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.
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.
If P is the center of gravity of the system, then as long as P is above the surface of the water, it will fall as the surface rises. When P reaches the surface, adding any more water will raise the water level and hence the level of P.
From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.
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?
We can answer this without knowing the car’s speed. If x is the number of signs that the car passes in one minute, then the car will pass 60x signs in an hour. We’re told that the car is traveling at 10x miles per hour, so in 10x miles it will pass 60x signs, and in one mile it will pass 60x/10x signs, or 6. So the signs are 1/6 mile, or 880 feet, apart.
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?
The first child stands in the center of the room, and the second stands at her side. Thereafter each entering child follows a rule: If she sees that the line of children are all wearing hats of the same color, then she stands at one end of the line. If they’re wearing both red and blue hats, then she inserts herself at the break between the two colors.
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.”)
WOMAN HERE HAS GIVEN BIRTH TO SEXTUPLETS. This appears in Dave Silverman’s article “Some Cryptographic Challenges” in the May 2008 issue of Word Ways; I’m not sure who the columnist was. Silverman’s own best guess was WISCONSIN MUSICAL MULTIMILLIONAIRE LEADS CONCERT UNDER PAVILION, which would have made for an entertaining comic.
Each point on a straight line is either red or blue. Show that it’s always possible to find three points of the same color in which one is the midpoint of the other two.
Choose two points of the same color, say blue. Call these A and B, and let C be their midpoint. If C is blue then we’re done. If C is red then mark points D and E as shown such that AD = AB = BE. Now if D or E is blue then we’re done, because these produce the blue triples DAB and ABE. And if D and E are both red then we have DCE, a red triple. So there’s no way to assign colors to these five points without producing some monochromatic triple.
John and Mary drive from Westville to Eastville. John drives the first 40 miles, and Mary drives the rest of the way. That afternoon they return by the same route, with John driving the first leg and Mary driving the last 50 miles. Who drives the farthest, and by what distance?
We don’t know the distance between Westville and Eastville, but suppose it’s only 50 miles. In that case John drives 40 miles altogether and Mary drives 60. If the distance is any greater than that, then the two drivers are dividing the extra miles equally between them, with Mary driving east and John west. So the total distance is immaterial — Mary drives 20 more miles than John.
Either you’ll toss more heads than I do or you’ll toss more tails than I do, but not both. (Right?) These outcomes are symmetric, so each has probability 1/2.
UPDATE: A number of readers questioned the validity of this solution, but I think that’s due to a bad writeup on my part. I don’t know who devised the problem originally; I found it in the 1995 collection Five Hundred Mathematical Challenges, by Edward Barbeau, Murray Klamkin, and William Moser. Here’s their presentation:
John tosses 6 fair coins, and Mary tosses 5 fair coins. What is the probability that John gets more “heads” than Mary?
More generally, suppose John tosses n + 1 coins and Mary tosses n coins. Then, either John tosses more heads than Mary does or John tosses more tails than Mary does — but not both. (Check this out!) Since these two outcomes are symmetric, each occurs with probability 1/2.
I think I confused the issue by failing to note that this reasoning addresses specifically the case of n coins vs. n + 1. I still haven’t had time to think about this carefully, but most of the people who’d objected wrote back later to change their minds. If you still think it’s incorrect, please do let me know, and I’ll post further updates here. Sorry for the confusion, and thanks to everyone who’s written in.
UPDATE #2: Yes, the solution appears to be valid with this change. Thanks to everyone who contributed!