# Find 2 heavy fakes among 11 coins on a 3-pan balance

I give you 11 coins, which look exactly the same. Among them are two fake coins. All genuine coins weigh the same. The fake coins are a little heavier than the genuine ones, and they may have a different weight from each other. You cannot distinguish their weight by hand.

You have no own utility to weigh coins, so I suggest you use my three-pan balance which, upon activation, lifts the unique pan with the lightest load. If there is no unique lightest pan, none of the pans will rise. You may only weigh the coins I gave you. Moreover, you are only allowed to perform three weighings.

As you may remember from a previous puzzle, under these conditions, the two fake coins can not always be identified.

To help you, and for my amusement, I give you one additional genuine coin, which looks and weighs the same as the other genuine coins, and which you may use in your weighings to come.

Can you identify the two fake coins under these circumstances? If so, how?

• I'd say it's not possible, but if you need one... well... Question: Can several coins be put in a pan at the same time? Jun 22, 2015 at 2:05
• Can we assume the weight difference is less than half the weight of a real coin? Jun 22, 2015 at 10:48
• @the4seasons Questions like these always allow you to put several coins in the pan at the same time.
– user88
Jun 22, 2015 at 13:31
• If I'm correct here, you have 64 possible outcomes, which you want to map to 55 results.
– user88
Jun 22, 2015 at 13:31
• @PrincessTrevor If there is no unique lightest pan, none of the pans will rise. This line is enough to explain what would happen. Jun 22, 2015 at 18:50

It is possible.

Label the coins A-L, with L being the known genuine coin.

Weigh ABCD vs EFGH vs IJKL.

If no pan rises, then both heavy coins are on the same pan. There are ${4\choose 2} = 6$ combinations where both heavy coins are on Pan 1, and 6 combinations where both are on Pan 2, but only ${3\choose 2} = 3$ combinations were both are on Pan 3, since we know coin L is genuine. That adds up to 15 possibilities.

If Pan 3 rises, then one heavy coin is on Pan 1 and another on Pan 2, for a total of $4\times 4 = 16$ combinations. If Pan 2 rises, then there are only $12$ combinations, again because we know coin L is genuine. Likewise for Pan 1.

If no pan rose on the first weighing, then the second weighing should be ABEF vs CGIJ vs DHKL. The results of this would be:

• None: AB, EF, IJ
• Pan 1: CD, GH, IK, JK
• Pan 2: AD, BD, EH, FH
• Pan 3: AC, BC, EG, FG

In the first case, for the third weighing, use AE vs BI vs EJ. The two heavy coins are now split and one of the three pans will rise.

In the other three cases, there are four patterns of which at least two intersect by one coin. Put the intersecting coin on Pan 1, and each of the others from the pairings on Pans 2 & 3. Pick one other pattern and split its coins between Pans 2 & 3. Put other coins aside and fill in with known genuine coins (of which you have several).

For example, if Pan 1 rose, you would weigh KA vs IC vs JD. If the pattern was GH, then no Pan would rise. CD would raise Pan 1, IK would raise Pan 3, and JK would raise Pan 2.

If Pan 3 rose on the first weighing, then one of the heavy coins is in ABCD, and the other in EFGH, for a total of $16$ possibilities. Coins I,J,K, & L are now all known to be genuine.

Weigh ABGH vs CDIJ vs EFKL.

• No Pan: AG, AH, BG, BH
• Pan 1: CE, CF, DE, DF
• Pan 2: AE, AF, BE, BF
• Pan 3: CG, CH, DG, DH

In all four of these cases, you have four patterns where one coin is from one pair and the other coin from the other pair. The weighing pattern for this is similar to above, but requiring 2 genuine coins.

No Pan: weigh AI vs BG vs HJ.

AG would raise Pan 3; AH would raise Pan 2; BG would raise no pan, and BH would raise Pan 1.

A similar pattern would work for any of the results of the second weighing.

If Pan 1 or 2 rose on the first weighing, follow the same logic for Pan 3 rising, with the different sets of coins. Any of the possibilities where coin L came out heavy would obviously not happen, but otherwise the logic would be the same.

QED