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.

First, realize that the total points awarded is 40. Given that we are dealing with integers that means the total number of Events (TE) times the total number of points awarded in an event (TP) must equal 40. We can quickly eliminate some possibilities.

So we now know it was either 4 events with 10 points awarded in an event or 5 events with 8 points awarded. Let’s look at the 4/10 situation first. The possible points for each place with 10 overall points available is (all other combinations are impossible because of the 1st > 2nd > 3rd > 0 constraint):

• 5, 3, 2 Not possible as there is no way to get to 22 points in 4 events (greatest possible points would be 20).
• 5, 4, 1 Same reason.
• 6, 3, 1 Not possible, since no combination of 4 numbers can get to 22 points (3 × 6 + 3 = 21, 4 × 6 = 24).
• 7, 2, 1 A possibility — let’s look further.

Can we get Adam to 22? Yes. Adam can finish 1st 3 times and 3rd once (in the Javelin). 3 × 7 + 1 = 22.

Can we get Bob to 9? Nope. Bob won the Javelin (7 points). There’s no way to get to 9 with the remaining event/point combinations. Sooooo …

There must be 5 events with a total of eight points (we’re getting close). Let’s look at the possible points (again, the other combinations are not possible because of the 1st > 2nd > 3rd > 0 constraint):

• 4, 3, 1 Not possible as there is no way to get to 22 points (4 × 5 = 20 is maximum)
• 5, 2, 1 Looks like this is the winner. Let’s check.

Can we get Adam to 22? Yes. Adam can finish 1st 4 times and 2nd once (4 × 5 + 2). Since we know Bob finished 1st in the Javelin, Adam must have finished 2nd.

Can we get Bob to 9? Yes. Bob finished 1st in the Javelin (5 points) and finished 3rd 4 times (4 ×1) for a total of 9 points.

Can we get Charlie to 9? Yes. Charlie finished 3rd in the Javelin (1 point) and must have finished 2nd in the 4 other events (4 × 2) for a total of 9 points.

Since Charlie finished 2nd in EVERY EVENT OTHER THAN THE JAVELIN, he MUST have finished 2nd in the 100-meter dash. Here is a little table:

1. Rc5+ unpins the black rook on b4, permitting the reply 1. … Rxf4 and thus avoiding mate.

From Dr. Crypton and His Problems: Mind Benders From Science Digest, 1982.

The mass of the bottle plus the mass of the gasoline is 1,000 grams. And the mass of the bottle plus twice the mass of the gasoline is 1,600 grams (because the acid is twice as dense as the gasoline). So the difference, 600 grams, is the mass of the bottle’s volume of gasoline. The mass of the bottle full of gasoline is 1,000 grams, so the mass of the bottle alone is 400 grams. (This checks out: It tells us that the mass of the acid is 1,600 – 400 = 1,200 grams, which is twice the mass of the gasoline.)

From the March-April 1993 issue of Quantum (page 32, “Kaleidoscope”).

This problem was proposed in the American Mathematical Monthly by A.L. Rubinoff of the University of Toronto in 1948 (Problem E773). It’s possible simply to count up the lines, but Leo Moser of the University of Manitoba hit on a simpler solution.

Encase the 8×8×8 cube in a “shell” of unit thickness, and extend each winning line so that it pierces the shell at either end. Each winning line pierces two cells in the shell, and each cell in the shell is pierced by only one winning line. Thus each winning line corresponds to a unique pair of cells in the shell, and the number of winning lines is just half the number of cells in the shell:

This solving strategy can be generalized to cubes of any size.

(Via Ross Honsberger, Mathematical Morsels, 1978.)