Can You Solve a High Schooler’s Favorite Puzzle?

Rivka Lipkovitz has excelled in mathematics research competitions. Do you have what it takes to solve one of her favorite math puzzles?

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “Extra Credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

Rivka Lipkovitz is a high school senior from San Francisco, California, whose research project earned her fifth place in the 2025 Regeneron Science Talent Search. For her research, Rivka modeled how U.S. voter identification laws affect voter turnout—very timely stuff!

An avid practitioner of the Doomsday Algorithm, Rivka took some time to share one of her favorite math puzzles with me:

A teacher is handing out candy to his students, of which there are at least four. He abides by the following rules:

• He hands out candy to groups of three students (i.e., “trios”) at a time. Each member of the trio gets one piece of candy.

• Each unique trio can ask for candy, but that same trio can’t come back for seconds. If students in the trio want more candy, they must return as part of a different trio.

• When a trio gets candy, the next trio can’t contain any students from that previous trio.

It turns out that every possible trio can get a helping of candy. What is the smallest class size for which this is possible?

This Week’s Extra Credit

Instead of trios of students, suppose now that groups of 10 students come up to get candy. This time, there are at least 11 students in the class. As before:

• Each member of the group of 10 gets one piece of candy per visit.

• Each unique group of 10 can ask for candy, but the exact same group of 10 can’t come back for seconds. If students in the group want more candy, they must return as part of a different group.

• When a group of 10 gets candy, the next group of 10 can’t contain any students from the previous group of 10.

Suppose the class size is the minimum that allows every possible group of 10 to get a helping of candy. How many pieces of candy does each student receive?

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. But this week, I’m sharing something a little outside the norm.

Eve Silberman, a former colleague of mine at Amplify, has a new project called playmix.ai, where you can describe a game you’d like to play to an AI, and it will make a playable version of your game for you, right then and there. It’s still in the early stages, but I think this happens to be one very cool application of AI!

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

I’m tracking submissions from paid subscribers and compiling a leaderboard, which I’ll reset every quarter. All timely correct solutions to Fiddlers and Extra Credits are worth 1 point each.

The results from Q1 are in, and the finest of Fiddlers is … 👑 Michael Schubmehl 👑, who finished with a perfect score! Tied for second were Eric Widdison and Peter Ji, who were each one off the lead. Congratulations to all who participated; the final standings are below. (If you think you see a mistake in the standings, kindly let me know.)

I will be reaching out to the top few finishers so I can send along a fancy-schmancy t-shirt prize!

Also, this week’s puzzles mark the beginning of the next quarter (Q2). If you’d like your shot at glory (and a t-shirt), become a paid subscriber today!

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Drew Mathieson 🎻 from San Francisco, California. I received 89 timely submissions, of which 85 were correct—good for a 96 percent solve rate.

Last week, there were four teams remaining in a college basketball bracket: the 1-seed, the 2-seed, the 3-seed, and the 4-seed. In the first round, the 1-seed faced the 4-seed, while the 2-seed faced the 3-seed. The winners of these two matches then faced each other in the regional final.

Also, each team possessed a “power index” equal to 5 minus that team’s seed. In other words:

• The 1-seed had a power index of 4.

• The 2-seed had a power index of 3.

• The 3-seed had a power index of 2.

• The 4-seed had a power index of 1.

In any given matchup, the team with the greater power index would have emerged victorious. However, the fans loved underdogs. As a result, the team with the lower power index got an effective “boost” B, where B was some positive non-integer. For example, B could have been 0.5, 133.7, or 2𝜋, but not 1 or 42. To be clear, B was a single constant throughout the tournament, for all matchups.

As an illustration, consider the matchup between the 2- and 3-seeds. The favored 2-seed had a power index of 3, while the underdog 3-seed had a power index of 2+B. When B was greater than 1, the 3-seed defeated the 2-seed in an upset.

Depending on the value of B, different teams could have won the tournament. Of the four teams, how many could never win, regardless of the value of B?

For convenience let’s do away with the confusing notion of “seeds,” and refer to teams simply by their power index. So the first round featured team 4 (i.e., the 1-seed) against 1 (the 4-seed) and 3 (the 2-seed) against 2 (the 3-seed), after which the winners faced each other in the finals.

One thing to notice was that all values of B between two particular whole numbers produced the same result. For example, suppose B was 1.5. In the semifinal rounds, team 4 defeated team 1 (which had an effective value of 1 + 1.5 = 2.5), while team 3 was defeated by team 2 (which had an effective value of 2 + 1.5 = 3.5). In the finals, team 4 defeated team 2 (which still had an effective value of 3.5). Instead of using 1.5 for B, if we used any other value between 1 and 2, the results would have been the same. That’s because all teams had whole number power indices, which in turn meant differences between power indices were whole numbers. So you only had to consider one value of B between each pair of consecutive whole numbers. The half-integers (i.e., numbers ending in “.5”) were a natural choice to consider.

As we just saw, when B was 1.5, team 4 was the victor. What happened with other half-integer values of B? Let’s take a look:

• When B was 0.5, 4 defeated 1 (effectively 1.5) and 3 defeated 2 (effectively 2.5). In the finals, 4 defeated 3 (effectively 3.5). The winner was 4.

• When B was 1.5, we already saw how the winner was 4.

• When B was 2.5, 4 defeated 1 (effectively 3.5) and 2 (effectively 4.5) defeated 3. In the finals, 2 (effectively 4.5) defeated 4. The winner was 2.

• When B was 3.5, 1 (effectively 4.5) defeated 4 and 2 (effectively 5.5) defeated 3. In the finals, 1 (effectively 4.5) defeated 3). The winner was 1.

At this point, we found values of B for which the winner was 4, 2, and 1. For even greater values of B we haven’t yet considered (i.e., B > 4), the winner remained team 1, since it was always the underdog but had a sufficient effective power index (1+B) to defeat any opponent.

So of the four teams, there was only one team (team 3, meaning the 2-seed) that could never win, regardless of the value of B. (If you simply submitted “2,” meaning only the 2-seed couldn’t win, I checked your explanation to confirm your intent and gave you credit.)

Another way to convince yourself that team 3 could never win was to look at how it did in each round. To defeat team 2 in the semifinals, B had to be less than 1. And if B was less than 1, that meant team 4 defeated team 1 in the other semifinal. Finally, for team 3 to defeat team 4 in the finals, that meant B had to be greater than 1. Of course, B couldn’t simultaneously be less than 1 and greater than 1. Because of this contradiction, team 3 could never win.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 David Slater 🎻 from Okemos, Michigan. I received 50 timely submissions, of which 42 were correct—good for an 84 percent solve rate.

Instead of four teams, now there were 2⁶, or 64, seeded from 1 through 64. The power index of each team was equal to 65 minus that team’s seed.

The teams played in a traditional seeded tournament format. That is, in the first round, the sum of opponents’ seeds was 2⁶+1, or 65. If the stronger team always advanced, then the sum of opponents’ seeds in the second round was 2⁵+1, or 33, and so on.

Once again, the underdog in every match got a power index boost B, where B was some positive non-integer. Depending on the value of B, different teams could have won the tournament. Of the 64 teams, how many could never win, regardless of the value of B?