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.)

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.)

1. Rxc6!

From Chess Problems, 1873.

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.)

If there were only 3 matches in the original pile, A would lose — he could withdraw 1 or 2 matches, but then B could immediately withdraw the remainder and win the game. If there were 4 or 5 matches in the original pile then A could saddle B with a losing 3-match pile, but if there were 6 in the original pile, then by the same principle B could saddle A with the 3-match pile.

The winning strategy, then, is always to leave your opponent a quantity that’s a multiple of 3. A can accomplish this on his first turn by taking 2 matches from the pile of 500, leaving 498 = 3 × 166 matches for B, who cannot now escape his fate.

Left: “White mates without moving simply by removing the Black Pawn from [d5], he having just taken that Pawn en passant with the White Pawn standing on [d6]. The position is impossible unless Black and White had thus played last, for in no other way could the White King be now standing in the check of the Pawn.”

Right: “The move of Castling may be divided into four parts. First, lifting up the King: second, placing him on the proper square; third, lifting up the Rook; and fourth, placing the Rook on the proper square. In this position White has made 3/4 of this move: he has taken up the King, and placed him on the Bishop’s square; he has also taken up the Rook, which completes the 3/4 of the move. All that remains, therefore, is to place the Rook on Queen’s square, the remaining fourth of the move, and checkmate is accomplished in a quarter of a move!”

Guinea pigs never really appreciate Beethoven.

(From the unpublished manuscript of a second volume of Carroll’s Symbolic Logic, discovered by philosopher William Bartley.)

In his 1994 book Mathematics: A Human Endeavor, Harold R. Jacobs gives a similar problem, by French puzzlist Pierre Berloquin. Around a table sit the four members of the street storekeepers’ committee: a grocer, a baker, a tobacconist, and a butcher, whose name is Andre. If Andre sits on Charmeil’s left, Berton sits at the grocer’s right, and Duclos, who faces Charmeil, is not the baker, then what are the professions of the other members?

That’s simple enough that it doesn’t require a solution, I think. But here’s a much harder one, composed by J.B. Parker for the January 1941 issue of Eureka:

Seven men, Messrs. Black, Blue, Brown, Gray, Green, Purple and White sat down at a circular table laid for eight. Each man was wearing a tie and a pair of socks — and also had a car, their colours being three of the names of the other six men. There was a tie, a pair of socks and a car of each of the seven colours.

Mr. Blue sat next to the man with the green tie; between Mr. Gray and the man with white socks there sat a man wearing a white tie, and opposite him sat Mr. Green.

Mr. Brown’s socks were of the same colour as the tie of the man who occupied the chair on his right, and Mr. Green wore a brown tie. The empty chair lay between Mr. Black and the man who wore a green tie. The man with black socks had a gray car, and the man with gray socks had a black car.

Mr. Purple’s car was the same colour as Mr. Gray’s tie, and this was of the same colour as Mr. White’s socks. The man with the name of the colour of Mr. White’s car wore socks of the colour of Mr. Black’s car, i.e. blue. Mr. Purple’s tie was of the colour of the car of the man who occupied the chair on his right, and Mr. Brown sat opposite the man with the white car. The colour of the socks of the man whose tie was the colour of Mr. Gray’s car was the same as that of the tie of the man whose socks were the colour of Mr. Black’s car, and this colour was not black.

Find the colours of the socks, tie, and car of each man.

The answer (but not the solution) appears on page 11 of the May 1941 issue. (The Eureka Digital Archive is published under a Creative Commons license.)

(Pier Square, “The Bridge Game,” Pi Mu Epsilon Journal 6:9 [Fall 1978], 530.)

05/24/2021 Reader Dharmesh Jain kindly offers this walkthrough of the Eureka puzzle.

Credit for the solution goes to Amy Kuiper of Michigan’s Alma College and to the Grand Valley State University Problem Solving Group of Allendale, Mich.

The face that receives our final scrutiny might be any one of the faces; all are equally likely. So the probability that it’s red is just

(Clayton W. Dodge, “Problem Department,” Pi Mu Epsilon Journal 11:1 [Fall 1999], 47-63. Gebhardt’s is Problem 948, on page 56.)