Walter Penney of Greenbelt, Md., offered this poser in the August 1969 issue of Word Ways: The Journal of Recreational Linguistics. Below are five groups of English words. Each group appears also in a foreign language. What are the languages?
- aloud, angel, hark, inner, lover, room, taken, wig
- alas, atlas, into, manner, pore, tie, vain, valve
- ail, ballot, enter, four, lent, lit, mire, taller
- banjo, chosen, hippo, pure, same, share, tempo, tendon
- ago, cur, dare, fur, limes, mane, probe, undo
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March 6, 2014 | Categories: Language, Puzzles
In projective geometry, every family of parallel straight lines intersects at an infinitely distant point. Chess problem composers in the former Yugoslavia have adapted this idea for the chessboard, adding four special squares “at infinity.”
Now a queen on a bare board, for example, can zoom off to the west (or east) and reach a square “at infinity” from which she attacks every rank on the board simultaneously from both directions. She might also zoom to the north (or south) to reach a different square at infinity; from this one she attacks every file simultaneously, again from both directions. Finally she can zoom to the northwest or southeast and attack all the diagonals parallel to a8-h1, or zoom to the northeast or southwest and attack all the diagonals parallel to a1-h8. These four “infinity squares,” plus the regular board, make up the field of play.
N. Petrovic created the problem below, published in Matematika Na Shahmatnoi Doske. White is to play and mate in at least two moves. Can you find the solution?
Click for solution …
March 5, 2014 | Categories: Puzzles
A Russian problem from the 1999 Mathematical Olympiad:
A father wants to take his two sons to visit their grandmother, who lives 33 kilometers away. His motorcycle will cover 25 kilometers per hour if he rides alone, but the speed drops to 20 kph if he carries one passenger, and he cannot carry two. Each brother walks at 5 kph. Can the three of them reach grandmother’s house in 3 hours?
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Yes. The father sets out with his first son, leaving the second boy to walk alone. Father and son drive 24 kilometers, which takes 6/5 hours. Then the father puts down the first son and rides back to meet the second son 9 kilometers from home, which takes 3/5 hours. Finally he and the second son ride 6/5 hours more and reach grandmother’s house just as the first son is arriving there on foot. Each son spends 6/5 hours riding 24 kilometers and 9/5 hours walking 9 kilometers, so the three of them reach grandmother’s in exactly 3 hours.
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March 3, 2014 | Categories: Puzzles
By Ottmar Weiß. White to mate in two moves.
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The surprising 1. Qh3! leaves Black with no options. If he moves his knight then White can take the rook with mate, and if he moves the rook then White’s knight will mate.
From Teplitz-Schönauer Anzeiger, 1921.
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February 28, 2014 | Categories: Puzzles
From Sam Loyd:
Two children, who were all tangled up in their reckoning of the days of the week, paused on their way to school to straighten matters out.
“When the day after tomorrow is yesterday,” said Priscilla, “then ‘today’ will be as far from Sunday as that day was which was ‘today’ when the day before yesterday was tomorrow!”
On which day of the week did this puzzling prattle occur?
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“The two children were so befogged over the calendar that they had started on their way to school on Sunday morning!”
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February 24, 2014 | Categories: Puzzles
Six boys are accused of stealing apples. Exactly two are guilty. Which two? When the boys are questioned, Harry names Charlie and George, James names Donald and Tom, Donald names Tom and Charlie, George names Harry and Charlie, and Charlie names Donald and James. Tom can’t be found. Four of the boys who were questioned named one guilty boy correctly and one incorrectly, and the fifth lied outright. Who stole the apples?
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In presenting this problem in Ingenuity in Mathematics, Ross Honsberger gave a complex analysis, but his friend Scott Vanstone showed a simple way to visualize it, building on an observation by John Annulis:
Each boy is represented by a point, and each pair of boys named in an accusation are joined by a line. Now we know that each line has a thief at one end and an innocent at the other, except for one line, which has an innocent at both ends (due to the outright liar). Since four boys named one guilty boy correctly, the two points that represent the thieves are attached to a total of four lines. And because no boy made two accurate accusations, the thieves themselves are not joined by a line. The only two points that satisfy all these requirements indicate Charlie and James.
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February 23, 2014 | Categories: Puzzles
A group of four missionaries are on one side of a river, and four cannibals are on the other side. The two groups would like to exchange places, but there’s only one rowboat, and it holds only three people, and only one missionary and one cannibal know how to row, and the cannibals will overpower the missionaries as soon as they outnumber them, either on land or in the boat. Can the crossing be accomplished?
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Yes. First three cannibals cross; then the two rowers return, bearing an extra missionary; then the rowing missionary crosses with a cannibal and returns with two missionaries; and finally the rowing cannibal crosses:
Mmmm Cccc
MmmmCcc c
mmcc MmCc
Mmmccc mC
ccc MmmmC
Cccc Mmmm
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February 22, 2014 | Categories: Puzzles
By Sam Loyd. White to mate in two moves.
February 21, 2014 | Categories: Puzzles
A brainteaser by Chris Maslanka:
A packet of sugar retails for 90 cents. Each packet includes a voucher, and nine vouchers can be redeemed for a free packet. What is the value of the contents of one packet? (Ignore the cost of the packaging.)
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80 cents. It’s tempting to say that the answer is 81 cents, since paying for nine 90-cent packets will give us 10 packets total, meaning we’ve paid ($0.90 × 9) / 10 = $0.81 for each packet. But the free packet itself includes a voucher, which is worth something.
Maslanka suggests this thought experiment: Borrow a packet from a friend, then buy eight additional packets for $0.90 × 8 = $7.20. Consume the nine packets, redeem the nine vouchers for a free packet, and use this to repay your friend. You’ve consumed nine packets of sugar for $7.20, so each packet is worth 80 cents.
From Maslanka’s “A Bouquet of Brainteasers,” in Tribute to a Mathemagician, 2005.
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February 19, 2014 | Categories: Puzzles
A variation on yesterday’s puzzle:
Suppose there are six bottles of pills, and more than one of them may contain defective pills that weigh 6 grams instead of 5. How can we identify the bad bottles with a single weighing?
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Take 1 pill from the first bottle, 2 from the second, 4 from the third, 8 from the fourth, 16 from the fifth, and 32 from the sixth. This gives us 63 pills that ought to weigh 315 grams. As before, the bad pills will add some excess weight, but now that weight can tell us which combination of bottles is bad. Suppose the 63 pills weigh 336 grams, or 21 grams too much. That extra 21 grams can only have come from the fifth bottle (16 grams), the third (4 grams), and the first (1 gram). The same is true for any overage: Because any integer is uniquely expressible as the sum of powers of 2, the excess weight will always identify uniquely any combination of bad bottles.
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February 16, 2014 | Categories: Puzzles
An efficiency-minded pharmacist has just received a shipment of 10 bottles of pills when the manufacturer calls to say that there’s been an error — nine of the bottles contain pills that weigh 5 grams apiece, which is correct, but the pills in the remaining bottle weigh 6 grams apiece. The pharmacist could find the bad batch by simply weighing one pill from each bottle, but he hits on a way to accomplish this with a single weighing. What does he do?
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He numbers the bottles from 1 to 10, then takes one pill from bottle one, two pills from bottle two, and so on. This gives him a pile of 55 pills that ought to weigh 275 grams. Its difference from this weight can tell him which bottle is bad: If the pile is 1 gram overweight then the bad pills are in bottle one, if 2 grams then they’re in bottle two, and so forth.
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February 15, 2014 | Categories: Puzzles
This week’s puzzle has a twist: Imagine that the board has been rolled into a cylinder so that the a- and h-files are joined and pieces can move across the boundary. How can White mate in two moves?
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1. Ra5! The rook tours the fifth rank and returns to its original square, forcing 1. … Kc1 2. Ra1#. The normal-seeming 1. Rxa6, which would ordinarily lead to the same mate, here fails because Black can take the rook.
By Kuznecov and Plaskin, from Miodrag Petković, Mathematics and Chess, 1997.
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February 14, 2014 | Categories: Puzzles
The first 10 letters of the alphabet, ABCDEFGHIJ, form a cipher that conceals the name of a number less than 100. What is the number?
February 10, 2014 | Categories: Puzzles
- How does a deaf man indicate to a hardware clerk that he wants to buy a saw?
- How can you aim your car north on a straight road, drive for a hundred yards, and find yourself a hundred yards south of where you started?
- What runs fore to aft on one side of a ship and aft to fore on the other?
- A very fast train travels from City A to City B in an hour and a quarter. But the return trip, made under identical conditions, requires 75 minutes. Why?
- Does Canada have a 4th of July?
- Exhausted, you go to bed at 8 p.m., but you don’t want to miss an appointment at 10 a.m. the next day, so you set your alarm clock for 9. How many hours do you sleep?
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- He says, “I want to buy a saw.”
- Drive backward.
- Its name.
- An hour and a quarter is 75 minutes.
- Yes.
- One.
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February 9, 2014 | Categories: Puzzles
By William Anthony Shinkman. White to mate in two moves.
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After the quiet 1. Qf2!, the queen will mate next move.
From the Dubuque Chess Journal, July 1871.
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February 7, 2014 | Categories: Puzzles
Imagine a die that exactly covers one square of a checkerboard. Place the die in the top left corner with the 6 uppermost. Now, by tipping the die over successively onto each new square, can you roll it through each of the board’s 64 squares once and arrive in the upper right, so that the 6 is exposed at the beginning and end but never elsewhere?
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The solution above is unique. I think this puzzle was devised by John Harris of Santa Barbara, Calif., in 1964; I found it in Miodrag Petkovic’s Mathematics and Chess, 1997.
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February 5, 2014 | Categories: Puzzles
In a photograph, is there a way to distinguish between a landscape and its reflection?
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Yes. Because the camera is above the reflective surface, the two images are not identical. If the camera lens is, say, 5 feet above the water, then the reflected image appears as if viewed from 5 feet below it.
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February 4, 2014 | Categories: Puzzles
By Viktor Holst. White to mate in two moves.
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1. Qe4! and, one way or another, the queen will mate.
From the Illustrated Family Journal, 1917.
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January 31, 2014 | Categories: Puzzles
manzil
n. the distance between two stopping places
Another puzzle by Sam Loyd: Two ferry boats ply the same route between ports on opposite sides of a river. They set out simultaneously from opposite ports, but one is faster than the other, so they meet at a point 720 yards from the nearest shore. When each boat reaches its destination, it waits 10 minutes to change passengers, then begins its return trip. Now the boats meet at a point 400 yards from the other shore. How wide is the river?
“The problem shows how the average person, who follows the cut-and-dried rules of mathematics, will be puzzled by a simple problem that requires only a slight knowledge of elementary arithmetic. It can be explained to a child, yet I hazard the opinion that ninety-nine out of every hundred of our shrewdest businessmen would fail to solve it in a week. So much for learning mathematics by rule instead of common sense which teaches the reason why!”
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At their first meeting, 720 yards from one shore, the boats together have traveled a distance equal to the width of the river. When each reaches its destination, the combined distance is twice the width of the river. At their second meeting, 400 yards from the other shore, the combined distance is three times the river’s width. So at the second meeting each boat has gone three times as far as it had at the first meeting.
At the first meeting one boat had traveled 720 yards, so at the second meeting it has gone three times this distance, or 2,160 yards. At this point it has covered 400 yards since doubling back, so the river is 2,160 – 400 = 1,760 yards, or one mile, wide.
“The amount of time each ship consumed at the landing does not affect the problem.”
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January 30, 2014 | Categories: Language, Puzzles
Sam Loyd devised this puzzle in 1898. Begin at the heart in the center and move three squares in any of the eight directions, north, south, east, west, northeast, northwest, southeast, or southwest. You’ll land on a number; take this as the length of your next “march,” which again can go in any of the eight directions. “Continue on in this manner until you come upon a square with a number which will carry you just one step beyond the border, thus solving the puzzle.”
Interestingly, Loyd devised this puzzle expressly to defeat Leonhard Euler’s method of solving mazes. “Euler, the great mathematician, discovered a rule for solving all manner of maze puzzles, which, as all good puzzlists know, depends chiefly upon working backwards. This puzzle, however, was built purposely to defeat Euler’s rule and out of many attempts is probably the only one which thwarts his method.” The original puzzle, as published in the New York Journal and Advertiser, contained a flaw that permitted multiple solutions. That’s been corrected here — there’s only one way out.
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Move:
- 3 squares southwest to 4.
- 4 squares southwest to 6.
- 6 squares northeast to 6.
- 6 squares northeast to 2.
- 2 squares northeast to 5.
- 5 squares southwest to 4.
- 4 squares southwest to 4.
- 4 squares southwest to 4.
- 4 squares southeast “to liberty.”
UPDATE: Wait, there’s at least one additional solution — in the last step you can go northwest as well as southeast. (Thanks, Carlos.) Now I wonder whether there’s ever been a version with a unique solution.
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January 27, 2014 | Categories: Puzzles
Two-move chess is just like regular chess except that each side makes two moves at a time. Prove that White, who goes first, can be sure of at least a draw.
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Well, suppose that isn’t the case. This means that Black has a strategy that’s guaranteed to win no matter what White plays. But imagine that, on his first pair of moves, White moves a knight out onto the board and then returns it to its home square. Now Black finds himself in White’s shoes: All the pieces are in their original positions and it’s his turn to move. Our hypothesis says that anyone in these circumstances is sure to lose, but it also says that Black is sure to win. This is a contradiction, so the hypothesis must be wrong: It can’t be the case that White is sure to lose. Hence he must be able to achieve at least a draw.
From Svetoslav Savchev, Mathematical Miniatures, 2003.
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January 25, 2014 | Categories: Puzzles
By George Edward Carpenter. White to mate in two moves.
January 23, 2014 | Categories: Puzzles
