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Find a stone which is not the lightest one

Which one is the lightest marble?Using a balance scale the minimum number of times, find the median weight personA balance with three pans, detecting the lightest pan (find the one lighter ball)A balance with three pans, detecting the lightest pan (find the one heavier ball)A balance with three pans, detecting the lightest pan (find the two heavier balls)A balance with three pans, detecting the lightest pan (find the one lighter/heavier ball, for a given number of balls)Find 2 heavy coins among 27 with a 3-pan balanceMinimum number of tries to find the balance!Lots of Gold Stacks and a Balance ScaleFive balls weighing

8

$begingroup$

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.

weighing

share|improve this question

edited 1 hour ago

Gilles

3,42531837

asked 10 hours ago

podloga123podloga123

411

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please give credit to the author of the puzzle.
  $endgroup$
  – Gilles
  1 hour ago
  

  
  
  
  

add a comment | 

8

$begingroup$

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.

weighing

share|improve this question

edited 1 hour ago

Gilles

3,42531837

asked 10 hours ago

podloga123podloga123

411

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please give credit to the author of the puzzle.
  $endgroup$
  – Gilles
  1 hour ago
  

  
  
  
  

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.

weighing

share|improve this question

edited 1 hour ago

Gilles

3,42531837

asked 10 hours ago

podloga123podloga123

411

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.

$endgroup$

share|improve this question

edited 1 hour ago

Gilles

3,42531837

asked 10 hours ago

podloga123podloga123

411

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.

share|improve this question

edited 1 hour ago

Gilles

3,42531837

asked 10 hours ago

podloga123podloga123

411

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 1 hour ago

Gilles

3,42531837

edited 1 hour ago

Gilles

3,42531837

asked 10 hours ago

podloga123podloga123

411

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 10 hours ago

podloga123podloga123

411

2 Answers
2

active

oldest

votes

8

$begingroup$

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.

share|improve this answer

answered 9 hours ago

HermesHermes

4407

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Doh!! You got in a little before me. :) Nice answer; have an upvote.
  $endgroup$
  – Brandon_J
  9 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
  9 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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!

share|improve this answer

answered 9 hours ago

Brandon_JBrandon_J

3,883447

$endgroup$

add a comment | 

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8

$begingroup$

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.

share|improve this answer

answered 9 hours ago

HermesHermes

4407

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Doh!! You got in a little before me. :) Nice answer; have an upvote.
  $endgroup$
  – Brandon_J
  9 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
  9 hours ago
  

  
  
  
  

add a comment | 

8

$begingroup$

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.

share|improve this answer

answered 9 hours ago

HermesHermes

4407

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Doh!! You got in a little before me. :) Nice answer; have an upvote.
  $endgroup$
  – Brandon_J
  9 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
  9 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.

share|improve this answer

answered 9 hours ago

HermesHermes

4407

$endgroup$

share|improve this answer

answered 9 hours ago

HermesHermes

4407

answered 9 hours ago

HermesHermes

4407

answered 9 hours ago

HermesHermes

4407

3

$begingroup$

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!

share|improve this answer

answered 9 hours ago

Brandon_JBrandon_J

3,883447

$endgroup$

add a comment | 

3

$begingroup$

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!

share|improve this answer

answered 9 hours ago

Brandon_JBrandon_J

3,883447

$endgroup$

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!

share|improve this answer

answered 9 hours ago

Brandon_JBrandon_J

3,883447

$endgroup$

share|improve this answer

answered 9 hours ago

Brandon_JBrandon_J

3,883447

answered 9 hours ago

Brandon_JBrandon_J

3,883447

answered 9 hours ago

Brandon_JBrandon_J

3,883447