More aerobics for your mind. First with the right answer wins.
There are three clay containers. One of them has only pennies from the year 2005 in it, one has only pennies from the year 1975 in it, and the third has an equal number of each. They are labeled "2005," "1975" and "mixed," however, you know that the labels have been switched, so that each container is marked incorrectly. Can you properly label the containers without looking in them, and only pulling one coin out from one container?
Yes.
Pull out one coin from the jar marked "mixed." You know that the correct label for that jar is either 1975 or 2005. If the coin you get is a 1975 penny, then the jar must be the true 1975 jar.
If the jar marked "Mixed" is actually the 1975 jar, then you know the jar marked 2005 is actually the mixed jar (because it can't be 2005 and it can't be 1975). Therefore, the jar marked 1975 is actually the 2005 jar.
Likewise, if the penny you pull from the jar marked "mixed" is a 2005 penny, then you know that the "mixed" jar is actually full of 2005 pennies. From there, you determine that the jar marked 1975 is actually the mixed jar, and thus, that the jar marked 2005 is actually the 1975 jar.
Posted by: Steve | September 20, 2007 at 14:00
You take one coin from the jar labelled mixed. Whatever date is on the coin determines how the jar should be labelled.
The two remaining jars are then the mixed and the alternate year. The jar labelled the alternate year is labelled incorrectly and thus must be the mixed jar.
By process of elimination, the third jar should be labelled the year you did not draw.
Posted by: Nephos | September 20, 2007 at 14:24
1) All the containers are marked incorrectly. Therefore the "mixed" container contains either all 1975 coins or all 2005 coins. Remove the "mixed" label from this container and pull out one coin to see which coins it contains. If you pull out a 1975 coin, remove the "1975" label from whatever container it is on and put it on this container instead. If you pull out a 2005 coin, remove the "2005" label from whatever container it is on and put it on this container instead. Now you have one correctly marked container. Set this container aside.
2) You have two containers remaining - one with a label and one without a label. You have not touched the container with a label, so it must still be marked incorrectly. Take the label from this container and put it on the container without a label. This container is now correctly marked. Set it aside also.
3) Finally, put the "mixed" label on the one container remaining. You have now marked all three containers correctly.
Posted by: Ray Fowler | September 20, 2007 at 15:28
You can accurately relabel the jars without looking at a single penny within. Weigh each jar and then, taking into account the change in metal content of pennies over the years, label them such that the heaviest is the 1975 jar, the middle-weight is the mixed jar, and the lightest is the 2005 jar.
Jim
Posted by: Jim A. | September 20, 2007 at 15:33
I scratched my head on this one until I realized a key statement in the puzzle, "each container is marked incorrectly". Given that all the containers are labeled wrong there are actually only two possible ways the container could be mislabeled and still fulfill that mislabeled constraint.
ConfigA
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1975 contains 2005
2005 contains mixed
mixed contains 1975
ConfigB
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1975 contains mixed
2005 contains 1975
mixed contains 2005
By drawing a coin out of the container labeled mixed we immediately know what that container contains. What remains is to determine what is remaining in the other containers.
When drawing a coin out of a reamining container we might now immediately the identity if the coin either matches the previous drawn coin or the label on its container. If the second container doesn't fulfill that the third will.
Here are the six possible outcomes, 3 each for the two possible Configurations
Scenario A1
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Draw 1975 from container labeled mixed, we know it is 1975.
Draw 1975 from container labeled 1975, we know it is mixed
The remaining container must be 2005.
Scenario A2
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Draw 1975 from container labeled mixed, we know it is 1975.
Draw 2005 from container labeled 1975, it is either 2005 or mixed.
Draw 1975 from container labeled 2005, it must be mixed and previous is 2005.
Scenario A3
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Draw 1975 from container labeled mixed, we know it is 1975.
Draw 2005 from container labeled 1975, it is either 2005 or mixed.
Draw 2005 from container labeled 2005, it must be mixed and previous is 2005.
Scenario B1
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Draw 2005 from container labeled mixed, we know it is 2005
Draw 2005 from container labeled 2005, we know it is mixed
The remaining container must be 1975
Scenario B2
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Draw 2005 from container labeled mixed, we know it is 2005
Draw 1975 from container labeled 2005, it is either either 1975 or mixed
Draw 2005 from container labeled 1975, it must be 2005 and the previous is mixed
Scenario B3
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Draw 2005 from container labeled mixed, we know it is 2005
Draw 1975 from container labeled 2005, it is either either 1975 or mixed
Draw 1975 from container labeled 1975, it must be mixed and the previous is 1975
It is interesting to note that there is a a 2 in 6 chance (33%) chance of being able to identify the containers in only 2 coins!
Posted by: Mark | September 20, 2007 at 16:08
Take the coin from the one marked mixed, which you know must be either 75 or 05.
If it comes up 75, then it is 75, the one marked 75 is 05, and the one marked 05 is mixed.
If it comes up 05, then it is 05, the on marked 75 is mixed, and the one marked 05 is 75.
Posted by: Paul | September 20, 2007 at 16:24
Just pull a coin from the jar labled Mixed. You know that it is not the mixed container so it must be whatever year penny you pulled. Do this again with the other jar and you have your solution.
Posted by: Carl Holmes | September 20, 2007 at 16:33
Fun stuff, Mr. D. Yes, I can.
Call the container mislabeled as 2005, "A".
Call the container mislabeled as 1975, "B".
Call the container mislabeled as mixed, "C".
If there are only three containers, and all are mislabeled, there are only two possible coin draw scenarios.
Scenario 1) If I pull either a 1975 or a 2005 coin out of A, a 2005 coin out of B, and a 1975 coin out of C, then A should be labeled mixed, B should be labeled 2005, and C should be labeled 1975.
Scenario 2) If I pull a 1975 coin out of A, either a 1975 or a 2005 coin out of B, and a 2005 coin out of C, then A should be labeled 1975, B should be labeled mixed, and C should be labeled 2005.
Do I need to show my work? :)
Posted by: Aaron Snell | September 20, 2007 at 17:56
Take a coin out of the jar that is labeled as being both 1975 and 2005 coins
If the coin is a 2005 coin, then you have found the jar containing ONLY 2005 coins (as they are all labeled incorrectly).
Then the jar that is labeled as only 1975 coins is the mixed jar, and the jar labeled 2005 is the 1975 jar
(and vice versa is you pick out a 1975 coin...
Posted by: Alan Grey | September 20, 2007 at 19:10
Sure. You pull one penny from the container incorrectly labeled "mixed." If it's a 2005, you know that the container is actually the 2005 container. The "1975" is actually mixed and the "2005" is actually 1975. If the penny pulled is a 1975, you know the container is actually the 1975, the "1975" is actually 2005 and the "2005" is actually mixed.
Posted by: Jason Clarke | September 20, 2007 at 19:39
Sure. Pull from the 'mixed'... Whatever you get obviously is that year's container. [say 2005]
The container marked (say) '1975' obviously must be the truly-MIXED container. [It can't be what it says it is; nor can it be the '2005' container.]
And the last container would be whatever's left. In this example, it'd be 1975.
Posted by: IndyChristian | September 20, 2007 at 22:15