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There are eight identical-looking coins. Six of them are genuine ones weighing 10 grams each; but the remaining two are fake and weigh 12 grams each. Using just a pan balance, what is the minimum number of trials needed to determine both the fake coins?

Click to show/hide hint

Note that a pan balance in each trial can give one of 3 possible outcomes; so 3 trials can lead to one of only 27 different outcomes.

Click to show/hide explanation

Let us label the eight coins as A,B,C...,H.

We proceed by the following method; which covers all the possible cases.

Thus, we need atleast

#### Puzzle No. 23

*Posted on 03 June 2016*There are eight identical-looking coins. Six of them are genuine ones weighing 10 grams each; but the remaining two are fake and weigh 12 grams each. Using just a pan balance, what is the minimum number of trials needed to determine both the fake coins?

Click to show/hide hint

**Hint:**

Note that a pan balance in each trial can give one of 3 possible outcomes; so 3 trials can lead to one of only 27 different outcomes.

Click to show/hide explanation

**Explanation:**

Let us label the eight coins as A,B,C...,H.

We proceed by the following method; which covers all the possible cases.

In the first trial, compare

**ABC vs DEF**### One side is heavier, say ABC > DEF.

So DEF are all real, and ABC contains at least one fake coin.

In the second trial, compare

In the second trial, compare

**G vs H**### One side is heavier, say G > H.

So G is one of the fake coins; and the other fake coin has to be located from ABC.

In the third trial, compare

In the third trial, compare

**A vs B**### One side is heavier, say A > B.

So A is the second fake coin.

### The sides are even, i.e. A = B.

So A and B are real; and C is the second fake coin.

### The sides are even, i.e. G = H.

So G and H are both real; and the ABC group contains both fake coins.

In the third trial, compare

In the third trial, compare

**A vs B**### One side is heavier, say A > B.

So B is a real coin; hence A and C are the two fake coins.

### The sides are even, i.e. A = B.

So A and B are the two fake coins.

### The sides are even, i.e. ABC = DEF

So either ABCDEF are all real (and G,H are both fake), or the ABC and DEF groups each contain one fake coin.

In the second trial, compare

In the second trial, compare

**A vs G**### A > G

So A is fake and G is real; and the other fake coin has to be located from DEF.

In the third trial, compare

In the third trial, compare

**D vs E**### One side is heavier, say D > E.

So D is the second fake coin.

### The sides are even, i.e. D = E.

So D and E are real; and F is the second fake coin.

### A = G

So A and G are both real; hence one fake coin is from BC, the other is from DEF.

In the third trial, compare

In the fourth trial, compare

In the third trial, compare

**B vs C**. Whichever is heavier, say B > C, means that B is one fake coin. The other fake coin has to be location from DEF.In the fourth trial, compare

**D vs E**.### One side is heavier, say D > E.

So D is the second fake coin.

### The sides are even, i.e. D = E.

So D and E are real; and F is the second fake coin.

### A < G

This means A is real; so G and H are the two fake coins.

Thus, we need atleast

**4**trials to determine the two fake coins.