Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.
Every Friday morning, I present mathematical puzzles intended to challenge and delight you. Most can be solved with careful thought, pencil and paper, and the aid of a calculator. The “Extra Credit” is where the analysis typically gets 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 puzzles are 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
This week’s puzzle was a collaboration with high-schooler Connor Hill, the First Place Winner from this year’s Regeneron Science Talent Search. Connor’s project involved an original, computer-assisted proof to find the complete list of noble polyhedra, in which all the faces are indistinguishable and all the vertices are indistinguishable. The five Platonic solids are perhaps the best known examples of noble polyhedra, but Connor documented all of them, including two infinite sets and 146 exceptional cases, among which were 85 new examples!
Here’s the puzzle from Connor—unrelated to his work—which is about a frog that makes very particular hops:
A frog is hopping around a chessboard, always from the center of one square to the center of another square. Each square has side length 1, but the board itself is not necessarily 8-by-8. Instead, it’s N-by-N, where N is some large whole number.
Every jump the frog makes must be the same distance, which we’ll call L. The frog wants to make four jumps such that:
After the fourth jump, the frog has returned to its starting square.
The frog visits a total of four distinct squares along the way, including the square on which it starts (and also stops).
The path the frog takes is not a square loop.
The frog is never on a square that is diagonal (i.e., a bishop’s move away) or horizontal/vertical (i.e., a rook’s move away) from the starting square.
What is the smallest jumping distance L for which this is possible?
This Week’s Extra Credit
From Connor also comes some Extra Credit:
The frog is jumping around the board with the same minimum distance L you just found.
But this time, the frog also wants to be able to hop to every location on the chessboard. What is the minimum value of N for which this is 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 an online game that came my way courtesy of Bernardo Recamán Santos. It’s called “The Snake and the Hunter” and it’s billed as a “a strategic combinatorial game.” Well, that’s exactly what it is. Can you beat the computer?
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 correct solutions to Fiddlers and Extra Credits are worth 1 point each. Solutions should be sent prior to 11:59 p.m. the Monday after puzzles are released. At the end of each quarter, I’ll 👑 crown 👑 the finest of Fiddlers. If you think you see a mistake in the standings, kindly let me know.
Last Week’s Fiddler
Congratulations to the (randomly selected) winner from last week: 🎻 Zach Donofrio 🎻 from Brooklyn, New York. I received 86 timely submissions, of which 79 were correct—good for a 92 percent solve rate.
Last week, you studied an analog clock that included an hour hand, a minute hand, and 60 minute markers, 12 of which were also hour markers.
At a certain time, the hour hand and minute hand were both pointing directly at minute markers (either of which could also have been an hour marker). The hour hand was 13 markers ahead (i.e., clockwise) of the minute hand.
At what time did this occur? (You didn’t have to worry about a.m. vs. p.m. for this puzzle.)
Suppose the time was m minutes after h o’clock, where m was at least 0 and less than 60. So, for example, if h was 3 and m was 47, then the time was 3:47. Let’s figure out just which markers the hour and minute hands were pointing in terms of m and h.
The minute hand was the easier of the two. Because the time was m minutes after the hour, the minute hand was pointing to minute marker m.
Now for the hour marker. There were five minute markers for every hour marker on the clock, so the hour hand pointed directly at a minute marker every fifth of an hour, or every 12 minutes. Thus, m had to be one of 0, 12, 24, 26, or 48. Moreover, hour marker h was equivalent to minute marker 5h. That meant the hour hand pointed to minute marker 5h + m/12. Sure enough, you can see that this expression was an integer only when m was a multiple of 12.
For convenience, let’s define k = m/12. Since m/12 had to be an integer, so did k. More specifically, k had to be 0, 1, 2, 3, or 4.
At this point, you knew the minute hand pointed to minute marker m = 12k, while the hour hand pointed to minute marker 5h + m/12 = 5h + k. For what values of h and k was the hour hand 13 markers ahead of the minute hand? In other words, when was 5h + k equal to 12k + 13?
Setting these two expressions equal to each other gave you 5h + k = 12k + 13, or 5h = 11k + 13. For there to be integer solutions, both sides of this last equation had to equal the same multiple of 5. Let’s check the five possible values of k to see when this occurred:
k = 0: Then 11k + 13 = 11·0 + 13 = 13, which was not a multiple of 5.
k = 1: Then 11k + 13 = 11·1 + 13 = 24, which was not a multiple of 5.
k = 2: Then 11k + 13 = 11·2 + 13 = 35, which was a multiple of 5 (huzzah!). That gave you 5h = 35, or h = 7.
k = 3: Then 11k + 13 = 11·3 + 13 = 46, which was not a multiple of 5.
k = 4: Then 11k + 13 = 11·4 + 13 = 57, which was not a multiple of 5.
The only time k and h were both integers was when k = 2 (in which case m = 24) and h = 7. Thus, the time on the clock—and the answer to the puzzle—was 7:24. Here’s an illustration of this time, courtesy of solver Michael Schubmehl:
Sure enough, you can see that the hour hand was exactly 13 markers ahead of the minute hand.
But we weren’t quite done with this puzzle yet. We had neglected the possibility where the hour hand was 13 markers clockwise of the minute hand, but they were on opposite sides of the 12-hour (or 0-minute) marker. For this to occur—and for the hour hand to still point to a minute marker—you needed m = 48. For the hour hand to be 13 markers ahead, it had to point to m = 1, which corresponded to a time of 12:12, and not 12:48. Thus, no such time was possible.
With this check complete, 7:24 was indeed the unique answer to the puzzle.
By the way, the most common incorrect answer I received was 4:36, which was when the hour hand was 13 markers behind (i.e., counterclockwise to) the minute hand.
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 Jason Shaw 🎻 from Decorah, Iowa. I received 39 timely submissions, of which 29 were correct—good for a 74 percent solve rate.
At various times of day, the minute and hour hands form a right angle. But was there a time of day when the hour hand, minute hand, and second hand—each of which rotated in a continuous manner—together formed two right angles, with any of the hands in the middle? If you could find such a time or times, what were they?
If you couldn’t find any such times, suppose the measures of the two nearly right angles formed by the three hands measured A and B degrees. What time or times of day minimized the square error function f(A, B) = (A − 90)2 + (B − 90)2?
Either way, your answer had to be precise to at least a thousandth of a second. (Again, you didn’t have to worry about a.m. vs. p.m. for this puzzle.)
