Tanya Khovanova's Math Blog

The data from annual surveys carried out by the American Mathematical Society shows the same picture year after year: the percentage of females in different categories decreases as the category level rises. For example, here is the data for 2006:

The high percentage of female math majors means that a lot of women do like mathematics. Why aren’t women becoming professors of mathematics? In the picture to the left, little Sanya fearlessly took her first integral. I hope, even as an adult, she will never be afraid of integrals.

I am one of the organizers of the Women and Mathematics Program at the Institute for Advanced Study at Princeton In 2009 we had a special seminar devoted to discussing this issue. Here is the report of our discussion based on the notes that Rajaa Al Talli took during the meeting.

Many of us felt, for the following three reasons, that the data doesn’t represent the full picture.

First, the different stages correspond to women of different ages; thus, the number of tenured faculty should be compared, not to the number of current math majors, but rather to women who majored in math many years ago. The percentage of female PhDs in mathematics has been increasing steadily for the past several years. As a result, we expect an eventual increase in the number of full-time female faculty.

Second, international women mathematicians might be having a great impact on the numbers. Let’s examine a hypothetical situation. If many female professors come to the US after completing their studies in other countries, it would be logical to assume that they would raise the numbers. But since the numbers are falling, we might be losing more females than we think. Or, it could be the opposite: international graduate students complete a PhD in mathematics in the USA and then go back to their own countries. In this case we would be losing fewer females to professorships than the numbers seem to suggest. Unfortunately, we can’t really say which case is true as we do not know the data on international students and professors.

Third, many women who major in mathematics also have second majors. For example, the women who have a second major in education probably plan to become teachers instead of pursuing an academic career. It would be interesting to find the data comparing women who never meant to have careers in science with those women who left because they were discouraged. If we are losing women from the sciences because they decide not to pursue scientific careers, then at least that is their choice.

It is also worth studying why so few women are interested in careers in mathematics in the first place. Changing our culture or applying peer pressure in a different direction might change the ambitions of a lot of people.

We discussed why the data in the table doesn’t represent the full picture. On the other hand, there are many reasons why women who can do mathematics and want to do mathematics might be discouraged from pursuing an academic career:

• Marriage and children distract from mathematics.
• The lack of legal protections for pregnant women, of required maternity leave and of childcare provision.
• The cultural skepticism that women can do math on a high level.
• An educational system that tends to tell students that math is very difficult, thus discouraging women from the early stages of their academic life.
• Boys tend to be more competitive than girls.
• The lack of job opportunities.
• A career in math often requires moving.

Our group proposed many solutions to help retain women in mathematics:

1. Find a way to get men pregnant as well.
2. Incorporate ideas from other countries (like Portugal), where they don’t have this problem.
3. Increase the level of social care for pregnant women and young children.
4. Create new laws to protect the rights of pregnant women.
5. Educate secondary, high school and college math teachers how to present math — such as through games — as an interesting subject, not as a difficult one.

At the end of our meeting, everyone accepted Ingrid Daubechies‘ proposal that we do the following:

Each woman in mathematics should take as her responsibility the improvement of the mathematical environment in which she works. If every woman helps change what’s going on in her university or the school where she teaches, that will help solve the problem on the larger scale.

The book How to Drive Your Man Wild in Bed by Graham Masterton has a chapter on how to choose a lover. It highlights red flags for men who need to be approached with caution. There is a whole list of potentially bad signs, including neglecting to shower in the previous week and talking only about himself.

The list of bad features also includes professions to avoid. Can you guess the first profession on the list? OK, I think you should be able to meta-guess given the fact that I am writing about it. Indeed, the list on page 64 starts:

Avoid, on the whole, mathematicians…

I am an expert on NOT avoiding mathematicians: in fact, I’ve married three of them and dated x number of them. That isn’t necessarily because I like mathematicians so much; I just do not meet anyone else.

When I was a student I had a theory that mathematicians are different from physicists. My theory was based on two conferences on mathematical physics I attended in a row. The first one was targeted for mathematicians and the second for physicists. The first one was very quiet, and the second one was all boozing and partying. So I decided that mathematicians are introverts and physicists are extroverts. I was sure then that my second husband chose a wrong field, because he liked booze and parties.

By now, years later, I’ve met many more mathematicians, and I have to tell you that they are varied. It is impossible and unfair to describe mathematicians as a type. One mathematician even became the star of an erotic movie. I write this essay for girls who are interested in dating mathematicians. I am not talking about math majors here, I am talking about mathematicians who do serious research. Do I have a word of advice?

I do have several words of caution. While they don’t apply to all mathematicians, it’s worth keeping them in mind.

First, there are many mathematicians who, like my first husband, are very devoted to mathematics. I admire that devotion, but it means that they plan to do mathematics on Saturday nights and prefer to spend vacation at their desks. If they can only fit in one music concert per year, it is not enough for me. Of course, this applies to anyone who is obsessed by his work.

Second, there are mathematicians who believe that they are very smart. Smarter than many other people. They expand their credibility in math to other fields. They start going into biology, politics and relationships with the charisma of an expert, when in fact they do not have a clue what they are talking about.

Third, there are mathematicians who enjoy their math world so much that they do not see much else around them. The jokes are made about this type of mathematician:

What is the difference between an extroverted mathematician and an introverted one? The extroverted one looks at your shoes, rather than at his own shoes.

Yes, I have met a lot of mathematicians like that. Do you think that their wives complain that their husbands do not notice their new haircuts? No. Such triviality is not worth mentioning. Their wives complain that their husbands didn’t notice that the furniture was repossessed or that their old cat died and was replaced by a dog. My third husband was like that. At some point in my marriage I discovered that he didn’t know the color of my eyes. He didn’t know the color of his eyes either. He wasn’t color-blind: he was just indifferent. I asked him as a personal favor to learn the color of my eyes by heart and he did. My friend Irene even suggested creating a support group for the wives of such mathematicians.

While you need to watch out for those traits, there are also things I like about mathematicians. Many mathematicians are indeed very smart. That means it is interesting to talk to them. Also, I like when people are driven by something, for it shows a capacity for passion.

Mathematicians are often open and direct. Many mathematicians, like me, have trouble making false statements. I stopped playing —Mafia— because of that. I prefer people who say what they think and do not hold back.

There is a certain innocence among some mathematicians, and that reminds me of the words of the Mozart character in Pushkin’s poetic drama, Mozart and Salieri: —And genius and villainy are two things incompatible, aren’t they?— I feel this relates to mathematicians as well. Many mathematicians are so busy understanding mathematics, they are not interested in plotting and playing games.

Would I ever date a mathematician again? Yes, I would.

33 horsemen are riding in the same direction along a circular road. Their speeds are constant and pairwise distinct. There is a single point on the road where the horsemen can pass one another. Can they ride in this fashion for an arbitrarily long time?

The puzzle appeared at the International Tournament of the Towns and at the Moscow Olympiad. Both competitions were held on the same day, which incidentally fell on Pi Day 2010. Just saying: at the Tournament the puzzle was for senior level competitors; at the Moscow Olympiad it was for 8th graders.

Warning: If you want to solve it yourself first, pause now, because here is the solution I propose.

First, consider two horsemen who meet at that single point. The faster horseman passes the slower one and gallops ahead and the slower one canters along. The next meeting point should be at the same place in the circle. Suppose the slower horseman rides n full circles before the next meeting, then the second horseman could not have passed the first in between, so he has to ride n+1 full circles. That means their speeds should have a ratio of (n+1)/n for an integer n. And vice versa, if their speeds have such a ratio, they will meet at the same location on the circle each time. That means that to solve the problem, we need to find 33 different speeds with such ratios.

As all speed ratios are rational numbers, we can scale speeds so that they are relatively prime integers. The condition that two integers have a ratio (n+1)/n is equivalent to the statement that two integers are divisible by their difference. So the equivalent request to the problem is to find a set of 33 positive integers (or prove non-existence), such that every two integers in the set are divisible by their difference.

Let’s look at examples with a small number of horsemen. For two riders we can use speeds 1 and 2. For three riders, speeds 2, 3 and 4.

Now the induction step. Suppose that we found positive integer speeds for k horsemen. We can add one more horseman with zero speed who quietly stays at the special point and everyone else passes him. The difference condition is satisfied. We just need to tweak the set of speeds so that the lazy horseman starts moving.

We can see that if we add the least common multiple to every integer in a set of integers such that every two numbers in a pair are divisible by their difference, then the condition stays satisfied. So by induction we can find 33 horsemen. Thus, the answer to the problem is: Yes they can.

Now I would like to apply that procedure from the solution to calculate what kind of speeds we get. If we start with one rider with the speed of 1, we add the second rider with speed 0, then we add 1 to both speeds, getting the solution for two riders: 1 and 2. Now that we have a solution for two riders, we add a third rider with speed 0 then add 2 to every speed, getting the solution for three horsemen: 2, 3 and 4. So the procedure gave us the solutions we already knew for two and three horsemen.

If we continue this, we’ll get speeds 12, 14, 15 and 16 for four riders. Similarly, 1680, 1692, 1694, 1695, and 1696 for five riders.

We get two interesting new sequences out of this. The sequence of the fastest rider’s speed for n horsemen is: 1, 2, 4, 16, 1696. And the sequence of the least common multiples for n−1 riders — which is the same as the lowest speed among n riders — is: 1, 1, 2, 12, 1680, 343319185440.

The solution above provides very large numbers. It is possible to find much smaller solutions. For example for four riders the speeds 6, 8, 9 and 12 will do. For five riders: 40, 45, 48, 50 and 60.

I wonder if my readers can help me calculate the minimal sequences of the fastest and slowest speeds. That is, to find examples where the integer speed for the fastest (slowest) horseman is the smallest possible.

One day we may all face the necessity of hiring a lawyer. If the case is tricky the lawyer must be smart and inventive. I am collecting puzzles to give to a potential lawyer during an interview. The following puzzle is one of them. It was given at the second Euler Olympiad in Russia:

At a local Toyota dealership, you are allowed to exchange brand new cars. You can exchange three Camrys for one Prius and one Avalon, and three Priuses for two Camrys and one Avalon. Assuming an unlimited supply of cars at the dealership, can collector Vasya exchange 700 Camrys for 400 Avalons?

The beauty of this puzzle is that the answer I may find acceptable from a mathematician is not the same as I want from my future lawyer.

Have I intrigued you? Get to work and send me your solutions.

I am sitting in front of my computer and scheming, or, more precisely, scamming. I am inventing scams as a way of raising awareness of how probability theory can be used for deception.

My first scam is my lottery project. Suppose I create and run a private lottery. I will award minor payments to some participants, while promising a grand prize of one hundred million dollars. However, there will be a very small probability that anyone will win the big payout. My plan is to live lavishly on my proceeds, hoping no one ever wins the big ticket.

The beauty of this scheme is that nobody will complain until someone scores the top prize. After all, everyone has been receiving what I promised, and no one realizes my fraud. If nobody wins the big award until I retire, I will have built my life style on deception without having been caught.

Suppose someone wins the hundred million dollars. Oops. I am in big trouble. On the other hand, maybe I can avoid jail time. I could tell the winner that the money is gone and if s/he complains to the police, I will declare bankruptcy and we will all lose. Alternatively, I can suggest a settlement in exchange for silence. For example, we could share future proceeds. Probability theory will help me run this lottery with only a small chance of being exposed.

But even a small chance of failure will cause me too much stress, so I have come up with an idea for another scam. I will write some complicated mathematical formulas with which to persuade everyone that global warming will necessarily produce earthquakes in Boston in the near future. Then I’ll open an insurance company and insure everyone against earthquakes. As I really do not expect earthquakes in my lifetime, I can spend the money. I’ll just need to keep everyone scared about earthquakes. This time I can be sure that I won’t be caught as no one will have a reason to complain. The only danger is that someone will check my formulas and prove that I used mathematics to lie.

Perhaps I need a scam that covers up the lie better. Instead of inventing an impossible catastrophe, I need to insure against a real but rare event. Think Katrina. I collect the money and put aside money for payouts and pocket the rest. But I actually tweak my formulas and put aside less than I should, boosting my bank account. I will be wealthy for many years, until this event happens. I might die rich but if this catastrophe happens while I’m still alive, I’ll declare bankruptcy.

Though I was lying to everyone, I might be able to avoid jail time. I might be able to prove that it was an honest mistake. Mathematical models include some subjective parameters; besides, everyone believes that nature is unpredictable. Who would ever know that I rigged my formulas in my favor? I can claim that the theory ended up being more optimistic than reality is. Who could punish me for optimism?

Maybe I can be accused of lying if someone proves that I knew that the optimistic model doesn’t quite match the reality. But it is very difficult for the courts to punish a person for a math mistake.

When I started writing this essay, I wanted to write about the financial crisis of 2008. I ended up inventing scams. In a way, I did write about the financial crisis. My scams are simplified versions of what banks and hedge funds did to us. Will we ever see someone punished?

Most movies related to mathematics irritate me because of simplifications. I especially do not like when a movie pretends to be intelligent and then dumbs it down. I recently watched the Spanish movie Fermat’s Room, which, as you may guess, annoyed me several times. In spite of that I enjoyed it very much.

The movie opens with people receiving invitations to attend a meeting for geniuses. To qualify for the meeting they need to solve a puzzle. Within ten days, they must guess the order underlying the following sequence: 5, 4, 2, 9, 8, 6, 7, 3, 1. Right away, at the start of the movie, I was already annoyed because of the simplicity of the question. You do not have to be a genius to figure out the order, not to mention how easy it would be to plug this sequence into the Online Encyclopedia of Integer Sequences to find the order in five minutes.

The participants were asked to hide their real names, which felt very strange to me. All famous puzzle solvers compete in puzzle championships and mystery hunts and consequently know each other.

The meeting presumably targets the brightest minds and promises to provide “the greatest enigma.” During the meeting they are given seven puzzles to solve. All of them are from children’s books and the so-called “greatest enigma” could easily be solved by kids. Though I have to admit that these were among the cutest puzzles I know. For example:

There are three boxes: one with mint sweets, the second with aniseed sweets, and the last with a mixture of the two. The boxes are labeled, but all the labels are wrong. What is the minimum number of sweets you need to taste to correctly re-label all the boxes?

Another of the film’s puzzles includes a light bulb in a room and three switches outside, where you have to correctly find the switch that corresponds to the bulb, but you can only enter the room once. In another puzzle you need to get out of prison by deciding which of two doors leads to freedom. You are allowed to ask exactly one question to one of the two guards, one of whom is a truth-teller and the other is a liar.

The other four puzzles are similar to these three I have just described. To mathematicians they are not the greatest enigmas. They are nice material for a children’s math club. For non-mathematicians, they may be fascinating. Certainly it’s a good thing that such tasteful puzzles are being promoted to a large audience. But they just look ridiculous as “the greatest enigmas.”

So what is it about this film that I so enjoyed?

The intensity of the movie comes from the fact that the people are trapped in a room that starts shrinking when they take more than one minute to solve a puzzle.

I well remember another shrinking room from Star Wars: A New Hope. When Princess Leia leads her rescuers to a room, it turns out to be a garbage compactor. The bad guys activate the compactor and two opposite walls start moving in. In contrast, Fermat’s room is shrinking in a much more sophisticated way: all four walls are closing in. Each of the walls in the rectangular room is being pressured by an industrial-strength press. The walls in the corners do not crumble, but rather one wall glides along another. I was more puzzled by this shrinking room than I was by the math puzzles. I recommend that you try to figure out how this can be done before seeing the movie or its poster.

However, the best puzzle in the movie is the plot itself. Though I knew all the individual puzzles, what happened in between grabbed me and I couldn’t wait to see what would happen next. I saw the movie twice. After the first time, I decided to write this review, so I needed to check it again. I enjoyed it the second time even better than the first time. The second time, I saw how nicely the plot twists were built.

Maybe I shouldn’t complain about the simplicity and the familiarity of the puzzles. If they were serious new puzzles I would have started solving them instead of enjoying the movie. The film’s weakness might be its strength.

I collect hats puzzles. A puzzle about hats that I hadn’t heard before appeared on the Konstantin Knop’s blog (in Russian):

The sultan decides to test his hundred wizards. Tomorrow at noon he will randomly put a red or a blue hat — for both of which he has an inexhaustible supply — on every wizard’s head. Each wizard will be able to see every hat but his own. The wizards will not be allowed to exchange any kind of information whatsoever. At the sultan’s signal, each wizard needs to write down the color of his own hat. Every wizard who guesses wrong will be executed. The wizards have one day to decide on a strategy to maximize the number of survivors. Suggest a strategy for them.

I’ll start the discussion with a rather simple idea: Each wizard writes down a color randomly. In this case the expected number of survivors is 50. Actually, if each wizard writes “red”, then the expected number of survivors is 50, too. Can you find a better strategy, with a greater expected number of survivors or prove that such a strategy doesn’t exist?

As a bonus question, can you suggest a strategy that guarantees 50 survivors?

Now that you’ve solved that issue, here’s my own variation of the problem.

The wizards are all very good friends with each other. They decide that executions are very sad events and they do not wish to witness their friends’ deaths. They would rather die themselves. They realize that they will only be happy if all of them survive together. Suggest a strategy that maximizes the probability of them being happy, that is, the probability that all of them will survive.

The following two problems appeared together in Martin Gardner’s Scientific American column in 1959.

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?

Many people, including me and Martin Gardner, wrote a lot about Mr. Smith. In his original column Martin Gardner argued that the answer to the first problem is 1/3. Later he wrote a column titled “Probability and Ambiguity,” where he corrected himself about Mr. Smith.

… the answer depends on the procedure by which the information “at least one is a boy” is obtained.

This time I would like to ignore Mr. Smith, as I wrote a whole paper about him that is now under consideration for publication at the College Mathematics Journal. I would rather get back to Mr. Jones.

Mr. Jones failed to stir a controversy from the start and was forgotten. Olivier Leguay asked me about Mr. Jones in a private email, reminding me that the answer to the problem about his children also depends on the procedure.

One of the reasons Mr. Jones was forgotten is that for many natural procedures the answer is 1/2. For example, the following procedures will produce an answer of 1/2:

• We ask Mr. Jones whether his older child is a daughter and he says “yes.”
• Mr. Jones flips a coin deciding which child to talk about. After that he has to tell us the gender and whether the child is the oldest.
• Mr. Jones is asked to say nothing if he doesn’t have a daughter, to select the daughter if has just one, or to pick one at random if he has two daughters. After that he has to tell us whether the daughter he has selected is the oldest.

There are many other procedures that lead to the answer 1/2. However, there are many procedures that lead to other answers.

Suppose I know Mr. Jones, and also know that he has two children. I meet Mr. Jones at a mall, and he tells me that he is buying a gift for his older daughter. Most probably I would assume that the other child is a daughter, too. In my experience, people who have a son and a daughter would say that they are buying a gift for “my daughter.” Only people with two daughters would bother to specify that they are buying a gift for “my older daughter.”

In some sense I didn’t forget about Mr. Jones. I wrote about him implicitly in my essay Two Coins Puzzle. His name was Carl and he had two coins instead of two children.

My blog is getting more famous. Now I don’t need to look around for nice problems, for my readers often send them to me. In response to my blog about him, Sergey Markelov’s Best, Markelov sent me more of his problems. Here is a cute tetrahedron problem that he designed:

Six segments are such that you can make a triangle out of any three of them. Is it true that you can build a tetrahedron out of all six of them?

Another reader, Alexander Shen, sent me a different tetrahedron problem from a competition after reading my post on Problem Design for Multiple Choice Questions:

Imagine the union of a pyramid based on a square whose faces are equilateral triangles and a regular tetrahedron that is glued to one of these faces. How many faces will this figure have?

Shen wrote that the right answer to this problem had been rumored to have a negative correlation with the result of the entire test.

86 is conjectured to be the largest power of 2 not containing a zero. This simply stated conjecture has proven itself to be proof-resistant. Let us see why.

What is the probability that the nth power of two will not have any zeroes? The first and the last digits are non-zeroes; suppose that other digits become zeroes randomly and independently of each other. This supposition allows us to estimate the probability of 2ⁿ not having zeroes as (9/10)^k-2, where k is the number of digits of 2ⁿ. The number of digits can be estimated as n log₁₀2. Thus, the probability is about cxⁿ, where c = (10/9)² ≈ 1.2 and x = (9/10)^log₁₀2 ≈ 0.97. The expected number of powers of 2 without zeroes starting from the power N is cx^N/(1-x) ≈ 40 ⋅ 0.97^N.

Let us look at A007377, the sequence of numbers such that their powers of 2 do not contain zeros: 1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 14, 15, 16, 18, 19, 24, 25, 27, 28, 31, 32, 33, 34, 35, 36, 37, 39, 49, 51, 67, 72, 76, 77, 81, 86. Our estimates predicts 32 members of this sequence starting from 6. In fact, the sequence has 30 conjectured members. Similarly, our estimate predicts 2.5 members starting from 86. It is easy to check that the sequence doesn’t contain any more numbers below 200 and our estimate predicts 0.07 members after 200. As we continue checking larger numbers and see that they do not belong to the sequence, the probability that the sequence contains more elements vanishes. With time we check more numbers and become more convinced that the conjecture is true. Currently, it has been checked up to the power 4.6 ⋅ 10⁷. The probability of finding something after that is about 1.764342396 ⋅10^-633620.

Let us try to approach the conjecture from another angle. Let us check the last K digits of powers of two. As the number of possibilities is finite, these last digits eventually will start cycling. If we can show that all the elements inside the period contain zeroes, then we need to check the finite number of powers of two until this period starts. If we can find such K, we can prove the conjecture.

Let us look at the last two digits of powers of two. The sequence starts as: 01, 02, 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 04. As we would anticipate, it starts cycling. The cycle length is 20, and 90% of numbers in the cycle don’t have zeroes.

Now let’s continue to the last three digits. The period length is 100, and 19 of them either start with zero or contain zero. The percentage of elements in the cycle that do not contain zero is 81%.

The cycle length for the last n digits is known. It is 4 ⋅ 5^n-1. In particular the cycle length grows by 5 every time. The number of zero-free elements in these cycles form a sequence A181610: 4, 18, 81, 364, 1638, 7371, 33170. If we continue with our supposition that the digits are random, and study the new digits that appear when we move from the cycle of the last n digits to the next cycle of the last n+1 digits, we can expect that 9/10 of those digits will be non-zero. Indeed, if we check the ratio of how many numbers do not contain zero in the next cycle compared to the previous cycle, we get: 4.5, 4.5, 4.49383, 4.5, 4.5, 4.50007. All of these numbers are quite close to our estimation of 4.5. If this trend continues the portion of the numbers in the cycle that don’t have zeroes tends to zero; however, the total of such numbers grows exponentially. We can even estimate that the expected growth is 4 ⋅ 4.5^n-1. From this estimation, we can derive the conjecture:

Conjecture. For any number N, there exists a power of two such that its last N digits are zero-free.

Indeed, the last N digits of powers of two cycle, and there are an increasing number of members inside that cycle that do not contain zeroes. The corresponding powers of two don’t have zeroes among N rightmost digits.

So, how do we combine the two results? First, the expected probability of finding the power of two larger than 86 that doesn’t contain zero is minuscule. And second, we most certainly can find a power of two that has as many zeroless digits at the end as we want.

To combine the two results, let us look at the sequences A031140 and A031141. We can deduce from them that for the power 103233492954 the first zero from the right occupies the 250th spot. The total number of digits of that power is 31076377936. So 250 is a tiny portion of the digits.

As time goes by we grow more and more convinced that 86 is the largest power of two without zeroes, but it is not at all clear how we can prove the conjecture or whether it can be proven at all.

My son, Sergei, suggested that I claim that I have a proof of this conjecture, but do not have enough space in the margin to fit my proof in. The probability that I will ever be shamed and disproven is lower than the probability of me winning a billion dollars in the lottery. Though then, if I do win the big bucks, I will still care about being shamed and disproven.