The Candy Thief

A problem by Wayne M. Delia and Bernadette D. Barnes:

Five children — Ivan, Sylvia, Ernie, Dennis, and Linda — entered a candy store, and one of them stole a box of candy from the shelf. Afterward each child made three statements:

Ivan:

1. I didn’t take the box of candy.
2. I have never stolen anything.
3. Dennis did it.

Sylvia:

4. I didn’t take the box of candy.
5. I’m rich and I can buy my own candy.
6. Linda knows who the crook is.

Ernie:

7. I didn’t take the box of candy.
8. I didn’t know Linda until this year.
9. Dennis did it.

Dennis:

10. I didn’t take the box of candy.
11. Linda did it.
12. Ivan is lying when he says I stole the candy.

Linda:

13. I didn’t take the box of candy.
14. Sylvia is guilty.
15. Ernie can vouch for me, because he has known me since I was a baby eight years ago.

If each child made two true and one false statement, who stole the candy?

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Not as hard as it sounds! If Ivan was the thief, then he told more than one lie. In the same way we can exclude Ernie, Dennis, and Linda. So Sylvia is the thief, and the false statements are #3, #4, #9, #11, and #15. From Pi Mu Epsilon Journal, Fall 1980 (PDF) (solution by John Wesley Emert of Knoxville, Tenn.).

February 25, 2017March 5, 2017 | Puzzles

The Windmill Algorithm

Suppose we have a finite set of points in the plane, no three of which are collinear. A line drawn through one of them pivots around that point until it encounters another point, when it adopts that point as the new pivot. Call this line a “windmill”; it continues indefinitely, always rotating in the same direction. Show that we can choose an initial point and line so that the resulting windmill uses each point as a pivot infinitely many times.

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Draw the line vertically through the centermost of the points, and color all the points to the left of it blue and all the points to the right red. After the line has turned 180 degrees, all the points to the left of it will now be red, and those to the right will be blue. Hence every point has changed color. Each time the pivot changes, the old pivot point returns to the same side of the line that the new one comes from. That means that the total number of blue points remains unchanged; so does the number of red points. Since a point can change color only by serving as a pivot, this means that every point must act as a pivot at some stage. And if the line turns forever, then each point will change color infinitely many times. This is laid out more rigorously here. By Geoffrey Smith of the United Kingdom for the 52st International Mathematical Olympiad in Amsterdam, 2011. The animation is from the excellent Symmetry tumblr.

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February 9, 2017February 19, 2017 | Puzzles

Black and White

weiß chess problem

By Berthold Weiß. White to mate in two moves.

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1. Qh8!, threatening 2. Qa8#. If 1. … Qb8 then 2. Qa1#! From The Good Companion, 1917.

February 3, 2017February 12, 2017 | Puzzles

The Mengenlehreuhr

Further to Saturday’s triangular clock post, reader Folkard Wohlgemuth points out that a “set theory clock” has been operating publicly in Berlin for more than 40 years. Since 1995 it has stood in Budapester Straße in front of Europa-Center.

The circular light at the top blinks on or off once per second. Each cell in the top row represents five hours; each in the second row represents one hour; each in the third row represents five minutes (for ease of reading, the cells denoting 15, 30, and 45 minutes past the hour are red); and each cell in the bottom row represents one minute. So the photo above was taken at (5 × 2) + (0 × 1) hours and (6 × 5) + (1 × 1) minutes past midnight, or 10:31 a.m.

Online simulators display the current time in the clock’s format in Flash and Javascript.

If that’s not interesting enough, apparently the clock is a key to the solution of Kryptos, the enigmatic sculpture that stands on the grounds of the CIA in Langley, Va. In 2010 and 2014 sculptor Jim Sanborn revealed to the New York Times that two adjacent words in the unsolved fourth section of the cipher there read BERLIN CLOCK.

When asked whether this was a reference to the Mengenlehreuhr, he said, “You’d better delve into that particular clock.”

January 17, 2017January 16, 2017 | Puzzles · Science & Math

Black and White

d'orville chess problem

By Pierre Auguste D’Orville. White to mate in two moves.

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1. Rd5! Kxd5 2. Qd4# From Problèmes d’Échecs, 1842.

January 15, 2017January 22, 2017 | Puzzles

The Perplexed Cellarman

dudeney cellarman puzzle

One last puzzle from Henry Dudeney’s Canterbury Puzzles:

Abbott Francis sends for his cellarman and complains that a particular bottling of wine is not to his taste. He asks how many bottles he had produced. The cellarman tells him that there had been 12 large and 12 small bottles, and that 5 of each have been drunk. The abbot replies that three men are waiting at the gate, and orders the cellarman to give each of them some combination of full and empty bottles so that each man receives the same quantity of wine and combination of bottles.

How can the cellarman do this? He has seven large and seven small bottles full of wine, and five large and five small bottles that are empty. A large bottle holds twice as much wine as a small one, but a large bottle when empty is not worth two small ones — hence the abbot’s order that each man must take away the same number of bottles of each size.

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“Brother John gave the first man three large bottles and one small bottleful of wine, and one large and three small empty bottles. To each of the other two men he gave two large and three small bottles of wine, and two large and one small empty bottles. Each of the three then receives the same quantity of wine, and the same number of each size of bottle.”

January 13, 2017January 22, 2017 | Puzzles

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