posted
So, my sister called me again with another puzzle. She does this every time one of her kids has a homework problem that neither they nor she nor her husband can solve.
So I had a solution for this one right away, but it required an assumption, and she said... well, here's the puzzle.
You have 10 stacks of coins. One stack is of counterfeits and the rest are genuine. You know how much a genuine coin weighs, and you know that a counterfeit coin weighs .5 grams more than a genuine one.
How can you determine which stack is of counterfeits with only one weighing?
So I asked her if we know how many coins are in the stacks, and she said "no". Which is a problem. Because there's an obvious solution, but it requires there to be a certain minimum number of coins there. For example, if each stack has 5 coins in it, you're SOL. But she said that they weren't even told that the stacks all have the same number of coins in them, let alone how many that is.
Did they leave something out of the problem? Should they have said that there were 10 stacks of 10 coins each? That'd be easy to solve. Or am I just missing something?
Posts: 12266 | Registered: Jul 2005
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posted
I thought the same thing as you. The obvious way to do it, if you have enough coins, is to do 10 from one stack, 9 from another, etc. and the solution is quite easy. Too bad they didn't say how many coins there are.
Posts: 2705 | Registered: Sep 2006
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posted
The original (which I knew I had seen before, and Google helped me find) is from Martin Gardner's 1994 book of puzzles, and it does have 10 coins in each stack. (Correction: it was earlier in a 1988 book of his, but I never read that one.)
They don't actually all have to be the same, but yeah, there needs to be at least one stack with 10 in it, for example.
I'm guessing there should be more to it.
Posts: 32919 | Registered: Mar 2003
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posted
I think it is safe to assume each stack has some substantial number of coins in it, in which case the answer is, as you note, obvious.
Posts: 15770 | Registered: Dec 2001
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posted
Off the top of my head... Couldn't you just take the total weight of each stack, divide by the number of coins then deduce the inconstant pile?
Posts: 1355 | Registered: Jul 2006
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Another question... The wording doesn't specify one weighing total or one weighing of each stack does it? I might be missing something from my reading, though.
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posted
Well, "one weighing" isn't the clearest phrasing, but I think the intended meaning is the simpler way to interpret it.
Posts: 32919 | Registered: Mar 2003
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quote:Originally posted by rivka: The original (which I knew I had seen before, and Google helped me find) is from Martin Gardner's 1994 book of puzzles, and it does have 10 coins in each stack. (Correction: it was earlier in a 1988 book of his, but I never read that one.)
I first saw it in a Games Magazine book of puzzles that my family had back in the early 80s.
Posts: 16551 | Registered: Feb 2003
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posted
I’ve heard a version of this puzzle where the solution was based on the definition of “one weighing”. Namely, the puzzle was so worded that only putting the same coin on the scales repeatedly was “more weightings”. Therefore, while there was at least one coin present, the others could be added and then removed in some order (never using one coin twice) to determine the properties of all stacks, all being considered ONE weighting.
Anyway, that’s twisting the definition so much that I suspect the intention here was to have at least 10 coins in each stack from the beginning, for which case the solution was given above.
quote:Originally posted by mr_porteiro_head: My book also had the bolts puzzle that's on the cover of your book.
Upon further consideration, I'm betting that both originally appeared in his Scientific American column, and from there made it into all of the books mentioned.
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