Before you are 22 coins, 2 of which are fake. The 2 fake coins have the same weight, but may be heavier
or lighter than the genuine coins. The genuine coins all have the same weight. There is a balance
scale with a weighing pan on each side. With proper strategy, at most, how many weighings are required
to split the coins into 2 groups of equal weight? Note that you are not required to identify the 2 fake
coins.

**ANSWER**:

*4.*
**EXPLANATION**: For the two groups to be of equal weight, each group must contain a
counterfeit coin. Here is one solution with the 22 coins labeled A to V.

**1st Weighing**:

ABCDE vs FGHIJ

**2nd Weighing**:

ABCDE vs KLMNO

**3rd Weighing**:

ABCDE vs PQRST

**4th Weighing**:

(see Scenario 3):

After three weighings, there are 4 scenarios.

**Scenario 1 (ALL THREE weighings are balanced)**:

In this scenario, the fake coins are the unweighed U and V coins and having one in each group of 11 coins,
for example, ABCDEFGHIJU and KLMNOPQRSTV will yield two groups of equal weight. No fourth weighing is
required.

**Scenario 2 (ONLY ONE of the three weighings is balanced)**:

In this scenario,

*consider the weighing that is balanced*. For example, if the 1st weighing is the
only balanced weighing, the fake coins are one of ABCDE and one of FGHIJ

**or** one of KLMNO
and one of PQRST in which case ABCDEKLMNOU and FGHIJPQRSTV, for example, will guarantee a fake coin in
each group and yield two groups of equal weight. No fourth weighing is required.

**Scenario 3 (TWO OF THE THREE weighings are balanced)**:

In this scenario,

*consider the two weighings that are balanced*. For
example, if the 2nd and 3rd weighing are balanced, then FGHIJ contains at least one fake coin, and the
second fake coin is either also in FGHIJ or it's in one of the remaining coins U or V. Now weigh FG vs HI.
If the 4TH WEIGHING IS BALANCED, then there is either one fake coin on each side or J is a fake coin and
either U or V are fake coins. Separating the coins FGJ and HIUV, for example ABCDEFGJKLM and HINOPQRSTUV
will guarantee a fake coin in each group and yield two groups of equal weight. If the 4TH WEIGHING IS NOT
BALANCED, then FGHI contains one fake coin and UV contains one fake coin. Separating the coins FGHI and UV,
for example ABCDEFGHIJK and LMNOPQRSTUVWXYZ will guarantee one fake coin in each group and yield two
groups of equal weight.

**Scenario 4 (NONE OF THE THREE weighings are balanced)**:

In this scenario, the fake coins are one of ABCDE and either U or V in which case ABCDEFGHIJK and
LMNOPQRSTUV will yield two groups of equal weight. No fourth weighing is required.

Do you have a

suggestion for this puzzle (e.g. something that should
be mentioned/clarified in the question or solution, bug, typo, etc.)?