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I have 9 coins; 3 coins each of 3 denominations.
Coins of the same denomination all weigh the same.
Each coin weighs an exact whole number of grams, and at most 9 grams.
I can weigh any subset of my coins together, to get their total weight accurate to the gram.

I want to determine how much a coin of each denomination weighs. Obviously I could do it in 3 weighings, just weighing one coin of each denomination separately.

How can I get the weights of the 3 denominations using only 2 weighings?

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    $\begingroup$ I hope this is original. I thought of it myself but it seems an obvious variation. $\endgroup$ Commented Jan 3, 2022 at 12:45

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Calling the denominations (or rather their weights) a,b,c one scheme would be weighing:

3a+b and 3b+c. Call the resulting readings X and Y. Then b=X mod 3 and consequently 3X+c=Y mod 9. Because of the limited range of admissible values this last equation determines c. b and a immediately follow via back substitution.

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  • $\begingroup$ My own solution is the same, though I had a less compact proof. Would it have been a better puzzle if there were for example 4 or 5 coins of each denomination? Clearly the same solution still works, as would a similar solution with larger numbers, but would that have made it less obvious? $\endgroup$ Commented Jan 3, 2022 at 14:31
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    $\begingroup$ @JaapScherphuis I think the 9 (grams) is more of a giveaway. Perhaps cutting that to e.g. 7 (assuming it doesn't create unwanted alternative solutions) could have made a good smoke screen. $\endgroup$
    – loopy walt
    Commented Jan 3, 2022 at 14:36
  • $\begingroup$ As is it the puzzle has 18 possible solutions according to my computer, so I don't think adding alternative solutions is a problem. Those alternative solutions are less neat than this one, though I think essentially the same proof works in each case. Maybe lowering the 9 introduces anomalous alternative solutions that don't have the same proof. I'll have to look into that. $\endgroup$ Commented Jan 3, 2022 at 14:56

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