Another coins weighing puzzle

Here is a simple solution that works. There are many combinations that you can use. The idea is to make sure you are always making 3 such groups and weighing them against each another so that all of them balance. Also any transfer should be done in a way that you cannot tell whether you transferred a fake or a real.

Jason makes 6 groups as below (there are many more possible solutions as you can understand after reading through my solution) -

G1 = 20 coins, G2 = 20 coins, G3 = 20 coins

G4 = 7 coins (1 fake coin), G5 = 7 coins (1 fake coin), G6 = 6 coins (1 fake coin)

He weighs G1 against G2 and G2 against G3. This shows to Mary and Christian that either G1, G2 and G3 all have 1 fake each or none of them have any fake.

Now Jason transfers 1 coin from G1 to G4, 1 from G2 to G5 and 2 from G3 to G6 (he can also take 2,2,3 or 3,3,4 or other counts as well making sure G4, G5 and G6 have equal number of coins after transfer).

So G4, G5 and G6 all have 8 coins each now after the transfer. Now he weighs G4 against G5 and G5 against G6. They all balance. This shows Mary and Christian that there are 3 fake coins as they know there are either 2 or 3 (they know zero or another multiple of 3 is not an option).

But what they cannot tell whether the fake coins were there in G4, G5 and G6 from before or the transferred coins were fake or fake one's are still in G1, G2 and G3.

I hope it is clear. Let me know if any questions.