A problem from the 2002 Moscow Mathematical Olympiad:
A group of recruits stand in a line facing their corporal. They are, unfortunately, rather poorly trained: At the command “Left turn!”, some of them turn left, some turn right, and some turn to face away from the corporal. Is it always possible for the corporal to insert himself in the line so that an equal number of recruits are facing him on his left and on his right?
Here are six new lateral thinking puzzles to test your wits! Solve along with us as we explore some strange scenarios using only yes-or-no questions. Many were submitted by listeners, and most are based on real events.
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A puzzle by A. Korshkov, from the Russian science magazine Kvant:
It’s easy to show that the five acute angles in the points of a regular star, like the one at left, total 180°.
Can you show that the sum of these angles in an irregular star, like the one at right, is also 180°?
A memorably phrased puzzle from The Graham Dial: “Consider a vertical girl whose waist is circular, not smooth, and temporarily at rest. Around the waist rotates a hula hoop of twice its diameter. Show that after one revolution of the hoop, the point originally in contact with the girl has traveled a distance equal to the perimeter of a square circumscribing the girl’s waist.”
University of Strathclyde mathematician Adam McBride recalls that in his student days a particular teacher used to present a weekly puzzle. One of these baffled him:
Find positive integers a, b, and c, all different, such that a3 + b3 = c4.
“The previous puzzles had been relatively easy but this one had me stumped,” he wrote later. He created three columns headed a3, b3, and c4 and spent hours looking for a sum that would work. On the night before the deadline, he found one: 703 + 1053 = 354.
“This shows how sad a person I was! However, I then realised also how stupid I had been. I had totally missed the necessary insight.” What was it?
1. A puzzle from J.A.H. Hunter’s Fun With Figures, 1956:
Tom and Tim are brothers; their combined ages make up seventeen years. When Tom was as old as Tim was when Tim was twice as old as Tom was when Tom was fifteen years younger than Tim will be when Tim is twice his present age, Tom was two years younger than Tim was when Tim was three years older than Tom was when Tom was a third as old as Tim was when Tim was a year older than Tom was seven years ago. So how old is Tim?
2. Another, by Sam Loyd:
“How fast those children grow!” remarked Grandpa. “Tommy is now twice as old as Maggie was when Tommy was six years older than Maggie is now, and when Maggie is six years older than Tommy is now their combined ages will equal their mother’s age then, although she is now but forty-six.” How old is Maggie?
3. According to Wirt Howe’s New York at the Turn of the Century, 1899-1916, this question inspired an ongoing national debate when it appeared in the New York Press in 1903:
Brooklyn, October 12
Mary is 24 years old. She is twice as old as Anne was when she was as old as Anne is now. How old is Anne now? A says the answer is 16; B says 12. Which is correct?
1. By A.F. Rockwell. White to mate in two moves. (Solutions are below.)
2. In 1866 Sam Loyd asked: Suppose you’re playing White against an opponent who’s required to mirror every move you make — if you play 1. Nf3 he must play 1. … Nf6, and so on. Can you design a game in which your eighth move forces your opponent to checkmate you with a nonmirror move?
3. An endgame study by J.A. Miles. “White to play and draw the game.”
A puzzle by Princeton mathematician John Horton Conway:
Last night I sat behind two wizards on a bus, and overheard the following:
A: I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is my own age.
B: How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?
B: Aha! AT LAST I know how old you are!
“This is an incredible puzzle,” writes MIT research affiliate Tanya Khovanova. “This is also an underappreciated puzzle. It is more interesting than it might seem. When someone announces the answer, it is not clear whether they have solved it completely.”
We can start by auditioning various bus numbers. For example, the number of the bus cannot have been 5, because in each possible case the wizard’s age and the number of his children would then uniquely determine their ages — if the wizard is 3 years old and has 3 children, then their ages must be 1, 1, and 3 and he cannot have said “No.” So the bus number cannot be 5.
As we work our way into higher bus numbers this uniqueness disappears, but it’s replaced by another problem — the second wizard must be able to deduce the first wizard’s age despite the ambiguity. For example, if the bus number is 21 and the first wizard tells us that he’s 96 years old and has three children, then it’s true that we can’t work out the children’s ages: They might be 1, 8, and 12 or 2, 3, and 16. But when the wizard informs us of this, we can’t declare triumphantly that at last we know how old he is, because we don’t — he might be 96, but he might also be 240, with children aged 4, 5, and 12 or 3, 8, and 10. So the dialogue above cannot have taken place.
But notice that if we increase the bus number by 1, to 22, then all the math above will still work if we give the wizard an extra 1-year-old child: He might now be 96 years old with four children ages 1, 1, 8, and 12 or 1, 2, 3, and 16; or he might be 240 with four children ages 1, 4, 5, and 12 or 1, 3, 8, and 10. The number of children increases by 1, the sum of their ages increases by 1, and the product remains the same. So if bus number b produces two possible ages for Wizard A, then so will bus number b + 1 — which means that we don’t have to check any bus numbers larger than 21.
This limits the problem to a manageable size, and it turns out that the bus number is 12 and Wizard A is 48 — that’s the only age for which the bus number and the number of children do not uniquely determine the children’s ages (they might be 2, 2, 2, and 6 or 1, 3, 4, and 4).
(Tanya Khovanova, “Conway’s Wizards,” The Mathematical Intelligencer, December 2013.)
I leave my front door, run on a level road for some distance, then run to the top of a hill and return home by the same route. I run 8 mph on level ground, 6 mph uphill, and 12 mph downhill. If my total trip took 2 hours, how far did I run?
A problem from the 2000 Moscow Mathematical Olympiad:
Some of the cards in a deck are face down and some are face up. From time to time Pete draws out a group of one or more contiguous cards in which the first and last are both face down. He turns over this group as a unit and returns it to the deck in the position from which he drew it. Prove that eventually all the cards in the deck will be face up, no matter how Pete proceeds.
In 1831 this riddle appeared in a British publication titled Drawing Room Scrap Sheet No. 17:
In the morn when I rise, / I open my eyes, / Tho’ I ne’er sleep a wink all night;
If I wake e’er so soon, / I still lie till noon, / And pay no regard to the light.
I have loss, I have gain, / I have pleasure, and pain; / And am punished with many a stripe;
To diminish my woe, / I burn friend and foe, / And my evenings I end with a pipe.
I travel abroad. / And ne’er miss my road, / Unless I am met by a stranger;
If you come in my way, / Which you very well may, / You will always be subject to danger.
I am chaste, I am young, / I am lusty, and strong, / And my habits oft change in a day;
To court I ne’er go, / Am no lady nor beau, / Yet as frail and fantastic as they.
I live a short time, / I die in my prime, / Lamented by all who possess me;
If I add any more, / To what’s said before / I’m afraid you will easily guess me.
It was headed “For Which a Solution Is Required,” perhaps meaning that the editor himself did not know the solution. I think he may have found the riddle in The Lady’s Magazine, which had published it anonymously in September 1780 without giving the answer. Unfortunately he seems to have been disappointed — the Drawing Room Scrap Sheet never printed a solution either.
A century and a half later, in 1981, Faith Eckler challenged the readers of Word Ways: The Journal of Recreational Linguistics to think of an answer, offering a year’s subscription to the journal as a reward. When no one had claimed the prize by February 2010, Ross Eckler renewed his wife’s challenge, noting that the National Puzzlers’ League had also failed to find a solution.
That’s understandable — it’s tricky. “The author of the riddle cleverly uses ambiguous phrases to mislead the solver,” Ross Eckler notes. “I still lie till noon (inert, or continue to?); evenings I end with a pipe (a tobacco holder, or a thin reedy sound?); to court I ne’er go (a royal venue, a legal venue, or courtship?).”
To date, so far as I know, the riddle remains unsolved. Answers proposed by Word Ways readers have included fame, gossip, chessmen, a hot air balloon, and the Star and Stripes, though none of these seems beyond question. I offer it here for what it’s worth.
UPDATE: A solution has been found! Apparently The Lady’s Diary published the solution in 1783, which Ronnie Kon intrepidly ran to earth in the University of Illinois Rare Book and Manuscript Library. He published it in Word Ways in November 2012 (PDF). I’ll omit the solution here in case you’d still like to guess; be warned that it’s not particularly compelling. (Thanks, Ronnie.)
Starting in Delaware, you must tour the 48 contiguous United States, visiting each state exactly once.
Where will you finish?
A puzzle from Oswald Jacoby’s Mathematics for Pleasure (1962):
The MacDonalds are planning a long car journey of 27,000 miles. If they use tires that last 12,000 miles each, how many tires will they need, and how can they make the best use of them?
A quickie submitted by John Astolfi to MIT Technology Review’s Puzzle Corner, July/August 2013:
Consider the expansion of π (3.14159 …) in base 2. Does it contain more 0s than 1s, more 1s than 0s, or an equal number of both? Or is it impossible to tell?
In March 2013, New Mexico art dealer Forrest Fenn announced that he had hidden a bronze treasure chest in the Rocky Mountains north of Santa Fe. In the chest, he says, are gold coins, artifacts, and jewelry worth more than $1 million.
Fenn said he’d conceived the idea when diagnosed with cancer in 1988, planning to bury the treasure as a legacy. The cancer went into remission, but he decided to bury the chest anyway. In a self-published memoir he offered the following poem, which he says contains nine clues:
As I have gone alone in there
And with my treasures bold,
I can keep my secret where,
And hint of riches new and old.
Begin it where warm waters halt
And take it in the canyon down,
Not far, but too far to walk.
Put in below the home of Brown.
From there it’s no place for the meek,
The end is ever drawing nigh;
There’ll be no paddle up your creek,
Just heavy loads and water high.
If you’ve been wise and found the blaze,
Look quickly down, your quest to cease,
But tarry scant with marvel gaze,
Just take the chest and go in peace.
So why is it that I must go
And leave my trove for all to seek?
The answers I already know,
I’ve done it tired and now I’m weak.
So hear me all and listen good,
Your effort will be worth the cold.
If you are brave and in the wood
I give you title to the gold.
Fenn has been releasing further clues periodically as he follows the search (“No need to dig up the old outhouses, the treasure is not associated with any structure”). A number of people claim to have found the chest, but none has provided evidence, and Fenn says that to the best of his knowledge it remains undiscovered. There’s much background and discussion about the treasure at ttotc.com.
If you find it, I figure you owe me 75%.
A water jug is empty, and its center of gravity is above the inside bottom of the jug. Water is poured into the jug until the center of gravity of the jug and water (considered together) is as low as possible. Explain why this center of gravity must lie at the surface of the water.
A puzzle from Martin Gardner’s column in Math Horizons, November 1995:
Driving along the highway, Mr. Smith notices that signs for Flatz beer appear to be spaced at regular intervals along the roadway. He counts the number of signs he passes in one minute and finds that this number multiplied by 10 gives the car’s speed in miles per hour. Assuming that the signs are equally spaced, that the car’s speed is constant, and that the timed minute began and ended with the car midway between two signs, what is the distance from one sign to the next?
A group of children are standing outside a room. Each wears a hat that’s either red or blue, and each child can see the other children’s hats but not her own. At a signal they enter the room one by one and arrange themselves in a line partitioned by hat color. How do they manage this without communicating?
Benedict Arnold encrypted his messages to the British Army using Blackstone’s Commentaries on the Laws of England. Arnold would replace each word in his message with a triplet of numbers representing the page number, line number, and word position where the word might be found in Blackstone. For example:
The 166.8.11 of the 191.9.16 are 129.19.21 266.9.14 of the .286.8.20, and 291.8.27 to be on 163.9.4 115.8.16 114.8.25ing — 263.9.14 are 207.8.17ed 125.8.15 103.8.60 from this 294.8.50 104.9.26 — If 84.8.9ed — 294.9.12 129.8.7 only to 193.8.3 and the 64.9.5 290.9.20 245.8.3 be at an 99.8.14.
British Army Major John André could then look up the words in his own copy of Blackstone to discover Arnold’s meaning:
The mass of the People are heartily tired of the War, and wish to be on their former footing — They are promised great events from this year’s exertion — If disappointed — you have only to persevere and the contest will soon be at an end.
The danger in using a book code is that the enemy can decode the messages if he can identify the book — and sometimes even if he can’t. In the comic strip Steve Roper, a reporter once excitedly telephoned the coded message 188-1-22 71-2-13 70-2-11 68-1-25 19-1-6 112-2-10 99-1-35. Reader Sean Reddick suspected that this message had been encoded using a dictionary, with each triplet of numbers denoting page, column, and word number. He never did discover the book that had been used, but by considering the ratios involved and consulting half a dozen dictionaries he managed to break the code anyway — he sent his solution to a nationally known columnist, who verified his feat when the comic strip bore out his solution. What was the message? (Hint: In the comic, the reporter mentions significantly that the plaintext message was given to him by “the delivery boy.”)