There are $10$ coins, one of which is a fake, and weighs either more or less than the others - you don't know which.

There are two sets of scales - one tells the truth, the other lies (always says equal weighings are unequal but either way, unequal weighings are either equal or wrong.) - you don't know which is which. The lying scales decisions are made by an AI unit - see comments.

Both scales only allow an equal number of coins on either side (and at least one) to be tested.

How many weighings are needed to find the fake coin?

  • $\begingroup$ Is it correct that the lying scale only lies 50% of the time given that the coins don't weigh the same? Is this random? I.e. is it possible that it tells the truth 50 times in a row (though unlikely)? Can it show that the lighter coin is heavier, or will it always be lighter or equal? $\endgroup$ Oct 14, 2019 at 8:12
  • $\begingroup$ The lying scale has a AI unit to choose between options as to hinder the tester as much as possible. The lying scale never gives a truthful result. $\endgroup$
    – JMP
    Oct 14, 2019 at 8:15
  • $\begingroup$ Then what's the 50/50? $\endgroup$ Oct 14, 2019 at 8:17
  • 1
    $\begingroup$ If I weigh two equal weight coins on the lying scale with one on both sides, and then I do it again, will the results from both weighings be the same i.e. will it always say that the coin on the right side, for example, is heavier than the one on the left side? $\endgroup$
    – HTM
    Oct 14, 2019 at 8:25
  • 3
    $\begingroup$ @PiIsNot3 I feel like this has been answered. It will choose between options to hinder the tester as much as possible. $\endgroup$
    – hdsdv
    Oct 14, 2019 at 8:27

1 Answer 1


In 5 moves :

Lets say coins are ABCDEFGHJK. Plus symbol means weighing.
First try AB+CD and EF+GH.
There are four possibilities (well 3, both equal, both unequal or one equal and second not):

A) if both are equal, it is simple J or K, since I used normal scales (total 3 moves)

B) if both are unequal, I used "AI" scales, so for the rest I will use normal one - and it is solvable in 3 more moves:
ABC+DEF - if unequal like ABC>DEF, I will weigh ABC+GHJ (since I know GHJ is surely normal) and will either found it is in ABC as heavier coin, or in DEF as lighter coin - in both cases I need only one more weigh, something like A+B - which is heavier is the one, or in equal case it would be C. Same for DEF, only it will be lighter one.
similar solution for ABC If I found that ABC=DEF I will weigh ABC+GHJ and I will either found that GHJ is lighter (*heavier) so I weigh G+H and it would be lighter one (or *heavier) or J. If equal - it would be K coin.

C) Either AB+CD is equal or EF+GH is equal (but not both, since it would be A) solution) so I weigh J+K:
If J+K is equal, I used normal scales so I identified normal scales - Now I know which 4 it is (for example ABCD).
If J+K is unequal, I used "AI" scales so I know which one are normal scales - Now I know which 4 it is anyway, since one weigh was equal (for example ABCD).
Now I use normal scales like: ABC + EFG (EFG can't be with fake coin). Again like in B) now I will know if it is lighter or heavier and one more weigh for A+B left, or it is D if it was equal.

  • $\begingroup$ Is this optimal? It needs $\ge 4$ weightings for sure, but I'm not sure if we can do $4$ or not. $\endgroup$
    – Vepir
    Nov 27, 2019 at 19:38

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