Tanya Khovanova's Math Blog

I gave the following problem as part of the entrance test for my STEP program.

Puzzle. What word would you use to describe a man who does not have all his fingers on one hand?

The test had 17 questions, and this one was the only trick question. My goal was to check whether the students were paying attention.

The standard answer is normal, or something equivalent: regular, average, two-handed, or just a man. Most people do not have all their fingers on one hand; they have some fingers on one hand and some on the other.

Some students gave correct answers with extra flair.

• A cautious answer: A regular person, as to my knowledge, has fingers on both hands.
• A logical answer: A person with multiple hands, if they don’t have all fingers on one hand, then they must have multiple hands. For example, a human would work in this case.
• A funny answer: The man who puts his eggs in two baskets.

I also got answers from people who fallen right into my trap: fingerless, handicapped, genetically-mutated, alien, asymmetrical, injured, one-handed, resourceful, five-fingered, disabilitized, and mono-hand.

Some students sympathized with the man and called him frugal, determined, and a super-hero.

One student misread the problem, but gave a technically correct answer.

• Human, because I don’t see any difference in the man whether he has fingers or not.

This is not the first time I have used this problem on a test. But this year, the variety of answers was awesome. Still, the funniest answer in the misreadings category was:

• A chef.

Here is a new puzzle by Nikolai Chernyatiev.

Puzzle. Five sages, who all know one another, are blindfolded, seated in a row in a dimly lit hall, and then have their blindfolds removed. Each sage can see both of their immediate neighbors, but no farther; the sages at the ends know that they are at the ends. After that, each sage writes down one of the numbers 1, 2, or 3. The complete information — who wrote which number, in seating order — is then announced to everyone.
Before being seated, the sages may agree on a rule for choosing 1, 2, or 3 based on what they see. After the five numbers are announced, each sage must reconstruct the full left-to-right order of all five sages.

I love puzzles related to information theory, and this is a lovely example. Let’s do a quick sanity check. There are 5! = 120 possible orders of the sages. The announced numbers form a ternary string of length 5, giving 3⁵ = 243 possible announcements. That is more than enough in principle; so far, so good.

AI can produce a possible table of answers, but the resulting strategy is not very inspiring. Fortunately, there is a much more elegant solution based on the following neat fact:

Among any three distinct residues modulo 5, exactly one is the average of the other two. Equivalently, any three vertices of a regular pentagon form an isosceles triangle: one of the three vertices lies on the axis of symmetry of the other two.

But wait: Konstantin Knop proved a much tighter result. Each sage can get away with writing down only one of two numbers. Wow!

Alexander Karabegov sends me new puzzles from time to time. This time, however, it is not a puzzle but a math joke.

Joke. If a woman gives birth to a child at the age of 30, then 60 years earlier, her child was twice as old as she was. Whatever that means.

I’ve been staring at my icosahedron, trying to solve the following puzzle by Konstantin Knop.

Puzzle. One face of an icosahedron is special. The numbers 2, 3, and 5 are written at its vertices in some order. All other vertices of the icosahedron are labeled with 1. In one query, we may ask an oracle for the product of the numbers assigned to any subset of icosahedron’s vertices. What is the minimum number of queries needed to determine the special face?

However, I misread the problem. I ended up solving a different puzzle instead—and had quite a bit of fun doing it.

Puzzle. One face of an icosahedron is special. The numbers 2, 3, and 5 are written at its vertices in some order. All other vertices of the icosahedron are labeled with 1. In one query, we may ask an oracle for the product of the numbers assigned to vertices of any one face of the icosahedron. What is the minimum number of queries needed to determine the special face?

I won’t post the solutions just yet, but let me begin with a simple observation: one question cannot possibly be enough. Indeed, with one question, the oracle’s answer must be a divisor of 30, and 30 has only 8 positive divisors. But an icosahedron has 20 faces, so a single question cannot distinguish among all possible choices for the special face.

Puzzle. A ball rolls forever on a frictionless billiard table with no pockets. Can you find a finite convex shape of the table for which no trajectory of the ball ever covers the entire surface?

I gave the following puzzle to my students.

Puzzle. A cowboy rides into town on Friday, stays for three days, then leaves on Friday. How come?

Most of them submitted the standard answer: his horse is named Friday.

One student suggested that the town was named Onfriday. This solution works well when the puzzle is given orally, but my homework was typed, so it feels less elegant.

Here are two more solutions that also work:

• He drank so much that he believed he stayed for three days, while in reality, the other four days were lost in a blackout.
• The cowboy left after three days, then later returned and left again on Friday.

And here is my favorite solution:

• The town has a quirky tradition: they celebrate fried foods and call the day Friday (get it?). The cowboy came for this special event, which happened to be held on a Tuesday.

I am so happy that I met Ivo David and Yogev Shpilman. Together, wrote the book Mathematical Puzzles and Curiosities, and had so much fun on the way. I have already mentioned one of the puzzles from the book on my blog—Two Points on a Cube. Here is another one.

Puzzle. Consider the sequence: O, T, R, F, I, S, ?
What letter should replace the question mark?

Puzzle. Two fathers gave money to their sons. The first father gave $200, and the second father gave $100. Yet the total amount received by the sons was only $200. How come?

Standard answer: There were three people: a son, his father, and his grandfather. The grandfather gave the father $200, and the father gave the son $100.

In many puzzles, my students come up with a surprising variety of alternative solutions—but not for this one. For many years of my teaching, this puzzle stayed untouched by new ideas. Perhaps the puzzle is simply too well known. But recently, I finally heard an alternative answer:

• The money was taxable.

In the homework for my STEP program, I gave the following challenge problem.

Puzzle. My sages each wear a hat of a different color. As in standard hat puzzles, they can see everyone else’s hat color. Unlike in many other hat puzzles, they know the color of their own hat as well. I announce which color each of them should end up wearing; this assignment is a permutation of the original colors. Each sage is allowed one swap of hats with another person per day. They have two days to rearrange the hats so that everyone ends up with the correct color. Can they do it?

Many students noticed that the permutation can be decomposed into disjoint cycles and suggested solving the problem cycle by cycle. A few of them even pushed this idea all the way to a complete solution. However, none of them connected the puzzle to a topic we had discussed in class: dihedral groups.

Here is an elegant way to finish the solution once the permutation is decomposed into cycles. A cyclic permutation on n elements can be viewed as a rotation of an n-gon. Any rotation of an n-gon can be written as a product of two reflections. Each reflection of an n-gon, viewed as a permutation, consists only of 1- and 2-cycles. Ta-da!

In a recent post, Each Point has Three Closest Neighbors, I mentioned the following conjecture.

Karabegov’s Conjecture. Any finite planar point configuration in which every point has exactly 3 closest neighbors must contain at least 16 points.

The conjecture was proposed by my dear friend Alexander Karabegov, whom I met in 1974. Wait. What?! I just realized that this was more than 50 years ago. How is that even possible?

After I posted the conjecture, we couldn’t resist working on it. We wandered through different types of graphs and found many cute definitions related to our problem.

A unit distance graph is formed from points in the plane by connecting two points whenever they are exactly distance 1 apart. A matchstick graph is a unit distance graph that can be drawn in the plane with edges of length 1 that do not cross. In other words, it is a unit distance graph that behaves nicely, by being planar. Think of laying matchsticks flat on a table: no overlaps, no chaos.

Here’s the difference visually: the left graph is a unit distance graph, while the right one is a matchstick graph.

Now for the star of the story. A penny graph connects two vertices if and only if their distance is the minimum distance among all pairs of vertices. The name is delightfully literal: imagine placing identical pennies at each vertex so that they do not overlap. Two pennies touch exactly when the corresponding vertices are connected by an edge. A penny graph is a special kind of matchstick graph: two vertices that are not connected are at a distance that is longer than the length of the matchstick.

Finally, a 3-regular graph is a graph where every vertex has degree 3. Three neighbors. No more, no less. They are also called cubic graphs. Not surprisingly, if the vertices of a cube are the vertices of our graph, and the edges of a cube are the edges of our graph, we get a 3-regular graph, as each vertex is incident to exactly 3 edges. Surprisingly, such graphs are not called tetrahedron graphs, as a tetrahedron, too, has each vertex incident to 3 edges. But the tetrahedron graph is special: it is a minimal 3-regular graph.

We wrote a paper Minimal 3-regular Penny Graph, in which we proved the conjecture. The conjecture has officially graduated to a theorem.

Theorem. The minimal 3-regular penny graph has 16 vertices.