pirates have found the chest simultaneously, no one can claim the chest as his own.

To protect the chest from the avarices of the pirates, the pirate leader Captain Jack Sparrow suggests a

scheme. “We must put a certain number of locks on this chest and distribute their keys amongst ourselves in

such a way that it can be opened only when 7 or more then 7 pirates dicide to open it. In other words, only

when a majority of the 13 pirates agree should the chest be able to be opened.”

How can this scheme be implemented? What would be the minimum number of locks required and how must

their keys be distributed?

]]>Let us rephrase our problem slightly.

We want that only a group of 7 or more pirates should be able to open the chest. This means that whenever

such a majority group decides to open the chest, they should have amongst themselves the keys to all the

locks on the chest.

Suppose on the other hand, that only 6 pirates decide to open the chest. Then there should be at least one

lock on the chest whose key(s) are not with anyone amongst that group of 6. Thus, that single lock will

prevent the minority group of 6 pirates from opening the chest.

This is the approach we now follow. For every possible group of 6 pirates, we put a lock on the chest and

distribute 7 keys of the lock amongst the remaining 7 pirates. That lock will prevent our group of 6 pirates

from opening the chest.

Such a lock will exist for every group of 6 pirates. Thus, whenever any group of 6 pirates decides to open the

chest, they will be prevented by one lock whose keys are with the other 7 pirates (the case when even less

than 6 pirates decide to open the chest is automatically solved because then there will be more than one lock

to prevent that group from opening the chest)

Also, whenever any group of 7 pirates decides to open the chest, there is no lock whose key is not amongst

one of the 7 members of that group. Thus, any group of 7 pirates (or more) will be able to open the chest.

Thus, what we need to do is put

^{13}C_{6}(=^{13}C_{7}) locks on the chest and for each lock, we select a group of 7pirates, make 7 keys of that lock and give one key to each member of this group.

A robot is placed on the square A1 of a standard chessboard and has to reach H8. It understands orders Up, Down, Left, Right. On some squares, there might be a cement block; is such a case, the robot does not execute the order and continues with the next one on your list.

There always is at least one possible path.

The list of directions is finite.

The robot might reach the destination somewhere in the middle of your list.

Give the answer in the form UURRU.

(This list will work i.e. on an empty 3x3 chessboard or with a single block on A3 but will fail with a single block on B3.)

]]>Arrange them in a 4 × 4 square so that every row, column and diagonal contains one card of each value (A,J,Q,K) and one card of each suit (Heart, Spade, Diamond, Club).]]>

To alleviate their boredom, the Angel and the Devil decide to play a game. The Angel's objective is to bring all the lanterns into complete order, winning immediately if either all lanterns are turned on or all lanterns are turned off. To ensure the game eventually concludes, the celestial beings agree that the game ends if the lanterns return to a previously encountered state. In this case, the Devil is declared the winner.

Now, the question is: Who will emerge victorious, and what strategy ensures the win?

]]>

*In a "seven-eleven" (7-11) store, a customer selected four items to buy. The check-out clerk says that he multiplied the costs of the items and obtained exactly $7.11, the very name of the store! The customer calmly tells the clerk that the costs of the items should be added, not multiplied. The clerk then informs the customer that the correct total is also $7.11. What are the exact costs of the 4 items? *

Reasoning is welcome as I am certainly not going to find the elegant solution. I see some brute force in my future.

Edit to add a dollar sign and italics

]]>I’ve created a little puzzle that follows the cryptographic principle of zero-knowledge proof.

Let P = xx, the age of Peter

To find xx, I will provide you with means to verify the statements of the puzzle, without giving you any direct informations about the ages of the characters.

The ages of the characters are not given but can be found.

Although there are an infinite number of answers that could verify the information I provide, there is one answer that can be verified to 99% assuming the puzzle is honest and verifiable, and that Peter has a realistic age and life.

How old is Peter ?

- Peter has 5 children, Matthew, Nancy, Phil, Quinlan and Ryan

- Peter’s age is the sum of the ages of all of his children

- The concatenation of his children’s ages forms a palindrom

- Peter’s age is a semi-prime number

- 2 of his children are the same age

- One of his children is half the age of one of his older siblings

- Quinlan is younger than Phil

- Only two of his children have a job

- At least 2 of his children have a palindrome age

- Matthew can’t read

- Peter didn’t have a child before the age of 30

- If x is the age of the child < 10, then we’ll write 0x, such that a 1 year-old child = 01

]]>

Bonus question: what % of coffee is in my last cups mixture after adding the 2 oz?

For added clarity, when I top it up, I mean that I am only adding my coffee to the mixture.

]]>

When I divide it by 2, the remainder is 1.

When I divide it by 3, the remainder is 2.

When I divide it by 4, the remainder is 3.

When I divide it by 5, the remainder is 4.

When I divide it by 6, the remainder is 5.

When I divide it by 7, the remainder is 6.

When I divide it by 8, the remainder is 7.

When I divide it by 9, the remainder is 8.

When I divide it by 10, the remainder is 9.

It's not a small number, but it's not really big, either.

When I looked for a smaller number with this property I couldn't find one.

Can you find it?

]]>CosA/a=CosB/b=CosC/c=CosD/d

What can be claimed about this quadrilateral?

]]>1. You begin the game with 10 US dollars

2. Every round, you may wager any of your current US dollars

3. After you have wagered, we toss a fair coin, which leads to one of two outcomes:

a. Heads - You lose your wager

b. Tails - You get your wager back AND win 200% of your wager's value

4. You have 100 rounds in which to maximise your profits as much as possible

5. It should go without saying, but if you ever end a round with 0 US dollars, you've lost

The game is obviously in your favour, but provide a betting strategy which uses these rules to win as many US dollars as possible.

]]>
Jill said: IF John is not involved and Joe told the truth THEN Mike is not involved

John said: IF Abby is involved or Sue is involved THEN Joe is involved

Abby said: IF Cindy is not involved or Joe is not involved THEN Mike is involved or Jill is not involved

Cindy said: IF Mike is involved and Joe told the truth THEN John is not involved or Jill is involved

Sue said: IF Joe is involved or Sue is not involved THEN Cindy is involved or Abby is not involved

Mike said: IF Cindy is not involved and Abby is involved THEN John is not involved or Jill is not involved

Joe said: IF Jill is telling the truth or Cindy is lying THEN Sue is telling the truth or Mike is telling the truth

https://sites.google.com/view/puzzlequizzes/home

]]>

]]>

(for example try computing a MacLauren series for (1+2x)/(1-x^3)

]]>

Find f(x) where f(x) is a polynomial.

]]>find f(x) =

]]>Multiple emojis placed on top of one another count as added together.

]]>