# You have 12 coins with the same appearance, but one is counterfeit.?

You don’t know if the counterfeit coin is heavier or lighter than the others, and you only have a pan balance. What’s the minimum number of weighings to determine the counterfeit, and how?

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It’s 3.

1. Divide the coins into three groups (A,B,C) of four.

2. Weigh groups A and B together.

- I. If they balance, then the counterfeit is in group C. Number the coins in group C 1, 2, 3, and 4. Take coins 1,2, and 3 and weigh them against three coins from group A (now known to be authentic). Obviously if they balance, coin 4 is counterfeit. If one pan is heavier, then 1,2 or 3 is the counterfeit. If the genuine coins are heavier, we know the counterfeit is lighter, and vice-versa. Assuming the counterfeit is lighter (switch all the following around if it is heavier), Make a group D of coin 1 and a genuine coin, and E of coin 2 and a genuine coin, and weigh them against each other. If they balance, then coin 3 is counterfeit; if they don’t, then if group D is lighter, then coin 1 is counterfeit; if group E is lighter, then coin 2 is counterfeit. [end]

- II.1 If they don’t balance, then group C is all genuine and we put it aside (Continue with step II.2).

- II.2 Make two new groups from groups A and B. We assume group A is heavier than group B (you may have to rename your groups at this point): group A’ consists of coins 1A, 2A, and 1B; group B’ consists of coins 3A, 4A, and 2B (you’ll have coins 3B and 4B leftover). We then weigh groups A’ and B’ against each other.

- II.3.I if they balance, then you know A’ and B’ are all genuine. Weighing the leftover coins 3B and 4B together will tell you which one is the counterfeit-whichever is lighter (based on the fact that group A was heavier than group B in step 4). [end]

- II.3.II If they don’t balance, then if A’ is heavier, coins 3A, 4A, and 1B are authentic. Place coins 1A and 2A in the balance; if they balance, then coin 2B is counterfeit; if they don’t, then the heavier one is counterfeit.

- II.3.III If they don’t balance and if B’ is heavier, the same reasoning as in the preceding step will tell us that one of 3A, 4A, or 1B is counterfeit.

I think I’ve managed to do this, while finding out if the counterfeit is heavier or lighter. Number balls from 1 to 12.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

WEIGH1: (1.2.3.4) and (5.6.7.8)

IF (1234)=(5678) then either 9,10,11,12 is BAD

WEIGH2: (1.2.3) & (9.10.11)

IF (1.2.3) = (9.10.11) then 12 is lighter or heavier

WEIGH3: (1) & (12)::

IF (1) > (12), 12 is LIGHTER: IF (1) < (12), 12 is heavier

IF (1.2.3)>(9.10.11) then either (9,10,11) is lighter

WEIGH3: (9) & (10)

IF (9) = 10 THEN 11 is LIGHTER

IF (9) > (10) THEN 10 is LIGHTER

IF (9) < (10) THEN 9 is LIGHTER

IF (1.2.3) < (9.10.11) {

WEIGH3: (9) & (10)

IF (9) > (10) THEN 9 is HEAVIER

IF (9) < (10) THEN 10 is HEAVIER

IF (9) = (10) THEN 11 is HEAVIER

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

IF (1.2.3.4) > (5.6.7.8), then either (1,2,3,4) is heavier or (5,6,7,8) is lighter

WEIGH2: (1.2.5) & (9.10.11)

IF (1.2.5) = (9.10.11), either 3 or 4 is heavier

WEIGH3: 3 & 4

IF (1,2,5) > (9,10,11), then either 1 or 2 is heavier

WEIGH3: 1 & 2 to determine which is heavier

IF (1.2.5) < (9,10,11), then either 1,2,5 is lighter

WEIGH3: 1 & 2

IF (1)=(2) THEN 5 IS LIGHTER

IF (1)>2 THEN 2 IS LIGHTER

IF (1)<2 THEN 1 IS LIGHTER

The case for:

IF (1.2.3.4) < (5.6.7.8) should be similar to the case for IF (1.2.3.4) < (5.6.7.8) above