Tanya Khovanova's Math Blog

2022 is abundant, composite, even, evil, square-free, and untouchable.

In addition, 2022 is the smallest number n such that n, n+1, n+2, and n+3 have the maximal exponents in prime factorization equal 1, 2, 3, and 4 correspondingly. Indeed, 2022 = 2·3·337, 2023 = 7·17², 2024 = 2³·11·23, and 2025 = 3⁴·5².

Problem. The numbers 2²⁰²¹ and 5²⁰²¹ are expanded, and their digits are written out consecutively on one page. How many total digits are on the page?

Drabble cartoon, Jun 15 1987, by Kevin Fagan.

Problem. For how many prime numbers p, the expression 2^p + p² is a prime?

* * *

My daughter was talking at her kindergarten about what her parents do for work. She said that her mom catches bugs, invokes demons, and talks to clods.

* * *

I have neither Twitter nor Instagram. I just go for a walk to tell strangers what I ate and drank and how things are at work and at home. I have three followers: a doctor and two policemen.

* * *

Life is like Rubik’s cube: fix one side, better not look at the rest.

* * *

My Roomba just devoured a piece of cheese I wanted to pick up and eat. The war between humans and robots is already here.

My friend, John Conway, had a trick to help him with tricky situations. Whenever he needed to make a non-trivial decision, he would ask himself, “What would John Conway do?” As he explained to me, he had in mind the public image he himself created. He liked the image and thought this mental trick helped him be a better, more productive, and not-to-forget, flashier person.

From time to time, I catch myself in need of a decision and ask myself, “What would John Conway do?” And he gave me the answer: I should change the question and ask myself, “What would Tanya Khovanova do?”

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?