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March 24th, 2004, 12:51 PM
#646
Registered User
Question 98 - RESOLVED
You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?
Multiple answers possible with this one.
__________
Find possible answers below.
Last edited by a d e p t; March 25th, 2004 at 12:55 PM.
Human beings, who are almost unique in having the ability to learn from the experience of others, are also remarkable for their apparent disinclination to do so. -Douglas Adams
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March 24th, 2004, 01:03 PM
#647
Registered User
Originally Posted by a d e p t
You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?
Multiple answers possible with this one.
10.
"Don't be so humble - you are not that great." - Golda Meir
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March 24th, 2004, 01:11 PM
#648
Registered User
Originally Posted by a d e p t
You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?
Multiple answers possible with this one.
6
If you were lucky enough to weigh each of the 6 coins of one type, you would then know that the 6 left were all of the other wight.
"You are never too smart to act stupid" - Christopher Lloyd in Camp Nowhere
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March 24th, 2004, 01:12 PM
#649
Registered User
Originally Posted by Camaro80z
6
If you were lucky enough to weigh each of the 6 coins of one type, you would then know that the 6 left were all of the other wight.
You could do it in 5 if you were lucky.
"Don't be so humble - you are not that great." - Golda Meir
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March 24th, 2004, 01:15 PM
#650
Registered User
Originally Posted by edball
10.
No, sorry.
And there seems to be something misssing...
..oh yeah - no explanation given!
Human beings, who are almost unique in having the ability to learn from the experience of others, are also remarkable for their apparent disinclination to do so. -Douglas Adams
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March 24th, 2004, 01:18 PM
#651
Registered User
Originally Posted by a d e p t
No, sorry.
And there seems to be something misssing...
..oh yeah - no explanation given!
If I'm wrong, why should I explain it ? This question is making my brain hurt anyway !
"Don't be so humble - you are not that great." - Golda Meir
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March 24th, 2004, 01:41 PM
#652
Registered User
Originally Posted by a d e p t
You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?
Multiple answers possible with this one.
11 assuming the first 2 coins were different and each coin weighed against one of the first 2 went into a different pile until you got to the last coin which would go into the pile with only 5 coins.
"Don't be so humble - you are not that great." - Golda Meir
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March 24th, 2004, 01:57 PM
#653
Registered User
Originally Posted by a d e p t
Yes. But if you can elaborate as the "why" part, that would give you another point.
Steel is denser than air, therefore it would take a higher volume of air to displace the weight of the steel in the boat over the surface of the water when the boat was upright.
"Don't be so humble - you are not that great." - Golda Meir
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March 24th, 2004, 02:47 PM
#654
Registered User
Originally Posted by a d e p t
You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?
Multiple answers possible with this one.
I like Ed's initial answer of 10. You pick one coin at random and place it on one side of the scale. You have 11 coins remaining. After measuring 10 of the coins against the initial coin and separating them based on your results, you will know what the weight of the last coin must be so there's no need to measure it. You will also know which pile the initial coin goes in so no need to weigh that one.
You could get lucky and do it in fewer tries but that sounds like the smallest number of measurements to guarantee they are all sorted.
If it's true that wherever you go, there you are: how come so many people look lost?
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March 24th, 2004, 02:49 PM
#655
The answer is C
Just had to post in a thread that had been going on for so long
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March 24th, 2004, 03:26 PM
#656
Registered User
Hrm, I now think the answer is eight. You place six coins on each side and note the level of each side of the scale. Swap one coin from each side to the other side. Worst case, five measurements each result in the two coins being identical so you now have six piles, each consisting of two identical coins.
Two more weighings of two of those piles against each other will result in one of three scenarios
1-two piles of four, each of which is known to contain heavy or light coins
2-two piles of four, neither of which is known to be heavy or light
3-one pile of four identical coins and two piles of two coins known to be heavy or light
Scenario 1, one more weighing of any weighed two coin pile against either unweighed pile will sort them all out
Scenario 2, same method as scenario 1
Scenario 3, the two unweighed piles must be identical. weigh either of them against either of the two separate weighed two coin piles that are known to be heavy or light.
So worst case with this method is five weighings plus two weighings plus one weighing equals eight.
If it's true that wherever you go, there you are: how come so many people look lost?
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March 24th, 2004, 03:59 PM
#657
Registered User
Originally Posted by a d e p t
You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?
Multiple answers possible with this one.
I say eight also, but I used a different weighing method.
Simple logic can eliminate some of these steps depending on the situation, but this is given worst case scenario.
Step 1: Split coins up into four groups of three, labeled A, B, C and D.
Step 2: Weigh groups A and B (1), this will determine either even or one side is heavy.
Step 3: Weigh 2 coins from group A (2), then weigh one of those and the third from group A (3). This should determine light and heavy for the three coins in group A, using logic on the three weighing results. Three down, 9 to go.
Step 4: Weigh 2 coins from group B (4), the weigh one of those and the third from group B (5). This should determine light and heavy for the three coins in group B, using the same logic as in step 3. 6 down, 6 to go.
Step 5: Weigh group C and group D (6), they will be either even or heavy.
Step 6: Weigh 2 coins from group C (7).
Step 7: At this point you will either weigh the third coin from group C against one of the other two, or you will weigh two coins from group D (8). Anyway it goes, you should be able to determine placement with a little logic.
I have used this method repeatedly and have come up with 5 to 8 weighings, depending on the situation. I'm not sure I've explained this properly, but it's difficult given the various possible situations. Using this method, however, 8 is the maximum number of weighings that guarantees separation.
A bored admin is a very dangerous person...
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March 25th, 2004, 12:53 PM
#658
Registered User
Originally Posted by edball
Steel is denser than air, therefore it would take a higher volume of air to displace the weight of the steel in the boat over the surface of the water when the boat was upright.
Extra point goes to you.
Human beings, who are almost unique in having the ability to learn from the experience of others, are also remarkable for their apparent disinclination to do so. -Douglas Adams
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March 25th, 2004, 12:56 PM
#659
Registered User
Originally Posted by cisco2
Hrm, I now think the answer is eight. You place six coins on each side and note the level of each side of the scale. Swap one coin from each side to the other side. Worst case, five measurements each result in the two coins being identical so you now have six piles, each consisting of two identical coins.
Two more weighings of two of those piles against each other will result in one of three scenarios
1-two piles of four, each of which is known to contain heavy or light coins
2-two piles of four, neither of which is known to be heavy or light
3-one pile of four identical coins and two piles of two coins known to be heavy or light
Scenario 1, one more weighing of any weighed two coin pile against either unweighed pile will sort them all out
Scenario 2, same method as scenario 1
Scenario 3, the two unweighed piles must be identical. weigh either of them against either of the two separate weighed two coin piles that are known to be heavy or light.
So worst case with this method is five weighings plus two weighings plus one weighing equals eight.
3 points.
Human beings, who are almost unique in having the ability to learn from the experience of others, are also remarkable for their apparent disinclination to do so. -Douglas Adams
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March 25th, 2004, 12:56 PM
#660
Registered User
Originally Posted by Deity
I say eight also, but I used a different weighing method.
Simple logic can eliminate some of these steps depending on the situation, but this is given worst case scenario.
Step 1: Split coins up into four groups of three, labeled A, B, C and D.
Step 2: Weigh groups A and B (1), this will determine either even or one side is heavy.
Step 3: Weigh 2 coins from group A (2), then weigh one of those and the third from group A (3). This should determine light and heavy for the three coins in group A, using logic on the three weighing results. Three down, 9 to go.
Step 4: Weigh 2 coins from group B (4), the weigh one of those and the third from group B (5). This should determine light and heavy for the three coins in group B, using the same logic as in step 3. 6 down, 6 to go.
Step 5: Weigh group C and group D (6), they will be either even or heavy.
Step 6: Weigh 2 coins from group C (7).
Step 7: At this point you will either weigh the third coin from group C against one of the other two, or you will weigh two coins from group D (8). Anyway it goes, you should be able to determine placement with a little logic.
I have used this method repeatedly and have come up with 5 to 8 weighings, depending on the situation. I'm not sure I've explained this properly, but it's difficult given the various possible situations. Using this method, however, 8 is the maximum number of weighings that guarantees separation.
3 points here too.
Human beings, who are almost unique in having the ability to learn from the experience of others, are also remarkable for their apparent disinclination to do so. -Douglas Adams
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