The weakest link Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraTrapped in my CellarFinding the number of ways of crossing a riverThirty genuine and seventy fake coinsEvaporating coins212 weights of 1 gramMinimum number of tries to find the balance!Lots of Gold Stacks and a Balance Scale30 fake coins out of 99 coins1 Fake Coin among N Amount of coinsWhich is heavier, the heaviest of the light or the lightest of the heavy?
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The weakest link
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraTrapped in my CellarFinding the number of ways of crossing a riverThirty genuine and seventy fake coinsEvaporating coins212 weights of 1 gramMinimum number of tries to find the balance!Lots of Gold Stacks and a Balance Scale30 fake coins out of 99 coins1 Fake Coin among N Amount of coinsWhich is heavier, the heaviest of the light or the lightest of the heavy?
$begingroup$
I've gotten this riddle and have been struggling with solving it:
Suppose you have 20 stones which have different weights [0 ... n] . You have no way of measuring the weight of any stone individually, instead you can only measure them by putting 10 on each side of a balance. After doing this you see which "group" of stones is heavier. You can do this only 10 times in total. You can also switch the stones from either side to the other as much as you want between the measurements.
Your task is to find a stone for which you can be certain is not the lightest of them all.
All the stones are marked, so you always know which is which. Their appearance or anything similar does not help you with to determine their weight, the only way you can do it is by putting it on the balance.
mathematics
edited 2 hours ago
Brandon_J
3,873447
asked 3 hours ago
podloga123podloga123
361
New contributor
podloga123 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 |
$begingroup$
I've gotten this riddle and have been struggling with solving it:
Suppose you have 20 stones which have different weights [0 ... n] . You have no way of measuring the weight of any stone individually, instead you can only measure them by putting 10 on each side of a balance. After doing this you see which "group" of stones is heavier. You can do this only 10 times in total. You can also switch the stones from either side to the other as much as you want between the measurements.
Your task is to find a stone for which you can be certain is not the lightest of them all.
All the stones are marked, so you always know which is which. Their appearance or anything similar does not help you with to determine their weight, the only way you can do it is by putting it on the balance.
mathematics
edited 2 hours ago
Brandon_J
3,873447
asked 3 hours ago
podloga123podloga123
361
New contributor
podloga123 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 |
$begingroup$
I've gotten this riddle and have been struggling with solving it:
Suppose you have 20 stones which have different weights [0 ... n] . You have no way of measuring the weight of any stone individually, instead you can only measure them by putting 10 on each side of a balance. After doing this you see which "group" of stones is heavier. You can do this only 10 times in total. You can also switch the stones from either side to the other as much as you want between the measurements.
Your task is to find a stone for which you can be certain is not the lightest of them all.
All the stones are marked, so you always know which is which. Their appearance or anything similar does not help you with to determine their weight, the only way you can do it is by putting it on the balance.
mathematics
edited 2 hours ago
Brandon_J
3,873447
asked 3 hours ago
podloga123podloga123
361
New contributor
podloga123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I've gotten this riddle and have been struggling with solving it:
Suppose you have 20 stones which have different weights [0 ... n] . You have no way of measuring the weight of any stone individually, instead you can only measure them by putting 10 on each side of a balance. After doing this you see which "group" of stones is heavier. You can do this only 10 times in total. You can also switch the stones from either side to the other as much as you want between the measurements.
Your task is to find a stone for which you can be certain is not the lightest of them all.
All the stones are marked, so you always know which is which. Their appearance or anything similar does not help you with to determine their weight, the only way you can do it is by putting it on the balance.
mathematics
mathematics
edited 2 hours ago
Brandon_J
3,873447
asked 3 hours ago
podloga123podloga123
361
New contributor
podloga123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Brandon_J
3,873447
asked 3 hours ago
podloga123podloga123
361
New contributor
podloga123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Brandon_J
3,873447
edited 2 hours ago
Brandon_J
3,873447
edited 2 hours ago
Brandon_J
3,873447
3,873447
asked 3 hours ago
podloga123podloga123
361
New contributor
podloga123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
podloga123podloga123
361
asked 3 hours ago
podloga123podloga123
361
361
New contributor
podloga123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
podloga123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
podloga123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Measure 10 and 10.
In each subsequent weighing you change one of the originals of one side for one of the originals of the other side.
At some point, the balance turns around.
The stone that you have put on the side that has come down is not the lightest, because the other is lighter than that.
If at the tenth weighing the sides have not been turned around, the only original stone left on the side that is below is not the lightest.
answered 2 hours ago
HermesHermes
4207
$endgroup$
$begingroup$
Doh!! You got in a little before me. :) Nice answer; have an upvote.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
2 hours ago
add a comment |
$begingroup$
Here's how to do it:
Step one:
make a measurement, and take note of which side is heavier.
Then,
Switch two stones. If the scale tips differently than before, then the stone that you switched to the side that tipped is guaranteed to not be the lightest stone.
If the scale does not tip differently, then
switch again.
Eventually, the scale will tip differently. If it does not do so after 10 trials,
then the remaining stone on the tipped-to side that has not yet been switched is guaranteed to not be the lightest stone.
and you have your answer!
answered 2 hours ago
Brandon_JBrandon_J
3,873447
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Measure 10 and 10.
In each subsequent weighing you change one of the originals of one side for one of the originals of the other side.
At some point, the balance turns around.
The stone that you have put on the side that has come down is not the lightest, because the other is lighter than that.
If at the tenth weighing the sides have not been turned around, the only original stone left on the side that is below is not the lightest.
answered 2 hours ago
HermesHermes
4207
$endgroup$
$begingroup$
Doh!! You got in a little before me. :) Nice answer; have an upvote.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
2 hours ago
add a comment |
$begingroup$
Measure 10 and 10.
In each subsequent weighing you change one of the originals of one side for one of the originals of the other side.
At some point, the balance turns around.
The stone that you have put on the side that has come down is not the lightest, because the other is lighter than that.
If at the tenth weighing the sides have not been turned around, the only original stone left on the side that is below is not the lightest.
answered 2 hours ago
HermesHermes
4207
$endgroup$
$begingroup$
Doh!! You got in a little before me. :) Nice answer; have an upvote.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
2 hours ago
add a comment |
$begingroup$
Measure 10 and 10.
In each subsequent weighing you change one of the originals of one side for one of the originals of the other side.
At some point, the balance turns around.
The stone that you have put on the side that has come down is not the lightest, because the other is lighter than that.
If at the tenth weighing the sides have not been turned around, the only original stone left on the side that is below is not the lightest.
answered 2 hours ago
HermesHermes
4207
$endgroup$
Measure 10 and 10.
In each subsequent weighing you change one of the originals of one side for one of the originals of the other side.
At some point, the balance turns around.
The stone that you have put on the side that has come down is not the lightest, because the other is lighter than that.
If at the tenth weighing the sides have not been turned around, the only original stone left on the side that is below is not the lightest.
answered 2 hours ago
HermesHermes
4207
answered 2 hours ago
HermesHermes
4207
answered 2 hours ago
HermesHermes
4207
answered 2 hours ago
HermesHermes
4207
4207
$begingroup$
Doh!! You got in a little before me. :) Nice answer; have an upvote.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
2 hours ago
add a comment |
$begingroup$
Doh!! You got in a little before me. :) Nice answer; have an upvote.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
2 hours ago
$begingroup$
Doh!! You got in a little before me. :) Nice answer; have an upvote.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
Doh!! You got in a little before me. :) Nice answer; have an upvote.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
2 hours ago
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
2 hours ago
add a comment |
$begingroup$
Here's how to do it:
Step one:
make a measurement, and take note of which side is heavier.
Then,
Switch two stones. If the scale tips differently than before, then the stone that you switched to the side that tipped is guaranteed to not be the lightest stone.
If the scale does not tip differently, then
switch again.
Eventually, the scale will tip differently. If it does not do so after 10 trials,
then the remaining stone on the tipped-to side that has not yet been switched is guaranteed to not be the lightest stone.
and you have your answer!
answered 2 hours ago
Brandon_JBrandon_J
3,873447
$endgroup$
add a comment |
$begingroup$
Here's how to do it:
Step one:
make a measurement, and take note of which side is heavier.
Then,
Switch two stones. If the scale tips differently than before, then the stone that you switched to the side that tipped is guaranteed to not be the lightest stone.
If the scale does not tip differently, then
switch again.
Eventually, the scale will tip differently. If it does not do so after 10 trials,
then the remaining stone on the tipped-to side that has not yet been switched is guaranteed to not be the lightest stone.
and you have your answer!
answered 2 hours ago
Brandon_JBrandon_J
3,873447
$endgroup$
add a comment |
$begingroup$
Here's how to do it:
Step one:
make a measurement, and take note of which side is heavier.
Then,
Switch two stones. If the scale tips differently than before, then the stone that you switched to the side that tipped is guaranteed to not be the lightest stone.
If the scale does not tip differently, then
switch again.
Eventually, the scale will tip differently. If it does not do so after 10 trials,
then the remaining stone on the tipped-to side that has not yet been switched is guaranteed to not be the lightest stone.
and you have your answer!
answered 2 hours ago
Brandon_JBrandon_J
3,873447
$endgroup$
Here's how to do it:
Step one:
make a measurement, and take note of which side is heavier.
Then,
Switch two stones. If the scale tips differently than before, then the stone that you switched to the side that tipped is guaranteed to not be the lightest stone.
If the scale does not tip differently, then
switch again.
Eventually, the scale will tip differently. If it does not do so after 10 trials,
then the remaining stone on the tipped-to side that has not yet been switched is guaranteed to not be the lightest stone.
and you have your answer!
answered 2 hours ago
Brandon_JBrandon_J
3,873447
answered 2 hours ago
Brandon_JBrandon_J
3,873447
answered 2 hours ago
Brandon_JBrandon_J
3,873447
answered 2 hours ago
Brandon_JBrandon_J
3,873447
3,873447
add a comment |
add a comment |
podloga123 is a new contributor. Be nice, and check out our Code of Conduct.
podloga123 is a new contributor. Be nice, and check out our Code of Conduct.
podloga123 is a new contributor. Be nice, and check out our Code of Conduct.
podloga123 is a new contributor. Be nice, and check out our Code of Conduct.
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