Welcome to the Riddler. Each week, I come up with problems related to things that are dear to us here: math, logic, and probability. Two puzzles are featured each week: the Riddler Express for those of you who want something small and the Riddler Classic for those of you who like the movement of slow puzzles. Submit a correct answer for either, and you might get a shoutout in the next column. Please wait until Monday to share your answers publicly! If you need a clue or if your favorite puzzle is gathering dust in your attic, find me on twitter.

## Riddler Express

As you may recall, around this time last year, I ran two puzzles by high school students who won the Regeneron Science talent search. This year I’m thrilled to have puzzles of two of *this* also the winners of the year.

Luke Robitaille is originally from Euless, Texas. As part of his research project, he proved that most simple braids – topological structures composed of intertwined strands – are ordered for a low number of strands. But as the number of strands increases, almost all single braids become messy. Luke also represented the United States three times at the International Mathematical Olympiad, winning three gold medals.

Now I don’t remember ever running a straight-up *word puzzle* when I was editor of The Riddler. But Luke’s puzzle was too good to pass up, so here it is:

Find a word in the English language to which you can add a vowel, resulting in another word that has *less of* syllables.

By “add a vowel”, I mean inserting an extra letter – a vowel – somewhere in the word (or at the beginning or at the end), while keeping the same order for all the other letters. For example, you can add a vowel to the word “TASTY” to get the word “TOASTY”. However, both words are two syllables, which means this is not the solution.

Submit your answer

## classic riddler

This week’s classic comes from Daniel Larsen of Bloomington, Indiana. For his research project, Daniel studied Carmichael numbers. Specifically, he proved that for a large enough number *NOT*it is guaranteed that there is at least one Carmichael number between *NOT* and 2*NOT* (similar to Bertand’s postulate for prime numbers).

It turns out that there is more than one way to define Carmichael numbers. For this puzzle, we will define *NOT* to be a Carmichael number if (1) it has no square factor, and (2) if one less than each prime factor of *NOT* is a factor of *NOT*−1.

That’s a lot to take in at once, so let’s look at an example. The smallest Carmichael number is 561. Effectively, it has no square factor (other than 1, which we don’t count here). Meanwhile, its prime factors are 3, 11, and 17. If we subtract one from each of these, 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 ABCABC in base 10. (**To note:** The numbers A, B and C are *do not* necessarily distinct.) Of course, you can look that up in seconds or even write some code to figure it out. But I assure you that this riddle can be done by hand, so give it a try!

Submit your answer

## Solution to last week’s Riddler Express

Congratulations to 👏 Aaron Cote 👏 from Los Angeles, CA, winner of last week’s Riddler Express.

Last week you saw how you could gauge the “real estate value” of a tic-tac-toe square based on the number of possible rows of threes that ran through it. So the center scored 4, the corners scored 3, and the edges scored 2. Here is a “bad drawing” of these scores, courtesy of the author (and the author of this puzzle) Ben Orlin:

This meant that 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 most and least valuable, and how much were they worth?

Before solving this problem, you can start by asking how many unique positions there were on the cube. Two positions were not unique if you could somehow rotate or mirror the cube so that the positions overlapped – those positions were guaranteed to have the same real estate value.

For two-dimensional play, there were three unique positions: center, edge, and corner. In three dimensions, there was the “center of the body” (in the very middle of the cube), a “center of the face” (the central cube for each of the six faces), an edge and a corner – thus four positions unique to the total .

As in the two-dimensional case, it seemed that the middle of the cube – the center of the body – must be the most valuable. There were many strategies to calculate its value. My personal favorite – used by solver Izumihara Ryoma – was to note that the 26 positions around the center of the body formed 13 pairs that were on opposite sides of the center. (Similarly, the eight positions in the two-dimensional game formed four pairs.) This meant that the **body center**which was indeed the most valuable, scored **13 dots**.

Meanwhile, each center of the face had the same 4 points of the two-dimensional tic-tac-toe, *more* one more in the direction orthogonal to this face and passing through the cube. This meant that 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 common edge of these two faces, bringing the score to 3. Each edge also had three in a row that ran diagonally across the center of the body, for a score total of 4.

Finally, each corner was part of three faces, worth up to 9 points. But again, each of the three edges it was on was counted twice, bringing the score to 6. Including the inner diagonal brought the final score to 7.

This meant that the last valid position was the **edge**Which side scored **4 tips**.

To see all these partitions arranged on a cube, watch this animation:

To get additional credit, you had to find the most and least valuable real estate again, but this time in *four*dimensional tic-tac-toe. Just as the center position scored (3^{2}−1)/2 in two dimensions and (3^{3}−1)/2 in three dimensions, he marked (3^{4}−1)/2 — or **40 dots** — in four dimensions. the *less* valuable position turned out to be worthwhile **7 points**, but it is not so simple to describe where it is on a four-dimensional cube. Instead, I 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, each player would secretly choose a number from 0 to 10. The numbers were revealed simultaneously, and the middle number won that number of points. (If two or more players picked the same number, the winner was randomly selected from among them.)

After five rounds, the winner was the one with the median number of points. (Again, if two or more players had the same score, the winner was randomly selected from among them.)

With one turn remaining, players A, B, and C had 6, 8, and 10 points respectively. Player A sighed and wrote “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 down their own numbers in secret. Assuming everyone played to win, which numbers B and C should they have chosen?

If A were to win the round and get those 3 points, then he would have 9 points in total and would be crowned the winner. It was up to players B and C to prevent that from happening – one way or another.

To have a chance of winning, C (who currently has the most total points) cannot win this final round. Therefore, C would have to choose an extreme number, i.e. 0 or 10. If C were to choose 0, then B could win the round by writing 1, in which case B would be the big winner with 9 points. Seeing this logic, suppose C writes 10. What would B do?

If B writes a number less than 3, then A will win the round and the whole 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 3 (like A), then the winner will be either A (if A wins the round) or C (if B wins the round). So B’s only logical remaining move is to write 10. In this case, B and C will flip a coin to see who wins the round. The big winner is the one who made *do not* win the round.

To summarize, **B and C should have chosen 10**giving them both a 50% chance of mediocre victory.

A number of readers have alternatively suggested that all three players choose 3, in which case the winner of the round would be chosen at random. 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 big winner. This gave each player a one in three chance of winning. As we have just seen, by both choosing 10, players B and C can increase their odds to one in two. So choosing 10 was the best strategy for B and C.

Several solvers, including Jenny Mitchell and Erin Sy, have also observed that B and C both choosing 10 represent the game’s only Nash equilibrium, where no individual player can do better by changing strategy.

And so, in this particular game of mediocrity, it paid off for being extreme.

## Want more puzzles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles in this column and some never-before-seen puzzles. It’s called “The Riddler”, and it’s in stores now!

## Want to submit a puzzle?

Email Zach Wissner-Gross at [email protected]