Tanya Khovanova's Math Blog

Here are some snowball sentences suggested by my students.

• I do not know about radon’s, osmium’s, polonium’s abilities.
• I am the only short person playing football.
• “I am not even smart,” mother remarks.
• I do not know where people acquire insanity.
• “A no,” Joe said while eating burgers mightily adultlike.
• I am not very happy during Mondays.
• I do not joke.
• I be—arr, mate—avast!
• I do not know super skates.
• I do not fear yucky cheese; however, kamikaze elephants jackhammer lumberjacks blackjacking backpedalling brontosauruses, artificializing territorializing icositetrahedrons.

Can you invent some other snowball sentences? But first, you need to figure out what they are.

Each year I look at the MIT Mystery hunt puzzles and pick ones related to mathematics, logic, and computer science. I usually give additional comments about the puzzles, but this year’s titles are quite descriptive. Let’s start with mathematics.

• Yes or No?
• Exactly.
• Numbers.
• Triangles.
• The Greatest Jigsaw.
• Clueless.
• Ignorance.
• The IMO Shortlist.

Now Nikoli-type logic puzzles. I really enjoyed “Fun with Sudoku” during the hunt.

• Better Bridges.
• ★.
• Bounce House.
• Slitheɹlᴉuʞs.
• Fun With Sudoku.

And computer science.

• Illiterate Programming.
• A Bit of Light.
• Complexity Evaluation.
• Analog Circuitry.
• Escape From Life.
• Extensive.
• In C.
• Befuddled.

I recently wrote a post, A Splashy Math Problem, with an interesting problem from the 2021 Moscow Math Olympiad.

Problem (by Dmitry Krekov). Does there exist a number A so that for any natural number n, there exists a square of a natural number that differs from the ceiling of Aⁿ by 2?

The problem is very difficult, but the solution is not long. It starts with a trick. Suppose A = t², then Aⁿ + 1/Aⁿ = t²ⁿ + 1/t²ⁿ = (tⁿ + 1/tⁿ)² − 2. If t < 1, then the ceiling of Aⁿ differs by 2 from a square as long as tⁿ + 1/tⁿ is an integer. A trivial induction shows that it is enough for t + 1/t to be an integer. What is left to do is to pick a suitable quadratic equation with the first and the last term equal to 1, say x² – 6x + 1, and declare t to be its largest root.

Today I present three problems from the 41-st Tournament of the Towns that I liked: an easy one, one that reminds me of the Collatz conjecture, and a hard one.

Problem 1 (by Aleksey Voropayev). A magician places all the cards from the standard 52-card deck face up in a row. He promises that the card left at the end will be the ace of clubs. At any moment, an audience member tells a number n that doesn’t exceed the number of cards left in the row. The magician counts the nth card from the left or right and removes it. Where does the magician need to put the ace of clubs to guarantee the success of his trick?

Problem 2 (by Vladislav Novikov). Number x on the blackboard can be replaced by either 3x + 1 or ⌊x/2⌋. Prove that you can use these operations to get to any natural number when starting with 1.

Problem 3 (by A. Gribalko). There are 2n consecutive integers written on a blackboard. In one move, you can split all the numbers into pairs and replace every pair a, b with two numbers: a + b and a − b. (The numbers can be subtracted in any order, and all pairs have to be replaced simultaneously.) Prove that no 2n consecutive integers will ever appear on the board after the first move.

Here is a cute old problem that Facebook recently reminded me of.

Puzzle. By mistake, a clock-maker made the hour hand and the minute hand on a clock exactly the same. How many times a day, you can’t tell the current time by looking at the clock? (It is implied that the hands move continuously, and you can pinpoint their exact location. Also, you are not allowed to watch how the hands move.)

Here is the solution by my son who was working on it together with my grandson.

The right way to think about it is to imagine a “shadow minute hand”, like this: Start at noon. As the true hour hand advances, the minute hand advances 12 times faster. If the true minute hand were the hour hand, there would have to be a minute hand somewhere; call that position the shadow minute hand. The shadow minute hand advances 12 times faster than the true minute hand. The situations that are potentially ambiguous are the ones where the shadow minute hand coincides with the hour hand. Since the former makes 144 circuits while the latter makes 1, they coincide 143 times. However, of those, 11 are positions where the true minute hand is also in the same place, so you can still tell the time after all. So there are 132 times where the time is ambiguous during the 12-hour period, which leads to the answer: 268.

I love the problem and gave it to my students; but, accidentally, I used CAN instead of CAN’T:

Puzzle. By mistake, a clock-maker made the hour hand and the minute hand on a clock exactly the same. How many times a day can you tell the current time by looking at the clock?

Obviously, the answer is infinitely many times. However, almost all of the students submitted the same wrong finite answer. Can you guess what it was? And can you explain to me why?

Puzzle. What is the probability of getting five Mondays in a 31-days month?

This is easy if we assume that the day of the week for the first day of the month is chosen at random. But we should know better. What is the actual probability? Bonus question: for which day of the week the probability of having this day five times in a 31-days month is the highest?

A cute puzzle found on Facebook:

Puzzle. Four wizards A, B, C, and D, were given three cards each. They were told that the cards had numbers from 1 to 12 written without repeats. The wizards only knew their own three numbers and had the following exchange.

• A: “I have number 8 on one of my cards.”
• B: “All my numbers are prime.”
• C: “All my numbers are composite. Moreover, they all have a common prime factor.”
• D: “Then I know the cards of each of you.”

Given that every wizard told the truth, what cards does A have?

* * *

I surveyed many people who had played Russian roulette. Seems like the probability of dying is actually 0%.

* * *

What has the probability of one in five million?
Zero: there’s no 1 in 5000000. Only a five and six zeros.

* * *

Two classmates:
—What did you think of our probability exam yesterday?
—All means to an end.

* * *

My classmate didn’t study for our test in probability.
“I’ll take my chances”, he said.

* * *

I saw my math teacher with a piece of graph paper yesterday. I think he must be plotting something.

* * *

Not all math puns are terrible. Just sum.

* * * (submitted by Sergei Bernstein)

A programmer walks into a bar, holds up two fingers, and says, “I’ll have three beers please.”

* * *

What is the similarity between me and an experiment involving a biased coin with two tails?
The probability of getting a head is zero.

I’ve been crocheting hyperbolic surfaces of constant curvature. The process is time-consuming, so while I am crocheting, I wonder about the mathematics of crocheting.

Hilbert’s theorem says that I can’t embed a hyperbolic plane in 3-dimensional space. The proof is rather involved. But here, I have an explanation from the point of view of a crochet hook. My hook starts with a tiny cycle of four stitches. Then for every x stitches the hook makes y stitches in the next row, where y is greater than x. The extra stitches should be evenly distributed to guarantee that locally every small area is approximately isomorphic to other areas, meaning that the surface has a constant curvature.

The ratio of stitches in the next row to the current row is r = y/x. Thus, the number of stitches in each row increases exponentially. But each row is a fixed height h. That means after k rows, my thingy has to fit inside a ball of radius kh. But the length of the last row is 4r^k-1. It becomes huge very fast. As the last row is a physical curve made out of stitches, there is a limit of how much of it I can fit into a given volume, creating a contradiction.

That means, if I start crocheting, something should happen that won’t allow me to continue. I decided to experiment and see what actually would happen. Being lazy, I preferred the disaster to happen sooner rather than later. So I chose the ratio of three: for each stitch on my perimeter, I added three new stitches. Shortly after I started to work, the process became more and more difficult. The ball was too tight. It was challenging to hold that thing in the place where I needed to insert the hook. And the loops were getting tighter, making it more exhausting to insert the hook into the proper hole. So each new stitch was taking more and more time to complete.

To my disappointment, the thing didn’t explode, as I was secretly hoping: I just couldn’t work on it anymore.

I like rewarding my students. Before covid, I used to give them star stickers for good ideas. When I started to teach remotely, I wondered what I should do instead. I could tell them that they had won a star, but it felt too weak. The next idea was to show them a star and tell them that it belonged to them. But that still felt insufficient. Then I had an epiphany. I would say to them they earned a star, show it to them, and stick it to my face. So they, and all the other students, would see it for the rest of the class. The photo shows how I looked at the end of a successful lesson.

Another picture shows what my MathRoots students posted on our Discord channel.

Now that I am back teaching in person, my students asked me to continue sticking their stars to my face. Sometimes I forget about the stars and, after my class, wander around MIT star-covered.