Logic Puzzles

Hint:

Based on what you know, why are the other princes unable to solve this puzzle?

Solution:

White.

The king would not select two white hats and one black hat. This would mean two princes would see one black hat and one white hat. You would be at a disadvantage if you were the only prince wearing a black hat.

If you were wearing the black hat, it would not take long for one of the other princes to deduce he was wearing a white hat.

If an intelligent prince saw a white hat and a black hat, he would eventually realize that the king would never select two black hats and one white hat. Any prince seeing two black hats would instantly know he was wearing a white hat. Therefore if a prince can see one black hat, he can work out he is wearing white.

Therefore the only fair test is for all three princes to be wearing white hats. After waiting some time just to be sure, you can safely assert you are wearing a white hat.

Hint:

Each pirate cares first about staying alive, then about getting as many coins as possible. A pirate will vote “Aye” only if the proposal gives them more than they would get after mutiny. Side deals, promises, and revenge votes do not count, because the pirates are selfish and treacherous.

What happens if there are two pirates? Who completely loses out? What happens if there are three pirates? Who completely loses out? What happens if there are four pirates? Which two pirates completely lose out?

Solution:

98

The captain says he will take 98 coins, and will give one coin to the third most senior pirate and another coin to the most junior pirate. He then explains his decision in a manner like this...

1. Pirate 1: most junior, keeps 1 coin.
2. Pirate 2: fourth most senior, keeps 0 coins.
3. Pirate 3: third most senior, keeps 1 coin.
4. Pirate 4: second most senior, keeps 0 coins.
5. Pirate 5: captain, most senior, keeps 98 coins.

If there were 2 pirates, because pirates 3,4 & 5 had walked the plank, then pirate 2 would be the most senior, and he would just vote for himself and that would be 50% of the vote, so he's obviously going to keep all the money for himself.

If there were 3 pirates, because pirates 4 & 5 had walked the plank, pirate 3 has to convince at least one other person to join in his plan. Pirate 3 would take 99 gold coins and give 1 coin to pirate 1. Pirate 1 knows if he does not vote for pirate 3, then he gets nothing, so obviously is going to vote for this plan.

If there were 4 pirates, because the captain had to walk the plank, pirate 4 would give 1 coin to pirate 2, and pirate 2 knows if he does not vote for pirate 4, then he gets nothing, so obviously is going to vote for this plan.

As there are 5 pirates, pirates 1 & 3 had obviously better vote for the captain, or they face choosing nothing or risking death.

Pirates left	Winning proposal
2	Senior pirate keeps 100.
3	Captain keeps 99, gives 1 to Pirate 1.
4	Captain keeps 99, gives 1 to Pirate 2.
5	Captain keeps 98, gives 1 each to Pirates 1 and 3.

Hint:

Use the same approach.

Solution:

The most senior pirate could give the coin to the least senior pirate. He can use the same logic in the previous puzzle to explain the futility of anyone trying to keep the coin for himself.

Pirates left	Captain needs	Result
2	1 vote	Pirate 2 keeps the coin. His own vote is enough.
3	2 votes	Pirate 3 gives the coin to Pirate 1. Pirate 1 would get 0 if Pirate 3 dies.
4	2 votes	Pirate 4 gives the coin to Pirate 2 or Pirate 3. Either would otherwise get 0.
5	3 votes	Pirate 5 cannot survive. He has only one coin and needs two extra votes.
6	3 votes	Pirate 6 gives the coin to Pirate 1. Pirate 5 votes yes because he dies if Pirate 6 dies. Pirate 1 votes yes for the coin.

Hint:

The one who stopped laughing, asked himself what the other philosophers were seeing that made them laugh.

Solution:

If he (the smartest philosopher) had nothing on his head, then he realized that the second smartest philosopher would have quickly worked out that the third smartest was laughing only at the second smartest philosopher, and thus the second smartest philosopher would have stopped laughing.

Hint:

You can weigh as few or as many of the ten coins from each set as you choose.

Solution:

One coin from the first set is placed on the scale along with two from the second set etc... If the weight is off by one hundredth of an ounce then it is the first set that is faulty, if the weight is off by two hundred of an ounce then it is the second set which is faulty, etc...

Hint:

 Look up the formula for the LCM.

Solution:

2519

The solution for the answer is the LCM (Lowest Common Multiple) of 10,9,8,7,6,5,4,3,2,1 -1. LCM would give the least number which is divisible by all of these number and subtracting one would give us the number of coconuts which were initially there.

Solution:

Create two sets of ten coins. Flip the coins in one of the sets over, and leave the coins in the other set alone. The first set of ten coins will have the same number of heads and tails as the other set of ten coins.

Hint: If I ask people at random if they have two children and if the youngest is a boy, then the probability that this person has two boys is 1/2.

If I asked people at random if they have two children and if one is a boy, then the probability that this person has two boys is 1/3.

Possibility	Has at least one boy?	Two boys?
BB	yes	yes
BG	yes	no
GB	yes	no
GG	no	no

My question is excluding anyone with two girls, so therefore there are only three cases left, only one of which has two boys.

Solution:

13/27. If you think the answer should be 1/2, you would be wrong. If you knew which child was a boy (say, the younger one), you would be closer to the truth. But since the boy could be either the younger or the older child, the analysis is more subtle. But what does Tuesday have to do with it?

Solution:

Person A writes a number that is her salary plus a random amount (AS + AR) and hands it to B, without showing C. B then adds his salary plus a random amount (BS + BR) and passes it to C (at each step, they write on a new paper and don't show the 3rd person). C adds CS + CR and passes it to A. Now A subtracts her random number (AR) and passes it to B. B and C each subtract their random number and pass. After C is done, he shows the result and they divide by 3.

As has been noted already, there's no way to liar-proof the scheme.

It's also worth noting that once they know the average, any of the three knows the sum of the other 2 salaries.

Solution contributed by Jeebok.

Solution:

It is impossible to cover the remaining 62 squares with 31 dominoes after cutting out two opposite corners of the chessboard. Each domino covers one black and one white square, and removing two opposite corners (which are the same color) leaves an imbalance of colors, resulting in 32 squares of one color and 30 of the other, making it impossible to cover the board completely.

Solution:

The man either started at the North Pole, or the man started at a point just over a mile north of the South Pole, or at any latitude where walking 1 mile south takes him to a circle of latitude with a circumference that is an exact factor of 1 mile (such as 1/2 mile, 1/3 mile, etc.). After walking 1 mile south, he reaches this latitude, then walks 1 mile east, completing a full circle back to his starting point on that latitude, and finally walks 1 mile north to return to his original position.

Solution:

The sage's plan was to drink a weak poison prior to the meeting with the king, and then he would drink the pharmacist's strong poison, which would neutralize the weak poison. As his own poison he would bring water, which will have no effect on him, but the pharmacist who would drink the water, and then his poison would surely die. When the pharmacist figured out this plan, he decided to bring water as well. So the sage who drank poison earlier, drank the pharmacist's water, then his own water, and died of the poison he drank before. The pharmacist would drink only water, so nothing will happen to him. And because both of them brought the king water, he didn't get a strong poison like he wanted.

Hint:

Among any five cards there must be at least two of the same suit. Use one of those as a “key” card. The order of the four cards can encode up to 24 different messages.

Solution:

Among any five cards, at least two cards must have the same suit, because there are only four suits. Choose two cards of the same suit. One of them will be hidden, and the other will be shown as the first card in the pile.

Use the rank order:

A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, then back to A.

Between two cards of the same suit, one direction around the cycle is always between 1 and 6 steps. Choose the visible card so that the hidden card is 1 to 6 ranks ahead of it.

The first visible card tells Peter the suit of the hidden card. The order of the other three visible cards tells Peter how many ranks to count forward.

Beforehand, Peter and the magician agree on a fixed sorting order for cards, for example:

Clubs, Diamonds, Hearts, Spades, and within each suit A, 2, 3, ..., K.

After the key card is placed first, sort the remaining three cards by that agreed order and call them A, B, and C.

Use this code:

ABC = 1
ACB = 2
BAC = 3
BCA = 4
CAB = 5
CBA = 6

So the magician puts the key card first, then arranges the other three cards in one of those six orders to encode the distance.

Example:

Suppose the five cards are:

♣9, ♥K, ♦4, ♣3, ♠A

The two clubs are ♣3 and ♣9.

Choose ♣3 as the key card and hide ♣9, because counting forward from 3 to 9 gives:

4, 5, 6, 7, 8, 9

That is 6 steps.

The remaining cards are:

♥K, ♦4, ♠A

Sorted by the agreed order, they are:

A = ♦4
B = ♥K
C = ♠A

Distance 6 is encoded by CBA, so the magician gives Peter the four cards in this order:

♣3, ♠A, ♥K, ♦4

Peter sees ♣3 first, so he knows the hidden card is a club. He sees the remaining three cards are in CBA order, so the distance is 6. He counts six ranks forward from 3 and gets 9.

Therefore Peter announces:

Nine of Clubs

The trick works every time because the first card gives the suit, and the six possible orders of the remaining three cards give the rank distance from 1 to 6.

Solution:

Original: New:

9 9 9 9 9 9
9 9 9 9 9 9
9 9 9 9 9 9

So the four moves are:
top-left moves left
top-right moves right
bottom-left moves left
bottom-right moves right

The 10 lines of three 9s are:

3 horizontal rows
1 vertical center line
2 long diagonals
4 shorter diagonals

Solution:

First, divide the 12 balls into three groups of four. Weigh two groups against each other. Depending on the outcome, you will know if the odd ball is in one of those groups or in the group not weighed. Continue to narrow it down using the balance scale until you identify the odd ball and determine if it is heavier or lighter.