Can You Squeeze the Sheets?

Two identical sheets have equally spaced ridges on one side. When the sheets are squeezed together with the ridges facing each other, what spacing maximizes the empty volume between them?

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: Starting next week, I’ll be tracking submissions from paid subscribers to compile a leaderboard, which I’ll reset every quarter. All timely correct solutions to Fiddlers and Extra Credits will be worth 1 point each. At the end of each quarter, I’ll 👑 crown 👑 the leader or leaders. Sign up for a paid subscription now to get in on the fun!

This Week’s Fiddler

From Q P Liu comes a bumpy way to start the new year:

Two large planar sheets have parallel semicircular cylindrical ridges with radius 1. Neighboring ridges are separated by a distance L ≥ 2. The sheets are placed so that the ridges extrude toward each other, and so that the sheets cannot shift relative to each other in the horizontal direction, as shown in the cross-section below:

Which value of L (again, that’s the spacing between ridges) maximizes the empty space between the sheets?

To be clear, you are maximizing the volume of empty space per unit area of one flat sheet. In the cross-section shown above, that’s equivalent to maximizing the area of empty space per unit length of one flat sheet.

This Week’s Extra Credit

From Q P Liu also comes some Extra Credit:

Instead of cylindrical ridges, now suppose the sheets have any number (greater than zero) of hemispherical deformations with radius 1 that extrude toward each other. This time, the sheets need not be the same as each other.

As before, the distance between the centers of any two deformations on the same sheet must be at least 2. What is the minimum empty space, again expressed as volume per unit area of one flat sheet?

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 an oldie but a goodie:

How can you turn a sphere inside out, without cutting it or creating any pinched creases along the way? Importantly, the sphere is allowed to “self-intersect,” or pass through itself.

The reason I ask is because an older video showing how to “evert” a sphere in this manner has amassed an impressive following online, which in turn has garnered some attention from the press.

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: 🎻 Hao Ran Wee 🎻 from Kuala Lumpur, Malaysia. I received 50 timely submissions, of which 44 were correct—good for an 88 percent solve rate.

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

You couldn’t make 2025 by adding two distinct primes. To do so, you had 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 could make 2025 by adding three distinct primes. For example, 661 + 673 + 691 = 2025. You could also make 2025 by adding four distinct primes: 2 + 659 + 673 + 691 = 2025.

What was the greatest number of distinct primes that added up to 2025?

As noted by reader Nic Warmenhoven (a former student of Dean Ballard, the puzzle’s author!), this was a variant of the knapsack problem in which the weights and costs of the “items” being placed in the knapsack were the same—they were the values of the numbers themselves. (The algorithm folks out there undoubtedly know that the knapsack problem is NP-complete.)

One way to begin was to determine an upper bound for the answer. For instance, instead of adding up primes, suppose we added up as many odd numbers as we could to make 2025. Since not every odd number is prime, this number had to be greater than our answer.

The sum of the first N odd numbers is always N². Setting this equal to 2025 gave you N² = 2025, or N = 45. That meant adding the first 45 odd numbers (i.e., from 1 to 99) resulted in a sum of 2025. That further meant that adding the first 45 primes had to be a value greater than 2025, so the answer had to be less than 45.

Solver Ricardo found a more aggressive upper bound by simply adding up primes (rather than odd numbers). The sum of the first 34 primes (i.e., from 2 to 139) was 2127. So the answer had to be less than 34.

Suppose you were adding 33 primes together. If the number 2 was among them, then the sum would be even, since you were adding one even number (2) and an even number (32) of odd primes. For the sum to be odd (as was the case here, since 2025 is odd), all 33 primes had to be odd. Alas, the sum of the smallest 33 odd primes (i.e., from 3 to 139) was 2127−2, or 2125. Since that too was greater than 2025, the answer had to be less than 33.

Now suppose you were adding 32 primes together. For the sum to be odd, 2 had to be among them. That meant the remaining 31 primes had to add up to 2023. As we already said, the sum of the 33 smallest odd primes (from 3 to 139) was 2125. So one way to hit the target sum of 2023 was to remove a pair of these 33 odd primes that added up to 102 (i.e., the difference between 2125 and 2023). As it turned out, there were eight such pairs: 5 and 97, 13 and 89, 19 and 83, 23 and 79, 29 and 73, 31 and 71, 41 and 61, and 43 and 59. That meant the answer was indeed 32.

Here were the eight sets of 32 primes for which the largest prime was 139:

• 2, 3, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 101, 103, 107, 109, 113, 127, 131, 137, 139

• 2, 3, 5, 7, 11, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139

• 2, 3, 5, 7, 11, 13, 17, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139

• 2, 3, 5, 7, 11, 13, 17, 19, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139

• 2, 3, 5, 7, 11, 13, 17, 19, 23, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139

• 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 37, 41, 43, 47, 53, 59, 61, 67, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139

• 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 47, 53, 59, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139

• 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139

Of course, there were other ways to find 32 primes that added up to 2025 when they were allowed to exceed 139. For example, solver Andrew Petersen added up the smallest 30 primes to get 1593. From there, he needed two primes whose sum was 432, such as 163 and 269.

A few brave solvers went a step further, venturing to find all such sets of 32 primes. There are 306 primes less than 2025, which meant there were 306 choose 32 cases to check, in theory. That was a mind-boggling 2.5×10⁴³ cases—far more than a computer could check.

That said, none of the primes could be close to 2025 (e.g., 2017), since adding another 31 distinct primes would have resulted in a sum greater than 2025. If 31 of the primes were as small as possible, meaning they were the 31 smallest primes, their sum would have been 1720. That meant the last and largest prime of the bunch was at most 305, or rather, 293 (the largest prime less than 305). Since 293 is the 62nd prime, the number of cases to check was now 62 choose 32, a more reasonable 4.5×10¹⁷ cases. You also knew one of the numbers had to be 2, which further reduced the figure to 2.3×10¹⁷ cases. Still—yikes!

Nevertheless, several solvers were able to determine this total. Peter Ji first used dynamic programming to identify all 59 trillion ways to write 2025 as a sum of distinct primes. Of these, Peter found that 4058 of them contained the maximum 32 primes. Eric Widdison found the same result, while Izumihara Ryoma was kind enough to list them all out.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Gary M. Gerken 🎻 from Littleton, Colorado. I received 9 timely submissions, of which 7 were correct—good for a 78 percent solve rate.

How could 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 added up to 2025?