Tanya Khovanova's Math Blog

Every year, I used to blog about math-related puzzles from that year’s MIT mystery hunt. However, I want to skip this year. I participated in the hunt, and a few of the puzzles I saw were not good enough. Moreover, the puzzles had boring steps or were inelegant. I am not motivated to sift through every math puzzle in the hunt and check its quality.

This year, instead of cataloging all the math puzzles, I will present a short list of puzzles from the hunt that were recommended to me by others. Most of these puzzles are unrelated to math, but I checked them out, and they seem cool.

After the hunt, I asked two people about their favorite puzzles. Both of them mentioned Showcase. Looking at it, I know why. I’ll give you a hint: this puzzle will appeal to people doing competitive programming.

Here are some other recommendations.

• Scicabulary, this is a word puzzle.
• Natural Transformation, you are presented with some very structured images.
• Collage, this is a funny farm.
• Cute Cats, you are presented with many images, some of them with cats.
• Hall of Archaeology, this is a Nikoli-type puzzle.
• Museum Rules, this puzzle looks like a children’s picture book.
• This Puzzle is Just Another Regular Cryptic, this is a cool cryptic puzzle.
• Parsley Garden, you are presented with a grid of words.
• Dispel the Bees, you are presented with crossword clues.

Puzzle. If you can move exactly two matches, what is the largest possible number you can get?

Many people assume that given that 508 has three digits, the answer also has three digits. They get the number 999 and stop there.

Some students realized they can make an extra 1 out of two matches and gave answers with four digits: 1503, 5051, and 5103.

Another great idea is to take out the two horizontal matches from 0, turning the 0 into 11. For example, one might use the two matches to turn 5 into 8 and get 8118. However, we can combine this idea with the previous one and use the two matches to make a new digit 1 to get 51181.

When I first saw this puzzle several years ago, 51181 was the official answer. But my students went further, much, much further.

Another elegant idea is to rotate the picture and get 81151.

Some students decided that having all the digits be the same height is not very important. One vertical match reads as 1, even if it is half the height. The largest number they got this way was 511811. Combining this idea with flipping the number allows us to get 811511.

However, small-font digits are often used in powers. This train of thought led to a brilliant answer, 5118¹¹. This number is way larger than everything we saw before: it has 41 digits. If we combine this idea with rotating the number, we can get 8115¹¹, a 43-digit number.

I recently decided to check the Internet again and discovered a chain of videos showing solutions to this puzzle by the same author. He started with a simple solution, 51181, then people left comments in the video with better solutions, so he remade the video. This happened several times. I probably should send him a link to this essay for yet another video.

Here is an interesting solution by one of my students that I haven’t seen anywhere else. The idea is to use scientific notation. The student took two matches from the right side of 0. He used one of them to convert the first digit from 5 to 9 and the second digit from 0 into the letter E. He got 9E8 = 900000000. If we combine this idea with using a small 1, we can get 5E81, an 82-digit number.

Another brilliant idea is using two matches to create a caret symbol representing an exponent. This way, we can get 5^118, which is an 83-digit number. If we combine this idea with the rotation idea, we can get 8^115, a 104-digit number.

If we ignore the relative sizes of the digits, we can have the small digits as the base of the exponent and the larger digits as the power. One of the students suggested 11⁵¹¹⁸, a 5330-digit number. And if we rotate the number, the answer we can get is 11⁸¹¹⁵: the number with 8451 digits. This is a humongous number compared to the mere 3-digit one we started with, to the pathetic 5-digit original solution, and to the pitiful 104-digit previous record. But my students didn’t stop there.

One student suggested snapping two matches and creating a double factorial 505!!, with 575 digits. Combining this with other ideas, we can get 8115!!, a 14102-digit number.

One of my students decided to use two matches from the last digit 8 to create a factorial sign. She glued one of the matches perpendicular to the plane to get 505! in the projection. She even made a picture that you can see below. This number has 1148 digits, and combining this with previous ideas, we can get 5118!, which has 16763 digits. Or even 8115!, which has 28202 digits.

For this puzzle, a factorial works better than a double factorial. Here is my own suggestion: use one of the matches to create a small digit one, and split the other match into a factorial. This way, we can get 81151!, a number with a whooping 363154 digits.

This is an excellent example of a thinking-outside-the-box puzzle. I love this puzzle because it has not one, but many boxes one can think outside off.

Sometimes I mess with my students. Recently, I gave them the following problem for homework.

Puzzle. Don’t read this sentence.

This problem is a paradox: students can’t know not to read it unless they read it! I expected my students to explain the paradox, and a couple of them did. But, most of them provided me with an infinite stream of entertainment. Here are some of their answers divided into categories, starting with the apologizing ones, and there were a lot of those:

• Oopsie.
• Too late?
• My apologies.

I wasn’t surprised by the apologies, as this problem is intended to be intimidating. However, other students tried to wiggle out of it:

• Next time, I will ask my parents to read my homework to me.
• A person named Don’t read, in the past tense, “this sentence”. I read this sentence, too!
• I happen to have conveniently forgotten English, and it cannot be proven otherwise.

Yet other students decided to rebel:

• Why not?
• I didn’t.
• You just lost your game.
• I will never read anything again, as it says, “Don’t read.”

Karma is a boomerang, and this problem got me into a pickle. One of the most brilliant and funny solutions possible was to leave the answer box blank, implying that the sentence wasn’t read. However, how would I grade this? What if they just skipped the problem? I decided to err on the students’ side and gave full credit for an empty answer box.

A couple of students made a point of pretending they didn’t read the sentence:

• Which sentence?
• I see an answer box, but there’s no problem.

But I saved the best for last. My favorite answer was the following:

• Don’t read this answer.

Puzzle. Alice and Bob play the following game on a regular chessboard, where all the pieces move according to the standard rules of chess. Alice has a king, and Bob has a knight. They would place their pieces on the chessboard without attacking each other. Alice starts, and they alternate moves. Alice loses if the knight checks or all available moves place her under the knight’s attack. For which initial positions of the pieces is Bob guaranteed a win?

Once, I was in Princeton and attended a lecture by Persi Diaconis. John Conway was also there. During his presentation, Persi mentioned John by referring to him as Wonderful John Conway. John couldn’t resist and said, “Thank you, you are one of not many people who remember my real first name.”

Every summer when I was little, my mom would take us on vacation to a rusty village far away from Moscow. I do not remember much, mostly just cows grazing on the grassy fields. However, one particular memory is really special and vivid.

I was five years old, tired of another day in the fields, lying in bed about to fall asleep. I started counting. I do not remember what. I am sure it wasn’t sheep; it could have been cows. Then, I got bored of small numbers and jumped to a thousand, counting from there. Then, I jumped to another even bigger number. After a few jumps, I realized that I could always add one to a previous number. The number of numbers must be infinite. Wow!

I will always remember the feeling I had. It was like touching eternity, being one with the whole universe.

You can imagine why I became a mathematician. From time to time, I am touching eternity and getting paid for the bliss.

I use Escher’s tessellations to teach wallpaper groups. Escher is the best painter among mathematicians and the best mathematician among painters. His fame helps energize my class. Plus, he has so many beautiful drawings to choose from.

However, there is another layer to his tessellations. Many paintings are not just a study in wallpaper groups but also in group-subgroup pairs. For example, consider these red/gray/black fishes on the left. There are three distinct points where three different reflection lines intersect. The first point is where three black fishes kiss each other. The other two points correspond to gray and red fish kisses. In orbifold notation, this symmetry group is *333.

But, the same drawing has another symmetry group. We just need to ignore color. That is, we consider all the fishes to be the same. In this case, our three distinct points where three reflection lines intersect become the same point: the point where fishes kiss, regardless of their color. The new symmetry group has an additional element: a 120-degree rotation where three fins of three different-colored fishes touch each other. Thus, the new symmetry group is 3*3.

Escher created a lot of examples of groups and their subgroups using color. But, sometimes, he was more subtle. In one of my previous posts, The Dark Secret of Escher’s Shells, I discussed my favorite Escher’s plane tessellation. In that drawing, the second group appears when we ignore the markings on one of the dark shells.

Here is another spectacular example of a group and subgroup, a tessellation of a hyperbolic plane with angels and devils. Do you see two different symmetry groups in the painting?

My PRIMES project last year, done with Rich Wang, was about violinists living in a hotel and being annoyed with each other. The project was suggested by Darij Grinberg and based on the following puzzle.

Puzzle. Consider a hotel with an infinite number of rooms arranged sequentially on the ground floor. The rooms are labeled from left to right with consecutive integers. A finite number of violinists are staying in the hotel with no more than one violinist in a room. Each night, two violinists staying in adjacent rooms get fed up with each other’s playing, and both request different rooms from the manager. The manager moves one of them to the nearest unoccupied room on the left and the other to the nearest unoccupied room on the right. This keeps happening every night as long as there are two violinists in adjacent rooms. Prove that this relocation will stop after a finite number of nights.

The project was not to solve the puzzle, and I won’t describe the solution, leaving it to my readers to ponder on their own. The project was to take this puzzle and see what we could discover. The relocation process in the puzzle is reminiscent of chip-firing, with a few differences.

Let me explain chip-firing in terms of violinists. The starting position would allow any number of violinists per room. Each night, two violinists in the same room will get annoyed with each other and request new rooms. There might be more violinists in that same room, but only two at a time are really annoyed. The manager will put one of them into the room on the left and the other into the room on the right, regardless of how crowded the rooms are.

In the chip-firing example, violinists always move to the next room. In our puzzle, they might need to go miles to a new room. And carry their luggage with them!

The original puzzle above implies that after several nights the relocations will stop, and violinists will enjoy the hotel in peace. This peaceful configuration is called a final state. Interestingly, depending on the order in which violinists are getting annoyed with each other, a starting configuration can result in different final states.

For example, consider the smallest interesting case, where only 3 violinists are staying in a hotel. We use 1 to describe a noisy room with a violinist and 0 to describe an empty room. Suppose the starting configuration is 0011100, where we ignore rooms far away from the noise. Depending on which two rooms have the complaining violinists on the first night, we can end up with a final state 0101001 or 1001010.

Initially, in the project, we looked at final states where we start with N violinists in consecutive rooms. We call such a configuration an N-clusteron. The example above describe the final states of a 3-clusteron. Here is an exercise for the reader: prove that in the final state, the gaps between occupied rooms have to be 1 or 2.

A final state can be described by the index of the leftmost occupied room and the sequence of gaps between occupied rooms. In the exercise above, it is easy to prove that the gaps must be size 1 or 2. In our paper, we proved a more refined statement: any final state has exactly one gap of size 2. Thus, a final state has two parameters: the index of the leftmost room and the location of the size 2 gap. For example, this means that in a final state, the span between the leftmost and the rightmost rooms, inclusive, is 2N.

As we continued exploring final states, we discovered something else. But first, I need to define the shadow of a final state. Let’s denote an occupied room with a one and an empty room with a zero. Then a final state corresponds to an infinite sequence of zeros and ones. But the interesting part of the final state is not infinite: it is a subsequence between the leftmost and rightmost ones, inclusive. We call this subsequence the shadow of the final state. In other words, the shadow describes the final state up to a translation and is completely defined by one parameter: the location of the 2-gap. In our example of a 3-clusteron, there are 2 final states with 2 distinct shadows. For larger clusterons, the situation is more interesting. There are 5 different possible final states for a 4-clusteron but only three distinct shadows. We discovered the following conjecture.

Conjecture. If we start from an N-clusteron and at each state uniformly select a move to perform from all possible moves, then the shadows of the final states are equiprobable.

You can find more details in our paper Ending States of a Special Variant of the Chip-Firing Algorithm. This conjecture is beautiful as it shows that there are hidden symmetries in this process of appeasing the fussy violinists. We didn’t prove this conjecture. Can you do it?

“Follow your heart!” This is the most common advice for young people contemplating their future career path. This is not good advice. At some point in my life, the most popular aspiration among my friends’ children was to become opera singers. But the world only has room for a few opera singers. All of these children ended up doing something completely unrelated.

Aspiring to be a mathematician is a much more practical dream. There are so many professions that are friendly to mathematicians: actuary, finance, economics, teaching, computer science, cryptography, and programming, to name a few. Unlike opera singers, skilled mathematicians can find a way to get paid for their mathematical gifts.

However, most of the youngsters around me want to be research mathematicians. This is a different story. My adviser, Israel Gelfand, told everyone that if they could survive without mathematics, they should drop it. I did drop mathematics for some time to care for my children, but I couldn’t live without mathematics. I fed math to my children for breakfast and pursued math hobbies that could fit a single mother’s lifestyle. Well, that means I was mostly working with sequences for the Online Encyclopedia of Integer Sequences and building the database for my Number Gossip website. But I digress.

I agree with Gelfand. Research in math as a career choice is non-trivial. Here are some of the big issues:

• Money. An entry-level salary for MIT undergraduate alums is about twice as much as graduate students’ stipends. This discrepancy in income continues for a long time.
• Location. It is challenging to find a professorship. People looking for academic jobs are expected to send hundreds of job applications and accept a position anywhere in the country. This might be unacceptable for people who value their family and roots.
• Time. Many research mathematicians I know work 24/7. Not always by choice. Some want to secure a future tenure and need to publish papers in addition to teaching innovative courses. Some do want to do research but are too distracted with their routine and administrative tasks. As a result, their research spills out into nights and weekends.
• Gender and such. Discrimination is a separate big problem for women and minorities, which I do not want to discuss today.

So what would I suggest for young people who love math?

Many people who love math do not really love math per se. They love the way of thinking that math encourages. They love logic, generating ideas, precision, innovation, and so on. This makes their potential job search much wider. Such people might enjoy programming, cryptography, data science, actuarial science, finance, economics, computer science, engineering, etc. I know students planned to become mathematicians but tried an internship in finance and found their real passion.

For those who want to be closer to mathematics, there is always teaching: the world needs way more math teachers than research mathematicians. Plus, teaching provides a strong feeling of making an impact.

So what do I suggest for young people who love opera singing?

Many skills are less in demand than mathematics. It is important to be realistic. So here is my advice:

• Expand your skill set. If you love opera singing, working as a voice coach might be a solution.
• Explore secondary interests. If you also enjoy programming, you might find your happiness by building music software.
• Have a backup plan. You might become a lawyer to support yourself and your family and keep your love of music as a hobby, for example, by singing in a choir.
• Meet yourself halfway. You might get a half-time job to feed yourself and work on your music business for the other half of the time.

I know a former Soviet mathematician who worked as a night guard and used his quiet work time to invent theorems. He was a good mathematician but couldn’t find a research position because he was Jewish. He later immigrated to the US and found a professorship. So sometimes my advice for opera singers works for mathematicians too.

Puzzle. “I guarantee,” said the pet-shop clerk, “that this parrot will repeat every word it hears.” A customer bought the parrot but found it wouldn’t speak a single word. Nevertheless, the clerk told the truth. Explain.

The official answer:

• The parrot is deaf.

Indeed, in this case, it is not a lie that the parrot will repeat every word it hears. My students had some other ideas. The following answer differs from the official one by one letter, but the spirit of the solution is the same.

• The parrot is dead.

Another idea my students had was to introduce a time component.

• The parrot wasn’t guaranteed to say the lines immediately; it would wait 30 years before repeating any words.
• The parrot only repeated the words in the customer’s absence.

And a couple of outside-of-the-box answers.

• The parrot was a toy and did not have a battery in it.
• The customer mistook the pet-shop owner to say the statement about the parrot when in fact, the parrot was saying it.