Four coins marked 1,2,3 and 5 respectively are given to you. The marks are supposed to be their weight, except that a fake coin weighs lighter than it should. There is one fake coin. Find it in 2 uses of the physical balance.

Cool! Now can you solve for it when there are 9 coins marked 1,2,3,4,5,6,7,8,9? Exactly one coin is fake, and the fake coin is lighter than it's marking. You are allowed to use the physical balance twice.

The idea is to eliminate 2/3rds of the coins in each weighing.

Consider the following:
G1=(1,9,5) G2=(2,6,7) G3=(3,4,8)
(Each sum to 15)

1. Weigh 2 of the groups. (2,6,7) vs (3,4,8)
a) The group that rises has the lighter (fake) coin.
b) If equal, the 3 coins that were not weighed contains the fake.

Without loss of generality let me assume the group G3=(3,4,8) rose up in the previous weighing.

We know that all other coins have true markings (1,2,5,6,7,9).

2. Weigh (4,1) vs (3,2)
a) If (4,1) rises, 4 is the fake coin.
b) If (3,2) rises, 3 is fake.
c) If both are equal, then 8 is fake.

This was with respect to G3=(3,4,8) being identified as the group with the fake coin. We can extend the same idea for the second weighing with other groups. e.g. If G1=(1,5,9) was identified as containing the fake coin, then second weighing can weigh (4,5) and (9).

Fake coin solution:

Step1: Weigh (2,3) vs (5)

If 5 rises, it is lighter, hence fake.

If both are equal, 1 is fake.

Otherwise, 2 or 3 is fake, go to step 2.

Step2: Weigh (1,2) vs 3

If 3 rises, 3 is lighter, hence fake.

Else, (1,2) should rise and 2 is fake.

They better not be equal.

[All under the assumption that exactly 1 coin is fake and that the fake coin is light]

Cool! Now can you solve for it when there are 9 coins marked 1,2,3,4,5,6,7,8,9? Exactly one coin is fake, and the fake coin is lighter than it's marking. You are allowed to use the physical balance twice.

The idea is to eliminate 2/3rds of the coins in each weighing.

Consider the following:

G1=(1,9,5) G2=(2,6,7) G3=(3,4,8)

(Each sum to 15)

1. Weigh 2 of the groups. (2,6,7) vs (3,4,8)

a) The group that rises has the lighter (fake) coin.

b) If equal, the 3 coins that were not weighed contains the fake.

Without loss of generality let me assume the group G3=(3,4,8) rose up in the previous weighing.

We know that all other coins have true markings (1,2,5,6,7,9).

2. Weigh (4,1) vs (3,2)

a) If (4,1) rises, 4 is the fake coin.

b) If (3,2) rises, 3 is fake.

c) If both are equal, then 8 is fake.

This was with respect to G3=(3,4,8) being identified as the group with the fake coin. We can extend the same idea for the second weighing with other groups. e.g. If G1=(1,5,9) was identified as containing the fake coin, then second weighing can weigh (4,5) and (9).