Tanya Khovanova's Math Blog

Konstantin Knop posted the following puzzle on Facebook.

Puzzle. An agent sends messages to the command center. The messages have to be encrypted as a stream of 512 characters that can only be zeros and ones. Unfortunately, his transmitter is malfunctioning and gobbles 16 characters of each message. The missing 16 characters are always in the same positions in any message. As a result, the command center receives a sequence of 496 bits. Neither the center nor the agent knows which 16 bits of the sequence are eaten up by the device.
They cannot replace the broken transmitter. However, they can agree ahead of time to send K test messages, the content of which they both know. Find the smallest possible K needed to determine the positions of the disappearing 16 bits.

A quine is a computer program which takes no input and produces a copy of its own source code as its only output.

Puzzle. Assuming English is a computer language, write a quine in English.

My students had many solutions on how to solve this puzzle. They were all variations on “Write this sentence.” This is a self-referential sentence which doesn’t quite work. I even tried it on ChatGPT with the following result, “Of course, I’d be happy to help! Please provide the sentence you’d like me to write, and I’ll assist you with it.”

However, the solution I originally had in mind worked. ChatGPT repeated my input. So, ChatGPT provides a simple way for you to check your answer to this puzzle.

My students had more ideas. One of them suggested screaming at a friend, forcing the friend person to scream back. In a similar vein, one might say hello to a person in order to hear hello back. I tried this with ChatGPT, but it didn’t work. The bot replied, “Hello! How can I assist you today?”

Another trivial idea is to write nothing. This certainly works perfectly with ChatGPT.

What’s a polyomino? A polyomino is a plane geometric figure formed by joining one or more equal squares edge to edge. Here, we have a puzzle about a polyomino that is almost a rectangle.

Puzzle. You are given a 5-by-7 rectangle with two corners cut out: A 1-by-1 tile is cut from the bottom left corner, and a 1-by-2 tile is cut out of the top right corner, as in the picture. The task is to cut the resulting shape into two congruent polyominoes.

Here’s another brainteaser from my friend, Alexander Karabegov.

Puzzle. Place the numbers from 1 to 8 at the vertices of a cube so that each face is balanced. On a balanced face, the sum of the numbers at the ends of one diagonal equals the sum of the numbers at the ends of the other diagonal.

My friend, Alexander Karabegov, sent me one of his puzzles. I love the mixture of algebra and calculus.

Puzzle. Describe a real quadratic function f such that the graph of its derivative f′ is tangent to the graph of f.

My former student, Xiaoyu He, invented this elegant puzzle and shared it with me.

Puzzle. We’ve got a murder mystery on our hands. There are four suspects, and it’s pretty clear that one of them is the actual murderer. But here’s the twist: there are also four witnesses who know who the killer is. Now, three of these witnesses are the honest type, always telling the truth, but the fourth one always lies.
You get to ask each of these witnesses a single yes-or-no question, and your question must be, “Is the murderer among this group of suspects?” You can choose any group of suspects you want. The challenge is to figure out who the murderer is.
Can you take it up a notch and determine the murderer if you have to list all your questions before getting any of the answers?

The Museum of Mathematics organizes a biannual conference in recreational mathematics called MOVES: Mathematics Of Various Entertaining Subjects. Being a math-recreatinalist myself, I attend all of them. The last one was held in August 2023 and was devoted to the mathematics of Fiber Arts. A few years ago, I wouldn’t have believed I had something to do with the arts. However, in recent years, I started crocheting mathematical objects for my classes, so I not only decided to attend, but also submitted a talk proposal.

The talk proposal was accepted. And, when I finished my slides, I realized that I created too many objects for a 25-minute talk. So, I divided my talk into seven sections and asked the audience to choose. I was lucky that they chose what I was most excited about, but I only had time to cover 4 out of 7 section.

Today I am posting a couple of pictures from my slides that showcase Brunnian links.

A Brunnian link is a set of loops that can’t be separated, but if any one loop is removed, they all fall apart. The most famous example of a Brunnian link is Borromean rings which I already wrote about. In the first picture, the Borromean rings are on the right. They are famous because they are the simplest Brunnian link.

There are two natural ways to generalize Borromean rings. One way is to use the same three loops, but make them twist around each other more. The second way is to increase the number of loops.

The left side of the first picture shows three tangled loops, which are more intertwined than Borromean rings. The crossing number of the link on the left is 12, while the crossing number of the Borromean rings is 6.

The second picture shows two Brunnian links with more than three loops. The left link has four loops, while the right link has five.

* * *

—What’s the best way to get a math tutor?
—An add!

* * *

—Why was the equal sign so humble?
—Because she knew she wasn’t greater than or less than anyone else.

* * *

—Where do mathematicians go on vacation?
—Times Square.

* * *

—Why do cheapskates make good math teachers?
—Because they make every penny count.

* * *

—Why was math class so long?
—The teacher kept going off on a tangent.

* * *

—What did the student say about the calculus equation she couldn’t solve?
—This is derive-ing me crazy!

* * *

—Did you hear about the statistician who drowned crossing a river?
—It was three feet deep, on average.

* * *

—What do organic mathematicians throw into their fireplaces?
—Natural logs.

* * *

—Why is the obtuse triangle always upset?
—It is never right.

* * *

—What is the integral of one divided by a cabin? A log cabin?
—No, houseboat — you forgot the C.

* * *

—What do you get when a bunch of sheep hang out in a circle?
—Shepherd’s pi.

* * *

—What do you call a metric cookie?
—A gram cracker.

* * *

—What state has the most math teachers?
—Math-achusetts.

* * *

—What does a hungry math teacher like to eat?
—A square meal.

* * *

—What is the mathematician’s favorite season?
—Sum-mer.

* * *

—What adds, subtracts, multiplies, divides, and bumps into light bulbs?
—A moth-ematician.

* * *

—What tools do you use for math?
—Multi-pliers.

* * *

—Why didn’t the quarter roll down the hill with the nickel?
—Because it had more cents!

* * *

—Which snakes are good at math?
—Adders.

* * *

—What is the butterfly’s favorite subject in school?
—Moth-ematics.

The Ural Math Battle in 2016 had several mafia-themed problems of various difficulty with the same initial setup.

Puzzle Setup. Among 100 residents of Saint-San, m are mafiosi, and the rest are civilians. A commissioner arrived to the town after getting this information. In an attempt to expose the mafia, this commissioner asked each of the residents to name s mafia suspects from among the other 99 residents. The commissioner knows that none of the mafiosi would name other mafiosi, but each civilian would name at least k mafia members. What is the maximum number of mafia members the commissioner can definitively identify after his survey?

1. The most difficult case was m = s = 3 and k = 2.
2. In the next case, where m = 3 and s = k = 2, the puzzle had a different task: prove that the commissioner can find at least one mafioso.
3. In the third case, where m = s = 10 and k = 6, the question was whether the commissioner can find at least three mafiosi.
4. In the fourth case, where m = s = 10 and k = 7, the question was whether the commissioner can find all the mafiosi.
5. The last case was for younger students with m = 6, s = 10, and k = 6. The question was whether the commissioner can find all the mafiosi.

When I asked ChatGPT to translate the first and the most difficult case of this puzzle from Russian, ChatGPT decided to solve it too. At the end of its ridiculous solution, it concluded that the commissioner could identify all 21 mafiosi out of the given 3. So, if you comment on this blog that the answer to the first case is 21, I will know that you are a bot.