Tanya Khovanova's Math Blog

Puzzle. Three brothers who are identical triplets live on the seventh, eighth, and ninth floors of the same apartment building. Their apartments are identical and vertically stacked. One day all three step onto their balconies, standing in the same upright posture. The brother on the eighth floor shouts “AAAA!” Which of the other two will hear him first?

I gave this puzzle to my students, and two of them offered the right answer for the wrong reasons. One said, “The seventh-floor brother, because air is warmer closer to the ground and sound travels faster in warmer air.” Another said, “The seventh-floor brother, because the air is denser at lower altitude and sound travels faster in denser air.”

What is the right reason?

Once, I gave a puzzle to my students as homework just to check their level of attention.

Puzzle. A family has two identical twins. One of them is a boy, what is the probability that the other one is a boy?

What can I say? Some of the students didn’t pay attention and gave weird answers like 1/2, 1/3, and 2/3.

The twins are IDENTICAL. The other one has to be a boy! Duh!

One of the students was well-educated and mentioned that it is theoretically possible for different twins to be different genders, though this is extremely rare. When one fertilized egg splits into two, producing two embryos, the genetic material of both eggs is the same, almost. Some errors during splitting are possible, and it seems that some very particular errors can lead to the identical twins being identified as a boy and a girl. I never knew that before!

However, one student thought outside the box. In his vision, a family adopted two identical twins who aren’t each other’s twins, just happen to be identical twins with someone else.

I once saw a TikTok video featuring a puzzle: four matches were arranged to form a plus sign, and the challenge was to move one match to create a square. The solution shown differed from what first came to my mind, so I decided to share the puzzle with my students. Instead of drawing a diagram, I described it to them in words.

Puzzle. Arrange four matches to form a plus sign. Move one match to form a square.

Most of my students gave the same solution shown in the video: moving one match slightly away from the center to create a square in the center, with the inner edges of the matches forming its borders. Not all the students were as lazy as I was; some drew pictures to illustrate. One example is shown below, where the resulting square is in green.

My solution, also discovered by a couple of students, was to move one match to form the number 4, which is a square.

I am glad I didn’t provide a picture because it led to two unexpected solutions. For the first one, imagine a 3D shape out of four matches where one projection forms three sides of a square, and the other one is a plus sign. We can take the match from the plus sign that is not a side of a square and use it to complete the square. The second solution is shown below. The arrangement already contains a small square, so you can take a match and put it back. Being lazy brings benefits!

Here is another riddle I discovered in a book and gave as homework to my students.

Puzzle. I can use the number 20 thrice to make 60: 20 + 20 + 20 = 60. Make it 60 again by using a different number three times.

The book’s answer was to use 5: 55 + 5 = 60.

My students were very inventive. All of them solved the puzzle, but only one out of ten students came up with the book’s answer.

• For most of them, the new number was 60, as in 60 + 60−60 = 60, or 60*60/60.
• Some used −60 or 1/60, as in −60 − (−60) − (−60) = 60, or ((1/60)/(1/60))/(1/60) = 60. Similarly, some multiplied the cube root of 60 three times.
• One student used 59 in a clever way, as in 59 + 59/59 = 60.
• Another student said the following. If you turn 60 upside-down, you will make 09, and now you can use the number 3 thrice: 3+3+3 = 09.
• And the last on my list is the student who said that 42 has to be the answer to the universe and everything. He summed up two instances of 42 to get 84 and then subtracted the third instance of 42 with the digits flipped to get 84 − 24 = 60.

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.”

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.

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.

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.

The homework I give to my students (who are in 6th through 9th grades) often starts with a math joke related to the topic. Once, I decided to let them be the comedians. One of the homework questions was to invent a math joke. Here are some of their creations. Two of my students decided to restrict themselves to the topic we studied that week: sorting algorithms. The algorithm jokes are at the end.

* * *

A binary integer asked if I could help to double its value for a special occasion. I thought it might want a lot of space, but it only needed a bit.

* * *

Everyone envies the circle. It is well-rounded and highly educated: after all, it has 360 degrees.

* * *

Why did Bob start dating a triangle? It was acute one.

* * *

Why is Bob scared of the square root of 2? Because he has irrational fears.

* * *

Are you my multiplicative inverse? Because together, we are one.

* * *

How do you know the number line is the most popular?
It has everyone’s number.

* * *

A study from MIT found that the top 100 richest people on Earth all own private jets and yachts. Therefore, if you want to be one of the richest people on Earth, you should first buy a private jet and yacht.

* * *

Why did the geometry student not use a graphing calculator? Because the cos was too high.

* * *

Which sorting algorithm rises above others when done underwater? Bubble sort!

* * *

Which sorting algorithm is the most relaxing? The bubble bath sort.