Piggy Banks
Three nickels and three dimes are distributed among three piggy banks such that each holds two coins:
one bank is labeled 10 cents
one bank is labeled 15 cents
one bank is labeled 20 cents
The banks are mislabeled according to the coin distribution ... is it possible to accurately re-label the banks by shaking only one of them until a single coin falls out? If so, explain!
one bank is labeled 10 cents
one bank is labeled 15 cents
one bank is labeled 20 cents
The banks are mislabeled according to the coin distribution ... is it possible to accurately re-label the banks by shaking only one of them until a single coin falls out? If so, explain!
Labels: logic
posted by Zaux at
6:06 AM
8 Comments:
Good one Zaux. I know it can be done, otherwise I doubt that you'd pose the problem. I'll post a non-trick reason soon.
The 15c labelled pig must really contain 5c+5c or 10c+10c. So shake a coin out of it. Two possibilities only:
1. If 5c comes out then it is actually the 10c pig. We cannot simply swap the 10c and 15c labels as that would leave the 20c pig with the wrong label. So we must move the 10c label onto the 15c labelled pig, the 15c label to the 20c labelled pig, and the 20c label to the originally labelled 10c pig.
2. If 10c comes out then it is actually the 20c pig. We cannot simply swap the 20c and 15c labels as that would leave the 10c pig with the wrong label. So we must move the 20c label onto the 15c labelled pig, the 15c label to the 10c labelled pig, and the 10c label to the originally labelled 20c pig.
i.e. we move the labels one position in the direction that puts the correct label on the originally labelled 15c pig.
10c=2 nickles
15c=1 nickle 1 dime
20c=2 dimes
since they are all mislabelled
10 cent label may contain-15c, or 20 c
15 cent label may contain 10c or 20c
20c label may contain- 15c, or 10c
shaking nickle or dime out of 15c label conclusively tells you its contents (shaking other is ambiguos). By knowing contents of this bank you know the others
e.g.
nickle falls out
15c label= 10 c
20 c label can only be 10c or 15c but we just eliminated 10c so it must contain 15c
10c label must contain 15c as it is the only remaining bank, and it is a valid solution
other possibility:
dime falls out
15c label= 20 c
10 c label can only be 20c or 15c but we just eliminated 20c so it must contain 15c
20c label must contain 10c as it is the only remaining bank, and it is a valid solution
Cam
Hi Cam, you seem to be hedging your bets: "...it is 'a' valid solution". The solution is the only solution. :)
This post has been removed by the author.
It should be pointed out that the piggy banks can only be accurately re-labelled with certainty by shaking a coin out of the one labelled "15c", as Cam has shown.
However, they can still be accurately re-labelled if a nickel falls out of the "10c" bank or if a dime falls out of the "20c" bank, but not otherwise.
So, the answer to the question as posed is "yes" if you are allowed to choose the piggy bank, then you would obviously choose the "15c" one. If you are not allowed to choose which bank, then the answer is "maybe"!
I've just noticed that there's only two ways to completely mislabel the piggy banks. The 10c bank would have had to be labelled with 15c or 20c. Having done that, you have no choice about labelling the other two. So you have a 50-50 chance of getting it right by rotating the labels round (with your eyes closed).
Chris,
"and it is a valid solution" was referring to the fact that the value matched one of the two possibilities for the piggy bank in question (i.e. an indication that something didn't go very wrong in my method), not an indication that other valid solutions existed.
The contents of the bank must, of course, be unique in order to obtain the desired result.
I can see how my choice of words may have been confusing. English is my first language, believe it or not, and I still struggle with it.
Cam
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