Three players enter a room, and a maroon or orange hat is placed on each one’s head. The color of each hat is determined by a coin toss, and the outcome of one toss has no effect on the others. Each player can see the other players’ hats but not his own.
The players can discuss strategy before the game begins, but after this they may not communicate. Each player considers the colors of the other players’ hats, and then simultaneously each player must either guess the color of his own hat or pass.
The group shares a $3 million prize if at least one player guesses correctly and no player guesses incorrectly. What strategy will raise their chance of winning above 50 percent?
I propose a card game. I’ll shuffle an ordinary deck of cards and turn up the cards in pairs. If both cards in a given pair are black, I’ll give them to you. If both are red, I’ll take them. And if one is black and one red, then we’ll put them aside, belonging to no one.
You pay a dollar for the privilege of playing the game, and then we’ll go through the whole deck. When the game is over, if you have no more cards than I do, you pay nothing. But for every card that you have more than I, I’ll pay you 3 dollars. Should you play this game?
By W.A. Shinkman. This is a self-mate in two moves: White makes a move, Black is allowed to make any legal reply, then White plays a second move that forces Black to checkmate him.
New Year’s Day normally falls one week after Christmas: If Christmas falls on a Thursday, then New Year’s Day will fall on a Thursday as well. What is the most recent year in which Christmas and New Year’s Day fell on different days of the week?
A prison warden greets 23 new prisoners with this challenge. They can meet now to plan a strategy, but then they’ll be placed in separate cells, with no means of communicating. Then the warden will take the prisoners one at a time to a room that contains two switches. Each switch has two positions, on and off, but they’re not connected to anything. The prisoners don’t know the initial positions of the switches. When a prisoner is led into the room, he must reverse the position of exactly one switch. Then he will be led back to his cell, and the switches will remain undisturbed until the warden brings the next prisoner. The warden chooses prisoners at his whim, and he may even choose one prisoner several times in a row, but at any time, each prisoner is guaranteed another visit to the switch room.
The warden continues doing this until a prisoner tells him, “We have all visited the switch room.” If this prisoner is right, then all the prisoner will be set free. But if he’s wrong then they’ll all remain prisoners for life.
What strategy can the prisoners use to ensure their freedom?
Draw an arbitrary triangle and build an equilateral triangle on each of its sides, as shown.
Now show that AP = BQ = CR.
Three hockey pucks, A, B, and C, lie in a plane. You make a move by hitting one puck so that it passes between the other two in a straight line. Is it possible to return all the pucks to their original positions with 1001 moves?
A problem from the 1994 Italian Mathematical Olympiad:
Every inhabitant on the island of knights and scoundrels is either a knight (who always tells the truth) or a scoundrel (who always lies). A visiting journalist interviews each inhabitant exactly once and gets the following answers:
A1: On this island there is at least one scoundrel.
A2: On this island there are at least two scoundrels.
An-1: On this island there are at least n – 1 scoundrels.
An: On this island everyone is a scoundrel.
Can the journalist decide whether the knights outnumber the scoundrels?
This pleasing cryptarithm, by Bob High, appears in the September/October 2014 issue of MIT Technology Review. If each letter stands for a digit, what arithmetic sum is enciphered here?
The Martian parliament consists of a single house. Every member has three enemies at most among the other members. Show that it’s possible to divide the parliament into two houses so that every member has one enemy at most in his house.
Take two decks of cards, minus the jokers, shuffle them together, and divide them into two piles of 52 cards. What is the probability that the number of red cards in the Pile A equals the number of black cards in Pile B? How many cards would you have to view to be certain of your answer?
This unit square is divided into four regions by a diagonal and a line that connects a vertex to the midpoint of an opposite side. What are the areas of the four regions?
A rather baroque problem by Émile Leonard Pradignat. White to mate in two moves.
Two lines divide this equilateral triangle into four sections. The shaded sections have the same area. What is the measure of the obtuse angle between the lines?
To pay for 1 frognab you’ll need at least four standard U.S. coins. To pay for 2, you’ll need at least six coins. But you can buy three with two coins. How much is 1 frognab?
The knight’s tour is a familiar task in chess: On a bare board, find a path by which a knight visits each of the 64 squares exactly once. There are many solutions, but finding them by hand can be tricky — the knight tends to get stuck in a backwater, surrounding by squares that it’s already visited. In 1823 H.C. von Warnsdorff suggested a simple rule: Always move the knight to a square from which it will have the fewest available subsequent moves.
This turns out to be remarkably effective: It produces a successful tour more than 85% of the time on boards smaller than 50×50, and more than 50% of the time on boards smaller than 100×100. (Strangely, on a 7×7 board its success rate drops to 75%; see this paper.) The video above shows a successful tour on a standard chessboard; here’s another on a 14×14 board:
While we’re at it: British puzzle expert Henry Dudeney once set himself the task of devising a complete knight’s tour of a cube each of whose sides is a chessboard. He came up with this:
If you cut out the figure, fold it into a cube and fasten it using the tabs provided, you’ll have a map of the knight’s path. It can start anywhere and make its way around the whole cube, visiting each of the 364 squares once and returning to its starting point.
Dudeney also came up with this puzzle. The square below contains 36 letters. Exchange each letter once with a letter that’s connected with it by a knight’s move so that you produce a word square — a square whose first row and first column comprise the same six-letter word, as do the second row and second column, and so on.
So, for example, starting with the top row you might exchange T with E, O with R, A with M, and so on. “A little thought will greatly simplify the task,” Dudeney writes. “Thus, as there is only one O, one L, and one N, these must clearly be transferred to the diagonal from the top left-hand corner to the bottom right-hand corner. Then, as the letters in the first row must be the same as in the first file, in the second row as in the second file, and so on, you are generally limited in your choice of making a pair. The puzzle can therefore be solved in a very few minutes.”
Canadian doctor Samuel Bean created a curious tombstone for his first two wives, Henrietta and Susanna, who died in succession in the 1860s and are buried side by side in Rushes Cemetery near Crosshill, Wellesley Township, Ontario. The original stone weathered badly and was replaced with this durable granite replica in 1982. What does it say?