Real Estate

A curious problem from the Stanford University Competitive Examination in Mathematics: Bob wants a piece of land that’s exactly level and has four boundary lines, two running precisely north-south and two precisely east-west. And he wants each boundary line to measure exactly 100 feet. Can he buy such a piece of land in the United States?

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Image: Wikimedia Commons Properly speaking, the plot that Bob is describing is bounded by two meridians and two parallel circles. But within each hemisphere the arc of each parallel circle between fixed meridians decreases as its latitude increases — in the Northern Hemisphere, the northern boundary of Bob’s “square” plot will be slightly shorter than the southern boundary. The center of the parcel he seeks must lie on the equator, and he can’t get that in the United States. From George Polya, The Stanford Mathematics Problem Book, 1974.

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December 14, 2023December 13, 2023 | Puzzles

Squaring Words

In his 1864 autobiography Passages From the Life of a Philosopher, Charles Babbage describes an “amusing puzzle.” The task is to write a given word in the first rank and file of a square and then fill the remaining blanks with letters so that the same four words appear in order both horizontally and vertically. He gives this example with the word DEAN:

D E A N
E A S E
A S K S
N E S T

“The various ranks of the church are easily squared,” he writes, “but it is stated, I know not on what authority, that no one has yet succeeded in squaring the word bishop.”

By an unlikely coincidence I’ve just found that Eureka put this problem to its readers in 1961, and they found three solutions:

B I S H O P    B I S H O P    B I S H O P
I L L U M E    I N H E R E    I M P A L E
S L I D E S    S H A R P S    S P I N E T
H U D D L E    H E R M I T    H A N G A R
O M E L E T    O R P I N E    O L E A T E
P E S E T A    P E S T E R    P E T R E L

The first was found by A.L. Cooil and J.M. Dagnese; the second by A.R.B. Thomas; and the third by R.W. Payne, J.D.E. Konhauser, and M. Rumney.

12/10/2023 UPDATE: Reader Giorgos Kalogeropoulos has enlisted a database of 235,000 words to produce more than 100 bishop squares (click to enlarge):

This is pleasing, because it’s a road that Babbage himself was trying to follow in the 19th century, laboriously cataloging the contents of physical dictionaries after an algorithm of his own devising — see page 238 in the book linked above. (Thanks, Giorgos.)

December 5, 2023December 10, 2023 | Puzzles

A Puzzle Forest

An unusual problem by Reddit user cgibbard, from a discussion in 2010:

Here’s a representation of 101010 for you to figure out.
      *             *      *
      |             |      |
   *  *  *   *  *   *  * * *
   |  |   \ /    \ /    \|/
*  *  *    *      *      *
As a bonus, here’s the corresponding representation for 42:
   * *   *
   |  \ /
*  *   *
The puzzle is to find the rule for this representation.

A commenter wrote, “This puzzle is really clever and very rewarding to figure out.”

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This thread contains two explanations: By user typographicalerror: Every prime is a tree. Every number is a forest whose trees correspond to the primes its prime factorization. The tree for 2 is a single vertex. The tree for any other prime is a root vertex connected to the forest representing your prime’s place in the list of all primes. (e.g., 7 is the 4th prime, so you create a root vertex and connect it to the representation for 4.) By user root45: Each prime number has it’s own tree and we simply put these trees together to represent multiplication. So, once we know what the trees look like for 3 and 7 we can put them next to each other to make 21. To find out what the tree looks like for a prime, we build it recursively using the number of primes less than or equal to our prime. To start off with, 2 is just a dot (). 2: To make the tree for 3 we note that there are 2 primes less than or equal to 3, so we take the tree for 2 and connect it to another * like so: 3: * \| * To make 4 we first prime factor 4 as 22 and then put the trees next to each other. 4: * To make 7 we note that there are 4 primes less than or equal to 7, so we take the forest for 4 and connect it to a root. 7: * * \ / * Now we can make the tree for 21 = 37. 21: * * \| \ / * * So basically you can recursively define each number using prime factorization and the number of primes less than or equal to a prime. Note that the function that gives this number is called π(x) (that is, ‘pi of x’).

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November 28, 2023November 26, 2023 | Puzzles

Cornered

“Here is an old favorite of mine that completely stumped me once when I was a student,” writes Jeff Hooper in Crux Mathematicorum. Suppose that each square in this 3×7 grid is painted red or black at random. Show that the board must contain a rectangle whose four corner squares are all the same color.

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Each column contains 3 squares, so each of the 7 columns must contain 2 squares of the same color. There are 3 ways to position 2 black squares in a column, and 3 ways to position 2 red squares, for 6 possibilities overall. Since there are 7 columns in the grid, some 2 of them must bear the same pattern, and thus there are 2 different columns that each contain 2 squares with the same color and position. These form the corners of a rectangle. (Jeff Hooper, “Pólya’s Paragon,” Crux Mathematicorum 33:3 [March 2007], 146-148.)

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November 27, 2023November 24, 2023 | Puzzles

Cutup

A problem by Russian mathematician Viktor Prasolov: Prove that it’s impossible to cut a 10×10 chessboard into T-shaped tiles of 4 squares each.

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Suppose that the task is possible, and that we’ve done it. Due to their shape, each of our 25 tiles contains either 1 or 3 black chessboard squares, always an odd number. So, taken together, the tiles contain an odd number of black squares. But the original board contained 100/2 = 50 black squares. That’s a contradiction, so the task is not possible. (From Prasolov’s Problems in Plane and Solid Geometry.)

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November 19, 2023November 18, 2023 | Puzzles

Profile

A problem from the Stanford University Competitive Examination in Mathematics:

How old is the captain, how many children has he, and how long is his boat? Given the product 32118 of the three desired numbers (integers). The length of the boat is given in feet (is several feet), the captain has both sons and daughters, he has more years than children, but he is not yet one hundred years old.

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The prime factors of 32,118 are 2, 3, 53, and 101. There are only six ways to decompose it into three factors that are different from 1: 6 × 53 × 101 3 × 101 × 106 3 × 53 × 202 2 × 101 × 159 2 × 53 × 303 2 × 3 × 5,353 Of these possibilities, only the first includes two factors in the range from 4 (the minimum number of children) to 99 (the captain’s maximum age). So the captain has 6 children, he’s 53 years old, and his boat is 101 feet long. From George Polya, The Stanford Mathematics Problem Book, 1974.

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November 17, 2023November 16, 2023 | Puzzles

A Tangle

A problem from the October 1960 issue of Eureka, the journal of the Cambridge University Mathematical Society:

Which of the following knots are knotted and which not?

https://www.archim.org.uk/eureka/archive/index.html — Image: Eureka

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(a) Not knot, (b) Knot, (c) Knot, (d) Not knot.

November 11, 2023November 10, 2023 | Puzzles

“Stock-Breeding”

From John Scott, The Puzzle King, 1899:

“A farmer, being asked what number of animals he kept, answered: ‘They’re all horses but two, all sheep but two, and all pigs but two.’ How many had he?”

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Three. 11/09/2023 UPDATE: There’s another solution: Possibly he has no horses, sheep, or pigs but two llamas (or any other sort of animal). That works too. (Thanks, Gareth and Bob.)

November 8, 2023November 9, 2023 | Puzzles

A Self-Descriptive Crossword Puzzle

From Lee Sallows:

Can you complete the ‘self-descriptive crossword puzzle’ at left below? As in the solution to a similar puzzle seen at right, each of its 13 entries, 6 horizontal, 7 vertical, consists of an English number name folowed by a space followed by a distinct letter. The number preceding each letter describes the total number of occurrences of the letter in the completed puzzle. Hence, in the example, E occurs thirteen times, G only once, and so on, as readers can check. Note that the self-description is complete; every distinct letter is counted.

Though far from easy, the self-descriptive property of the crossword enables its solution to be inferred from its empty grid using reasoning based on orthography only.

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Further information about the construction of such crosswords can be found at https://www.leesallows.com/files/Reflexicons%20NEW(4c).pdf (Thanks, Lee!)

November 7, 2023November 9, 2023 | Puzzles

Snow Manipulation

A puzzle by James M., an operations researcher at the National Security Agency:

Frosty the Snowman wants to create a small snowman friend for himself. The new snowman needs a base, torso, and a head, all three of which should be spheres. The torso should be no larger than the base and the head should be no larger than the torso.

For building material, Frosty has a spherical snowball with a 6-inch radius. Since Frosty likes to keep things simple, he also wants the radius of each of the three pieces to be a positive integer. Can Frosty accomplish this?

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Yes! The formula for the volume of a sphere with radius r is $\frac{4}{3}\pi r^{3}$ . Let a, b and c be the radii of the base, torso, and head, respectively, which are integers greater than 0. The problem then amounts to Frosty solving the equation $\displaystyle \frac{4}{3}\pi 6^{3} = \frac{4}{3}\pi a^{3} + \frac{4}{3}\pi b^{3} + \frac{4}{3}\pi c^{3} = \frac{4}{3}\pi \left ( a^{3} + b^{3} + c^{3}\right ).$ By canceling out the factor of $\frac{4}{3}\pi$ from both sides, we’re left with $\displaystyle 216 = a^{3} + b^{3} + c^{3}$ As stated in the problem, a ≥ b ≥ c. The biggest that a can be is 5, so let’s try that first. Subtracting 5³ = 125 from both sides gives us $\displaystyle 91 = b^{3} + c^{3}$ The biggest that b can be is 4, so let’s try that next. Subtracting 4³ = 64 from both sides gives us $\displaystyle 27 = c^{3}$ Frosty is in luck, since 3³ = 27. Thus we have that 6³ = 5³ + 4³ + 3³. Now Frosty can build his new snowman friend to his specifications. In fact, (5, 4, 3) is the only combination of numbers that will work. From NSA’s Puzzle Periodical series.

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November 3, 2023November 2, 2023 | Puzzles