A puzzle by Pierre Berloquin:
In my house are a number of rooms. (A hall separated from the rest of the house by one or more doors counts as a room.) Each room has an even number of doors, including doors that lead outside. Is the total number of outside doors even or odd?
Click for solution …
July 26, 2014 | Categories: Puzzles
From The Booke of Meery Riddles, 1629:
A soldier that to Black-heath-field went,
Prayed an astronomer of his judgment,
Which wrote these words to him plainly,–
Thou shalt goe thither well and safely
And from thence come home alive againe
Never at that field shalt thou be slaine.
The soldier was slaine there at that field,
And yet the astronomer his promise held.
How?
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“The pointing of the words is the solution, as follows: Thou shalt goe thither well and safely, and from thence come home alive againe never: at that field shalt thou be slaine.”
See “Ambiguous Lines.”
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July 15, 2014 | Categories: Puzzles
By Alexander Yarosh. The position above was reached in a legal game, except that one piece has been knocked off the board. What was it?
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Black’s pawns have made 14 captures altogether, which accounts for every one of the missing white pieces. So the missing piece cannot be white. Both kings are in check, which is illegal, so the missing black piece must be on a2, blocking the rook’s check. The missing piece cannot be a queen or rook, as this would put the white king in an illegal double check. So it’s a black knight or a black bishop. And it can’t be a bishop — Black’s light-squared bishop never left its home square (the pawns on b7 and d7 have not moved), and all his pawns are still on the board, so Black has made no promotions. So the missing piece is a black knight on a2.
From Schachmatnaja komposizija, 2007.
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July 3, 2014 | Categories: Puzzles
A problem from the 2004 Moscow Mathematical Olympiad:
An arithmetic progression consists of integers. The sum of its first n terms is a power of two. Prove that n is also a power of two.
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If a and b are the first and nth terms of the progression, then the sum of the first n terms is S = (a + b)n/2. So 2S is divisible by n, and since 2S is a power of 2, so is n.
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June 29, 2014 | Categories: Puzzles
My wife and I walk up an ascending escalator. I climb 20 steps and reach the top in 60 seconds. My wife climbs 16 steps and reaches the top in 72 seconds. If the escalator broke tomorrow, how many steps would we have to climb?
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Let x be the number of steps on the escalator. I climb x – 20 steps in 60 seconds, and my wife climbs x – 16 steps in 72 seconds. This means the escalator ascends at a rate of 4 steps every 12 seconds, or 20 steps in 60 seconds. Its length is the sum of these 20 steps and the 20 steps I climb, or 40 steps.
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June 28, 2014 | Categories: Puzzles
By Rudolf L’hermet. White to mate in two moves.
June 20, 2014 | Categories: Puzzles
Only one of these statements is true. Which is it?
A. All of the below
B. None of the below
C. One of the above
D. All of the above
E. None of the above
F. None of the above
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Consider them in turn. A can’t be true because some of the other statements contradict one another (for example, C and D). B can’t be true because it implies that C is also true. Since A and B are false, C is also false, and likewise D. But E is true, and this makes F false. So the answer is E.
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June 17, 2014 | Categories: Puzzles
A conundrum by French puzzle maven Pierre Berloquin:
Caroline leaves town driving at a constant speed. After some time she passes a milestone displaying a two-digit number. An hour later she passes a milestone displaying the same two digits but in reverse order. In another hour she passes a third milestone with the same two digits (in some order) separated by a zero.
What is the speed of Caroline’s car?
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The number on the first milestone can be expressed as 10A + B, and the number on the second milestone is 10B + A. The number on the third milestone is either 100A + B or 100B + A.
Caroline’s speed is constant, so the distances between the milestones are equal. What’s the greatest this distance could be? Well, setting the two digits as far apart as possible would mean that one digit is 1 and the other is 9; then the first milestone would be at 19, the second at 91, and the third at 163. This doesn’t satisfy the puzzle, but it tells us that the number in the hundreds place must be 1 — Caroline doesn’t get as far as the 200-mile mark.
That means that either 100A or 100B must equal 100. Which is it? It must be 100A, since the number on the first milestone, where A stands for tens, is smaller than the number on the second milestone, where B stands for tens. So A is 1, the first milestone is 10 + B, the second milestone is 10B + 1, and the third milestone is 100 + B. Since the distances between milestones are equal, we have:
(10B + 1) – (10 + B) = (100 + B) – (10B + 1)
18B = 108
B = 6
A = 1 and B = 6, so the milestones carry the numbers 16, 61, and 106. Caroline’s speed is 45 mph.
(From Berloquin’s 1976 book 100 Numerical Games.)
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June 15, 2014 | Categories: Puzzles
1. A friend gives you a bottle that contains seven olives. Two of them are green and five are black. He bets that if you remove three olives at random from the bottle, they’ll include a green one. Should you take the bet?
2. Agnes has a tin of olives. It originally contained both black and green ones, but someone has been eating them, so she’s not sure of the colors of the 14 olives that remain. She removes 7 at random and finds that they’re all green. If the odds of this happening were exactly 50-50, what are the colors of the remaining 7?
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1. No: (5/7) × (4/6) × (3/5) = 0.28571428571. The odds are slightly against you even if you draw only two olives.
From John Fisher, Never Give a Sucker an Even Break, 1976.
2. One is black and six are green. You can work this out mathematically as above, or make an observation: If there’s a single black olive, it has an equal chance of lying among the first seven olives or the second seven.
From Chris Maslanka, “A Bouquet of Brainteasers,” in Barry Cipra’s Tribute to a Mathemagician, 2005.
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May 21, 2014 | Categories: Puzzles
In 2000, the residents of Luppitt, East Devon, installed a granite bench decorated with a variety of puzzles and curiosities that “it is hoped will be practical and entertaining for most of the next millennium.”
Among the puzzles is this “railway maze,” contributed by Roger Penrose. Make your way from Start in the upper left to Finish in the lower right. The catch is that your train has no reverse gear — you must move continually forward, following the natural curve of the track and making no sharp turns.
May 20, 2014 | Categories: Puzzles
Print out two copies of this pattern, cut them out, and fold each along the dotted lines, making two identical solids. Then fit these two pieces together to make a regular tetrahedron.
This sounds dead simple, but it stumped me for a while. See if you can do it. (There’s no trick — the task is just what it seems.)
May 18, 2014 | Categories: Puzzles
In S.S. Van Dine’s The Bishop Murder Case (1929), someone is killing chessplayers and leaving a black bishop at each crime scene. The prime suspect is John Pardee, promoter of a chess opening called the Pardee Gambit, which he hopes to establish in master play. But Pardee kills himself, despondent after losing to Akiba Rubinstein at the Manhattan Chess Club. It turns out that the real killer was only using the chess angle to throw suspicion onto others.
Van Dine based Pardee on a real person, Isaac Leopold Rice, who sponsored numerous tournaments in which his Rice Gambit was the required opening. But practice showed that the best White could hope for was a draw, and the line was abandoned after World War I. In 1979 Larry Evans wrote, “One of the most heavily analyzed openings in history is now never played, interred in a footnote of the latest opening manual.”
In the book, investigators determine that Pardee had faced the position above against Rubinstein shortly before his suicide. White has just realized that Black has a forced win in four moves. How does Black play?
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1. … b1=Q+! 2. Kxb1 Kd3 3. Ka1 Kc2 4. d4 (or d3) 5. Bb2#
Markham was frowning in deep perplexity.
‘You say it’s unusual for a bishop alone to mate?’ he asked Arnesson.
‘Never happens — almost unique situation. And that it should happen to Pardee of all people! Incomprehensible!’ he gave a short ironic laugh. ‘Inclines one to believe in a nemesis. You know, the bishop has been Pardee’s bête noir for twenty years — it’s ruined his life. Poor beggar! The black bishop is the symbol of his sorrow. Fate, by Gad! It’s the one chessman that defeated the Pardee gambit. Bishop-to-Knight-5 always broke up his calculations — disqualified his pet theory — made a hissing and a mocking of his life’s work. And now, with a chance to break even with the great Rubinstein, the bishop crops up again and drives him back into obscurity.’
(Thanks, Ray.)
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May 16, 2014 | Categories: Puzzles
By Dezso Elekes. White to mate in two moves.
May 9, 2014 | Categories: Puzzles
By George Edward Carpenter. White to mate in two moves.
May 3, 2014 | Categories: Puzzles
Nine wolves share a square enclosure at the zoo. Build two more square enclosures to give each wolf a pen of its own.
May 1, 2014 | Categories: Puzzles
A conundrum by the late brilliant Japanese puzzle maven Nob Yoshigahara.
Lee Sallows writes, “You have to solve this yourself, otherwise you won’t see how beautiful it is.”
April 30, 2014 | Categories: Puzzles
A cylindrical problem by Miodrag Petkovic: Imagine that the board is rolled into a cylinder, with the a-file adjoining the h-file. How can White mate in two moves?
April 25, 2014 | Categories: Puzzles
By John Griswold White. White to mate in two moves.
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The quiet 1. Kg5! forces Black to move a knight, permitting White to mate from e6 or g6. (Note that 1. Ke5 won’t work because of 1. … Nc6+!)
From Canadian Illustrated News, 1880.
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April 18, 2014 | Categories: Puzzles
Kepler’s second law holds that a line segment connecting an orbiting planet to its sun sweeps out equal areas in equal periods of time: In the diagram above, if the time intervals t are equal, then so are the areas A.
If gravity were turned off, would this still be true?
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Yes. The planet would pass into space in a straight line, and each time interval would define a triangle with height h and base t. Since a triangle’s area is half the product of its height and base, they’re all equal.
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April 12, 2014 | Categories: Puzzles, Science & Math
By Sam Loyd. White to mate in two moves.
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Typically for Loyd, the answer is the least likely move on the board:
1. Nd2! (threatening 2. Rxe2#)
If 1…Kxf2 then 2.Qg3#
If 1…Kxd2 then 2.Qc3#
If 1…Rxe3 then 2.Qxe3#
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April 11, 2014 | Categories: Puzzles
Prove that log10 5 is irrational.
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Suppose it’s rational. Then log10 5 = a/b, where a and b are positive integers. Then 10a/b = 5 and 10a = 5b. But these two values can never be equal — any positive integral power of 10 will end in 0, and any such power of 5 will end in 5. So log10 5 cannot be rational.
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April 10, 2014 | Categories: Puzzles
By George Edward Carpenter. White to mate in two moves.
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1. Qh2! threatens 2. Nf5#. If 1. … Kf4 then 2. Ne4#, and if 1. … Kd4 then the queen mates from h8.
From The Chess Players’ Chronicle, 1874.
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April 4, 2014 | Categories: Puzzles
Imagine a 1000 x 1000 chessboard on which a white king and 499 black rooks are placed at random such that no rook threatens the king. And suppose the king goes bonkers and wants to kill himself. Can he reach a threatened square in a finite number of moves if Black is trying actively to avoid this?
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Yes. Imagine that the king is on the lower left square and that on each move he takes a step along the main diagonal toward the upper right. After his first move and Black’s response, there must be no rooks on the lowest three ranks or the leftmost three files, or the king could immediately reach a threatened square. After his 998th move, the king arrives at a similar situation in the upper right: At this point there must be no rooks on the uppermost three ranks or the rightmost three files, for the same reason. This means that when the king begins his journey, all the rooks must be northeast of him, and when he finishes, all must be southwest. During the trip, then, each of Black’s 499 rooks must change both its rank and its file, in order to keep out of his way. This requires 499 x 2 = 998 moves. So, at the latest, Black will run out of time just a half-move too soon: White will squeeze Black’s maneuvering room down to zero on his own 998th move, just before Black evacuates the last rook.
From Miodrag Petkovic, Mathematics and Chess, 1997.
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March 26, 2014 | Categories: Puzzles
