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
I recently introduced my children to a game called “Buzz” (also known as “Fizz buzz”), in which players take turns reciting whole numbers in order. However, in one particular variant of the game, anytime a number is a multiple of 7 or at least one of its digits is a 7, the player must say “buzz” instead of that number.
For example, here is how the first 20 turns of the game should proceed: 1, 2, 3, 4, 5, 6, buzz, 8, 9, 10, 11, 12, 13, buzz, 15, 16, buzz, 18, 19, 20.
How many times should “buzz” be said in the first 100 turns of the game (including those mentioned above in the first 20 turns)?
This Week’s Extra Credit
As we just saw, in the first 20 turns of the game, 15 percent of the numbers were “buzzed.” But as the game proceeds, an increasing frequency of numbers get buzzed.
There is a certain minimum number N such that, for the Nth turn in the game and for every turn thereafter, at least half the numbers up to that point have been buzzed. 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 that The Daily Baffle, whose Arithmeglyphs puzzle I previously shared, is now available for download on various devices. Yay, more puzzling fun on the go!
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: 🎻 Jason Donofrio 🎻 from Brooklyn, New York. I received 63 timely submissions, of which 56 were correct—good for an 89 percent solve rate.
Last week, you had four number cubes, where each face of each cube could display one numeric digit from 0 to 9. You could make various numbers by picking three faces on three distinct cubes and lining them up. For example, you could make the number “123” by lining up three cubes to show 1-2-3, and you could make the number “7” by lining up three cubes to show 0-0-7.
Importantly, any face with a “6” could also be used to display a “9” by flipping the cube around, and vice versa. However, no other pairs of digits were interchangeable in this way.
You could choose which digits to place on the various faces of the four cubes. Your goal was to be able to make all the whole numbers from 1 to N, without skipping any numbers in between. With optimal design, what was the greatest possible value of N?
Solver Nis Jørgensen noted that it was not possible to make all the whole numbers up to and including 777. Why? Because then you would have needed:
two 0s ( for 007, 070, 700, etc.)
three 1s (for 111)
three 2s (for 222)
three 3s (for 333)
three 4s (for 444)
three 5s (for 555)
three 6s (for 666)
three 7s (for 777)
two 8s (for 088, 188, etc.)
If you tallied all these up, that was 25 digits. But with four number cubes, each of which had six faces, you could only place a total of 24 digits on the cubes.
So while 777 was impossible, 776 was still in play. If you didn’t need to generate 777, then you only needed two 7s on the dice, reducing the total number of required digits to 24. Thus, 776 seemed likely to be the answer.
But to prove that 776 was indeed possible, you still had to assign a set of six digits to each of the four cubes in some valid way. Solver Jerome Socolof came up with one such assignment:
{0, 1, 3, 4, 5, 7}
{0, 2, 3, 4, 6, 8}
{1, 2, 3, 5, 6, 7}
{1, 2, 4, 5, 6, 8}
Sure enough, with these digits on the four dice, you could generate every whole number from 1 (i.e., 0-0-1) to 776.
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 Joe Kong 🎻 from Mississauga, Ontario, Canada. I received 33 timely submissions, of which 23 were correct—good for a 70 percent solve rate.
Now that you had determined (in last week’s Fiddler) that the greatest possible value of N was 776, how many distinct ways could you have assigned numbers to the four cubes that allowed you to generate all the whole numbers from 001 to 776?
Note that the cubes were not ordered in any way. And, very importantly, swapping digits between two faces on a single cube did not count as producing a distinct arrangement. In other words, you didn’t have to worry about the various ways to assign six given digits to the faces of a cube. (Also, “6” and “9” counted as the same digit here.)
