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?
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.
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?
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.
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?
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.”
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.
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.
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?
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.
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 + 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).
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.
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?
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.)
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?
Cemetery caretaker John Hammond decoded the inscription in 1947, some 80 years after the women were interred. The message begins at the I just west of the center; it starts in a counterclockwise spiral but then begins to zigzag:
IN MEMORIAM
HENRIETTA 1ST WIFE OF S BEAN M. D.
WHO DIED 27TH SEP 1865 AGED 23 YEARS 2 MONTHS & 17 DAYS
& SUSANNA HIS 2ND WIFE
WHO DIED 27TH APRIL 1867 AGED 26 YEARS 10 MONTHS & 15 DAYS
2 BETTER WIVES 1 MAN NEVER HAD
THEY WERE GIFTS FROM GOD BUT ARE NOW IN HEAVEN
MAY GOD HELP ME S. B. TO MEET THEM THERE
You’re a logician who wants to know which of two roads leads to a village. Standing nearby, inevitably, are three natives: one always lies, one always tells the truth, and one answers randomly. You don’t know which is which, and you can ask only two yes-or-no questions, each directed to a single native. How can you get the information you need?
Ask native A, “Is B more likely than C to tell the truth?” Then go to the native he indicates is less trustworthy and ask, “If I were to ask you whether this road goes to the village, would you say yes?”
This procedure ensures that you won’t put the second question to the random answerer. Once that’s sure, then you’re just facing the classic version of the puzzle, with one liar, one truth-teller, and one yes-no question.
Suppose you have three identical Dewar flasks labeled A, B, and C. (A Thermos is a Dewar flask.) You also have an empty container labeled D, which has thermally perfect conducting walls and which fits inside a Dewar flask.
Pour 1 liter of 80°C water into flask A and 1 liter of 20°C water into flask B. Now, using all four containers, is it possible to use the hot water to heat the cold water so that the final temperature of the cold water is higher than the final temperature of the hot water? How? (You can’t actually mix the hot water with the cold.)
Pour half the cold water into D and put D into the hot flask (A). The final temperature in both A and D will be about 60°C. Empty D into C and the repeat the procedure, pouring the remaining half of the cold water in B into D and immersing it in the hot water. Now the final temperature of the water in A and D will be about 47°C. Empty D again into C; the final temperature of the mixture there will be about 53°C. So in two operations we’ve warmed the cold water to 53°C and cooled the hot water to 47°C.
From Christopher Jargodzki’s books Science Brain-Twisters (1976) and Mad About Physics (2002).
Timothy and Urban are playing a game with two six-sided dice. The dice are unusual: Rather than bearing a number, each face is painted either red or blue.
The two take turns throwing the dice. Timothy wins if the two top faces are the same color, and Urban wins if they’re different. Their chances of winning are equal.
The first die has 5 red faces and 1 blue face. What are the colors on the second die?
Throwing two six-sided dice produces 36 possible outcomes. If Timothy and Urban have equal chances of winning, then there must be 18 outcomes in which both dice display the same color. If x is the number of red faces on the second die, then:
18 = 5x + 1(6 – x)
x = 3
The second die must have 3 red faces and 3 blue faces.
(From Pierre Berloquin, 100 Numerical Games, 1994.)
From the Second All Soviet Union Mathematical Competition, Leningrad 1968:
On a teacher’s desk sits a balance scale, on which are a set of weights. On each weight is the name of at least one student. As each student enters the classroom, she moves all the weights that bear her name to the other side of the scale.
Before any students enter, the scale is tipped to the right. Prove that there’s some set of students that you can let into the room that will will tip the scale to the left.
Make a list of all possible subsets of the students, including the empty set and the full set, and suppose you let in all of these groups. Each weight would spend half its time in the left pan and half in the right, so the total weight for each pan is the same. Since the empty set produces a tip to the right, there must be some other set that produces a tip to the left.
Two circles intersect. A line AC is drawn through one of the intersection points, B. AC can pivot around point B — what position will maximize its length?
Call the other intersection point D and draw AD and DC. All angles inscribed in a circle and subtended by the same chord are equal, so angle BAD retains the same measure as A travels around its circle. Similarly, angle BCD remains the same as C travels around its circle. This means that triangle ADC will always have the same shape: As line AC pivots around B, triangle ADC will vary in size but remain self-similar.
So which position will maximize its size? AD and DC are chords of their respective circles, and the longest chord is a diameter. So turn the triangle until both of these lines are diameters (this will happen simultaneously); at that point triangle ADC will reach its maximum size and line AC its maximum length.
From Mogens Larsen in Richard Guy and Robert Woodrow, eds., The Lighter Side of Mathematics, 1994.
