Tanya Khovanova's Math Blog

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.

When I was in 8th grade, I was selected to be part of the Moscow math team and went to Yerevan, Armenia, to participate in the All-Soviet Math Olympiad. A group of us boarded a bus, and Alexander Karabegov paid for all of our bus tickets. He was from Yerevan himself and wanted to be a gracious host. I was impressed. The next time I met him was when I started studying at the Moscow State University. We have been friends ever since. He was even the best man at one of my weddings. Now, he lives in Texas and sends me his original puzzles from time to time. Today, he sent me a new one.

WARNING. His solution to the puzzle is also included. So if you want to solve it yourself, stop reading after the next paragraph.

Puzzle. A number c is called a fixed point of a function f, if it is a solution of the equation f(x) = x; that is, if f(c) = c. Find all solutions of the equation g(g(x)) = x, where g(x) = x² + 2x − 1; that is, find all fixed points of the function f(x) = g(g(x)). (We can assume that x is a real number.)

I gave the puzzle to my students, and they converted it to a fourth-order equation, which they solved using various methods. What I liked about Alexander’s solution is it only uses quadratic equations. I am too lazy to give his full solution. Here is just his solve path.

Solve path. If c is a fixed point of the function g(x), then it is a fixed point of f(x) = g(g(x)). Solving the equation g(c) = c gives us two fixed points. We need two more, as our equation is quartic. Suppose a is another fixed point. Let b = g(a). It follows that g(b) = a. Moreover, we can assume that a is not b, as we covered this case before. We get two equations a² + 2a − 1 = b and b² + 2b − 1 = a. Subtracting one equation from another, we get a quadratic equation that has to be divisible by a −b. As b is not a, by our assumption, we can divide the result by a − b, expressing b as a linear function of a. We plug this back into one of the two equations and get a quadratic equation for a, supplying us with the remaining two solutions. TADA!

Here’s one by Sergei Luchinin, designed for 7th graders.

Puzzle. We have an 8-by-8 chessboard, but it’s not colored in the usual checkerboard pattern. Instead, all cells in odd-numbered columns are black, and all cells in even-numbered columns are white. A limping rook is placed in the lower-left corner and can only move one cell to the right or one cell up. The rook’s goal is to reach the upper-right corner.
The question is: Are there more paths that pass through more white cells, or more that pass through more black cells?

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.