Tanya Khovanova's Math Blog

Here is a new report of interesting homework solutions from my students.

Puzzle. One day, two sisters decided to clean the old shed at the bottom of their garden. When they finished cleaning, one had a dirty face and the other had a clean face. The sister with the clean face went and washed her face, but the girl with the dirty face did not wash. Why should this be so?

The expected answer: The sister with the clean face saw her sister’s dirty face and assumed her own face must be dirty as well, so she washed it. The sister with the dirty face saw her sister’s clean face and assumed her own face must also be clean, so she didn’t feel the need to wash.

Another student suggested a different but quite realistic answer.

The realistic answer: The sisters’ home ran out of water after the clean sister washed her face, preventing the dirty sister from washing her own.

The other student watched too many sitcoms.

The sitcom answer: The sister with the dirty face purposefully kept her face dirty, so she could show her parents that she did all the work, as she was the only one with dirt on her face.

I asked ChatGPT to solve the puzzle, and, unsurprisingly, it came up with the standard answer. I pushed and got the following.

The ChatGPT answer: The sister with the clean face washed up because she was an Instagram influencer and couldn’t risk being seen dirty, even in her own garden. Meanwhile, the sister with the dirty face was a carefree adventurer who believed dirt was “nature’s makeup.” Plus, she figured that if she waited long enough, the dirt would either blow away or blend into a trendy new skincare routine—”Exfoliation by Shed Dust.”

Here’s a fresh challenge from the recent Tournament of the Towns, crafted by Alexander Shapovalov.

Puzzle. A mother and her son are playing a game involving cheese and butter. The son starts by cutting a 300-gram block of cheese into 4 pieces. Then, the mother divides 280 grams of butter between two plates. Afterward, the son places the cheese pieces onto these same plates. The son wins if, on both plates, there is at least as much cheese as butter. If not, the mother wins. Can either the mother or the son guarantee a win, regardless of the other’s moves?

A while ago I took writing lessons with Sue Katz. Below is my homework from 2010 (lightly edited). If I remember correctly, this piece was inspired by Sam Steingold.

—My friend Sam installed six locks on his door to protect himself from burglars.
—I know. I visited your friend. He has six very cheap locks. Any professional could open one in a second, so Sam’s door will only resist for six seconds.
—Yeah, but those locks aren’t completely identical. Three of them unlock with a clockwise motion, and three with a counterclockwise motion.
—So what? The thieves will just turn the lock mechanism whichever way it can be turned.
—Not so fast. Sam never locks all of them. Every time, he randomly picks which ones to lock.
—That might work, but what if he forgets which ones he locked?
—That’s okay, He remembers which way to turn every lock to unlock.

I love hat puzzles, and this one, posted on Facebook by Konstantin Knop, is no exception.

Puzzle. The sultan decided to test his three sages once again. This time, he showed them five hats: three red and two green. Each sage was blindfolded and had one hat placed on their head. When the sages removed their blindfolds, they could see the hats on the other sages but not their own. The twist in this puzzle is that one of the sages is color-blind and cannot distinguish red from green. The sages are all friends and are aware of each other’s perception of color. The sages are then asked, in order, if they know the color of their hats. Here’s how the conversation unfolded:

• Alice: I do not know the color of my hat.
• Bob: Me too, I do not know the color of my hat.
• Carol: Me too, I do not know the color of my hat.
• Alice: I still do not know the color of my hat?

The question is: Who is color-blind?

From time to time, the homework for my PRIMES STEP students includes questions that are not exactly mathematical. Last week, we had the following physics puzzle.

Puzzle. A fisherman needed to move a heavy iron thingy from one river’s shore to another. When he put the thingy in his boat, the boat lowered so much that it wasn’t safe to operate. What should he do?

The expected answer: He should attach the thingy to the bottom of the boat. When the object is inside the boat, the boat needs to displace enough water to account for the entire weight of the boat and the thingy. When the thingy is attached to the bottom of the boat, the thingy experiences its own buoyancy. Thus, the water level rises less because the thingy displaces some water directly, reducing the boat’s need to displace extra water. Thus, the amount of weight the fisherman saves is equal to the amount of water that would fit into the shape of this thingy.

As usual, my students were more inventive. Here are some of their answers.

• The fisherman could cut the iron thingy and transport it piece by piece.
• He can swim across and drag the boat with a rope with the thingy inside.
• He can use a second boat to pull the first boat with the thingy in it.
• It is another river’s shore, so he can just take the iron with him to a different river without going over water.
• If the fisherman has extra boat material, heightening the boat’s walls would keep it from sinking.

Also, some funny answers.

• He could fast for a few days, making him lighter.
• He could tie helium balloons to the boat to keep it afloat even after he gets in.
• Wait until winter and slide the boat on ice.

And my favorite answer reminded me of a movie I recently re-watched.

• You’re gonna need a bigger boat.

Before moving to the US, I attended the Gelfand seminar and took some pictures. I was a regular participant and have some bittersweet memories from that time. I’ve written about my experiences in several blog posts related to Gelfand, who was my advisor.

The year was 1990, and I just acquired my first camera. I was about to leave the USSR for the US and took a bunch of pictures of family, friends, and other moments. I wasn’t happy with the photos I captured at the seminar due to the poor quality of the camera and the dim lighting in the lecture hall. For context, the seminar was held on Mondays from 7 to 10 pm. So I put the pictures away and forgot about them.

Recently, I decided to digitize all of my old pictures. While the seminar photos are still grainy, they feel more precious now. Perhaps it’s the fact that they’ve survived for over 30 years, or maybe I’ve just grown more sentimental.

A big part of the seminar was the networking that happened beforehand. Although the seminar was scheduled to start at 7 pm, it often began at random times, anywhere between 7 and 8:30 pm. Gelfand disliked tardiness, so everyone would arrive by 7 and hang. All of my photos were taken before the seminar: some in the hallway and some in the seminar room.

In the last three pictures, the socializing had ended, and the seminar was about to start.

The term fibonometry was coined by John Conway and Alex Ryba in their paper titled, you guessed it, “Fibonometry”. The term describes a freaky parallel between trigonometric formulas and formulas with Fibonacci (F_n) and Lucas (L_n) numbers. For example, the formula sin(2a) = 2sin(a)cos(a) is very similar to the formula F_2n = F_nL_n. The rule is simple: replace angles with indices, replace sin with F (Fibonacci) and cosine with L (Lucas), and adjust coefficients according to some other rule, which is not too complicated, but I am too lazy to reproduce it. For example, the Pythegorian identity sin²a + cos²a = 1 corresponds to the famous identity L_n² – 5F_n² = 4(-1)ⁿ.

My last year’s PRIMES STEP senior group, students in grades 7 to 9, decided to generalize fibonometry to general Lucas sequences for their research. When the paper was almost ready, we discovered that this generalization is known. Our paper was well-written, and we decided to rearrange it as an expository paper, Fibonometry and Beyond. We posted it at the arXiv and submitted it to a journal. I hope the journal likes it too.

Consider the following Fibonacci trick. Ask your friends to choose any two integers, a and b, and then, starting with a and b, ask them to write down 10 terms of a Fibonacci-like sequence by summing up the previous two terms. To start, the next (third) term will be a+b, followed by a+2b. Before your friends even finish, shout out the sum of the ten terms, impressing them with your lightning-fast addition skills. The secret is that the seventh term is 5a+8b, and the sum of the ten terms is 55a+88b. Thus, to calculate the sum, you just need to multiply the 7th term of their sequence by 11.

If you remember, I run a program for students in grades 7 through 9 called PRIMES STEP, where we do research in mathematics. Last year, my STEP senior group decided to generalize the Fibonacci trick for their research and were able to extend it. If n=4k+2, then the sum of the first n terms of any Fibonacci-like sequence is divisible by the term number 2k+3, and the result of this division is the Lucas number with index 2k+1. For example, the sum of the first 10 terms is the 7th term times 11. Wait, this is the original trick. Okay, something else: the sum of the first 6 terms is the 5th term times 4. For a more difficult example, the sum of the first 14 terms of a Fibonacci-like sequence is the 9th term times 29.

My students decided to look at the sum of the first n Fibonacci numbers and find the largest Fibonacci number that divides the sum. We know that the sum of the first n Fibonacci numbers is F_n+2 – 1. Finding a Fibonacci number that divides the sum is easy. There are tons of cute formulas to help. For example, we have a famous inequality F_4k+3 – 1 = F_2k+2L_2k+1. Thus, the sum of the first 4k+1 Fibonacci numbers is divisible by F_2k+2. The difficult part was to prove that this was the largest Fibonacci number that divides the sum. My students found the largest Fibonacci number that divides the sum of the first n Fibonacci numbers for any n. Then, they showed that the divisibility can be extended to any Fibonacci-like sequence if and only if n = 3 or n has remainder 2 when divided by 4. The case of n=3 is trivial; the rest corresponds to the abovementioned trick.

They also studied other Lucas sequences. For example, they showed that a common trick for all Jacobsthal-like sequences does not exist. However, there is a trick for Pell-like sequences: the sum of the first 4k terms (starting from index 1) of such a sequence is the (2k + 1)st term times 2P_2k, where P_n denotes an nth Pell number.

You can check out all the tricks in our paper Fibonacci Partial Sums Tricks posted at the arXiv.

Have you heard of Grigori Perelman? If you like math, you probably have. He is one of the most renowned mathematicians in the world. I recently got a book on the Leningrad Mathematical Olympiads (scheduled for publication in English in 2025) and found Grigori’s name there. He authored one of the Olympiad problems from 1984. For context, he was born in 1966. Here it is.

Puzzle. You are given ten numbers: one “1” and nine “0”s. You are allowed to replace any two numbers with their arithmetic mean. What is the smallest number that can appear in place of the “1” after a series of such operations?

The famous 5-card trick begins with the audience choosing 5 cards from a standard deck. The magician’s assistant then hides one of the chosen cards and arranges the remaining four cards in a row, face up. Upon entering the room, the magician can deduce the hidden card by inspecting the arrangement. To eliminate the possibility of any secret signals between the assistant and the magician, the magician doesn’t even have to enter the room — an audience member read out the row of cards.

The trick was introduced by Fitch Cheney in 1950. Here is the strategy. With five cards, you are guaranteed to have at least two of the same suit. Suppose this suit is spades. The assistant then hides one of the spades and starts the row with the other one, thus signaling that the suit of the hidden card is spades. Now, the assistant needs to signal the value of the card. The assistant has three other cards than can be arranged in 6 different ways. So, the magician and the assistant can agree on how to signal any number from 1 to 6. This is not enough to signal any random card.

But wait! There is another beautiful idea in this strategy — the assistant can choose which spade to hide. Suppose the two spades have values X and Y. We can assume that these are distinct numbers from 1 to 13. Suppose, for example, Y = X+5. In that case, the assistant hides card Y and signals the number 5, meaning that the magician needs to add 5 to the value of the leftmost card X. To ensure that this method always works, we assume that the cards’ values wrap around. For example, king (number 13) plus 1 is ace. You can check that given any two spades, we can always find one that is at most 6 away from the other. Say, the assistant gets a queen of spades and a 3 of spades. The 3 of spades is 4 away from the queen (king, ace, two, three). So the assistant would hide the 3 and use the remaining three cards to signal the number 4.

I skipped some details about how permutations of three cards correspond to numbers. But it doesn’t matter — the assistant and the magician just need to agree on some correspondence. Magically, the standard deck of cards is the largest deck with which one can perform this trick with the above strategy.

Later, a more advanced strategy for the same trick was introduced by Michael Kleber in his paper The Best Card Trick. The new strategy allows the magician and the assistant to perform this trick with a much larger deck, namely a deck of 124 cards. But this particular essay is not about the best strategy, it is about the Cheney strategy. So I won’t discuss the advanced strategy, but I will redirect you to my essay The 5-Card Trick and Information, jointly with Alexey Radul.

63 years later, the 4-card trick appeared in Colm Mulcahy’s book Mathematical Card Magic: Fifty-Two New Effects. Here the audience chooses not 5 but 4 cards from the standard deck and gives them to the magician’s assistant. The assistant hides one of them and arranges the rest in a row. Unlike in the 5-card trick, in the 4-card trick, the assistant is allowed to put some cards face down. As before, the magician uses the description of how the cards are placed in a row to guess the hidden card.

The strategy for this trick is similar to Cheney’s strategy. First, we assign one particular card that the magician would guess if all the cards are face down. We now can assume that the deck consists of 51 cards and at least one of the cards in the row is face up. We can imagine that our 51-card deck consists of three suits with 17 cards in each suit. Then, the assistant is guaranteed to receive at least two cards of the same imaginary suit. Similar to Cheney’s strategy, the leftmost face-up card will signal the imaginary suit, and the rest of the cards will signal a number. I will leave it to the reader to check that signaling a number from 1 to 8 is possible. Similar to Cheney’s strategy, the assistant has an extra choice: which card of the two cards of the same imaginary suit to hide. As before, the assistant chooses to hide the card so that the value of the hidden card is not more than the value of the leftmost face-up card plus 8. It follows that the maximum number of cards the imaginary suit can have is 17. Magically, the largest possible deck size for performing this trick is 52, the standard deck of cards.

Last academic year, my PRIMES STEP junior group decided to dive deeper into these tricks. We invented many new tricks and calculated their maximum deck sizes. Our cutest trick is a 3-card trick. It is similar to both the 5-card trick and the 4-card trick. In our trick, the audience chooses not 5, not 4, but 3 cards from the standard deck and gives them to the magician’s assistant. The assistant hides one of them and arranges the other two in a row. The assistant is allowed to put some cards face down, as in the 4-card trick, and, on top of that, is also allowed to rotate the cards in two ways: by putting each card vertically or horizontally.

We calculated the maximum deck size for the 3-card trick, which is not 52, as for the 5- and 4-card trick, but rather 54. Still, this means the 3-card trick can be performed with the standard deck. The details of this trick and other tricks, as well as some theory, can be found in our paper Card Tricks and Information.