, and the other two pans will be equal to each other but move opposite the lone pan. Given that there are TWO IDENTICAL counterfeit coins, this may be less informative than you might otherwise hope for. My shorthand for the 7 outcomes is as follows:
- A = B = C
- A > B = C
- A < B = C
- B > A = C
- B < A = C
- C > A = B
- C < A = B
For sake of convention, I consider the coins to be numbered 1 through whatever, and the pans labeled A, B and C. The 4th location, D, is used to denote a coin not being the scale.
Warning: I do not believe there is a "cute" solution and it took me weeks to solve it... and even then it isn't quite a "proof". I really enjoyed going through it, but seriously... this isn't for everybody. And for those that do get it... there's a bunch of much harder variants awaiting you!
weighing physics
asked 8 mins ago
Dark ThunderDark Thunder
436
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add a comment |
$begingroup$
For a starting number of otherwise identical coins there are among them TWO IDENTICAL counterfeit coins which are either heavier or lighter than the rest. Using a three-pan balance (described in detail below), and three weighings, what is the most number of starting coins (including the counterfeits) you can have while being able to always determine the counterfeits and whether they are both heavier or both lighter?
This should not be confused with the balance detailed in Find 2 heavy coins among 27 with a 3-pan balance as it has a bunch of different rules.
So there isn't confusion, I want to explicitly state how this thing works. It has three equal-length arms supporting three pans (120deg apart), and is balancing on a point. It works like other balances in that the center-of-gravity is below the balance point and thus will tilt to counteract imbalance in the pans.
There are exactly 7 outcomes the balance will give you. The first is the "balanced" outcome, where all pans stay at the same height. The other six are variations on the following:
One "lone" pan will go up or down, and the other two pans will be equal to each other but move opposite the lone pan. Given that there are TWO IDENTICAL counterfeit coins, this may be less informative than you might otherwise hope for. My shorthand for the 7 outcomes is as follows:
- A = B = C
- A > B = C
- A < B = C
- B > A = C
- B < A = C
- C > A = B
- C < A = B
For sake of convention, I consider the coins to be numbered 1 through whatever, and the pans labeled A, B and C. The 4th location, D, is used to denote a coin not being the scale.
Warning: I do not believe there is a "cute" solution and it took me weeks to solve it... and even then it isn't quite a "proof". I really enjoyed going through it, but seriously... this isn't for everybody. And for those that do get it... there's a bunch of much harder variants awaiting you!
weighing physics
asked 8 mins ago
Dark ThunderDark Thunder
436
New contributor
Dark Thunder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
For a starting number of otherwise identical coins there are among them TWO IDENTICAL counterfeit coins which are either heavier or lighter than the rest. Using a three-pan balance (described in detail below), and three weighings, what is the most number of starting coins (including the counterfeits) you can have while being able to always determine the counterfeits and whether they are both heavier or both lighter?
This should not be confused with the balance detailed in Find 2 heavy coins among 27 with a 3-pan balance as it has a bunch of different rules.
So there isn't confusion, I want to explicitly state how this thing works. It has three equal-length arms supporting three pans (120deg apart), and is balancing on a point. It works like other balances in that the center-of-gravity is below the balance point and thus will tilt to counteract imbalance in the pans.
There are exactly 7 outcomes the balance will give you. The first is the "balanced" outcome, where all pans stay at the same height. The other six are variations on the following:
One "lone" pan will go up or down, and the other two pans will be equal to each other but move opposite the lone pan. Given that there are TWO IDENTICAL counterfeit coins, this may be less informative than you might otherwise hope for. My shorthand for the 7 outcomes is as follows:
- A = B = C
- A > B = C
- A < B = C
- B > A = C
- B < A = C
- C > A = B
- C < A = B
For sake of convention, I consider the coins to be numbered 1 through whatever, and the pans labeled A, B and C. The 4th location, D, is used to denote a coin not being the scale.
Warning: I do not believe there is a "cute" solution and it took me weeks to solve it... and even then it isn't quite a "proof". I really enjoyed going through it, but seriously... this isn't for everybody. And for those that do get it... there's a bunch of much harder variants awaiting you!
weighing physics
asked 8 mins ago
Dark ThunderDark Thunder
436
New contributor
Dark Thunder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
For a starting number of otherwise identical coins there are among them TWO IDENTICAL counterfeit coins which are either heavier or lighter than the rest. Using a three-pan balance (described in detail below), and three weighings, what is the most number of starting coins (including the counterfeits) you can have while being able to always determine the counterfeits and whether they are both heavier or both lighter?
This should not be confused with the balance detailed in Find 2 heavy coins among 27 with a 3-pan balance as it has a bunch of different rules.
So there isn't confusion, I want to explicitly state how this thing works. It has three equal-length arms supporting three pans (120deg apart), and is balancing on a point. It works like other balances in that the center-of-gravity is below the balance point and thus will tilt to counteract imbalance in the pans.
There are exactly 7 outcomes the balance will give you. The first is the "balanced" outcome, where all pans stay at the same height. The other six are variations on the following:
One "lone" pan will go up or down, and the other two pans will be equal to each other but move opposite the lone pan. Given that there are TWO IDENTICAL counterfeit coins, this may be less informative than you might otherwise hope for. My shorthand for the 7 outcomes is as follows:
- A = B = C
- A > B = C
- A < B = C
- B > A = C
- B < A = C
- C > A = B
- C < A = B
For sake of convention, I consider the coins to be numbered 1 through whatever, and the pans labeled A, B and C. The 4th location, D, is used to denote a coin not being the scale.
Warning: I do not believe there is a "cute" solution and it took me weeks to solve it... and even then it isn't quite a "proof". I really enjoyed going through it, but seriously... this isn't for everybody. And for those that do get it... there's a bunch of much harder variants awaiting you!
weighing physics
weighing physics
asked 8 mins ago
Dark ThunderDark Thunder
436
New contributor
Dark Thunder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 mins ago
Dark ThunderDark Thunder
436
New contributor
Dark Thunder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 mins ago
Dark ThunderDark Thunder
436
New contributor
Dark Thunder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 mins ago
Dark ThunderDark Thunder
436
asked 8 mins ago
Dark ThunderDark Thunder
436
436
New contributor
Dark Thunder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dark Thunder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dark Thunder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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