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Worth the Weight



You are shown a pile of dimes all of which have one of two distinct weights differing by a small amount not detectable by feel.  Forty eight dimes are separated from this pile and you are told of these forty eight, light ones are a dime a dozen (literally - i.e. 44 heavy dimes and 4 light dimes).  Using a balance scale twice, find seven heavy dimes.


EDIT:  for clarification

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4 light ones/48

when 14(A) = 14(B)

The remaining 20 coins = C

A = 7 (A1) + 1(A2) + 6(A3)

Compare (B+A1) and ( A2 + C )

If .... = .....

=> have 2 light coins each side, in B(1) A1(1) C(2)

=>  A2 +A3 = 7 heavy coins


If .... >….

A,B have 0 light coin ,

Or B have 1 light coin, C have 2, A1 have 0

=> get 7 coins in A1


If ....<….

C have 0 light coins

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Let's take the OP to be a riddle of sorts that implies precisely 4 light coins in the group of 48 coins.

That means in the first weighing, the heavier side, if there is one, has at most 1 light coin.
If neither side is heavier, both sides can have 0-2 light coins.
That gets us close, but I think it's not quite enough.

Compare any two groups of 14 coins.

If there is a heavier group (at most 1 light coin), compare any 7 of that group with the other 7.
The heavier group of 7, if there is one, otherwise both groups of 7, are free of light coins. Done

If the 14-14 comparison balances, both groups of 14 contain 0-2 light coins.
Compare 7 of either group of 14 with the other 7 of that group.

  • If there is a heavier side, it contains 0 light coins. Done.
  • Otherwise both sides contain 0 or 1 light coins. Fail.

Alternative scheme

If the 14-14 comparison balances, Take any 7 (A) from either group of 14.
Compare (A) with any 7 (B) from the remaining 20 coins and take the heavier group if there is one, else either group.
Three cases:

  1. (A) has two light coins. (B) must have 0 light coins. Done.
  2. (A) has one light coin. (B) can have 0-2 light coins. Fail.
  3. (A) has no light coins. Done.


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hey bn - Thanks for getting this one started.  You are correct in all you say above including your initial summary of the problem.  Have edited the OP to include that explanation.  I assure you there is a scheme in which one can guarantee finding seven heavy coins.  Perhaps start off by finding one guaranteed heavy coin in two uses of a balance scale.  Or not.  I have no doubt you or others here will discover the method for finding seven heavy dimes.

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This is a nice puzzle.


The solution seems to consist of (1) choosing the right number of coins for the first comparison and (2) what to do if weights are equal. (1) depends on how many heavy coins you want to find. I thought I had (2) figured out, and I think it's on the right track, but it's still not quite definitive. I'll edit this post when I get it.


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I think I've got it.  Nah, I definitely don't have it.  I can't account for one particular case.

First, I'd like to say that any dealer who shorts a dime is a criminal.  It's just a dime!

My method can't tell the difference between the first weighing having 2 light vs 2 light and then a second weighing having 1 vs 1 compared to no light dimes being present during any weighing.  I'll detail my answer anyway, though, just to get it out of my head.

First weigh 16 vs 16.  If they balance, the least optimal possibility is 2 vs 2 or 0 vs 0, which will be indistinguishable* with only one weighing remaining.  1 vs 1 would be doable and any unbalanced amount would also be doable.

*Note: If I can come up with a way to make this distinction, I would have a solution.

I think 16 is a good number to begin, though.  It's 1/3 of the pile and allows itself to be split in half and be left with more than the required number of coins.  It's also a solid square number and a power of 2, which likely have no bearing on this puzzle, but it does please me.

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