Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
As you may recall, around this time last year, I ran two puzzles by high school students who were winners of the Regeneron Science Talent Search. This year, I am delighted to have puzzles from two of this year’s winners as well.
Luke Robitaille is from Euless, Texas. As part of his research project, he proved that most simple braids — topological structures composed on intertwining strands — are orderly for low numbers of strands. But as the number of strands increases, nearly all simple braids become chaotic. Luke also represented the United States three times in the International Math Olympiad, taking home three gold medals.
Now, I can’t recall ever running a straight-up word puzzle in my days as the editor of The Riddler. But Luke’s puzzle was too good to pass up, so here goes:
Find a word in the English language to which you can add a vowel, resulting in another word that has fewer syllables.
By “add a vowel,” I mean insert one additional letter — a vowel — somewhere in the word (or at the beginning or end), while keeping the ordering of all the other letters the same. For example, you could add a vowel to the word “TASTY” to get the word “TOASTY.” However, both words are two syllables, meaning this is not the solution.
The solution to this Riddler Express can be found in the following column.
This week’s Classic comes from Daniel Larsen of Bloomington, Indiana. For his research project, Daniel studied Carmichael numbers. More specifically, he proved that for a sufficiently large number N, there is guaranteed to be at least one Carmichael number between N and 2N (resembling Bertand’s postulate for prime numbers).
As it turns out, there’s more than one way to define Carmichael numbers. For this riddle, we’ll define N as being a Carmichael number if (1) it is composite and has no square factors, and (2) if one less than every prime factor of N is a factor of N−1.
That’s a lot to take in at once, so let’s look at an example. The smallest Carmichael number is 561. Sure enough, it has no square factors (other than 1, which we’re not counting here). Meanwhile, its prime factors are 3, 11 and 17. If we subtract one from each of those, we get 2, 10 and 16, all of which divide 560 (one less than the original number).
Daniel’s puzzle asks you to find a six-digit Carmichael number that can be written as ABCABC in base 10. (Note: The digits A, B and C are not necessarily distinct.) Of course, you could look this up in a matter of seconds or even write some code to figure it out. But I assure you, this riddle can be done by hand, so please give it a shot!
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to 👏 Aaron Cote 👏 of Los Angeles, California, winner of last week’s Riddler Express.
Last week, you saw how you could rate a square’s “real estate value” in tic-tac-toe based on how many possible three-in-a-rows passed through it. Thus, the center scored 4, the corners scored 3 and the edges scored 2. Here’s a “bad drawing” of these scores, courtesy of author (and this puzzle’s submitter) Ben Orlin:
That meant the center was the “most valuable” real estate, while the edges were the “least valuable.”
Your challenge was to try your hand at three-dimensional tic-tac-toe, played on a 3-by-3-by-3 board. Which positions were the most and least valuable, and how much were they worth?
Before working this out, you could start by asking how many unique positions there were on the cube. Two positions were not unique if you could somehow rotate or reflect the cube so the positions were superimposed — such positions were guaranteed to have the same real estate value.
For the two-dimensional game, there were three unique positions: the center, the edge and the corner. In three dimensions, there was the “body center” (in the very middle of the cube), a “face center” (the center cube for each of the six faces), an edge and a corner — so four unique positions in total.
As in the two-dimensional case, it seemed like the middle of the cube — the body center — should be the most valuable. There were many strategies for calculating its value. My personal favorite — used by solver Izumihara Ryoma — was noting that the 26 positions around the body center formed 13 pairs that were on opposite sides of the center. (Similarly, the eight positions in the two-dimensional game formed four pairs.) That meant the body center, which indeed was the most valuable, scored 13 points.
Meanwhile, each face center had the same 4 points from two-dimensional tic-tac-toe, plus one more in the direction orthogonal to that face and going through the cube. That meant each face center scored 5.
Now each edge scored 2 points from one of the faces it was on, plus 2 points from the other face it was on. However, we just double-counted the three in a row along the shared edge of those two faces, bringing the score back down to 3. Each edge also had three in a row that went diagonally through the body center, for a total score of 4.
Finally, each corner was part of three faces, worth up to 9 points. But once again, each of the three edges it was on was double counted, bringing the score down to 6. Including the internal diagonal brought the final score up to 7.
That meant the last valuable position was the edge, which scored 4 points.
To see all these scores laid out on a cube, check out this animation:
For extra credit, you had to find the most and least valuable real estate once again, but this time in four-dimensional tic-tac-toe. Just as the central position scored (32−1)/2 in two dimensions and (33−1)/2 in three dimensions, it scored (34−1)/2 — or 40 points — in four dimensions. The least valuable position turned out to be worth 7 points, but it’s not so simple to describe where it is on a four-dimensional cube. Instead, I’ll refer you to solver Dean Ballard, who also analyzed five- and six-dimensional tic-tac-toe!
Solution to last week’s Riddler Classic
Congratulations to 👏 Sam Bergman 👏 of Evanston, Illinois, winner of last week’s Riddler Classic.
Last week, you studied a three-player Game of Mediocrity, where you won by not winning too much.
Each round, every player secretly picked a number from 0 to 10. The numbers were simultaneously revealed, and the median number won that number of points. (If two or more players picked the same number, then the winner was randomly selected from among them.)
After five rounds, the winner was whoever had the median number of points. (Again, if two or more players had the same score, then the winner was randomly selected from among them.)
With one round remaining, players A, B and C had 6, 8 and 10 points, respectively. Player A sighed and wrote down “3,” but failed to do so in secret. Players B and C both saw player A’s number (and both saw that the other saw A’s number) and planned to take care to write their own numbers in secret. Assuming everyone played to win, what numbers should B and C have chosen?
If A were to win the round and get those 3 points, then they would have 9 points in total and be crowned the winner. It was up to players B and C to prevent that from happening — somehow.
To have any shot at winning, C (who currently has the most overall points) cannot win this final round. Therefore, C should pick a number that is extreme — that is, either 0 or 10. If C were to pick 0, then B could win the round by writing down 1, in which case B would be the overall winner with 9 points. Seeing that logic, let’s suppose C were to write down 10. What would B do?
If B writes a number less than 3, then A will win the round and the entire game. If B writes a number greater than 3 but less than 10, then B will win the round but C will win the overall game. If B writes down 3 (like A), then the winner will either be A (if A wins the round) or C (if B wins the round). So B’s only logical remaining move is to write down 10. In this case, B and C will flip a coin to see who wins the round. The overall winner is whoever between them did not win the round.
To summarize, both B and C should have chosen 10, giving them both a 50 percent chance of mediocre victory.
A number of readers alternatively suggested that all three players should pick 3, in which case the winner of the round would be picked randomly. If A won the round, then A would be the overall winner. If B won the round, then C would be the overall winner. And if C won the round, B would be the overall winner. That gave each player a one-in-three chance of winning. As we just saw, by both choosing 10, players B and C can improve their chances to one in two. So choosing 10 was the better strategy for both B and C.
Several solvers, including Jenny Mitchell and Erin Sy, also observed that B and C both choosing 10 represented the lone Nash equilibrium of the game, where no individual player can do better by switching strategies.
And so, in this particular Game of Mediocrity, it paid to be extreme.
Want more puzzles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
Want to submit a riddle?
Email Zach Wissner-Gross at firstname.lastname@example.org.