# Can You Solve a High Schooler’s Favorite Puzzle?

### Linus Tang has excelled in research competitions and math olympiads. Do you have what it takes to solve his favorite math puzzle?

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. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

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

Suppose you have infinitely many 3-by-3 cm tiles and infinitely many 5-by-5 cm tiles. You want to use some of these tiles to precisely cover a square whose side length is a whole number of centimeters. Tiles may not overlap, and they must completely cover the larger square, without jutting beyond its borders.

What is the smallest side length this larger square can have, such that it can be precisely covered using at least one 3-by-3 tile *and *at least one 5-by-5 tile?

## This Week’s Extra Credit

This week’s Extra Credit is a personal favorite of high schooler Linus Tang. Linus was a Finalist of the 2024 Regeneron Science Talent Search, where he presented his research on measures of peripherality (the opposite of centrality) in graphs. He was also a Gold Award winner at the 2024 USA Math Olympiad. Outside of his impressive academic accomplishments, Linus is also an avid player of Go.

Here’s is Linus’s puzzle:

This time, you have an infinite supply of square tiles for *each* odd whole number side length (as measured in centimeters) greater than 1 cm. In other words, you have infinitely many 3-by-3 cm tiles, infinitely many 5-by-5 cm tiles, infinitely many 7-by-7 cm tiles, and so on.

You want to use one or more of these tiles to precisely cover a square whose side length is *N* cm, where *N* is an integer. Once again, tiles may not overlap, and they must completely cover the larger square without jutting beyond its borders. (Before you ask—yes, all odd values of *N* result in squares that can be covered with a single tile.)

What is the *largest* integer *N* for which this task is *not* possible?

## 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. This week, I’m sharing a puzzle called “Bug Byte,” which was apparently concocted over at Jane Street Capital. (Note: I have no ties to Jane Street—this just genuinely looked like a great puzzle to me!) For the brief but complete set of rules, follow the link to the puzzle.

Here’s a screenshot of the rather snazzy-looking challenge:

I spent some time trying to solve it by hand, which soon led to frustration. Then I wrote code that used the conditions to considerably reduce the number of potential solutions. After checking the last few cases by hand, I’m proud to report that the prize for solving this puzzle is digital confetti:

If you have the time, give it a try, and get some of that digital confetti for yourself!

## 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.

## Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 John 🎻 from Washington, DC. I received 80 timely submissions, of which 68 were correct—good for an 85 percent solve rate.

Last week, there were 25 sprinters at a meet, and there was a well-defined order from fastest to slowest among the runners. That is, the fastest sprinter would always beat the second-fastest, who would in turn always beat the third-fastest, and so on. However, this order was not known to you in advance.

To reveal this order, you could choose any 10 sprinters at a time to run in a heat. For each heat, you only knew the *ordinal* results, so which runner finished first, second, third, and so on up to 10th. You did *not* know the specific finishing time for any runner, making it somewhat difficult to compare performances across heats.

Your goal was to determine with absolute certainty which of the 25 sprinters was fastest, which was second-fastest, and which was third-fastest. What was the fewest number of heats needed to guarantee you could identify these three sprinters?

First off, for convenience, let’s label the sprinters using the numbers 1 through 25, noting that these assignments have been randomized (i.e., “1” is not necessarily the fastest).

Now, it wasn’t possible to identify the fastest three sprinters in just *two* heats. Why? Because with two heats you could only test the mettle of at most 20 sprinters. That meant you couldn’t observe the remaining five sprinters, and if any of them had been one of the fastest three runners you simply couldn’t know about it.

At the same time, there were plenty of ways to identify the three fastest sprinters via *four* heats. One such way was to have Heat 1 consist of sprinters 1-10, Heat 2 consist of sprinters 11-20, and Heat 3 consist of sprinters 21-25. If a sprinter was among the fastest three overall, then they had to come in the top three positions in one of these heats (or any potential heat, for that matter). That meant you could take the top three finishers from Heats 1, 2, and 3, for a total of nine sprinters, and have them all square off in a final heat. The winner of this heat was the fastest overall, second-place was the second-fastest overall, and third-place was third-fastest overall. Done and … done?

So the real question here was this: Was it possible to identify the three fastest sprinters in just *three* heats? A few readers decided this was impossible, offering an answer of four. But let’s take a closer look, together.

For Heat 1, you wanted to glean as much information as possible, and there was no reason to race any fewer than 10 sprinters. Any 10 would have done, so let’s again have sprinters 1-10 square off. As we already said, only the top three in this heat had any chance of being the top three overall, so let’s keep track of them. Let’s label first place in Heat 1 as “A,” second place as “B,” and third place as “C.”

For Heat 2, it was tempting to carry over A, B, and C, and have them race against a fresh set of seven more sprinters, such as 11-17. The fastest three in this heat would still have a shot at being the fastest three overall. However, Heat 3 then required the three winners of Heat 2 to square off against the eight remaining sprinters, 18-25. Because you couldn’t squeeze these 11 runners into a single heat, this strategy was doomed to failure.

It looked like you had to be a little more economical in terms of whom you carried over from Heat 1 to Heat 2. Suppose you had two of the top three finishers from Heat 1 also compete in Heat 2, along with eight other sprinters (e.g., 11-18). (The two sprinters you carried over could have been A and B, A and C, or B and C.) If these two sprinters finished in the same positions in Heat 2 as they did in Heat 1, then you had a total of four candidates—A, B, C, and the other top-three finisher from Heat 2—that you had to bring into Heat 3. Unfortunately, Heat 3 also had the remaining seven sprinters (i.e., 19-25), which meant you again had to somehow squeeze 11 runners into a single heat.

At this point, you knew you could only carry over *one* of A, B, or C into Heat 2, along with nine other sprinters (e.g., 11-19). Let’s start with A. If A finished first in Heat 2, then you had five candidates for top-three overall: A, B, C, second place from Heat 2, and third place from Heat 2. Those five had to compete against the remaining 6 sprinters (20-25), which meant Heat 3 had to accommodate up to 11 runners. Nope.

What about having C run again in Heat 2? If C finished third in Heat 2, then you again had five candidates for top-three overall: A, B, C, first place from Heat 2, and second place from Heat 2.

The only remaining option was having B run again in Heat 2, but it was hard to see how that could have fared any better. As long as there were at most four candidates for top-three overall in the first two heats to compete against the remaining six sprinters (19-25), then this strategy was viable. Let’s consider all the cases:

If B won Heat 2, then you had four candidates for top-three overall who would advance to Heat 3: A, B, C, and second place from Heat 2. You weren’t sure which of the last two here was faster, but you were able to rule out everyone else who had run.

If B came in second in Heat 2, then you had three candidates for top-three overall: A, B, and first place from Heat 2. All three of these sprinters were faster than both C and the second-place sprinter from Heat 2.

If B came in third in Heat 2, then you had three candidates for top-three overall: A, first place from Heat 2, and second place from Heat 2. All three of these sprinters were faster than B.

If B came in fourth or later in Heat 2, then you had four candidates for top-three overall: A, first place from Heat 2, second place from Heat 2, and third place from Heat 2. You weren’t sure where A stacked up against the top three finishers of Heat 2, but you were able to rule out everyone else who had run.

And so indeed it was possible to determine the fastest three overall runners in just **three heats**. The strategy was to have any 10 sprinters in Heat 1, have second place from Heat 1 compete against another nine sprinters in Heat 2. Then, depending on how that second-place runner finished in Heat 2, have the three or four candidates for top-three overall race against the remaining six sprinters in Heat 3. The first three across the finish line in Heat 3 were precisely the fastest, second-fastest, and third-fastest sprinters among the 25.

## Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Paige Kester 🎻 from Southlake, Texas. I received 25 timely submissions, of which 18 were correct—good for a 72 percent solve rate.

At a different meet, there were six sprinters that could race head-to-head. (In other words, there were only two sprinters per heat.) Again, they finished races in a consistent order that was not known to you in advance.

This time, your goal was to determine the *entire* order, from the fastest to the slowest and everywhere in between. What was the fewest number of head-to-head races needed to guarantee you could identify this ordering?

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