Tanya Khovanova's Math Blog

My goal is to expand my students’ minds. So, though my STEP program is about mathematics, I sometimes give problems from other areas for homework. Here is a recent physics one.

Puzzle. You have a brick of 1 kilogram. How does the weight of the brick change during the year?

As always, my students generated tons of ideas about what can influence the weight.

• The Sun’s gravitational pull can influence the weight.
• The Moon’s gravitational pull can influence the weight.
• It would possibly decrease slightly due to factors such as weathering.
• During rainy days, it could absorb moisture and become heavier.
• Due to thermal expansion, the brick will be larger in the summer than in the winter, which means it displaces more air. As a result, it will weigh less in the summer.

I am not into physics. So, when I got these replies, I contacted a real physicist friend, Levy Ulanovsky. He referred me to Wikipedia: The first operational definition of weight was given by Euclid, who defined weight as: “the heaviness or lightness of one thing, compared to another, as measured by a balance.” This implies that when we talk of “weight”, we need to specify how we measure it. He continued by saying that the above ideas all make sense if we measure the weight force, e.g., by using a spring or pendulum frequency. Yet if our measurement is relative, e.g., by using a lever-like scale, then, for example, the sun’s gravitational pull is not a valid answer.

For example, if we measure the weight using a lever-like scale, with our brick on one side and a known weight combination on the other, then the weight reading on the moon will be the same as on Earth. If you use a spring, the weight will be different.

He added: People often use the words “weight” and “mass” interchangeably. But for teaching, you may wish to clarify that weight force is mg. A change in pendulum frequency shows a change in g, the acceleration due to gravity. A balance (lever-like) can show a change in m, the mass on one of its two plates relative to the mass on the other plate, with g equal at both ends. A spring, a rubber band, or a springboard can show a change in the weight force mg, whether caused by a change in m, in g, or in both m and g. Of the four student answers, the moon and sun change g; weathering and moisture change m; thermal expansion has several effects that interplay in a complicated way, so we’re better off forgetting about it.

I also asked Levy which effect is the strongest. His reply was: assuming a spring, a pendulum, or the like, the strongest effect is due to the Moon.

Here is a probability puzzle I heard from my son Sergei. We even included this puzzle in our book Mathematical Puzzles and Curiosities. Our book includes the answer but omits the details. So, this blog post is devoted to said details.

Puzzle. Alice rolls a die until she gets 6. Then Bob observes that she never rolled a 5.
Question. What is the expected number of times that Alice rolled the die?

The answer depends on Bob’s strategy. Many people assume that Bob loves 5 and is only looking for 5. In this case, the answer is 3. Here is the argument: the expected number of rolls to get 5 or 6 is 3: this is equivalent to rolling a three-sided die and waiting to one side to appear. Only on the rolls without 5 will Bob say something.

However, there are other natural assumptions. In the book, we have two suggestions, where Bob treats every digit that is not 6 equally.

Modeling assumption 1. Suppose Bob lists all the numbers that are missing. Then, when he says that 5 is missing, we are guaranteed that Alice rolled 1, 2, 3, and 4 before 6. Such a strategy by Bob noticeably increases the expected number of rolls, and the answer is 8.7. Let us prove this.

This version of the problem is related to the coupon collector’s problem. Suppose we randomly get coupons, where the total number of coupons is 5, and we get each one with probability 1/5. How many coupons will we need to collect to get 4 different coupons? The first coupon appears immediately after one draw; after that, a different coupon appears with probability 4/5, which means the expected additional wait is 5/4. After we get the second coupon, the expected wait for the third coupon is 5/3. Continuing, the total wait for four different coupons to appear is 5/5 + 5/4 + 5/3 + 5/2 = 77/12.

However, we actually need 4 different coupons, not out of 5, but out of 6 to appear. That means that we need to multiply the answer by 6/5 to get 7.7. Then we add one extra roll for the final 6. The answer is 8.7.

Modeling assumption 2. Suppose Bob randomly chooses one number out of the ones that are missing. For example, if Alice rolled 1, 2, 3, 2, 1, 6, then Bob notices that 4 and 5 are missing, and mentions 5 with probability 1/2. In this case, the number of expected rolls is 4.26.

By using coupon-collecting ideas, we know, for each k, the expected number of rolls until k+1 distinct dice faces appear. To wit, for each k=0,1,2,3,4, the expected number of rolls is 1, 2.2, 3.7, 5.7, and 8.7, respectively.

Now we need to condition on the event that Bob actually says 5. By symmetry among the non-6 faces, the probability that Bob’s announcement is 5, given that he says something at all, is the same for each of the five digits. This conditioning does not bias the waiting time toward any particular missing digit, so the conditional distribution of the stopping time is obtained by averaging these expectations over the five possible values of k. Therefore, the expected number of rolls is (1 + 2.2 + 3.7 + 5.7 + 8.7)/5 = 4.26.

I am grateful to my other son, Alexey, for discussing this problem with me. Probability is a tricky subject, and it is nice to have experts in the family.

I wrote a book. This is my first book, so I am very proud. I wrote it together with two brilliant puzzle lovers, Ivo Fagundes David de Oliveira and Yogev Shpilman. The book is published by World Scientific and is available for pre-order: Mathematical Puzzles and Curiosities. Here is one sample puzzle from the book.

Puzzle. The centers of two opposite faces of a cube are connected by four distinct shortest paths, shown in the picture in different colors. Can you find two points on the surface of a cube such that there are exactly three shortest paths connecting them?

This puzzle appeared in the latest issue of SLMath’s newsletter, 17 Gauss Way. The issue has a puzzle column that I coauthored with Joe Buhler and Pavlo Pylyavskyy. The coolest images in the column were done by Tracy Hicks, and this image is no exception. The picture is better than our original one in the book.

I recently gave my STEP students a question from our old 2014 PRIMES entrance test.

Puzzle. John’s secret number is between 1 and 2¹⁶ inclusive, and you can ask him yes-or-no questions, but he may lie in response to one of the questions. Explain how to determine his number in 21 questions.

Here is the standard solution. We start by asking John to convert his number into binary and add zeros at the beginning if needed to make the result a binary string of length 16. For the first 15 questions, we do the following. For question i, we ask: “Is the i-th digit of your string zero?” For question 16, we ask, “Have you lied in response to a previous question?” If he lied on a previous question, he must say YES. If he didn’t, he might lie on question 16 and also say YES. In any case, if the answer is NO, he didn’t lie on the first 15 questions and we know the first 15 digits of the number. Then, we ask about the last digit three times, and the answer given at least twice is correct, so we know the number.

If the answer to question 16 is YES, then he lied on one of the questions 1–16. From now on, he has to tell the truth since he already lied. We use binary search (4 questions) to determine on which question he lied. This will tell us the first 15 digits, and we can use the 21st question to find the last digit.

One of my students, Tanish, invented an out-of-the-box solution that uses 18 questions. The idea is to force John to lie in the first two questions, and then safely proceed with the binary search.

He suggested asking the following two questions: “Are you going to answer NO in response to the next question?” and “Did you respond YES to the previous question?” The reader can check that whatever John replies, he is forced to lie exactly once.

Another student, Vivek, had a similar idea but used only one question to force John to lie: “Will you say NO to this question?”

My team, Death and Mayhem, organized the 2025 MIT Mystery Hunt. The hunt was a great success. Many people commented that it was the best mystery hunt ever.

This year, we added a new and interesting feature. Not only were teams allowed to choose which puzzles to unlock, but they were also given a short description of each puzzle in addition to its title. So, small teams who liked crosswords could choose to work only on crosswords.

As usual, I will list the mathy puzzles, including our official puzzle descriptions. All the puzzles can be found at the hunt’s All puzzles page.

We had a special round called Stakeout, with easy puzzles. My team isn’t too nerdy, so we didn’t have too many mathematical puzzles overall, and just two puzzles with a math flavor in the Stakeout round, incidentally coauthored by me. Somehow, I like designing easy puzzles. There were two additional puzzles in this round that I enjoyed during testing. I loved the popsicle puzzle so much that I brought it to my grandchildren to solve.

• A Math Quiz: A list of math questions.
• A Sudoku?: Grids with numbers.
• Anything Is Popsicle: Physical Puzzle — popsicle sticks.
• Some Assembly Required: A list of phrases and a word bank.

The first round wasn’t too difficult either. Several people praised the ChatGPT puzzle, though it’s not mathy.

• ChatGPT: A blank textbox with a text entry field below it.

Now, moving to more difficult puzzles, Denis Auroux is famous for designing fantastic logic puzzles. His puzzles below aren’t easy, but many people loved them. I even heard magnificent as praise.

• Maze of Lies: A text adventure exploring a maze.
• Bermuda Triangle: Triangular battleship puzzles.
• This is Just a Test: A bunch of grid logic puzzles.
• Good Fences Make Otherwise Incompatible Neighbors: A hexagonal logic puzzle.

Here are two puzzles I test-solved and enjoyed. The first one is a logic puzzle, while the second one isn’t math-related.

• Unreal Islands: A grid filled with letters and numbers.
• To Do: Tile That Rectangle: Rectangles of different lengths with words on the center and edges.

Here are two puzzles that I edited and highly recommend. The first puzzle was initially called Gin and Tonic; I wonder if anyone can guess why.

• Follow The Rules: An interactive interface with a grid of toggle switches and a grid of lights.
• Incognito: Cryptic crossword.

These are math-related puzzles that people liked.

• The Inspectre: Physical Puzzle — A bag of acrylic pieces and a sheet of paper.
• Knights of the Square Table: A 9×9 grid logic puzzle.
• Do The Manual Calculations (Don’t Try Monte Carlo): Some unusual chessboards and graphs with rules.

I asked only a few people for recommendations. These are math-related puzzles that weren’t mentioned but seem cool. The fourth puzzle was an invitation to the Mystery Hunt, which, not surprisingly, was a puzzle.

• The 10000-Sheet Excel File: An Excel file with 10000 sheets.
• Eponymous Forensic Accountant: Physical Puzzle — Bring your own bag (The PDF with receipts is available).
• _land: A thin, multicolored horizontal line.
• Engagements and Other Crimes: A printed invitation.
• A Map and a Shade (or Four): A chart with letters, colors, ticks, crosses, and a series of pie charts below.
• Garden Anecdotes: Post-it notes and some gibberish text.

I also got a recommendation for a non-math puzzle, which I would definitely have enjoyed watching solved. I’m not sure I’d enjoy solving it alone.

• Men’s at My Nose: A series of quotes.

Finally, here is the list of non-math puzzles that seem cool. A warning about the first puzzle: It’s rated R. The first three puzzles are relatively easy; they are from the Stakeout round.

• A Recipe For Success: Several completely normal and reasonable pickup lines.
• Charged: Interactive interface with text entry boxes and lines between them.
• Magic i: Two columns of text.
• Star Crossed: A large isometric grid of letters in the shape of a circle.
• Beyond a Shadow of a Doubt: A series of colorful dropquote puzzles.

Here is a video from Cracking the Cryptic, joined in this episode by Matt Parker, titled Matt Parker Sets Us A Challenge!. The video is devoted to the second part of the puzzle Maze of Lies, mentioned above, by Denis Auroux and Becca Chang.

Take a look at a card one of my students gave me last December. You can spot the Koch snowflake, the Sierpiński triangle, and the Sierpiński carpet on it. I guess my fractal class was a hit.

I am a proud member of the Death and Mayhem team, which participates in the MIT Mystery Hunt every year. This year, our team had the honor of running the hunt.

Here is a puzzle I contributed, titled A Math Quiz. It consists of a list of math problems. I am especially happy that I was able to turn a collection of cute math puzzles into a puzzle-hunt challenge with a word or phrase as its final answer.

Imagine you’re watching a magician. She pulls out two perfectly ordinary boxes — or so it seems. One box is inside the other, like a set of nesting dolls. So far, nothing suspicious.

Then she removes the smaller box, closes the larger one, and slides the larger box inside the smaller one. Ta-da!

The name comes from the way one box goes into the other. Would you like to know the secret? The two boxes are actually identical. Moreover, they are not cubes but cuboids. The inner box is fully closed, while the outerbox is slightly expanded, and the inner box is rotated relative to the outer one.

I first heard about Gozinta Boxes at the Gathering for Gardner conference in 2024. Ivo David gave a talk and presented his new trick: Triple Gozinta Boxes, which you can now buy at TCC Magic. He can place three boxes inside one another — and then repeat the trick in the reverse order.

During his presentation, David mentioned that he knew how to prove that you cannot have more than ten Gozinta Boxes. My immediate reaction was that ten must be overkill. So I decided to give the problem to my STEP students as a project.

We proved that in three or higher dimensions, the maximum number of boxes is three. We also showed that in two dimensions, the maximum is four. You can find all the details in our paper Mathematics of Gozinta Boxes, posted on the arXiv. But we didn’t stop there. We invented a new trick. We constructed three boxes such that not only can they be nested in one order — say, ABC — and in the reverse order, CBA, but they can also be nested in three additional orders, for example ACB, BAC, and BCA. We also proved that achieving all six possible orders is impossible. You can see the trick by following the link for A New Gozinta Boxes Trick.

I met Alexander Karabegov during the All-Soviet Math Olympiad in Yerevan. He was one year older than me. By then, when I was still competing in 1976, he was already a freshman at Moscow State University. He proposed the following two related puzzles for the Moscow Olympiad, which I had to solve.

Puzzle 1. You are given a finite number of points on a plane. Prove that there exists a point with not more than 3 closest neighbors.

Just in case, by closest neighbors I mean all points at the minimal distance from a given point. I am sure I solved both puzzles at the time. I leave the solution to the first one to the reader.

Puzzle 2. Can you place a finite number of points on the plane in such a way that each point has exactly 3 closest neighbors?

The last problem has an elegant solution with 24 points chosen from a triangular grid. The story continued almost 40 years later, when Alexander sent me an image (below) of such a configuration with 16 points. He conjectures that this is the minimal configuration.

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

Can you prove it?

Initially, I didn’t want to give the 24-points solution, but the image above is a big hint, so here you go.

Both constructions reveal the same underlying pattern. The constructions consist of rhombuses formed by two equilateral triangles, and the rhombuses are connected to each other. The 24-point construction consists of 6 rhombuses, while the 16-point construction consists of 4 rhombuses. What will happen if we try the construction with 3 rhombuses? The image below shows such a configuration, which now has extra edges with the shortest distance. We now see 3 points with more than three closest neighbors each, violating the condition. So the conjecture doesn’t break.

So far, every smaller attempt failed — can you prove that 16 is minimal?

I was teaching my students the Knaster method of dividing an estate, which I learned from my friend Ingrid Daubechies. Let’s look at an example.

Problem. Alice and Bob are divorcing. They have the portrait of Alice’s grandpa and $10,000. Alice values the portrait at $10,000 because of its sentimental value. Bob values it at a market price of $2,000. How do they divide their estate?

Here is what my students initially suggest.

1. Give Alice the portrait, and give Bob all the cash.
2. Give Alice the portrait, and divide the cash in half.

Let’s look at these suggestions in greater detail. Alice values the whole estate at $20,000; Bob values it at $12,000. In the first version, Alice gets half of the estate from her point of view; Bob gets the rest, which is more than half in his point of view. The students are obviously rooting for Bob. In the second version, Bob gets half of the estate in his estimate, while Alice gets the rest, which is more than half in her estimate. The students are obviously rooting for Alice. After some discussion, the students agree that there should be a number between $5,000 and $10,000 that Bob gets, which would be a fairer division than the two initial examples. But how do we find such a number?

This is where the Knaster algorithm comes in. The main idea is that each gets the same amount of money on top of their perceived half. In other words, the Knaster method treats the estate like a sealed-bid auction and equalizes bonuses. Alice thinks that her fair half is $10,000, while Bob thinks his half is $6,000. To equalize bonuses, we want Alice and Bob to each receive their perceived half plus the same amount — call it x. Solving gives x = $2,000. Alice gets the portrait and $2,000, while Bob gets $8,000.

This is a beautiful algorithm that allows each person to be very happy, receiving more than one half. The bigger the taste difference, the more each person gets on top of their portion. The next question is: how can people cheat if they know this algorithm is used?

Alice can cheat by claiming that she values the portrait at $2,000 plus epsilon. Epsilon is needed to guarantee she gets the portrait. This way, they both value the estate the same. Alice gets the portrait and $4,000, which is $2,000 more than the honest way. Symmetrically, Bob can cheat by claiming he values the portrait at $10,000 minus epsilon. This way, he gets $10,000, which is $2,000 more than the honest way.

I’ve been teaching this topic several times now. This year, my student Ben had an out-of-the-box idea on how Alice can cheat. Alice declares that she values the portrait at $0. Bob thinks the estate is worth $12,000, while Alice pretends that she values it at $10,000. After the calculation, Bob gets the portrait and $4,500, which is $500 more than his half of the estate in his view. Alice gets $5,500, which is $500 more than half of the estate in her declaration. Then she buys the portrait from Bob for $2,000. In the end, Alice gets the portrait and $3,500, way more than she would get after an honest use of the algorithm.

The first cheating method seems more profitable than the new one. But still, I love it when my students suggest unexpected ideas.