Discover more from Fiddler on the Proof
Can You Light Up the Pinball Machine?
Try lighting up all four lanes on a pinball machine. The catch? Hit a lane twice, and the light turns off again.
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
From Bowen Kerins comes a puzzle about pinball:
You’re playing a game of pinball that includes four lanes, each of which is initially unlit. Every time you flip the pinball, it passes through exactly one of the four lanes (chosen at random) and toggles that lane’s state. So if that lane is unlit, it becomes lit after the ball passes through. But if the lane is lit, it becomes unlit after the ball passes through.
On average, how many times will you have to flip the pinball until all four lanes are lit?
Instead of four lanes, now suppose your pinball game has N lanes. And let’s say that E(N) represents the average number of pinball flips it takes until all N lanes are lit up.
Now, each time you increase the number of lanes by one, you find that it takes you approximately twice as long to light up all the lanes. In other words, E(N+1) seems to be about double E(N).
But upon closer examination, you find that it’s not quite double. Moreover, there’s a particular value of N where the ratio E(N+1)/E(N) is at a minimum. What is this value of N?
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 recent Headscratcher puzzle from New Scientist that was “set” by Colin Wright.
You can find a screenshot of the puzzle here. Simply stated, it’s this: How can you partition a square floor into an even number of congruent tiles (not necessarily square themselves), such that the center of the square floor is contained entirely within one of the tiles (i.e., it’s not the edge of a tile)?
Last Week’s Fiddler
Congratulations to the (randomly selected) winner from last week: 🎻 Greg Tanner 🎻, from Moorhead, Minnesota. I received 78 timely submissions, of which 73 were correct—good for a 94 percent solve rate, a new record for a Fiddler. Great work, everyone! (Yes, a few Extra Credits have had higher solve rates, likely due to survivorship bias.)
Earlier this month, I watched La Vuelta, one of the three grand tours of cycling (a trio that includes the Tour de France). I noticed the peloton—the main group of riders—took on an aerodynamic profile that sometimes looked like a triangle, sometimes looked like a rhombus, and sometimes looked somewhere in between.
For example, the figure below shows the four possible formations between a triangle and a rhombus when the peloton’s maximum width was four riders:
For certain numbers of riders, multiple formations like these were possible. In particular, there were two formations with 15 riders: a triangle that was five riders wide at the base and an almost-rhombus that was four riders wide in the middle, but missing the bottommost rider, as shown below.
After 15, what was the next smallest number of riders that similarly had two distinct formations between a triangle and a rhombus?
Several solvers took to calling numbers with this property “peloton numbers.” More specifically, let’s call these “2-peloton numbers,” since we wanted at least two distinct formations, as opposed to 1-peloton numbers (at least one such formation), 3-peloton numbers (at least three such formations), etc. Some solvers found the next 2-peloton number by guess-and-check, others with computer code, and others with spreadsheets.
But no matter what strategy you used, the next 2-peloton number turned out to be 36, as illustrated by solver Rohan Lewis:
Interestingly, the two formations with 36 riders were the extremes allowed by the puzzle—a rhombus and a triangle—rather than one of the in-between formations. A few solvers noted that 36 is a special number: It’s what’s known as a square triangular number, because it’s both square (62 = 36) and triangular, as shown above. Indeed, the peloton’s “rhombus” formation was nothing more than a slightly deformed square, as illustrated by 🎬 Michael Schubmehl 🎬 below. There happen to be infinitely many square-triangular numbers (sequence A001110 on OEIS), each of which is also a 2-peloton number.
But those weren’t the only 2-peloton numbers. After all, 15, the smallest 2-peloton number, isn’t a square. This led a number of solvers to hunt for more 2-peloton numbers. Thomas Stone found the next one after 36, which turned out to be 43:
To find more 2-peloton numbers, it was helpful to have a systematic (i.e., algebraic) approach. As highlighted in Michael Schubmehl’s animation, each formation consisted of a square with a triangle removed from the bottom. If the square had A riders on a side, while the triangle being removed had a base that was B riders wide (noting that B had to be less than A), the total number of riders was A2 − B(B+1)/2. With any whole number values of A and B (again, with B < A), you can find a 1-peloton number. Plugging this formula into a computer and checking for duplicates gave you a list of 2-peloton numbers.
The sequence of 2-peloton numbers turned out to be 15, 36, 43, 49, 64, 66, 78, 85, 99, 100, 118, 120, 134, 141, 151, 159, 168, 169, 190, and so on. As I’m writing this column, this sequence of numbers still hasn’t been published on OEIS. Has anyone submitted it yet? If I don’t see it soon, I’ll just have to submit the sequence myself.
While the 2-peloton sequence hasn’t been published, I did want to give a shout-out to solver Greg Tanner, who found the 1-peloton sequence, which was equivalently described as “offsets i such that i + n(n+1)/2 is a perfect square for some positive integer n.” (Can you reason out why this would give you the 1-peloton numbers?)
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 John O'Donoghue 🎻, from Eugene, Oregon. I received 58 timely submissions, of which 53 were correct—good for a 91 percent solve rate. This set a record for most correct submissions for an Extra Credit, besting the previous record by more than 35 percent!
In last week’s Fiddler, you’re looking for the second-smallest 2-peloton number (as defined above), which turned out to be 36. But for Extra Credit, you had to find the smallest 3-peloton number. Stated algebraically, this meant you wanted to find a number that could be written as A2 − B(B+1)/2, with whole number B less than whole number A, for at least three distinct ordered pairs (A, B).
Again, most solvers had their computers generate the arrangements for various A and B, and then looked for numbers that appeared in triplicate. Solver Clem Lelievre correctly identified the smallest 3-peloton number as 141, as demonstrated below:
None of these three arrangements was a rhombus or a triangle; they were all somewhere in between.
But the fun didn’t stop there. There was a whole sequence of 3-peloton numbers to explore: 141, 190, 253, 288, 435, 463, 484, 610, 624, 631, 729, 778, 831, 841, 925, 946, and so on. This was another sequence that I still (as I’m writing this) can’t find on OEIS.
So far, we’ve looked at 1-peloton numbers, 2-peloton numbers, and 3-peloton numbers. What about 4-peloton numbers and higher? Does an N-peloton number exist for every N? These were great questions that a few bold readers explored.
The smallest 1-, 2-, 3-, 4-, and 5-peloton numbers were respectively, 1, 15, 141, 610, and 2,395. Interestingly, 2,395 was also the smallest 6-peloton number. The smallest 5-peloton number that wasn’t also a 6-peloton number was 6,903. In any case, this was yet another sequence I still can’t find on OEIS.
Finally, solver 🎬 Steve Curry 🎬 plotted how the number of N-peloton numbers increased for different values of N, as shown below. (Note that Steve defined an N-peloton number as meaning there were precisely N formations, rather than at least N formations.) Early on, there were more 2-peloton numbers, followed by 3-peloton numbers, 4-peloton numbers, 6-peloton numbers, 5-peloton numbers, and 8-peloton numbers. The rarest were the 7-peloton numbers, the smallest of which was 338,241. It was also intriguing that the 4-peloton line caught up to and surpassed the 3-peloton line, while the 8-peloton line similarly surpassed the 5-peloton line.
I know I sound like a broken record, but I really do hope that a few readers will post some of the sequences they found to OEIS. Let future generations of number theorists bask in your brilliance!
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.
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