You toss 6 fair coins, and I toss 5 fair coins. What is the probability that you get more heads than I do?
Click for solution …
February 23, 2015 | Categories: Puzzles
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?
|
SelectClick for Answer |
A player who sees two hats of the same color (say, orange) guesses the opposite color (maroon). A player who sees hats of two different colors passes. There are 8 possible arrangements of hats:
OOO
OOM
OMM
MOO
MMO
MMM
MOM
OMO
In six of these, two of the players see hats of different colors and so will pass, and the third player sees two hats of the same color and guesses the opposite color, winning. In the other two cases, all three players are wearing hats of the same color, so all guess the wrong color and lose. This strategy produces a win in six of the eight cases, so the players will win 3/4 of the time.
From “Applications of Recursive Operators to Randomness and Complexity,” the 1998 UCSB doctoral thesis of computer scientist Todd Ebert.
|
February 17, 2015 | Categories: Puzzles
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?
|
SelectClick for Answer |
No. Because the cards that are put aside are paired by color, the remaining cards must be divided equally between red and black. So at the end of the game you and I will always have the same number of cards. You’ll lose a dollar every time you play.
From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.
|
February 14, 2015 | Categories: Puzzles
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.
|
SelectClick for Answer |
1. Qd6! protects both white rooks and keeps Black’s bishop pinned. Black has only two options:
1. … Kc3 Rf3+ 2. Bxf3#
1. … Ke3 Rb3+ 2. Bxb3#
From Benjamin Glover Laws, The Two-Move Chess Problem, 1890.
|
February 8, 2015 | Categories: Puzzles
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?
|
SelectClick for Answer |
Trick question. Within any calendar year, Christmas and New Year’s Day always fall on different days:
|
February 6, 2015 | Categories: Puzzles
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?
|
SelectClick for Answer |
If the prisoners knew that both switches were off at the start, then the solution would be fairly straightforward. One switch (say, the left one) will be used for counting, and the other is a meaningless dummy. One prisoner is chosen as the Counter. Each other prisoner is instructed to flip the left switch from off to on one time, at the first opportunity, but otherwise to flip the dummy switch on each visit. When the Counter enters the room and finds the left switch on, he turns it off; otherwise he too flips the dummy switch. When the Counter has found the left switch turned on 22 times, he knows that all prisoners have visited the room and can safely tell the warden so.
Because the prisoners don’t know the initial states of the switches, this plan must be enhanced (otherwise, if the left switch starts in the on position, then the Counter will make his announcement prematurely). The solution is to ask each prisoner to flip the left switch from off to on twice in total, rather than once. The Counter will wait until he’s seen the left switch turned on 44 times altogether, and then tell the warden that all prisoners have visited the room.
(At first I wondered why they couldn’t just stick with the first scenario but add 1 to the expected count to allow for the possibility that the left switch had started in the on position. But this won’t work: If the Counter is waiting until he’s found the left switch turned on 23 times, and if it started in the off position, then he’ll wait forever, because each of the 22 other prisoners will turn it on only once.)
|
February 4, 2015 | Categories: Puzzles
By J. Keeble. White to mate in two moves.
|
SelectClick for Answer |
1. Ng5!, and the other knight will discover mate.
From O.A. Brownson Jr., Chess Problems, 1876.
|
January 28, 2015 | Categories: Puzzles
Draw an arbitrary triangle and build an equilateral triangle on each of its sides, as shown.
Now show that AP = BQ = CR.
|
SelectClick for Answer |
Rotating the figure 60° counterclockwise around B carries A to R and P to C. And rotating it 60° clockwise around A carries B to R and Q to C. So AP = BQ = CR.
From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.
|
January 23, 2015 | Categories: Puzzles
By Theophilus A. Thompson. White to mate in two moves.
January 19, 2015 | Categories: Puzzles
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?
|
SelectClick for Answer |
No. When viewed from above the three pucks form triangle ABC. With each hit the orientation of the triangle changes, so that sometimes “ABC” reads clockwise, sometimes counterclockwise. The original position was reached after zero hits, so it can be restored only after an even number of hits.
From Arthur Engel, Problem-Solving Strategies, 1998.
|
January 15, 2015 | Categories: Puzzles
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?
|
SelectClick for Answer |
I don’t have the olympiad solution, but it seems to me that there must be an equal number of knights and scoundrels. The first half of the interviewees are knights, and the second half are scoundrels.
If there were more than n/2 knights, then they’d crowd out the scoundrels and their greater claims could not be true.
If there were more than n/2 scoundrels, then they’d crowd out the knights and their lesser claims could not be false.
UPDATE: Also, n must be even. If it’s odd, then the middle interviewee can be neither a knight nor a scoundrel, which is impossible. Thanks to those who wrote in to point this out.
|
January 10, 2015 | Categories: Puzzles
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?
|
SelectClick for Answer |
POCO is 4595 and MUCHO is 68925.
There are 15 POCOs, so the sum of each column is a multiple of 15, and hence (without the carry from the column to its right) ends in a 5 or a 0. The rightmost column has no carry, so the O in MUCHO stands for 0 or 5. Which is it?
- If it’s 0, that means that the sum of all the Os in the rightmost column is 0, and that there’s no carry into the tens column. If there’s no carry, we need to find a digit for C that when multiplied by 15 produces a two-digit answer (the CH in MUCHO) that starts with that digit (since the hundreds column in the sum is otherwise zero). The only digit that will do that is 1 (1 × 15 = 15). But if CH is 15 then there’s no carry into the thousands column, which means that U must be 0 or 5 (15 × P). And we’ve already assigned 0 to O and 5 to H. So this can’t be right.
- If it’s 5, then the carry into the tens column is 7 (15 × 5 = 75). What is C? Every possibility from 0 to 8 produces conflicts or requires C to have two different values, but C = 9 produces a sum in the tens column of 142 (15 × 9 plus the 7 carry). In that case H = 2 and 14 is carried into the hundreds column. We know that O is 5, so that column now totals 89 (5 × 15 + 14). C is 9 and 8 is carried into the thousands column, where it’s added to 15 Ps. Here every possibility except P = 4 creates duplications or conflicts, so U is 8 and M is 6.
“Because O must equal 5 and C must equal 9 and P must equal 4, I believe the solution is unique,” writes NYU computer scientist Allan Gottlieb, who conducts the puzzle column. “I like these alphabetical number problems; but, cannot recall seeing this one before.”
|
January 7, 2015 | Categories: Puzzles
By H.R. Agnel. White to mate in two moves.
January 6, 2015 | Categories: Puzzles
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.
|
SelectClick for Answer |
Consider all the possible divisions, and in each case count up the total number of cases E in which a member has an enemy in his own house. The division in which E is minimized must have the required property. If it didn’t, that would mean that some member has at least two enemies in his own house. In that case he’d have one enemy at most in the other house, and could reduce E further by moving to the opposite house. That’s a contradiction, so the required task must be possible.
|
January 2, 2015 | Categories: Puzzles
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?
|
SelectClick for Answer |
The probability is 100 percent. First suppose that Pile A is all red and Pile B all black. This fulfills the condition. Now move one red card from A to B. In order to keep 52 cards in each pile, we must also move a black card from B to A, so this new arrangement also fulfills the condition. And so on: So long as there are 52 cards of each color and 52 cards in each pile, the number of red cards in one pile must equal the number of black cards in the other. So you don’t have to view any cards at all.
See Alcohol Problem.
|
December 27, 2014 | Categories: Puzzles
By S.C. Dickinson. White to mate in two moves.
December 26, 2014 | Categories: Puzzles
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?
|
SelectClick for Answer |
a + b = 1/4, b + c = 1/2, and a + d = 1/2. Triangles a and c are similar, and c has twice the linear dimensions of a, so c = 4a. That’s enough to work out the areas: a = 1/12, b = 1/6, c = 1/3, and d = 5/12.
From University of Toronto mathematician Ed Barbeau’s After Math (1995).
|
December 15, 2014 | Categories: Puzzles
A rather baroque problem by Émile Leonard Pradignat. White to mate in two moves.
|
SelectClick for Answer |
The surprising 1. Qxh8! leaves Black with no options. If he captures on c4 or d3 with his knight, then White recaptures with his king, giving mate. Any other response allows mate along the first rank or first file.
From Prager Tagblatt, Jan. 17, 1904.
|
December 14, 2014 | Categories: Puzzles
By Nikolai Maximov. White to mate in two moves.
November 30, 2014 | Categories: Puzzles
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?
|
SelectClick for Answer |
Triangles CAM and ABN have the same area. Rotating triangle ABC 120° takes CAM into ABM1. Since ABN and ABM1 have the same area and both N and M1 fall on BC, N = M1. Since rotating CM produces AN, the angle between them is 120°, or 180° – 120° = 60°, which is the same.
(Posed by V. Proizvolov in Math Horizons, Spring 1994.)
|
November 25, 2014 | Categories: Puzzles
By Sam Loyd. White to mate in two moves.
|
SelectClick for Answer |
1. Re3! wins at once: If the king takes the rook then 2. Qd2# is mate; if the pawn moves then 2. Re4#; and if the knight moves then 2. Qe5#.
From Alain C. White, Sam Loyd and His Chess Problems, 1913.
|
November 16, 2014 | Categories: Puzzles
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?
|
SelectClick for Answer |
Seventeen cents:
17 cents = 1 dime + 1 nickel + 2 pennies
34 cents = 1 quarter + 1 nickel + 4 pennies
51 cents = 1 half dollar + 1 penny
From Oswald Jacoby’s Mathematics for Pleasure (1962). See Cash and Carry.
|
November 13, 2014 | Categories: Puzzles
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.”
November 10, 2014 | Categories: Puzzles, Science & Math
