Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,¹ and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter or send me an email.

Riddler Express

For Easter, you and your family decide to decorate 10 beautiful eggs. You pull a fresh carton of eggs out of your fridge and remove 10 eggs. There are two eggs remaining in the carton, which you return to the fridge.

The next day, you open the carton again to find that the positions of the eggs have somehow changed — or so you think. Perhaps the Easter Bunny was snooping around your fridge?

The 12 slots in the carton are arranged in a six-by-two array that is symmetric upon a 180-degree rotation, and the eggs are indistinguishable from each other. How many distinct ways are there to place two eggs in this carton? (Note: Putting two eggs in the two leftmost slots should be considered the same as putting them in the two rightmost slots, since you can switch between these arrangements with a 180-degree rotation of the carton.)

Extra credit: Instead of two eggs remaining, suppose you have other numbers of indistinguishable eggs between zero and 12. How many distinct ways are there to place these eggs in the carton?

Submit your answer

Riddler Classic

From Nis Jørgensen comes a picaresque puzzle of a captain and crew:

You are the captain of a three-member crew (not including yourself): Geordi, Sidney and Alandra. Your ship has been captured by a previously unknown foe, who has decided to return your ship if you can win a simple game.

Each of the three crew members is to be issued a number between zero and one, randomly and uniformly picked within that range. Each crew member knows their own number, but not those of their fellow crew mates. As the captain, your objective is to guess who has the highest number.

The catch is that you can only ask one yes-or-no question to each crew member. Based on the answer to the question you ask the first crew member, you can update the question you’d ask the second. Similarly, based on the answers to the first two questions, you can update the third question you’d ask. But in the end, you still have to guess which crew member has the highest number.

What is your optimal strategy, and what are your chances of regaining your ship?

Submit your answer

Solution to the last Riddler Express

Congratulations to 👏 Sweet Tea Dorminy 👏 of Greenville, South Carolina, winner of last week’s Riddler Express.

Last week’s Express was submitted by high schooler Max Misterka, a winner of the 2023 Regeneron Science Talent Search. Max and I were playing a game in which we both picked a number in secret. Let’s call Max’s number m and my number z. After we both revealed our numbers, Max’s score was m^z, while my score was z^m. Whoever had the greater score won.

When we played most recently, Max and I selected distinct whole numbers. Surprisingly, we tied — there was no winner! Which numbers did we pick?

Because Max and I had tied, that meant the whole numbers m and z satisfied the equality m^z = z^m. By taking the m-th and z-th roots of both sides, this gave you m^1/m = z^1/z. At this point, it was worth taking a closer look at the function f(x) = x^1/x. After all, having m^1/m = z^1/z meant the same as having f(m) = f(z).

This function increased for small values of x, reaching a maximum value when x was approximately 2.718 (i.e., e). Beyond this maximum, the function forever decreased, asymptotically approaching 1. Because the function was increasing and then decreasing, with no other change of direction in between, that meant either m or z had to be less than e, while the other number had to be greater than e. Let’s suppose that m was the smaller number.

At this point, there weren’t many options: m had to be either 1 or 2. If m had been 1, then you needed 1^z = z¹, which meant z was also equal to 1. Since the puzzle said m and z were distinct, this wasn’t a viable solution. If m had instead been 2, then you needed 2^z = z². Sure enough, this equation had two solutions: z = 2 (which again did not result in distinct numbers) and z = 4. Thus, the only two whole numbers Max and I could have picked were 2 and 4, as 2⁴ = 4².

For extra credit, you had to analyze another round of the game in which Max and I both picked positive numbers that weren’t necessarily whole numbers. I told Max my number without knowing his, at which point he told me the game was once again a tie. “Ah,” I replied, “that meant we must have picked the same number!” Which number did we both pick?

Mathematically, this meant that f(m) = f(z) implied that m and z were equal. For any value of m between 1 and e, there was a corresponding z greater than e such that f(m) = f(z). So for f(m) = f(z) to imply m = z, given that they were both at least 1, both m and z had to be e. Alternatively, as noted by solver Fernando Mendez, they both could have been any positive number less than or equal to 1.

Going up against a high schooler who’s in the top of his class in math and science, all I can say is I’m glad to have tied (rather than lost) both times we played this game.

Solution to the last Riddler Classic

Congratulations to 👏Jason Winerip 👏 of Phoenix, Arizona, winner of last week’s Riddler Classic.

Last week, you were introduced to the sudoku-like game of Star Battle. In the five-star variant of the game, you were trying to fill a 21-by-21 grid with stars according to certain rules:

• Every row had to contain exactly five stars.
• Every column had to contain exactly five stars.
• Every bold outlined region had to contain exactly five stars.
• No two stars could be horizontally, vertically or diagonally adjacent.

For example, here was a solved game board:

In this example, the stars seemed to be rather evenly distributed throughout the board, although there were some gaps. In particular, this board had 20 empty two-by-two squares, highlighted below:

Some of these two-by-two regions overlapped — even so, they still counted as distinct.

In a solved board of Star Battle, what were the minimum and maximum possible numbers of empty two-by-two squares?

At first glance, this looked like a rather complicated combinatorial puzzle, or perhaps something that required lots of simulation. But as it turned out, you could figure this out with some relatively straightforward algebra!

Solver N. Scott Cardell started by getting a lay of the land. Each of the 21 rows had five stars, which meant there were 105 stars in total. Meanwhile, there 20², or 400, total two-by-two squares in the grid. Because stars couldn’t be adjacent, that meant any given two-by-two square had at most one star on it.

Now a star in one of the four corners occurred on exactly one of these two-by-two squares, while a star on one of the edges occurred on two such squares and a star in the interior of the grid occurred on four such squares. If there were C corner stars, E edge stars and I interior stars, the number of two-by-two squares with a star on them was C + 2E + 4I. Since there were 400 two-by-two squares in all, the number of squares without a star was 400 − (C + 2E + 4I).

Since the total number of stars was 105, that meant C + E + I = 105, or I = 105 − E − C. Moreover, because each edge (like any other row or column) had five stars, with corner stars being counted for two edges, you had E + 2C = 20, or E = 20 − 2C.

At this point, you could algebraically eliminate variables from the expression for the number of empty two-by-two squares, 400 − (C + 2E + 4I). Plugging in 105 − E − C for I gave you 3C + 2E − 20. Finally, plugging in 20 − 2C for E gave you 20 − C.

After all that work, this was a surprisingly simple result. To find the number of empty two-by-two squares, all you had to do was count up the number of stars that were in the four corners and subtract that from 20. Sure enough, this was consistent with the solved game of Star Battle in the original puzzle: No stars were in a corner, and there were 20 empty two-by-two squares.

So what was the answer? The minimum number of empty two-by-two squares was 16, when all four corners had stars. The maximum was 20, when all four corners were devoid of stars. (In my opinion, this riddle turned out to be simpler than it seemed at first — as opposed to Star Battle itself, which is much trickier than it seems.)

Want more puzzles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

Want to submit a riddle?

Email Zach Wissner-Gross at riddlercolumn@gmail.com.