Happy (Almost) New Year from The Fiddler!

Can you solve two puzzles about the number 2025, which happens to be a perfect square?

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 🎬.

Note: Due to the upcoming holidays, the next edition of Fiddler on the Proof will be coming out on January 3, 2025. See you next year!

This Week’s Fiddler

From Dean Ballard comes the perfect puzzle to usher in 2025:

The number 2025 is not prime. As a matter of fact, it’s a perfect square: 2025 = 45².

You cannot make 2025 by adding two distinct primes. To do so, you’d have to add an even prime and an odd prime. The only even prime is 2, but 2025 − 2 = 2023, which is not prime (it’s equal to 7∙17²).

But you can make 2025 by adding three distinct primes. For example, 661 + 673 + 691 = 2025.

You can also make 2025 by adding four distinct primes: 2 + 659 + 673 + 691 = 2025.

What is the greatest number of distinct primes that add up to 2025?

This Week’s Extra Credit

From Dean Ballard also comes some Extra Credit:

How can you assign a set of 20 distinct prime numbers to the 20 vertices of a dodecahedron, so that the numbers on the five vertices of each face add up to 2025?

(I realize it’s tricky to describe how you’re placing the numbers, so please just do your best.)

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 that the 85th William Lowell Putnam Mathematical Competition (known colloquially as “The Putnam”), which was administered earlier this month, has been publicly posted courtesy of Kiran Kedlaya.

You are welcome to spoil yourself with some solutions. You might also be interested to compare these with the responses from OpenAI o1 pro, which took about 36 minutes to generate.

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: 🎻 Michael Nolte 🎻 from Cologne, Germany. I received 33 timely submissions, of which 32 were correct—good for a 97 percent solve rate.

Last week, you considered a simplified version of the game show, The Floor, with nine contestants on a 3×3 square grid. Each round of the game consisted of the following steps:

• One of the remaining contestants was chosen at random. (Note that each contestant was equally likely to be chosen, regardless of how many squares they currently controlled.)

• The set of eligible opponents for this contestant included anyone whose territory shared a common edge with the contestant. One of these eligible opponents was chosen at random. (Again, all eligible opponents were equally likely to be chosen, regardless of how many squares they controlled or how many edges they had in common with the opponent.)

• The contestant and their selected opponent had a trivia duel, each with a 50 percent chance of winning. The loser was eliminated, and their territory was added to that of the winner.

These rounds repeated until one contestant remained, and that contestant was the overall winner.

You were a contestant on a new season of this 3×3 version of The Floor. The nine positions on the grid are shown below:

Which position would you have chosen? That is, which position or positions gave you the best chance of being the overall winner?

Since your chances of winning any given duel were always the same (50 percent), you could maximize your chances of winning by participating in as few duels as possible. Duels were always between neighbors, so to have as few duels as possible, you wanted as few neighbors as possible.

The central square (E) had four neighbors: B, D, F, and H. These four side squares each had three neighbors in turn: E and two corner squares. Finally, the corner squares (A, C, G, and I) each had just two neighbors—the minimum to be found among the squares in the grid.

Therefore, logic dictated that squares A, C, G, and I gave you the best chance of winning.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Lowell Vaughn 🎻 from Bellevue, Washington. I received 20 timely submissions, of which all 20 were correct—good for a perfect 100 percent solve rate. That’s just the fifth time among all 144 Fiddlers and Extra Credits (holy cow, has it been that many?) this has ever happened. Way to go, everyone!

Again, a new season of the 3×3 version of The Floor (as described above) was about to begin.

What was your probability of winning for each of the nine starting positions?