Place 4R+3F in left and 3R+4F in right pan. Left pan must be heavier.

Second Weighing.

Remove 2R from left pan and 2F from right pan. This will lead 2R+3F in left and 3R+2F in right pan. Right pan must be heavier.

This proves 2R and 2F removed after first weighing are true.

Third Weighing.

Remove 3F from left and 3R from right pan and add now proven 2F in left and 2R in right pan. The pans must balance.

This will prove all 14 coins true, 7 real and 7 fake.

]]>Weigh R1 against F1; R1 is heavier. Everyone now knows R1 is real and F1 is fake.

Weigh F1+R2+R3 against R1+F2+F3; F1+R2+R3 is heavier. Everyone now knows R1, R2, R3 are real and F1, F2, F3 are fake.

Weigh F1+F2+F3+R4+R5+R6+R7 against R1+R2+R3+F4+F5+F6+F7. That settles it.

This can be generalized to handle 2^n-1 real coins and 2^n-1 fake coins in n weighings. But the question is, is that the greatest number of coins you can handle for any number of weighings? My guess is, probably not. There’s probably some surprises lurking here.

]]>Weigh R1 against F1; R1 is heavier. Everyone now knows R1 is real and F1 is fake.

Weigh F1+R2+R3 against R1+F2+F3; F1+R2+R3 is heavier. Everyone now knows R1, R2, R3 are real and F1, F2, F3 are fake.

Weigh F1+F2+F3+R4+R5+R6+R7 against R1+R2+R3+F4+F5+F6+F7. That settles everything.

This can naturally be generalized to handle 2^n-1 real coins and 2^n-1 fake coins in n weighings. But the question is, is this the greatest number of coins you can handle, regardless of the number of weighings? My guess is, probably not.

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