Tanya Khovanova's Math Blog

One hundred people play the following game. Their names are written on pieces of paper and put into 100 labeled boxes at random. Each box is labeled with a number from 1 to 100 and one name has been placed inside each box. The boxes are placed on a table in a separate room. The players go into the room one by one and each has to open 99 boxes one after another. After each player finishes and leaves the room, the boxes are closed again. The players are not allowed to communicate with each other in any way, although they have been given one day before the event to discuss their strategies. They only win if every one of the one hundred players avoids opening the box with his or her own name. What is the optimal strategy?

Let me first discuss a simpler version of the game. Each player has to open exactly one box and they win if each one of them finds their name. After each player finishes and leaves the room, the boxes are closed again and the room is re-set.

If all of them decide to open box number 42, they are guaranteed to lose. They can try to open random boxes, then they win with probability (1/100)¹⁰⁰. Can they use a joint strategy that is better than random?

Yes, they can. Clearly, two people shouldn’t open the same box. So on the day before, if each agrees to open a box with a different assigned number, their probability to win is one over 100!. I leave it to my readers to prove that this is the best strategy.

What is the difference between this problem and the original problem? Isn’t choosing the last box the same as choosing the first? Aha! When they open 99 boxes they see the names, so they can use this information as part of their strategy.

I hope that this new version is so intriguing that you will start solving this puzzle right away.

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Decimals have a point.

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During the show “Are You Smarter Than a 5th Grader?” the following question was asked:

What is superfluous in the following list: a carrot, an onion, a potato, a Lexus?

A smart 5th grader answered: a carrot, an onion, and a potato.

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If you buy 3 DVDs for the price of 4, you will get one more as a bonus.

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Only yesterday, today was tomorrow.

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By definition, one divided by zero is undefined.

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Finally artificial intelligence has caught up with humans: when filling out electronic forms, many humans need several tries to prove they are not robots.

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Be back in 5 minutes. If I am late, reread this sms.

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— We’ll split the money 50-50.
— I want 70.
— Okay, 70-70!

I recently published my new favorite math problem:

A deck of 36 playing cards (four suits of nine cards each) lies in front of a psychic with their faces down. The psychic names the suit of the upper card; after that the card is turned over and shown to him. Then the psychic names the suit of the next card, and so on. The psychic’s goal is to guess the suit correctly as many times as possible.
The backs of the cards are asymmetric, so each card can be placed in the deck in two ways, and the psychic can see which way the top card is oriented. The psychic’s assistant knows the order of the cards in the deck; he is not allowed to change the order, but he may orient any card in either of the two ways.
Is it possible for the psychic to make arrangements with his assistant in advance, before the latter learns the order of the cards, so as to ensure that the suits of at least (a) 19 cards, (b) 23 cards will be guessed correctly?
If you devise a guessing strategy for another number of cards greater than 19, explain that too.

If the psychic is only allowed to look at the backs of the cards, then the amount of transmitted information is 2³⁶, which is the same amount of information as suits for 18 cards. This number of guesses is achievable: the backs of every two cards can clue in the suit of the second card in the pair. This way the psychic can guess the suits of all even-numbered cards in the deck. So the problem is to improve on that. Using the info from the cards that the psychic is permitted to turn over can help too.

The problem is from the book Moscow Mathematical Olympiads, 2000-2005. The book and Russian blog discussions provide many different ideas on how to guess more than half of the deck.

Here is the list of ideas.

Idea 1. Counting cards. If you count cards you will know the suits of the last cards.

Idea 2. Trading. As we discussed before, the psychic can correctly guess the suits of even-numbered cards. By randomly guessing the odd-numbered cards she can correctly guess on average the suits of 4.5 additional cards. Unfortunately, this is not guaranteed. But wait. What if we trade the knowledge of the second card’s suit for the majority suit among odd-numbered cards?

Idea 3. Three cards. Suppose we have three cards. Three bits can provide the following knowledge: the majority color, plus the suit of the first and of the second cards in the majority color. Thus, three bits of information will allow the psychic to guess the suits of two cards out of three.

Idea 4. Which card. Suppose the assistant signals the suits of even-numbered cards. With no loss, the psychic can guess the even-numbered card and repeat the same suit for the next card. If this is the plan, the assistant can choose which of the two cards to describe. Which card of the two matches the psychic’s guess provides an additional bit of information.

Idea 5. Surprise. Suppose we have a strategy to inform the psychic about some cards. Suppose the assistant deliberately fails on one of the cards. Then the index of this card provides info to the psychic.

I leave it to my readers to use these ideas to find the solution for 19, 23, 24 and maybe even for 26 cards.

Imagine a slice of buttered white bread with a heap of sugar on top. That was my favorite lunch when I was a kid. My mom was working very hard, I was the oldest sister, and this was what I would make for myself almost every day.

Later someone told me that sugar is brain food. I believed that sugar and chocolate helped me do mathematics, so my love for sugar got theoretical support. I finally figured out the source of this love when my first son was born. To teach my son to stop requesting milk at night, my mother pushed me to give him sugar-water instead. At that moment, I realized that I developed my love of sugar with my mother’s milk. Or, more precisely, instead of my mother’s milk.

Now there is more and more evidence that the love of my life is a mistake. See for example Is Sugar Toxic?. Will I ever be able to break my oldest bad habit, the one I developed before I can remember myself doing it?

This is how my ex-husband Joseph Bernstein used to start his courses in representation theory.

Suppose there is a four-armed dragon on every face of a cube. Each dragon has a bowl of kasha in front of him. Dragons are very greedy, so instead of eating their own kasha they try to steal kasha from their neighbors. Every minute every dragon extends four arms to the neighboring cube’s faces and tries to get the kasha from the bowls there. As four arms are fighting for every bowl of kasha, each arm manages to steal one-fourth of what is in the bowl. Thus each dragon steals one-fourth of of the kasha of each of his neighbors, while all of his own kasha is stolen too. Given the initial amounts of kasha in every bowl, what is the asymptotic behavior of the amounts of kasha?

You might ask how this relates to representation theory. First, it relates to linear algebra. We can consider the amounts of kasha as a six-dimensional vector space and the stealing process as a linear operator. As mathematicians, we can easily assume that a negative amount of kasha is allowed.

Now to representation theory. The group of rotations of the cube naturally acts on the 6-dimensional vector space of kashas. And the stealing operator is an intertwining operator of this representation. Now for a spoiler alert: I’m about to finish the solution, so stop here if you want to try it on your own.

The intertwining operator acts as a scalar on irreducible representations of the group. Thus we should decompose our representation into irreducible ones. The group has five irreducible representations with dimensions 1, 1, 2, 3, and 3.

We can decompose the kasha into the following three representations:

• One-dimensional. Every dragon has the same amounts of kasha. The stealing operator acts as identity.
• Three-dimensional. Dragons on opposite sides have the opposite amount of kasha. The stealing operator acts as zero.
• Two-dimensional. Dragons on opposite sides have the same amount of kasha and the total amount of kasha is zero. The stealing operator acts as −1/2.

We see that asymptotically every dragon will have the same amount of kasha.

Now it is your turn to use this method to solve a similar problem, where there are n dragons sitting on the sides of an n-gon. Each dragon has two arms, and steals half of the kasha from his neighbors. Hey, wait a minute! Why dragons? There are people around the table stealing each other’s kasha. But the question is still the same: What is the asymptotic behavior of the amounts of kasha?

Students should use a different strategy for the AIME than for the AMC. So students who are approaching the AIME for the first time need to question the habits they have developed after years of doing multiple choice tests. Here are some suggestions.

Checking. I’ve noticed that the accuracy level of students who take the AIME for the first time drops significantly. It seems that they are so used to multiple choice questions that they rely on multiple choices as a confirmation that they are right. So when someone solves a problem, they compare their answer to the given choices and if the answer is on the list they assume that the answer must be correct. Their pattern is broken when there are no choices. So they arrive at an answer and since there is no way to check it against choices, they just submit it. Because of this lack of confirmation, checking their answer in other ways becomes more important.

Timing. At the AMC we have 3 minutes per problem. At the AIME — 12. That means the timing strategies need to be different. Indeed, the AMC is so fast-paced that it is reasonable to save time by not reading a problem twice. If you read it, you either solve it or skip it and go on. The student who is not trying to achieve a perfect score can decide in advance not to read those final, highly-difficult problems.
For the AIME it is not expensive, in relative terms of time, to read all the problems. The student can read the problems and choose the most promising ones to start with, knowing that if there is time they can always come back to other problems.

Guessing. Guessing at the AMC is very profitable if you can exclude three choices out of the given five. Guessing for the AIME is a waste of time because the answers are integers between 000 and 999. So the probability of a random guess is one in a thousand. Actually, this is not quite right, because the problem writers are human and it is much easier to write a problem with an answer of 10 than one with an answer of 731. But the AIME designers are trying very hard to make answers that are randomly distributed. So the probability of a random guess is not one in a thousand, but it is very close. You can improve your chances by an intelligent guess. For example, you might notice that the answer must be divisible by 10. But guessing is still a waste of time. Thinking about a problem for two minutes in order to increase the probability of a correct guess to one in a 100 means that your expected gain is 1/200 points per minute. Which is usually much less than the gain for checking your answers. You can play the guessing game if you have exhausted your other options.

What saddens me is that the students who are not trained in checking use their first guess to make their life choices. But this is a subject for a separate discussion.

I have already written about how American math competition are illogically structured, for the early rounds do not prepare students for the later rounds. The first time mathletes encounter proofs is in the third level, USAMO. How can they prepare for problems with proofs? My suggestion is to look East. All rounds of Russian math Olympiads — from the local to the regional to the national — are structured in the same way: they have a few problems that require proofs. This is similar to the USAMO. At the national All-Russian Olympiad, the difficulty level is the same as USAMO, while the regionals are easier. That makes the problems from the regionals an excellent way to practice for the USAMO. The best regional Olympiad in Russia is the Moscow Olympiad. Here is the problem from the 1995 Moscow Olympiad:

We start with four identical right triangles. In one move we can cut one of the triangles along the altitude perpendicular to the hypotenuse into two triangles. Prove that, after any number of moves, there are two identical triangles among the whole lot.

This style of problems is very different from those you find in the AMC and the AIME. The answer is not a number; rather, the problem requires proofs and inventiveness, and guessing cannot help.

Here is another problem from the 2002 Olympiad. In this particular case, the problem cannot be adapted for multiple choice:

The tangents of a triangle’s angles are positive integers. What are possible values for these tangents?

The problems are taken from two books: Moscow Mathematical Olympiads, 1993-1999, and Moscow Mathematical Olympiads, 2000-2005. I love these books and the problems they present from past Moscow Olympiads. The solutions are nicely written and the books often contain alternative solutions, extended discussion, and interesting remarks. In addition, some problems are indexed by topics, which is very useful for teachers like me. But the best thing about these books are the problems themselves. Look at the following gem from 2004, which can be used as a magic trick or an idea for a research paper:

A deck of 36 playing cards (four suits of nine cards each) lies in front of a psychic with their faces down. The psychic names the suit of the upper card; after that the card is turned over and shown to him. Then the psychic names the suit of the next card, and so on. The psychic’s goal is to guess the suit correctly as many times as possible.
The backs of the cards are asymmetric, so each card can be placed in the deck in two ways, and the psychic can see which way the top card is oriented. The psychic’s assistant knows the order of the cards in the deck; he is not allowed to change the order, but he may orient any card in either of the two ways.
Is it possible for the psychic to make arrangements with his assistant in advance, before the latter learns the order of the cards, so as to ensure that the suits of at least (a) 19 cards, (b) 23 cards will be guessed correctly?
If you devise a guessing strategy for another number of cards greater than 19, explain that too.

Do you remember how to divide three apples among four people? Make apple sauce, of course. In the following two puzzles you are not allowed to cut apples. Here is an old riddle:

There are four apples in a basket. How can you divide them among four people, so that one apple remains in the basket?

Here is a variation from Konstantin Knop’s blog:

There are four apples in a basket. How can you divide them among three people, so that no one has more than the others and one apple remains in the basket?

Imagine a magician who comes on stage and performs the following magic trick:

He asks someone in the audience to think of a two-digit number, then subtract the sum of its digits. He waves his wand and guesses that the result is divisible by nine. Ta-Da!

This is not magic. This is a theorem. To make it magical we need to disguise the theorem.

First, there are many ways to hide the fact that we subtract the sum of the digits. For example, we can ask to subtract the digits one by one, while chatting in between. It is better to start with subtracting the first digit. Indeed, if we start with subtracting the second digit, the audience might notice that the result is divisible by 10 and start suspecting that some math is involved here. You can be more elaborate in how you achieve the subtraction of the sum of digits. For example, subtract twice the first digit, then the second, then add back the original number divided by 10.

Second, we need to disguise that the result is divisible by 9. A nice way to do this is implemented in the online version of this trick. The website matches the resulting number to a gift that is described on the page in pale letters. Paleness of letters is important as it is difficult to see that the same gift reappears in a pattern. In my work with students I use the picture on the left. At the end I tell them, “Ta-Da! the resulting number is blue.” Here is the full sized version of the same picture that you can download.

My students are too smart. They see through me and guess what is going on. Then I ask them the real question, “Why do I have some cells with question marks and other symbols?” To give you a hint, I can tell you that the symbols are there for the same reason some blue numbers are not divisible by 9.

by Andrew Buchanan, Tanya Khovanova, and Alex Ryba

We got this problem from Rados Radoicic:

A 7 by 7 board is covered with 38 dominoes such that each covers exactly 2 squares of the board. Prove that it is possible to remove one domino so that the remaining 37 still cover the board.

Let us call a domino covering of an n by n board saturated if the removal of any domino leaves an uncovered square. Let d(n) be the number of dominoes in the largest saturated covering of an n by n board. Rados’ problem asks us to prove that d(7) < 38.

Let’s begin with smaller boards. First we prove that d(2) = 2. Suppose that 3 dominoes are placed on a 2 × 2 board. Let us rotate the board so that at least two of the dominoes are horizontal. If they coincide, then we can remove one of them. If not, they completely cover the board and we can remove the third one. Similarly, you can check all the cases and show that d(3) = 6.

Now consider a saturated domino covering of an n × n board. We can view the dominoes as vertices of a graph, joining two if they share a cell of the board. No domino can share both cells with other dominoes, or we could remove it. Hence, each domino contains at most one shared cell. This means that all the dominoes in a connected component of the graph must overlap on a single shared cell. Hence, the only possible connected components must have the following shapes:

The largest shape in the picture is the X-pentomino. We can describe the other shapes as fragments of an X-pentomino, where the center and at least one more cell is intact. We call these shapes fragments.

A saturated covering by D dominoes corresponds to a decomposition of the n × n board into F fragments. Note that a fragment with k cells is made from k − 1 dominoes. Summing over the dominoes gives: D = n² − F. Thus, in order to make D as large as possible, we should make F as small as possible. Let f(n) be the minimal number of fragments that are required to cover an n by n board without overlap. Then d(n) = n² − f(n).

Consider the line graph of the n by n board. The vertices of the line graph correspond to cells in the original board and the edges connect vertices corresponding to neighboring cells. Notice that in the line graph our fragments become all star graphs formed by spokes coming out from a single central node. Thus a decomposition of a rectangular board into fragments corresponds to a covering of its line graph by star graphs. Consider an independent set in the line graph. The smallest independent set has the same number of elements as the smallest number of stars that can cover the graph. This number is called a domination number.

Now let’s present a theorem connecting domino coverings with X-pentomino coverings.

Theorem. f(n) equals the smallest number of X-pentominoes that can cover an n by n board allowing overlaps and tiles that poke outside, which is the same as the domination number of the corresponding line graph.

The proof of this theorem and the solution to the original puzzle is available in our paper: “Saturated Domino Coverings.” The paper also contains other theorems and discussions of other boards, not to mention a lot of pictures.

The practical applications of star graph coverings are well-known and widely discussed. We predict a similar future for saturated domino coverings and its practical applications, two examples of which follow:

First, imagine a party host arranging a plate of cookies. The cookies must cover the whole plate, but to prevent the kids sneaking a bite before the party, the cookies need to be placed so that removal of just one cookie is bound to expose a chink of plate. This means the cookies must form a saturated covering of the plate. Of course the generous host will want to use a maximal saturated covering.

For the second application, beam yourself to an art museum to consider the guards. Each guard sits on a chair in a doorway, from where it is possible to watch a pair of adjacent rooms. All rooms have to be observed. It would be a mistake to have a redundant guard, that is, one who can be removed without compromising any room. Such a guard might feel demotivated and then who knows what might happen. This means that a placement of guards must be a saturated domino covering of the museum. To keep the guards’ Union happy, we need to use a maximal saturated covering.

We would welcome your own ideas for applications of saturated coverings.