Tanya Khovanova's Math Blog

I am so tired of spam emails. I keep thinking about how we can fight spam, and here is an idea.

Gmail should change its system: every email you send to me would cost 1 dollar, payable to me. We can add an exception for people on my contact list. Everyone else, pay up!

I do not often contact strangers. But if I do, it is always important. So paying 1 dollar seems more than fair. On the other hand, this system will immediately discourage mass emails to strangers. Spam would go down, and I would stop receiving emails inviting me to buy a pill to increase the size of a body part I do not have.

This idea of getting paid for reading an email is not new. It was implemented by Jim Sanborn, the creator of the famous Kryptos sculpture. Kryptos is located at the CIA headquarters and has four encrypted messages. People tried to decrypt them and would send Jim their wrong solutions. Jim got tired of all the emails and administered Kryptos fees. Anyone who wants Jim to check their solution, can do so by paying him 50 dollars. I wonder if Jim would still charge the fee if someone sent him the correct solution.

Thinking about it, I would like the payable email system to be customizable, so I can charge whatever I want. After all, I do value my time.

Gmail could get a small percentage. Either Gmail, together with me, gets rich, or spam goes away. Both outcomes would make my life easier.

Puzzle. A goat was on a 10-meter leash. Yet it managed to go 300 meters away from the post. How come?

The standard answer. The leash wasn’t attached to the post.

My students scrutinized the puzzle and found some other possible ambiguities. For example, there might be two posts: the goat was leashed to one and was far away from the other. In another example, the timing is not given. It is possible that the goat was on the leash at one time and unleashed and far away from the post at another time. Here is my favorite answer.

My favorite answer. The goat ate the leash.

Puzzle. Find the largest and the smallest 4-digit numbers n such that when you erase the first two digits of n, you get the sum of the digits of n.

I recently posted a gnome puzzle by Alexander Gribalko.

Puzzle. Nine gnomes repeat the following procedure three times. They arrange themselves on a 3 by 3 chessboard with one gnome per cell and greet all of their orthogonal neighbors with handshakes. Prove that not all pairs of gnomes greet each other.

As often happens with my blog puzzles, I used this puzzle as homework for my students. They calculated that the total number of handshakes needed for all nine gnomes to greet each other is 36. On the other hand, each arrangement of gnomes creates 12 handshakes. This means that the numbers are tight: no greeting can be wasted, and every pair of gnomes need to greet each other exactly once. The students then studied different cases to prove this was impossible.

In each arrangement, a gnome can have either 2, 3, or 4 handshakes. Hence, we can distribute handshakes over three placements as 2+2+4 or 2+3+3. It follows that if a gnome is ever in the center of the grid, he has to be in a corner for the other two arrangements. Therefore, the three gnomes who end up in the center for one of the arrangements never greet each other.

However, I always love solutions that involve coloring the board in a checkerboard manner. Here is my solution.

Solution. Let’s color the board in a checkerboard manner. We assign each gnome a binary string, of length 3, describing the colors of the cells where the gnome was in each placement. There are 8 different possible strings. It follows that at least two gnomes are assigned the same string. But they can’t greet each other if they are standing on the same colors in each arrangement!

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I hate getting into debates about Möbius strips. They’re always one-sided.

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North Korea’s ballistic missile test failed due to a bug in Windows. The next missile containing a bug report has been automatically sent to Microsoft.

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4 out of 3 people have trouble with fractions.

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Do you know what seems odd to me?
Numbers that aren’t divisible by two.

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Why was algebra so easy for the Romans? X was always 10.

My readers know that I love hat puzzles. This is why I decided to turn a number trick by Konstantin Knop (in Russian) into a hat trick.

Hat Trick. The audience has a bottomless supply of hats in ten different colors. They arrange ten people in a line and put one of the hats on each person. Then the magician’s assistant comes in and removes a hat from one of the ten people. After that, the magician appears and, abracadabra, guesses the color of the removed hat. The magician and the assistant agreed on a strategy beforehand. What is it?

Keep in mind that this trick won’t work with fewer than ten colors. As a bonus, can you explain why?

Sep 18, 2022 Correction: I meant “with fewer than ten people.”

As my readers know, I am devoted to my students. When I need something I can’t buy, I try to make it. That is why I crocheted a lot of mathematical objects. One day, I resolved to have in my possession a Seifert surface bounded by Borromean rings (a two-sided surface that has Borromean rings as its border).

However, my crocheting skills were not advanced enough, so I signed up for a wet and needle felting workshop. When I showed up, Linda, our teacher, revealed her lesson plan: a felted soap with a nice pink heart on top. It looked cool to have soap inside a sponge, not to mention that wool is anti-bacterial. But I had bigger plans than soap and eagerly waited for no one else to show up.

When my dream materialized, and, as I had hoped, no one else was interested in felt, I asked Linda if we could drop the hearty soap and make my dream thingy. She agreed, but my plan didn’t survive for long. As soon as Linda saw a picture of what I wanted, she got scared. Seifert surfaces were not in the cards, so soap it was. I told her that there was no way I was going to needle-felt a pink heart onto my felted soap. I ended up with a blue Sierpiński gasket.

We had a great time. Linda was teaching me felting, and I was teaching her math. I am a good teacher, so even felters working on a farm enjoy my lessons.

After the workshop, I went online and found my dream surface on Shapeways. In the end, I was happy to just buy it and not have to make it.

But my felting workshop wasn’t a waste of time: tomorrow I will wash myself with a gasket.

A combinatorial games section? At a math competition? I have never heard of it before. But here it is, at the Lomonosov Tournament. The first problem is from the 2012 Tournament (the link is to a Russian version). The problem is by Alexander Shapovalov. I picked it for its elegance and simplicity.

A rectangular band. You have a paper rectangle of size m by n with grid lines, where both m and n are greater than 1. Two players play the following game. The first player cuts the rectangle along any grid line into two rectangles. The next player picks one of the rectangles and cuts it again along a grid line, and so on. The winner is the player who, after their move, can arrange all the rectangles into a band of width 1. Which player is guaranteed a win regardless of the moves of their partner? Consider two cases.

1. One of the numbers, m or n, is even;
2. Both m and n are odd.

The next problem, by I. Raskin, is from the 2015 Tournament (in Russian). The problem is about fruits, and I know a lot of great puzzles with bananas and oranges, so I immediately got attracted to this one.

The original version had three types of fruit starting with the letters a, b, and c in Russian. They were oranges, bananas, and plums. I reused bananas and replaced oranges with apples, but I got stuck on the letter c. The players in this game eat their fruits, so using cantaloupes seemed like overkill. Plus, I am on a diet, so I decided on cherries.

Fruits. There are a apples, b bananas, and c cherries on the table. Two players play a game where one move consists of eating two different fruits. The person who can’t move loses. Assuming the players use their best strategies, would the first or the second player win in the following cases?

1. a = 1 (the answer might depend on the values of b and c);
2. a = 6, b = 8, c = 10;
3. a = 7, b = 9, c = 15;
4. a = 19, b = 20, c = 21.

The last problem is from the 2012 Tournament (in Russian). It has great sentimental value for me, as it was created by John Conway. It also reminds me of the famous Frobenius (chicken McNugget) problem.

Coin mintage. Once upon a time, in a faraway kingdom, two treasurers were minting coins. They decided to make it into a game, taking turns minting coins. Each turn, the player chooses a particular integral denomination and mints an infinite supply of coins of this denomination. The rules of the game forbid choosing a new denomination that can be paid with the existing coins. The treasurer who is forced to mint a coin of denomination 1 loses.

1. Prove that if the first treasurer starts with denomination 2 or 3, s/he loses.
2. Is it profitable for the first treasurer to start with denomination 4?
3. Is it profitable for the first treasurer to start with denomination 6?
4. Suppose the first treasurer minted coins of denomination 5, and the second treasurer of denomination 6. What is the winning strategy for the first treasurer after that?
5. Suppose the first treasurer minted coins of denomination 5, and the second treasurer of denomination k. Prove that the largest denomination available for minting is 4k − 5.
6. Prove that the first treasurer can win by starting with denomination 5. (Hint: Suppose the second treasurer replied with denomination k, and the first treasurer minted 4k − 5 after that. If this strategy wins, the problem is solved. However, if after that, the second treasurer wins by minting denomination m, then minting denomination 4k − 5 was the wrong move. What was the right move?)

Here is another exciting puzzle posted on Facebook by Konstantin Knop.

Puzzle. Eight glasses of wine are placed in a circle on a round table. Three sages are invited to take the following challenge. In the presence of the first sage, five glasses are filled with good wine and the other three with poisoned wine, indistinguishable from the good wine. After drinking the poisoned wine, the person will die a terrible tormented death. Each sage has to drink one full glass. The first sage is not allowed to give any hints to the other sages, but they can see which glass he chooses before making their own selection. The three sages can agree on their strategy beforehand. What is the strategy to keep them all alive?
An extra question. Does a strategy exist if fewer than eight glasses are placed around the table?

Let’s start with two trivial variations. If there is only one sage, s/he knows which glass to drink. Now, suppose there is only one poisoned glass and any number of sages. If the total number of good glasses is not less than the number of sages, the solution is obvious. The first sage drinks the glass clockwise from the poisoned one, and the other sages continue clockwise.

The next slightly less trivial case involves two sages and two poisoned glasses. If the total number of glasses is at least five, the sages are safe. The reason: there are at least two good glasses in a row. So the sages can agree that the first one drinks a good glass followed clockwise by another good glass. If the total number of glasses is less than 5, there is no reliable strategy, as the reader can check.

The above idea works if the number of sages is S, the number of poisoned glasses is P, and the total number of glasses is T, where T is greater than SP. Then, the strategy is the same since there is a guaranteed continuous stretch of S good glasses.

On the other hand, one can argue that for S and P more than 1, and T = S + P, it is impossible to find a strategy.

Our original problem corresponds to the case S = P = 3, and T = 8. Presumably, the strategy doesn’t exist if S = P = 3, and T < 8. If the 8-glasses problem is difficult, here is a much easier version, corresponding to the case S = 2, P = 3, and T = 6.

An Easier Puzzle. Six glasses of wine are placed in a circle on a round table. Two sages are invited to take the following challenge. In the presence of the first sage, three glasses are filled with good wine and the other three with poisoned wine, indistinguishable from the good wine. After drinking the poisoned wine, the person will die a terrible tormented death. Each sage has to drink one full glass. The first sage is not allowed to give any hints to the other sages, but they can see which glass he chooses before making their own selection. The two sages can agree on their strategy beforehand. What is the strategy to keep them all alive?

I am trying to make a point that, mathematically, the Sleeping Beauty problem is resolved, and the Wikipedia article about it should stop assuming that there is an ongoing debate.

The Sleeping Beauty problem. Sleeping Beauty participates in the following experiment. On Sunday, she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and asked whether she thinks the coin was heads or tails. Regardless of her answer, if the coin was heads, the experiment ends. However, if the coin was tails, she is put back to sleep with her memory erased and awakened again on Tuesday and asked the same question. In this case, the experiment stops on Tuesday. She knows the protocol. She is awakened one morning. From her point of view, what is the probability that the coin was heads?

Here is the solution. If it is Monday, then the probability that the coin is heads is one half. So the probability of Monday/heads is the same as Monday/tails. If the coin is tails, Sleeping Beauty can’t distinguish between Monday and Tuesday. So the probability of Monday/tails is the same as Tuesday/tails. Thus, the three cases Monday/heads, Monday/tails, and Tuesday/tails are equally probable. It follows that when she is awakened, the probability of heads is one third.

However, there are still people — called halfers — who think that the probability of heads is one half.

But suppose we ask a different question.

Different question. She is awakened one morning. From her point of view, what is the probability that the day is Tuesday?

As I explained before, when she is awakened, the probability of it being Tuesday is one third. Let us calculate what the halfers think. Suppose they think that the probability of the day being Tuesday is x, then the probability of Monday is 1 − x. Let’s calculate the probability of the coin being heads from here. The probability of heads, if today is Tuesday, is zero. The probability of heads, if today is Monday, is 1/2. Therefore, the probability of heads equals 0 · x + (1 − x)/2 = (1 − x)/2. In halfers’ view, the resulting calculation equals 1/2. In other words, (1 − x)/2 = 1/2. It follows that x is zero. Doesn’t make much sense that the probability of the day being Tuesday is zero, does it?

Halfers are wrong. Wikipedia should update the article.