Tanya Khovanova's Math Blog

Here is a palindrome problem by Nazar Agakhanov from the All-Russian Mathematical Olympiad, 1996:

Can the number obtained by writing the numbers from 1 to n, one after another, be a palindrome?

Of course, we should assume that n > 1.

When I give this problem to my friends, they immediately answer, “No, of course not.” The reasoning is simple. The last 9 digits of this palindrome should be 987654321 — all different digits. The earliest you can get these digits at the end of your string 1 to n, is when your number n is actually 987654321. By that time your string of digits is really huge. If we take a random number with 2k digits, then the probability of it being a palindrome is 10^(-k). There is no reason to think that writing consecutive integers increases the probability of finding a palindrome. So the probability of hitting a palindrome is so low, that you can safely answer, “No, of course not.”

After confidently saying “no”, my friends usually stop thinking about the problem altogether. Only my friend David Bernstein suggested a simple solution for this problem. You can try to find the proof, but I do not insist that you do. I am about to give you many other fun problems to solve, and you can choose which ones you want to think about.

Naturally, we can replace the sequence of natural numbers in the Agakhanov’s problem with any sequence. So, one problem becomes 160,000 different problems when you plug all current sequences from the Online Encyclopedia of Integer Sequences into it.

For some periodic sequences every concatenation you create is a palindrome. For example, for the sequence of all zeroes, every set of the first n terms is a palindrome. More interestingly, you can think of an increasing sequence such that any first n terms comprise a palindrome, as we see in the sequence of repunits: 1, 11, 111, 1111, etc. Can you think of other sequences like this?

For some sequences, not every concatenation you create is a palindrome, but you can obtain an infinite number of palindromes. One example, suggested by Sergei Bernstein, is the sequence a(n) of the last digit of the greatest power of 2 that divides n. Can you think of other sequences like that?

For some sequences only the initial term itself is a palindrome. Beyond that you can’t obtain a palindrome by concatenating the first n terms. For a few of those sequences, this fact is easy to prove. Take, for example, the sequence of powers of ten, or the sequence of squares, or the sequence of prime numbers. Can you think of other similar sequences?

There are many sequences that do not produce a palindrome beyond the first term, even if you try many times. I suspect that they do not, in fact, ever produce a palindrome, but I have no clue how to prove that. I have in mind the sequence of the digits of π. Can you suggest other sequences like that?

Instead of solving the initial problem, I gave you a variety of other problems. My last challenge for you is to find other sequences that can replace the sequence of natural numbers in the Agakhanov’s problem so that the problem becomes solvable and the solution is interesting.

In 2007 Alexander Shapovalov suggested a very interesting coin problem. Here is the kindergarten version:

You present 100 identical coins to a judge, who knows that among them are either one or two fake coins. All the real coins weigh the same and all the fake coins weigh the same, but the fake coins are lighter than the real ones.

You yourself know that there are exactly two fake coins and you know which ones they are. Can you use a balance scale to convince the judge that there are exactly two fake coins without revealing any information about any particular coin?

To solve this problem, divide the coins into two piles of 50 so that each pile contains exactly one fake coin. Put the piles in the different cups of the scale. The scale will balance, which means that you can’t have the total of exactly one fake coin. Moreover, this proves that each group contains exactly one fake coin. But for any particular coin, the judge won’t have a clue whether it is real or fake.

The puzzle is solved, and though you do not reveal any information about a particular coin, you still give out some information. I would like to introduce the notion of a revealing coefficient. The revealing coefficient is a portion of information you reveal, in addition to proving that there are exactly two fake coins. Before you weighed them all, any two coins out of 100 could have been the two fakes, so the number of equally probable possibilities was 100 choose 2, which is 5050 4950. After you’ve weighed them, it became clear that there was one fake in each pile, so the number of possibilities was reduced to 2500. The revealing coefficient shows the portion by which your possibilities have been reduced. In our case, it is (5050 − 2500)/5050 (4950-2500)/4950, slightly more less than one half.

Now that I’ve explained the kindergarten version, it’s time for you to try the elementary version. This problem is the same as above, except that this time you have 99 coins, instead of 100.

Hopefully you’ve finished that warm-up problem and we can move on to the original Shapovalov’s problem, which was designed for high schoolers.

A judge is presented with 100 coins that look the same, knowing that there are two or three fake coins among them. All the real coins weigh the same and all the fake coins weigh the same, but the fake coins are lighter than the real ones.

You yourself know that there are exactly three fake coins and you know which ones they are. Can you use a balance scale to convince the judge that there are exactly three fake coins, without revealing any information about any particular coin?

If you are lazy and do not want to solve this problem, but not too lazy to learn Russian, you can find several solutions to this problem in Russian in an essay by Konstantin Knop.

Your challenge is to solve the original Shapovalov puzzle, and for each solution to calculate the revealing coefficient. The best solution will be the one with the smallest revealing coefficient.

My son, Sergei Bernstein, got accepted to MIT through early action. Because the financial costs of studying at MIT worried me, I insisted that Sergei also apply to Princeton and Harvard, as I had heard they give generous financial packages. In the end, Sergei was rejected by Princeton and wait-listed and finally rejected by Harvard. Though many people have been rejected by Princeton and Harvard, not too many of them have won places on US teams for two different international competitions — one in mathematics and the other in linguistics. To be fair, Sergei was accepted by these teams after Princeton had already rejected him. Nonetheless, Sergei has an impressive mathematical resume:

• In 2005 he was the National MathCounts Written Test Champion.
• In 2005 he was the National MathCounts Master’s Round Champion.
• In 2007 and 2009 he was a USAMO winner.
• In 2008 he passed Math 55a at Harvard taught by Dennis Gaitsgory, which is considered to be the hardest freshman math course in the country. More than 30 students started it and less than 10 finished. Sergei was one of the finishers, and he was only a high school junior.
• In 2007, 2008 and 2009 he competed at a 12th grade level at the Math Kangaroo, while he actually was in 10th, 11th and 12th grade. He placed first all three times.
• In 2009 he was on the US team at the Romanian Masters in Mathematics competition, which might be a harder competition than the International Mathematical Olympiad. He got a silver medal and was second on the US team.
• In 2009 he placed 5th in the North American Computational Linguistics Olympiad, making it to the Alternate US Team for the International Linguistics Olympiad.

I am trying to analyze why he was rejected and here are my thoughts.

1. His application forms to Harvard and Princeton were different from MIT. Yes, MIT was his first choice and he wrote a customized essay for MIT. For other places he had a common essay. But as he was supposed to be flagged as a top math student, his essay should have been irrelevant, in my opinion.
2. Admissions offices made a mistake. I can imagine that admissions offices never heard of the Romanian Masters in Mathematics competition, because it is a relatively new competition and the USA only joined it in 2009 for the first time. On its own, though, it should have sounded impressive. Also, they might not have known about the Math 55 course at Harvard, as usually high-schoolers do not take it. But that still leaves many other achievements. Many people told me that admissions offices know what they are doing, so I assume that I can disregard this point.
3. Princeton and Harvard knew that he wanted to go to MIT and didn’t want to spoil their admission rate. I do not know if colleges communicate with each other and whether Princeton and Harvard knew that he was admitted early to MIT. Because he had sent them a common application essay, they may have been suspicious that they weren’t his first choice.
4. Harvard and Princeton didn’t want him. I always heard that Harvard and Princeton want to have well-rounded people, whereas MIT likes geeks. I consider Sergei quite well-rounded as he has many other interests and achievements beyond mathematics. Perhaps his other accomplishments aren’t sufficiently impressive, making him less round than I thought he was.
5. Harvard and Princeton are not interested in mathematicians. Many people say that they want future world leaders. I think it is beneficial for a world leader to have a degree in math, but that’s just my personal opinion. And of course, to support their Putnam teams, it is enough to have one exceptional math student a year.
6. Sergei couldn’t pay. Yes, we marked on the application that we need financial help. In the current financial crisis it could be that even though Harvard and Princeton do not have enough money to support students, they do not want to go back and denounce their highly publicized generosity.

Many people told me of surprising decisions by Ivy League schools this year. The surprises were in both directions: students admitted to Ivy League colleges who didn’t feel they had much of a chance and students not admitted that had every right to expect a positive outcome. I should mention that I personally know some very deserving kids who were admitted.

I wonder if there has been a change in the financial demographics of the students Harvard and Princeton have accepted this year. If so, this will be reflected in the data very soon. We will be able to see if the average SAT scores of students go down relative to the population and previous years.

I do not know why Sergei wasn’t accepted; perhaps I’m missing something significant. But if it was because of our finances, it would be ironic: Sergei wasn’t admitted to Princeton and Harvard for the same reason he applied there.

There is interesting data to show that MIT takes math students more seriously than Harvard and Princeton. By Michael Sipser’s suggestion I looked at the Putnam Competition results. Out of the top 74 scorers of 2007, 21 were from MIT, 9 from Harvard and 7 from Princeton. Keep in mind that the total freshman enrollment at MIT is much lower than at Harvard or Princeton. This story repeated itself in 2008: out of top 79 scorers 23 were from MIT, 11 from Harvard and 11 from Princeton.

Ironically, MIT’s team didn’t win Putnam in those years. MIT’s team won the third place after Harvard and Princeton. If you look at the results more closely, you will notice that had MIT arranged teams differently, MIT would have won.

It appears that MIT put their three top scorers from the previous year on their lead team. MIT shouldn’t assume that those three continue to be their strongest competitors. Instead they should probably test their students right before the Putnam competition, because if you look at MIT’s top individual performers, had they been on a team together, they would have won.

Maybe MIT should rethink its algorithm for creating teams, or maybe we should just wait. As it is obvious that MIT is more serious about math, all top math students may want to go to MIT in coming years. If this happens, the mathematics field will be absolutely dominated by MIT.

A long ago my son Sergei went to the Streamline School Olympiad. Some of the problems were really nice and I asked the organizer, Oleg Kryzhanovsky, where he took the problems from. It seems that he himself supplied all the problems, many of which are his original creations. He told me that he can invent a math Olympiad problem on demand for any level of difficulty on any math topic. No wonder that he is the author of almost all math problems at the Ukraine Olympiad.

The following is a sample of his problems from the Streamline Olympiad. For my own convenience I have chosen problems without figures and equations. Note: I edited some of them.

1998 (8th – 9th grade). Find three numbers such that each of them is a square of the difference of the two others.

1999 (9th – 10th grade). The positive integers 30, 72, and N have a property that the product of any two of them is divisible by the third. What is the smallest possible value of N?

1999 (9th – 10th grade). You have 6 coins weighing 1, 2, 3, 4, 5 and 6 grams that look the same. The number (1, 2, 3, 4, 5, 6) on the top of each coin should correspond to its weight. How can you determine whether all the numbers are correct, using the balance scale only twice?

1999 (11th – 12th grade). In how many ways can the numbers 1, 2, 3, 4, 5, 6 be ordered such that no two consecutive terms have a sum that is divisible by 2 or 3.

2000 (6th – 7th grade). Let A be the least integer such that the sum of all its digits is equal to 2000. Find the left-most digit of A.

2000 (8th grade). You have six bags with coins that look the same. Each bag has an infinite number of coins and all coins in the same bag weigh the same amount. Coins in different bags weigh 1, 2, 3, 4, 5 and 6 grams exactly. There is a label (1, 2, 3, 4, 5, 6) attached to each bag that is supposed to correspond to the weight of the coins in that bag. You have only a balance scale. What is the least number of times do you need to weigh coins in order to confirm that the labels are correct?

Until the introduction of the Abel prize, the Fields medal was the most prestigious prize in mathematics. The medal has been awarded 48 times and all of the recipients have been men. Can we conclude that women are inferior to men when it comes to very advanced mathematics? I do not think so.

The Fields medal was designed for men; it is very female-unfriendly. It is the prize for outstanding achievement made by people under age 40. Most people start their research after graduate school, meaning that people have 10-15 years to reach this outstanding achievement. If a woman wants to have children and devote some time to them, she needs to do it before she is 40. That puts her at a big disadvantage for winning the medal.

Recently the Abel prize for mathematics was introduced. This is the math equivalent of a Nobel prize and nine people have received the prize, all of them male. The Wolf prize is another famous award: 48 people have received it so far and they too have all been male.

On the grand scale of things, women have only recently had the option of having a career in mathematics. Not so long ago it was considered quite exceptional for a woman to work in mathematics. The number of female mathematicians is increasing, but as this is a new trend, they are younger people. At the same time, Abel prizes and Wolf prizes are given to highly accomplished and not-so-young people. That means the increase in the percentage of women PhDs in mathematics might affect the percentage of females getting the prize, but with a delay of several dozen years.

There are other data covering extreme math ability. I refer to the International Math Olympiad. The ability that is needed to succeed in the IMO is very different from the ability required to succeed in math research. But still they are quite similar. The IMO data is more interesting in the sense that the girls who participate are usually not yet distracted by motherhood. So in some sense, the IMO data better represents potential in women’s math ability than medals and prizes.

Each important math medal or prize is given to one person a year on average. So the IMO champion would be the equivalent of the Fields medal or the Wolf prize winner. While no girl was the clear best in any particular year, there were several years when girls tied for the best IMO score with several other kids. For example:

• In 1995, 14 students tied for the perfect score; two of them were girls. (Maryam Mirzakhani and Chenchang Zhu)
• In 1994, 22 students tied for the perfect score; two of them were girls. (Theresia Eisenkölbl and Catriona Maclean)
• In 1991, 9 students tied for the perfect score; one of them was a girl. (Evgenia Malinnikova)
• In 1990, 4 students tied for the perfect score; one of them was a girl. (Evgenia Malinnikova)
• In 1987, 22 students tied for the perfect score; one of them was a girl. (Jun Teng)
• In 1984, 8 students tied for the perfect score; one of them was a girl. (Karin Gröger)

In one of those years, a girl might have been the best, but because the problems were too easy, she didn’t have a chance to prove it. Evgenia Malinnikova was an outstanding contender who twice had a perfect score. In 1990, she was one out of four people, and she was younger than two of them, as evidenced by the fact they they were not present in 1991. Only one other person — Vincent Lafforgue — got a perfect score in 1990 and 1991. We can safely conclude that Evgenia was one of two best people in 1990, because she was not yet a high school senior.

This might be a good place to boast about my own ranking as IMO Number Two, but frankly, older rankings are not as good as modern ones. Fewer countries were participating 30 years ago, and China, currently the best team, was not yet competing.

Girls came so close to winning the IMO that there is no doubt in my mind that very soon we will see a girl champion. The Fields medal is likely to take more time.

Today I have my first invited guest blogger, J.B. He is a 2006, 2007, 2008 and 2009 USAMO qualifier. He was also selected to be on the US team at the Romanian Masters in Mathematics competition. Also, he placed 6th at the North American Computational Linguistics Olympiad. Here is his piece:

The analysis is based on the list of 2009 USAMO qualifiers.

There is a rule that if nobody naturally qualifies for the USAMO from a state, then the highest scoring individual will qualify. Unfortunately, this means that we must remove those states with only one USAMO qualifier. We have 33 states remaining. If we sort these strictly by number of USAMO qualifiers, then we find the following result.

States with at least 4 USAMO qualifiers (24 total) voted for Obama, with the following exceptions: Georgia, Texas, South Carolina, and Missouri. In addition, of the two states with 3 USAMO qualifiers, one voted for Obama and one for McCain. The remaining states with 2 qualifiers (5 total) voted Republican.

Now this is not really unexpected. States with very large populations tend to be democratic and also produce more USAMO qualifiers. The most notable exceptions are Georgia and Texas, both of which were indeed exceptions (major outliers, in fact) above. This prompts the following consideration.

States with at least 8 USAMO qualifiers per 10 million residents (25 total) voted for Obama, with the following exceptions: Florida, Wisconsin, South Carolina, Missouri, and Georgia. Of these, all but Georgia fall within 50% of the target 8 USAMO qualifiers per 10 million residents. Georgia has 18 qualifiers per 10 million residents. Note also that the entire USA has 16 qualifiers per 10 million residents.

Furthermore, if USAMO qualifiers had been used instead of population for determining electoral votes, Obama would have won with 86% of the vote rather than 68%. In general, if the Democrat can secure all those states with at least 1 qualifier per million residents (plus DC), he will win with 303 votes. He can even lose the three red states in that category (Georgia, Missouri, and South Carolina) for exactly 269.

USAMO qualifiers per 10 million residents (for states with more than one qualifier) are:

• NH — 122 (all of their qualifiers are from Phillips Exeter Academy)
• MA — 55
• ME — 30
• CA — 29
• NJ — 29
• CT — 29
• VA — 28
• MD — 25
• WA — 21
• IN — 19
• OR — 19
• NY — 18
• GA — 18
• IA — 17
• NM — 15
• MI — 15
• PA — 14
• MO — 14
• SC — 13
• IL — 13
• NC — 13
• OH — 9
• CO — 8
• MN — 8
• UT — 7
• KS — 7
• WI — 7
• KY — 7
• TX — 7
• FL — 6
• LA — 5
• AL — 4
• TN — 3

The states with only one USAMO qualifier are WY, VT, ND, AK, SD, DE, MT, RI, HI, ID, NE, WV, NV, AR, MS, OK, and AZ. The only blue one of these which falls below 8 qualifiers per 10 million is Nevada (we would expect it to have at least 2 qualifiers to fit the expected pattern). Otherwise, it is at least possible that each state fits the pattern of 8 qualifiers per 10 million residents if and only if it votes Democratic.

Testing in the US is dominated by multiple-choice questions. Together with the time limit, this encourages students to stop thinking and go for guessing. I recently wrote an essay AMC, AIME, USAMO Contradiction, in which I complained about the lack of proofs in the first two rounds of math competitions.

Is there a way to improve the situation? I grew up in the USSR, where each round of the math competition had the same format: you were given several hours to write proofs for three or four difficult problems. There are two concerns with organizing a competition in this way. First, the Russian system is much more expensive, whereas the US’s multiple choice tests can be inexpensively checked by a computer. Second, the Russian system is prone to unfairness. You need many math teachers to check all these papers on the highest level. Some of these teachers might not be fully qualified, and it is difficult to ensure uniform checking. This system can’t easily be adopted in the US. I am surprised I haven’t heard of lawsuits challenging USAMO results, but if we were to start having proofs at the AMC level with several hundred thousand participants, we would get into lots of trouble.

An interesting compromise was introduced at the Streamline Olympiad. The problems were multiple choice, but students were also requested to write proofs. Students got two points for a correct multiple choice answer, and if the choice was correct the proof was checked. Students could get up to three points for a correct proof. This idea solves two issues. The writing of proofs is rewarded at an early stage and the work of the judges is not as overwhelming as it would have been, had they needed to check every proof. However, there is one problem that I discussed in previous posts that this method doesn’t solve: with multiple choice, minor mistakes cost you the whole problem, even though you might have been very close to a solution. If we want to reward thinking more than accuracy, the proof system allows us to give credit for partial solutions.

I can suggest another approach. If the Russians require proofs for all problems and the Americans don’t require proofs for any problem, why not compromise and require a proof for one problem out of the set.

But I actually have a bigger idea in mind. I think that current development in artificial intelligence may soon help us to check the proofs with the aid of a computer. Artificial intelligence is still far from ready to validate that a mathematical text a human has produced constitutes a proof. But in this particular case, we have two things working for us. First, we can use humans and computers together. Second, we do not need to check the validity of any random proof; we need to check the validity of a specific proof of a simple problem that we know in advance, thus allowing us to prepare the computers.

Let us assume that we already can convert student handwriting into computer-legible text or that students write directly in LaTeX.

Here is the plan. Suppose for every problem, we create a database of some sample right, wrong and partial solutions with corresponding scores. The computer checks the students’ solutions against the given sample. Hopefully, the computer can recognize small typos and deviations that shouldn’t change the point value. If the computer encounters a solution that is significantly different from the ones in the sample, it sends the solution to human judges. Humans decide how to score the solution and the solution and its score is added to the sample database.

For this system to work, computers should be smart enough not to send too many solutions to humans. So how many is too many? My estimate is based on the idea that we wouldn’t want the budget of AMC to go too much higher than the USAMO budget. Since USAMO has 500 participants, judges check just a few hundred solutions to any particular problem. With several hundred thousand participants in AMC, the computer would have to be able to cluster all the solutions into not more than a few hundred groups. The judges only have to check one solution in each group.

As a bonus, we can create a system where for a given solution that is not in the database, the computer finds the closest solution and highlights the difference, thus simplifying the human’s job.

In order to improve math education, we need to add proofs when teaching math. My idea might also work for SATs and for other tests.

Now that there is more money available for education research, would anyone like to explore this?

To get to the national swimming championship, you need to win the state running championships.

What? Is that a joke? Perhaps you’re having the same reaction. Because this is exactly what is happening with math competitions. The official USA math competition has three rounds: AMC, AIME and USAMO.

AMC is a multiple-choice competition with 25 problems in 75 minutes. To be good at it, you need speed, accuracy and the ability to guess well.

AIME is 3 hours long and has 15 problems. The problems are a different level of difficulty and guessing will not help you. Though AIME is also multiple-choice, unlike AMC where you choose out of 5, in AIME you choose out of 1,000. But you still need speed and accuracy. A small arithmetic mistake will cost you the whole problem.

USAMO is a competition that runs for 9 hours and has 6 problems. The problems are much harder and you have to write proofs. Proofs? What proofs? Where are the proofs coming from? It is like you got to the national swimming championship because you are a great runner, but you do not know how to swim.

As the result of this system of selection, the USA team at the International Math Olympiad has diverse skills: these kids are good at all three types of the math competitions. It is like taking an Olympic triathlon team to the Olympic swimming event.

However, the US loses by selecting in this way. There are many kids who are great mathematicians: they may be good at difficult problems but not that good at speed-racing problems. An arithmetic mistake costs you only one point at IMO, but a whole problem at AIME. There are kids who are deep mathematicians prone to small arithmetic mistakes. They could get a gold medal at IMO, but they can’t pass AMC or AIME.

Not only that. As many AMC problems are boring and do not require ideas, AMC might discourage kids from all math competitions at an early stage.

I will write later with my ideas about how to change AMC. Right now I would like to offer a solution to a smaller problem. I am sure that the US math team organizers know many cases in which a non-senior kid does great at USAMO and is potentially a team member for the next year’s US IMO team, but, oops, next year he can’t pass AMC.

I suggest the following: USAMO participants are allowed to go to next year’s AIME no matter what their AMC scores are. USAMO winners are allowed to go to the next year’s USAMO no matter what their AIME results are. This way kids who have proved that they are great swimmers do not need to compete in running again.

I have forgotten all the problems from the math Olympiads I participated in, except for one. That problem cost me one point, and as result I didn’t get a perfect score. Here is Problem Number 4 from the 1976 IMO:

Determine the greatest number that is the product of some positive integers, and the sum of these integers is 1976.

The solution goes like this. Consider a partition of 1976. If there is an integer x in the partition greater than 5, then replacing it with two integers x-2 and 2 gives a bigger product. Replacing 4 in the partition by 2 and 2 doesn’t change the product. On the other hand, if there is 1 in the partition, it is profitable to attach it to any other number: to replace 1 and x by x+1.

That means the maximum-product partition of a number greater than one has to consist only of twos and threes. If we have three twos in the partition, we can replace them by two threes, thus increasing the product. That means that our partition should consist of twos and threes, and the number of twos shouldn’t be greater than three.

I lost my point when I made an elementary arithmetic mistake while calculating the remainder of 1976 modulo 3.

Let’s generalize this problem for any number n. If we define the maximum product of partitions a(n) as a function of the number n, we get the sequence A000792 in the Encyclopedia of Integer Sequences. This sequence has other interesting definitions, which appear if we add some structure to partitions of n.

For example, we can suppose that our n is not just a number, but it also represents n objects that are being permuted by the symmetric group S_n. In this case, we can associate some Abelian subgroups of the symmetric group with every partition. Namely, we can divide the set of n elements into disjoint subsets of sizes corresponding to our partition and cycle the elements in every subset. The product of these cycles is an Abelian subgroup, the order of which doesn’t depend on how exactly we partition or how we cycle. This order is the product of the partition numbers. It appears that we can get all maximal Abelian subgroups of the symmetric group in this manner. Thus the sequence a(n) is the maximum size of the maximal Abelian subgroup in a symmetric group S_n.

We can do something else with our n objects. Suppose they are represented by vertices of a graph. We take any partition of n and divide our graph into groups of vertices according to this partition. Next we connect vertices of the graph that belong to different groups. If we take a vertex from every group, then the corresponding subgraph is a clique. We can see that it is a maximal clique as two vertices from the same group are never connected and we can’t add any more vertices. The number of different cliques of this size is the product of the number of elements in every group. We can prove that a(n) is the maximum possible number of maximal cliques in a graph with n vertices.

Even though I can’t return to 1976 to correct my stupid mistake, I love this problem and the corresponding sequence.