Crying Wolf

A puzzle from reader Paul Sophocleous:

Van Helsing, who is of course famous for his part in the destruction of Dracula, has had many other encounters with supernatural creatures. In the early hours of one morning, he was woken by a loud knock at the door. “Come quickly!” cried the chief of police. “There’s been a ghastly attack at the manor house on the hill!”

Van Helsing dressed hurriedly and followed the chief. A grisly sight met him when he arrived. The front door of the house was open, and the beam of light that came from within shone on the body of a young man lying on the path. His throat had been torn out viciously, as though he had been attacked by some kind of hideous wild beast. Van Helsing looked around, but the grounds were dark, since the moon had set some time before, and he could see nothing else.

He stepped inside and found that several officers of the local constabulary were comforting a woman who appeared to be the maid. “It was horrible!” she cried. “I came down here after hearing some racket outside, and I found the young master at the door. ‘There’s something out there,’ he told me, ‘some beast, and I mean to drive it off.’ And he had in his hand the poker from the fireplace as a weapon. But when he opened the door, it was on him in a flash, a great beast, all hairy and shaggy, bigger than a man it was!”

Van Helsing stepped forward. “What was it?” he demanded.

The maid let out a little scream and gasped, “It was a werewolf!” And with that she fainted dead away.

“Could it be, Van Helsing?” said the chief, sounding worried.

Van Helsing shook his head. “Not a chance.”

Why not?

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Van Helsing knew it couldn’t be a werewolf because the moon was not full. A full moon is always directly opposite the sun in the sky — it sets when the sun rises and rises when the sun sets. However, if it was the early hours of the morning, the full moon would still be up, lighting up the land. Since the moon had already set, it could not possibly be full. And if the moon wasn’t full, it could not have transformed anyone into a werewolf.

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January 7, 2017January 15, 2017 | Puzzles

A Compensatory Harmonica

https://www.flickr.com/photos/soppyfrog/3931666786 — Image: Flickr

A problem from the American Mathematical Monthly, March 1930:

Two men jointly own x cows. They sell these for x dollars per head and use the proceeds to buy some sheep at $12 per head. Their income from the cows isn’t divisible by 12, so they buy a lamb with the remainder. Later they divide the flock so that each man has the same number of animals. This leaves the man with the lamb somewhat short-changed, so the other man gives him a harmonica. What’s the harmonica worth?

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The men sold x cows for $x, so their income from that sale is $x². x² isn’t divisible by 12, so x can’t be a multiple of 6 (if it were, then x² would be divisible by 36 and hence by 12). So x is 12n ± y, where n is any integer and y is an integer from 1 to 5. Each man ends up with the same number of animals, so the total number of animals must be even, which means there’s an odd number of sheep and one lamb. So (12n ± y)² when divided by 12 should leave a remainder and have an odd integer for its quotient. $\displaystyle \frac{\left ( 12n \pm y \right )^{2}}{12}=\frac{144n^{2}\pm 24ny+y^{2}}{12}=12n^{2} \pm 2ny + \frac{y^{2}}{12}.$ 12n² ± 2ny will always be even, so that last term, y²/12, must provide an odd contribution toward this quotient. In order to do this, y² must exceed 12, so y must be 4 or 5. If it’s 5, we get $\displaystyle \frac{y^{2}}{12}=\frac{25}{12}=2+\frac{1}{12},$ which provides an even contribution (2) to the quotient. So y must be 4 and y² = 16: $\displaystyle \frac{y^{2}}{12}=\frac{16}{12}=1+\frac{4}{12}.$ That’s what we need; the integer part of this quotient, 1, is odd. Since the remainder when 4² is divided by 12 is 4, a lamb costs $4. Consequently, in dividing the flock one man got a $4 lamb and the other got a $12 sheep. To make this fair, the traded harmonica must be worth $4. (The problem was proposed by J.H. Neelley, T.L. Smith, and P.S. Ganesa Sastri in AMM 37:3, 162; Ross Honsberger introduced the harmonica in recounting it in his Mathematical Morsels, 1978. This solution is by P.S. Ganesa Sastri of Trichinopoly, South India.)

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January 5, 2017January 15, 2017 | Puzzles

Podcast Episode 135: Lateral Thinking Puzzles

Here are six new lateral thinking puzzles to test your wits and stump your friends — play along with us as we try to untangle some perplexing situations using yes-or-no questions.

See full show notes …

December 26, 2016October 1, 2017 | Podcast · Puzzles

Graft

graft puzzle

You’re a venal king who’s considering bribes from two different courtiers.

Courtier A gives you an infinite number of envelopes. The first envelope contains 1 dollar, the second contains 2 dollars, the third contains 3, and so on: The nth envelope contains n dollars.

Courtier B also gives you an infinite number of envelopes. The first envelope contains 2 dollars, the second contains 4 dollars, the third contains 6, and so on: The nth envelope contains 2n dollars.

Now, who’s been more generous? Courtier B argues that he’s given you twice as much as A — after all, for any n, B’s nth envelope contains twice as much money as A’s.

But Courtier A argues that he’s given you twice as much as B — A’s offerings include a gift of every integer size, but the odd dollar amounts are missing from B’s.

So who has given you more money?

December 23, 2016December 22, 2016 | Puzzles

The Two Errand Boys

dudeney errand boys

Another conundrum from Henry Dudeney’s Canterbury Puzzles:

A country baker sent off his boy with a message to the butcher in the next village, and at the same time the butcher sent his boy to the baker. One ran faster than the other, and they were seen to pass at a spot 720 yards from the baker’s shop. Each stopped ten minutes at his destination and then started on the return journey, when it was found that they passed each other at a spot 400 yards from the butcher’s. How far apart are the two tradesmen’s shops? Of course each boy went at a uniform pace throughout.

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All that is necessary is to add the two distances at which they meet to twice their difference. Thus 720 + 400 + 640 = 1760 yards, or one mile, which is the distance required. A more general solution can be found on page 416 of Pi Mu Epsilon Journal, Spring 1982 (PDF).

December 19, 2016January 1, 2017 | Puzzles

Growth Potential

Suppose you’re working on an algebraic expression that involves variables, addition, multiplication, and parentheses. You try repeatedly to expand it using the distributive law. How do you know that the expression won’t continue to expand forever?

For example, expanding

(x + y)(s(u + v) + t)

gives

x(s(u + v) + t) + y(s(u + v) + t),

which has more parentheses than the original expression.

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Set all the variables equal to 2! “The point of the distributive law is that its application doesn’t change the value of the expression,” writes Dartmouth mathematician Peter Winkler. “The value of the initial expression limits the size of anything you can get from it by expansion.” Winkler received the puzzle from Dick Lipton of the College of Computing at Georgia Tech and included it in his “Seven Puzzles You Think You Must Not Have Heard Correctly” at the seventh Gathering for Gardner, in March 2006.

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December 15, 2016December 25, 2016 | Puzzles

Black and White

hochberg puzzle

To kill some time before a meeting of chess grandmasters, Burt Hochberg offered this anonymous puzzle from the 15th century. White must place four white rooks on the board, one at a time, giving check with each one. After each placement the black king can respond with any normal legal move. How can White plan his moves so that the fourth rook reliably gives checkmate?

There’s no trick, and in fact there are several solutions, but Hochberg says the grandmasters studied the position for several minutes before Paul Keres came up with an answer. What was it?

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The first rook goes on d2. Now if the king flees to the left, check him from the a-file, and if he flees to the right, check him from the g-file. So, for example, if he goes to e4, then play 2. g4+: That’s the key — this plan traps the king in a narrow pair of files, and now it’s an easy matter to attack those with the remaining rooks (above, for example: 2. … Kf3 3. Rf4+ 4. Ke3 Re2#). From Chess Braintwisters, 1999.

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December 14, 2016December 25, 2016 | Puzzles

8 Is Enough

8 Is Enough - problem

This is said to have been the most popular problem presented in the American Mathematical Monthly. It was proposed by P.L. Chessin of Westinghouse in the April 1954 issue. Each of the digits in this long division problem has been replaced with an x — except for a single 8 in the quotient. Can you reconstruct the problem?

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In long division, whenever two digits are brought down instead of one — as happens twice in this problem — a zero must appear in the quotient. That tells us that the quotient is x080x. The visible 8 multiplied by the divisor produces a three-digit number. But when the last number in the quotient is multiplied by the divisor, it produces a four-digit number. So the last digit in the quotient must be 9. The divisor has three digits and produces a three-digit number when multiplied by the visible 8. So the divisor must be less than 125, because 8 × 125 = 1000. That tells us something further. When the divisor is multiplied by the first digit of the quotient, it produces a three-digit number, and when this is subtracted from the first four digits of the dividend it yields a two-digit difference. That means that the first digit of the quotient must be greater than 7, because 7 × 124 = 868, and that would yield a three-digit difference when subtracted from even the smallest possible four-digit number, 1000 (1000 – 868 = 132). An 8 or a 9 could produce a two-digit difference, but a 9 would produce a four-digit number when multiplied by the divisor. So the first digit of the quotient is 8, and the full quotient is 80809. Now, 80809 × 123 = 9939507, which has only seven digits, and the dividend has eight. So the divisor is less than 125 and greater than 123, which means it’s 124. That gives us everything we need:

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December 13, 2016December 25, 2016 | Puzzles

The Squire’s Puzzle

dudeney squire puzzle

Another conundrum from Henry Dudeney’s The Canterbury Puzzles:

A squire has drawn a portrait of King Edward III with a single continuous stroke of his pen. “‘Tis a riddle to find where the stroke doth begin and where it doth also end. To him who shall first show it unto me will I give the portraiture.” What is the answer?

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“The portrait may be drawn in a single line because it contains only two points at which an odd number of lines meet, but it is absolutely necessary to begin at one of these points and end at the other. One point is near the outer extremity of the King’s left eye; the other is below it on the left cheek.”

December 2, 2016December 11, 2016 | Puzzles

The Cornish Cliff Mystery

dudeney cliff puzzle

Police have chased two thieves to the Cornish coast. There they find that two sets of footprints depart the hard road and cross soft soil to the edge of a cliff, where they end. The cliff juts out 200 feet above sea-washed boulders. No one could survive a fall from this height, there is no way to descend the cliff, and there are no other footprints.

The police have proven that the footprints match those of the fleeing criminals. The small foot belongs to Marsh, who apparently takes relatively long strides, walking heavily on his heels. Lamson takes shorter strides, treading more on his toes and evidently following behind Marsh, as he sometimes treads over the smaller man’s footprints.

The two men did not walk to the cliff edge and then return to the road by walking backward in their own footprints — such precision over a course of 200 yards is impossible. Accordingly the inspector says he will report that the criminals, hopeless to escape justice, have hurled themselves to their death.

“Then you will make a fatal mistake,” says Henry Melville, a visiting member of the Puzzle Club. “The men are alive and in hiding in the district.” He proves to be right, but how did the men get away from the edge of the cliff?

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From the road, Marsh alone walked to the edge of the cliff, carrying Lamson’s boots in his hands. At the edge of the cliff he changed boots and walked backward to the road, carrying his own boots. “This little maneuver accounts for the smaller footprints showing a deeper impression at the heel, and the larger prints a deeper impression at the toe, for a man will walk more heavily on his heels when going forward, but will make a deeper impression with the toes in walking backwards. It will also account for the fact that the large footprints were sometimes impressed over the smaller ones, but never the reverse; also for the circumstance that the larger footprints showed a shorter stride, for a man will necessarily take a smaller stride when walking backwards.” From Henry Dudeney’s The Canterbury Puzzles, 1907.

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November 27, 2016December 4, 2016 | Puzzles