a3. The game runs:

1. b4 a5
2. h4 axb4
3. Rh3 Rxa2
4. Ra3 Rxa3
5. Rxa3 bxa3

The moves are illustrated here (line b). (Thanks, Éric.)

Choose three delegates at random. Two of them must share a language. Put those two together into a room. Now 998 delegates remain. Again choose three and put the two who can communicate together into a room. Continue until four delegates remain. Call these A, B, C, and D. If all four speak the same language then assign them arbitrarily to the last two rooms. And if (say) A and B can’t communicate directly then both C and D can serve as their interpreters, so we can lodge A with C and B with D.

From D.O. Shklyarsky, N.N. Chentsov, and I.M. Yaglom, Selected Problems and Theorems in Elementary Mathematics, 1979.

This solution is by Leon Bankoff. Denote each man by his initial and write x → y to indicate that x’s hat was taken by y. We know that B → x → A, where x is the man who took Baily’s hat, and that D → z → y → C, where z is the man who took Dunlop’s hat and y’s hat was taken by Caldwell. These two sequences comprise seven symbols altogether, so x, y, or z must be duplicating A, B, C, or D in one of them. There are two ways to combine the sequences:

1. B → D → A → y → C (where x = D and z = A)
2. D → z → B → C → A (where x = C and y = B)

Each of these new sequences has five symbols, so each lacks one. Add s to the first sequence and t to the second to provide this missing link and to make each sequence into an unbroken cycle:

1. s → B → D → A → y → C → s
2. t → D → z → B → C → A → t

The first cycle fits the problem statement if we let y = F and s = E, and the second matches if we let t = F and z = E. In either case, Fort took Aitkins’ hat.

(“Problem Department,” Pi Mu Epsilon Journal 1:8 [April 1953], 329-330.)

“Make the cut as shown in the illustration and fit the four triangular pieces into the places enclosed by the dotted lines.”

06/02/2024 UPDATE: Reader Mark Schachner found a more conventional solution (click to enlarge):

“I wanted to share this because I noticed humorously that it supports Dudeney’s claim that ‘the easiest dissection puzzles are the prettiest’, but in contrapositive: my solution is not very pretty at all, and it took me the better part of a morning to come up with it!”

Yes. Let E stand for an even number and O for an odd and consider the sequence OEO OEO OEO OEO OEO OEO OEO OEO O. Any group of three consecutive terms contains one even number and two odd, giving an even sum. But 17 of the numbers are odd, so the sum of the whole group is odd.

From Vaderlind’s excellent 2002 book The Inquisitive Problem Solver.

Impressively, he can get 63 gold pieces. He starts by proposing that just over half of the 65 citizens (33 voters) have their salaries doubled to 2 gold pieces, at the expense of the other 33 voters, including himself. Then he does the same thing again, proposing that just over half (17) of the 33 salaried voters receive an increase to 3 or 4 gold pieces, while the remaining 16 in that group are reduced to zero. By continuing in this manner he can reduce the number of salaried voters to 9, 5, 3, and finally 2, with each receiving 33 gold pieces. Then the king can propose that three of the unsalaried citizens receive a salary of 1 gold piece if these two large salaries are now reassigned to himself.

From Peter Winkler’s excellent Mathematical Puzzles, 2021.

Cut the square into four smaller squares, each of which has a side length of 1/2. Now one of these smaller squares must contain at least three of the points. And the largest triangle that these points can form has an area half that of the small square — or 1/8.

05/16/2024 This is an old Martin Gardner puzzle, intended to illustrate the pigeonhole principle. Reader Jon Jerome points out that the three points in the small square might be collinear, but if we accept a degenerate triangle with zero area, then the proof holds.

06/02/2024 Reader Drake Thomas writes:

“An alternate approach to nine points in the square: sort the points by y coordinate. There are four disjoint intervals [y₁, y₃], [y₃, y₅], [y₅, y₇], [y₇, y₉], so at least one of those intervals [y_i, y_i+2] is of length at most 1/4. Then the triangle y_i, y_i+1, y_i+2 occupies at most half of the rectangle of dimensions 1 × 1/4, so is area at most 1/8.”

(Thanks, Drake.)

This solution is by V. Dubrovsky. The size of the product shows that the factors must have three digits each. So let one of them be abc (or 100a + 10b + c) and the other be cba. The product ends in 5, so either a or c must be 5. Say that’s a. The other factor starts with 5, and 92,565 / 500 < 200, so c must be 1. As to b, we can see that the 6 in 92,565 is the last digit of 5b + b, or 6b, so b must be either 1 or 6, and we can test these candidates to learn that it’s 6. The numbers we seek are 165 and 561.

05/17/2024 UPDATE: Reader Robert Filman points out that we can solve this without actually having to multiply the numbers together. The sum of the digits of any number is the remainder of that number divided by 9, and the sum of the digits of 92,565 mod 9 = 0. So once we’ve established that the end digits of the factors we’re seeking are 1 and 5 and that the middle digit is 1 or 6, as above, we can notice that none of the resulting candidates (115, 165, 511, 561) has a digit sum divisible by 9 and hence each factor must be divisible by 3 — which means that the digit sum of each factor must be a multiple of 3. The only possibilities are 165 and 561. (Thanks, Robert.)

No. Impose coordinates so that the frogs sit at (0,0), (1,0), and (0,1). Now when a frog at (x, y) jumps over a frog at (a, b), it lands at (2a – x, 2b – y). Thus the parities of each frog’s coordinates never change. At the start each frog had at least one even coordinate, so none of them can ever reach a point with two odd coordinates, such as (1,1).

See the issue, Solution M14 on page 59, for a visual proof.