Evolution

I just ran across this anecdote by Jason Rosenhouse in Notices of the American Mathematical Society. In a middle-school algebra class Rosenhouse’s brother was given this problem:

There are some horses and chickens in a barn, fifty animals in all. Horses have four legs while chickens have two. If there are 130 legs in the barn, then how many horses and how many chickens are there?

The normal solution is straightforward, but Rosenhouse’s brother found an alternative that’s even easier: “You just tell the horses to stand on their hind legs. Now there are fifty animals each with two legs on the ground, accounting for one hundred legs. That means there are thirty legs in the air. Since every horse has two legs in the air, we find that there are fifteen horses, and therefore thirty-five chickens.”

(Jason Rosenhouse, “Book Review: Bicycle or Unicycle?: A Collection of Intriguing Mathematical Puzzles,” Notices of the American Mathematical Society, 67:9 [October 2020], 1382-1385.)

June 23, 2021 | Puzzles · Science & Math

“Fermat’s Last Theorem”

A puzzle by H.A. Thurston, from the April 1947 issue of Eureka, the journal of recreational mathematics published at Cambridge University:

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The answer appears in the March 1948 issue: “The statements of 2 and 3 are true, those of 1, 4 and 5 are false.” But no full solution is given. (The Eureka archive is published under a Creative Commons license.)

June 23, 2021June 23, 2021 | Puzzles

State House

A quickie from Peter Winkler’s Mathematical Puzzles, 2021: Can West Virginia be inscribed in a square? That is, is it possible to draw some square each of whose four sides is tangent to this shape?

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Yes. Start by drawing an arbitrary rectangle that hugs the state. Now rotate that rectangle, adjusting its dimensions as it turns to keep each side in contact with the shape. If you turn it through 90 degrees, then the “height” and “width” will have traded places: What had been the shorter side of the rectangle is now the longer. So at some point the two must have had the same value. (This is a two-dimensional variation on a classic puzzle.)

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June 15, 2021June 14, 2021 | Puzzles

Black and White

carpenter chess problem

By George E. Carpenter. White to mate in two moves.

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1. Qh8! and the queen will mate. From Carpenter’s Chess Problems, 1886.

June 14, 2021June 12, 2021 | Puzzles

Hooper’s Paradox

https://commons.wikimedia.org/wiki/File:Hooper_paradox.svg — Image: Wikimedia Commons

William Hooper published the oddity in 1774. The rectangle at the top measures 10 units by 3, giving an area of 30. But its dissected pieces seem to produce two other rectangles, with areas 12 and 20. Where did the two extra units come from?

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Image: Wikimedia Commons The pieces don’t quite match. If the pieces in the second figure are arranged carefully back into the first, it will be seen that they overlap. The indicated parallelogram measures exactly 2 square units.

June 12, 2021 | Puzzles

“Imitative Chess”

dudeney imitative chess

A puzzle by Henry Dudeney:

A chessboard was on the table with the pieces all set up for a game. So I asked Dr. Bates to play a game with the Major on these conditions: Whatever move Bates made throughout, with the white pieces, the Major must exactly imitate with the black, and Bates must give checkmate on the fourth move. As an experiment, Bates started off with 1. e4, and Rackford replied with 1. e5. Then Bates played 2. Qh5, and the Major had to reply with 2. Qh4. This gave me a good opportunity to explain that White cannot now play 3. QxQ, because it would be impossible for Black then to imitate the move. Neither could he play 3. Qxf7+, because Black cannot do the same thing, as he would have to get out of check. White must always make a move that Black can copy, until the checkmate is actually given on the fourth move.

“This puzzle caused great interest, and it was some time before somebody (I think it was Strangways) hit on a solution.”

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1. c4 c5 2. Qa4 Qa5 3. Qc6 Qc3 4. Qxc8#: (Henry E. Dudeney, “A Budget of Christmas Puzzles,” Strand 64:12 [December 1922], 619-622.) (If you want to solve the other puzzles, the solutions are here.)

June 8, 2021June 6, 2021 | Puzzles

The Hidden Element

The name of one chemical element appears as an unbroken string in the names of four other elements. What is the element, and what are the four?

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Tin (actinium, astatine, platinum, protactinium). (Dave Morice, “Kickshaws,” Word Ways 28:3 [August 1995], 170-180.) 06/09/2021 UPDATE: In French, the name or (gold) appears as an unbroken string in the names of nine elements: bore (boron), californium, chlore (chlorine), fluor (fluorine), livermorium, phosphore (phosphorus), rutherfordium, seaborgium, and thorium. (Thanks, Jean-Claude.)

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June 7, 2021June 9, 2021 | Puzzles

Stonework

A problem from the Leningrad Mathematical Olympiad: You have 32 stones, each of a different weight. How can you find the two heaviest in 35 weighings with an equal-arm balance?

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Think of the stones as tennis players and put them through a singles tournament. There are no upsets: In each match (weighing), the better (heavier) player always wins and goes on to the next round. With 32 players, this will produce a tournament with five rounds, comprising 16, 8, 4, 2, and 1 matches, successively, enough to produce 31 losers and identify the best (heaviest) stone. Now, the second-heaviest stone must have “lost” to the heaviest at some point in the tournament (since that was the only “player” strong enough to beat it). So take the five players beaten by the heaviest stone and put them through a little tournament of their own. Four matches will be enough to find the strongest of these five, and that must be the second-heaviest stone overall. (Via Ross Honsberger’s Mathematical Delights, 2004.)

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June 4, 2021 | Puzzles

Black and White

thompson chess problem

By Theophilus A. Thompson. White to mate in two moves.

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1. Rxc6! From Chess Problems, 1873.

May 24, 2021May 23, 2021 | Puzzles

Quickie

Express 1,000,000 as the product of two numbers, neither of which contains any zeroes.

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1,000,000 = 2⁶ × 5⁶. To avoid zeroes we need to collect the 2s into one factor and the 5s into the other: 64 × 15625. (Thanks, Nick.)

May 23, 2021 | Puzzles