# Logical problem -- AMC 2013 (Australia) [closed]

In 3013, King Warren of Australia is finally deposed. The five remaining earls argue about which one of them will be king, and which one of the others will be treasurer.

• Akaroa will be satisfied only if Darlinghurst or Erina is treasurer.
• Bairnsdale will be satisfied only if Claremont is treasurer.
• Claremont will be satisfied only if Darlinghurst is either king or treasurer.
• Darlinghurst will be satisfied only if Akaroa is either king or treasurer.
• Erina will be satisfied only if Akaroa is not king.

It is not possible for all five to be satisfied, so in the end they appoint king and treasurer so that the other three earls are satisfied.
Who becomes king? And who becomes treasurer?

I list the conditions in the image:

My thinking is as the following:
The combinations of 3 people to be satisfied are to choose 3 from 5, that are (5*4*3)/(1*2*3)=10 combinations:
ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE
if ABC,ABD,ABE then: (C is T) and (D or E is t), contradicts;
if ACD then: D is t, A is k, C is t, contradicts;
if BCD,BDE then: C is t, A is k, E is not satisfied, contradicts;
if BCE then: C is t, D is k, A is not satisfied, contradicts;
if CDE then: A,D is k,t, B is not satisfied, contradicts;
if ACE then: D is t, B is k. It works:

• What have you tried? Jan 5, 2020 at 4:28
• Think about what happens if both Darlinghurst and Akaroa are satisfied. There is only one outcome from there. Jan 5, 2020 at 4:37

The trick is to find three heirs satisfied after not being elected, leaving the two elected heirs in an unknown state of satisfaction.

If D is treasurer then A and C are satisfied. So A and C won’t count for election of the king, being already satisfied. D is obviously dissatisfied, but he’s elected and his satisfaction won’t count.

Now only B and E are left. If E was king, do you think B would be satisfied as to make 3 satisfied yet unelected heirs? I think B won’t be satisfied, leaving only two heirs unelected but satisfied.

If B was elected king (after D was first elected treasurer) would that make E satisfied? I’ll let you guess the answer.

Bairnsdale is the King and Darlinghurst is the Treasurer. There are only twenty possibilities which are easy to figure out. All of them strike out except the one I have mentioned in my answer. Try yourself.

Heres a hint: The classic way to approach this type of problem is with a matrix of possibilities. Make a five by five grid, label each row and column with each of the five names. Let the row represent the king and column represent the treasurer. If you like, you can black out the main diagonal (since the same person can't be both king and treasurer).

$$\left[\begin{array}{c|c|c|c|c|c} & A & B & C & D & E\\ \hline A & \blacksquare & \\ \hline B & & \blacksquare \\ \hline C & & & \blacksquare\\ \hline D & & & & \blacksquare\\ \hline E & & & & & \blacksquare\\ \end{array}\right]$$

Now you only need five steps to solve. For each condition, cross out the matrix elements that represent dissatisfying outcomes. Remember that it's good enough to just appoint a dissatisfied person as king or treasurer, so don't cross out Akaroa's row or column when considering Akaroa's condition.

Akaroa will be satisfied only if Darlinghurst or Erina is treasurer

$$\left[\begin{array}{c|c|c|c|c|c} & A & B & C & D & E\\ \hline A & \blacksquare & & \\ \hline B & & \blacksquare & \times \\ \hline C & & \times & \blacksquare\\ \hline D & &\times & \times & \blacksquare\\ \hline E & & \times & \times & & \blacksquare\\ \end{array}\right]$$

When you finish, the uncrossed elements will represent the possible outcomes.