Skip to main content
Can You Eat All The Chocolates?

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 next week’s 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.

Riddler Express

From Benjamin Dickman comes proof that it’s hip to be (a difference of) squares:

Benjamin likes numbers that can be written as the difference between two perfect squares. He thinks they’re hip. For example, the number 40 is hip, since it equals 72−32, or 49−9. But hold the phone, 40 is doubly hip, because it also equals 112−92, or 121−81.

With apologies to Douglas Adams, 42 is not particularly hip. Go ahead and try finding two perfect squares whose difference is 42. I’ll wait.

Now, Benjamin is upping the stakes. He wants to know just how hip 1,400 might be. Can you do him a favor, and figure out how many ways 1,400 can be written as the difference of two perfect squares? Benjamin will really appreciate it.

Extra credit: Can you find a general formula or approach for counting the number of ways any whole number can be written as the difference between two perfect squares? (Your approach might be a function of whether the number is even or odd, its prime factorization, etc.)

The solution to this Riddler Express can be found in the following week’s column.

Riddler Classic

From mathematician (and author of Basic Probability: What Every Math Student Should Know) Henk Tijms comes a choice matter of chance and chocolate:

I have 10 chocolates in a bag: Two are milk chocolate, while the other eight are dark chocolate. One at a time, I randomly pull chocolates from the bag and eat them — that is, until I pick a chocolate of the other kind. When I get to the other type of chocolate, I put it back in the bag and start drawing again with the remaining chocolates. I keep going until I have eaten all 10 chocolates.

For example, if I first pull out a dark chocolate, I will eat it. (I’ll always eat the first chocolate I pull out.) If I pull out a second dark chocolate, I will eat that as well. If the third one is milk chocolate, I will not eat it (yet), and instead place it back in the bag. Then I will start again, eating the first chocolate I pull out.

What are the chances that the last chocolate I eat is milk chocolate?

The solution to this Riddler Classic can be found in the following week’s column.

Solution to last week’s Riddler Express

Congratulations to 👏 Karen Campe 👏 of New Canaan, Connecticut, winner of last week’s Riddler Express.

Last week, you took a swing at a riddle about golf. (See what I did there?)

A typical hole is about 400 yards long, while the cup measures a mere 4.25 inches in diameter. Suppose that, with every swing, you hit the ball X percent closer to the center of the hole. For example, if X were 75, then with every swing the ball would be four times closer to the hole than it was previously.

For a 400-yard hole with no hazards (water, sand or otherwise) in the way, what was the minimum value of X so that you shot par, meaning you hit the ball into the cup in exactly four strokes?

First, it was a good idea to work with a single set of units. One yard equals 3 feet, or 36 inches. That meant the total length of the hole — 400 yards — was 14,400 inches. Meanwhile, the cup’s diameter was 4.25 inches, which meant its radius was 2.125 inches. So once the ball came within 2.125 inches of the hole’s center, it fell in. (Some readers used the diameter instead of the radius here, which led to an answer that was off by a few percentage points.)

Most solvers assumed left the value of X as a variable. As we said, the initial distance to the hole was 14,400 inches. After one swing, it was 14,400·(1−X). After two swings, it was 14,400·(1−X)2. After three swings, it was 14,400·(1−X)3. Finally, after four swings, it was 14,400·(1−X)4.

You shot par when this distance, 14,400·(1−X)4, was equal to 2.125 inches (or, rather, just a shade less than 2.125 inches). Mathematically, that meant 14,400·(1−X)4 = 2.125, or (1−X)4 = 17/115,200. Taking the fourth root of both sides and solving gave a solution of approximately 0.88978. In other words, X had to be at least about 89 for you to make par.

Some solvers took this problem even further, calculating the value of X for holes that were longer or shorter and also for different numbers of strokes. For example, what would it take to shoot a birdie (a score of one less than par — in this case, three strokes) on this hole? To answer this question, you had to solve the equation (1−X)3 = 17/115,200, with an exponent of three rather than four. That gave a value of X that is approximately 95. In other words, to go from a par to a birdie, you would need to hit the ball about twice as close to the hole with each shot.

Anyway, the next time you’re out on a golf course and you’re hitting the ball 89 to 95 percent of the way to the hole on each shot, be respectful to those playing ahead of you and shout, “Four!”

Golf humor. Apologies.

Solution to last week’s Riddler Classic

Congratulations to 👏 Jez Schaa 👏 of Singapore, winner of last week’s Riddler Classic.

Last week, you had half of a 10-inch pizza that you were cramming into your fridge. You had circular plates of all different sizes, so to save space in your fridge, you wanted to place the pizza on the smallest plate that held the entire semicircle (i.e., with no pizza hanging off the plate).

Unfortunately, the smallest plate that could hold half of a 10-inch pizza was just a circle with a 10-inch diameter. So much for saving space.

But there was a catch — you were allowed to make a single straight cut and then rearrange the two resulting pieces on a circular plate, provided the pieces didn’t overlap and no pizza was hanging off the plate.

What was the diameter of the smallest circular plate onto which you could fit the two resulting pieces?

Reader Pierre Bierre put it best: “This [was] a humdinger of a geometry Riddler.” What made this problem so difficult? Consider all the dimensions at play here:

  • First, should both endpoints of the cut have been on the circumference, or should one have been along the circumference and the other on the diameter? The latter was a better bet.
  • Where along the circumference should the first endpoint have been?
  • Where along the diameter should the second endpoint have been?
  • Finally, how should you have arranged the two pieces so that they were enclosed by the smallest possible circle?

Ideally, you would have turned those last three questions into three parameters. From there, you had to find the radius of the enclosing circle as a function of those three parameters and then minimize it (either with calculus or computationally). So yes, this was certainly a “humdinger.”

Some readers cheated and had the two pieces overlapping. In this case, the best you could do was to cut the semicircle in half, forming two quarter-circles that could be stacked directly on top of each other. The smallest plate each quarter-circle could fit on had a diameter of 5√2, or about 7.07 inches. While this was not the correct answer, it served as a good lower bound — the answer could be no less than 7.07 inches.

Tom Singer of Melbourne, Florida was able to properly fit his semicircular pizza onto a plate with a diameter of approximately 8.79 inches by assuming the cut was perpendicular to the slice’s diameter. Here’s a similar arrangement, courtesy of the puzzle’s submitter, Dean Ballard:

cutting a semicircle into two pieces, such that the cut is perpendicular to the diameter

But if you made a cut that was not necessarily perpendicular to the diameter, it was possible to do even better. The best anyone was able to do was about 8.16 inches. Again, here’s an illustration from Dean:

cutting a semicircle into two pieces so that they fit within a small circle

And if that wasn’t hard enough, last week’s extra credit had you making two straight cuts, resulting in three pieces to fit on your plate. Whereas the two-cut problem involved optimizing across three parameters, the three-cut version had those same parameters, plus many more. In particular, you had to choose which piece had the second cut, the particular edges or curves that were the endpoints of this second cut and how you could arrange all three pieces as tightly as possible.

Dean made his first cut perpendicular to the diameter, achieving a minimum plate that was about 7.737 inches in diameter (shown below). Meanwhile, Steven Trautmann of Aurora, Colorado wrote some C++ code to solve this for him, which gave an answer of about 7.584 inches.

cutting a semicircle into three pieces so that they fit within a small circle

As far as I’m concerned, this remains an open problem. If you find a plate size that works and that’s smaller than these answers, let me know (preferably with an image included!).

Want more riddles?

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


  1. Important small print: In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.