Breaking Balance (Part C-minor: easy version!)
$begingroup$
Identical to Breaking Balance (Part C), except with two weighings.
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 in the previous question), and two 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?
The previous question comes with a Warning: I do not believe there is a "cute" solution and it took me weeks to solve it... this isn't for everybody. I think the opposite is true here, I think the answer is cute, not trivial, but not too hard.
I request @DarkThunder, OP of the previous question, to delay answering for a while :-)
weighing
asked 14 mins ago
deep thoughtdeep thought
3,2531738
$endgroup$
add a comment |
$begingroup$
Identical to Breaking Balance (Part C), except with two weighings.
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 in the previous question), and two 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?
The previous question comes with a Warning: I do not believe there is a "cute" solution and it took me weeks to solve it... this isn't for everybody. I think the opposite is true here, I think the answer is cute, not trivial, but not too hard.
I request @DarkThunder, OP of the previous question, to delay answering for a while :-)
weighing
asked 14 mins ago
deep thoughtdeep thought
3,2531738
$endgroup$
add a comment |
$begingroup$
Identical to Breaking Balance (Part C), except with two weighings.
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 in the previous question), and two 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?
The previous question comes with a Warning: I do not believe there is a "cute" solution and it took me weeks to solve it... this isn't for everybody. I think the opposite is true here, I think the answer is cute, not trivial, but not too hard.
I request @DarkThunder, OP of the previous question, to delay answering for a while :-)
weighing
asked 14 mins ago
deep thoughtdeep thought
3,2531738
$endgroup$
Identical to Breaking Balance (Part C), except with two weighings.
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 in the previous question), and two 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?
The previous question comes with a Warning: I do not believe there is a "cute" solution and it took me weeks to solve it... this isn't for everybody. I think the opposite is true here, I think the answer is cute, not trivial, but not too hard.
I request @DarkThunder, OP of the previous question, to delay answering for a while :-)
weighing
weighing
asked 14 mins ago
deep thoughtdeep thought
3,2531738
asked 14 mins ago
deep thoughtdeep thought
3,2531738
asked 14 mins ago
deep thoughtdeep thought
3,2531738
asked 14 mins ago
deep thoughtdeep thought
3,2531738
asked 14 mins ago
deep thoughtdeep thought
3,2531738
3,2531738
add a comment |
add a comment |
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