Tanya Khovanova's Math Blog

I promised to discuss how to improve the accuracy of your guessing at AMC 10/12, or other tests for that matter. There are two types of guessing. First, meta-guessing is when you do not look at the problem, but rather guess from just looking at the choices. Second, in the one I refer to as an educated-guess you do look at the problem. Instead of completely solving it, you try to deduce some information from the problem that will help you eliminate some of the choices. In this essay, I discuss both methods.

But first let me emphasize that solving problems is a more important skill than meta-guessing from a list of choices. Solving problems not only teaches you to think mathematically, but also increases your brain power. Spending time improving your meta-guessing skills can help you at multiple-choice tests and may give you insight into how test designers think, but this will not increase your math knowledge in the long run.

On the other hand, educated-guessing is a very useful skill to obtain. Not only can it improve your score during a test, the same methods can be applied to speed up the process of re-checking your answers before handing in the test. This skill will also come in handy when you start your research. Some problems in research are so difficult that even minor progress in estimating or describing your answer is beneficial.

Before discussing particular methods, let me remind you that AMC 10/12 is a multiple-choice competition with five choices for each question. The correct answer brings you 6 points. A wrong answer brings you 0 points, and not answering brings you 1.5 points. So if you randomly guess one of the five choices, your expected average score is 1.2 points, which is 0.3 less than your score for an unanswered question. Thus guessing is unprofitable on average.

However, if you can eliminate one choice, your expected average score becomes 1.5 points. In this case guessing doesn’t bring you points on average, but it does create some randomness in your results. For strategic reasons, you might prefer guessing, as I discussed in my earlier piece.

If you can eliminate two wrong choices, then guessing becomes profitable. A random guess out of three possibilities brings you 2 points, a better result than 1.5 points for an unanswered question. Even more, if you can eliminate three choices, then guessing will increase your score by 3 points on average.

Now that we’ve covered the benefits of excluding choices before guessing, I would like to discuss how to exclude choices by just looking at them. Let us take one of the problems from the 2002 AMC10B. Here are the choices: (-2,1), (-1,2), (1,-2), (2,-1), (4,4). The pair (4,4) is a clear outlier. I suggest that an outlier can’t be a correct choice. If (4,4) were the correct answer, then it would have been enough, instead of solving the problem, to use some intermediate arguments to choose it. For example, if you can argue that both numbers in the answer must be at least 2, or must be positive or be even, then you can get the correct answer without solving the problem. Any problem for which you can easily pick the correct answer without solving it is an unacceptably poor problem design. Thus, (4,4) can’t be the correct answer, and should be eliminated during guessing.

Let us look at a 2002 AMC10A problem with the following choices: 4/9, 2/3, 5/6, 3/2, 9/4. Test designers want to create choices that are plausible. They try to anticipate possible mistakes. In this set of choices, we can deduce that one of the mistakes that they anticipate is that students will confuse a number with its inverse. In this case 5/6 can’t be the correct answer. Otherwise, 6/5 would have been included as a choice. In another similar example from the 2000 AMC10 with choices -2, -1/2, 1/3, 1/2, 2, the designers probably hope that students will confuse numbers with their inverses and negations. Hence, we can exclude 1/3.

Sometimes the outlier might hint at the correct answer. Suppose you have the following list of choices: 2, 1/2π, π, 2π, 4π. The number 2 is an outlier here. Probably, the problem designers were contemplating that students might forget to multiply by π. In this case the likely correct answer would be 2π, because only from this answer can you get 2 by forgetting to multiply by π.

As an exercise, try to eliminate the wrong choices from the following set from a problem given at 2000 AMC10: 1/(2m+1), m, 1-m, 1/4m, 1/8m².

AMC designers know all of these guessing tricks, so they attempt to confuse the competitors from time to time by going against common sense. For example, in a 2002 AMC10A problem the choices were: -5, -10/3, -7/3, 5/3, 5. I would argue that -7/3 is a clear outlier because all the others are divisible by 5. Furthermore, there is no point in including so many answers with 3 in the denominator unless there is a 3 in the denominator of the correct answer. So I would suggest that one of -10/3 and 5/3 must be the answer. My choice would be 5/3, as there is a choice of 5 which I would assume is there for students who forgot to divide by 3. As I said the designers are smart and the actual answer is -10/3. They would have tricked me on this problem.

One of the best ways to design a multiple choice question is to have an arithmetic progression as a list of choices. There is no good way to eliminate some choices from 112, 113, 114, 115, 116. Unfortunately for people who want to get an advantage by guessing, many of the problems at AMC have an arithmetic progression as their choices.

Now that we’ve seen the methods of meta-guessing, let’s look at how to make an educated guess. Let us look at problem 1 in 2003 AMC10A.

What is the difference between the sum of the first 2003 even counting numbers and the sum of the first 2003 odd counting numbers?

Without calculations we know that the answer must be odd. Thus, we can immediately exclude three out of five choices from the given choices of 0, 1, 2, 2003, 4006. Parity consideration is a powerful tool in eliminating wrong answers. Almost always you can decide the parity of the answer much faster than you can calculate the answer. Similar to using parity, you can use divisibility by other numbers to have a fast elimination. Here is a problem from 2000 AMC10:

Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71, 76, 80, 82, and 91. What was the last score Mrs. Walter entered?

The first four numbers entered must be divisible by 4. The total of the given numbers is divisible by 4. Hence, the last number must also be divisible by 4. This reasoning eliminates three out of five choices.

Another powerful method is a rough estimate. Let us look at the next problem in 2003 AMC10A.

Members of the Rockham Soccer League buy socks and T-shirts. Socks cost $4 per pair and each T-shirt costs $5 more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is $2366, how many members are in the League?

If we notice that the cost per person is more than $25, we can conclude that there are less than a hundred members in the League. Given the choices of 77, 91, 143, 182, and 286, we immediately can eliminate three of them.

Another method is to use any partial knowledge that you may have. Consider this problem from 2003 AMC10A:

Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?

You might remember that there is a formula for this. Even if you do not remember the exact formula, you might still have a vague memory that the answer must be a binomial coefficient that somehow uses the number of cookies and the number of flavors. Looking at the choices — 22, 25, 27, 28, 29 — you can see that the only choice that appears in the first 10 rows of the Pascal’s triangle is 28. So you should go with 28.

It is easy to talk about easy problems; let us see what we can do about difficult ones. Consider the last problem on 2003 AMC10A:

Let n be a 5-digit number, and let q and r be the quotient and the remainder, respectively, when n is divided by 100. For how many values of n is q+r divisible by 11?

The choices are 8180, 8181, 8182, 9000, 9090. They can be naturally split into two groups: three choices below 9000 and the rest. By my rules of removing outliers the group of numbers below 9000 seems the more promising group. But I would like to discuss how to approximate the answer. There is no reason to believe that there is much correlation between remainders by 100 and divisibility by 11. There is a total of 90,000 5-digit numbers; among those numbers, approximately 90,000/11 = 8182 is divisible by 11, so we should go with the group of answers close to 8182.

Another way of thinking about this problem is the following. There are 900 different quotients by 100 to which we add numbers between 0 and 99. Thus for every quotient our sums are a set of 100 consecutive numbers. Out of 100 consecutive numbers usually 9, and rarely 10, are divisible by 11. Hence, the answer has to be less than 9000.

Sometimes methods you use for guessing can bring you the answer. Here is a problem from 2001 AMC12:

What is the product of all odd positive integers less than 10000? (A) 10000!/5000!², (B) 10000!/2⁵⁰⁰⁰, (C) 9999!/2⁵⁰⁰⁰, (D) 10000!/(2⁵⁰⁰⁰5000!), (E) 5000!/2⁵⁰⁰⁰.

For a rough estimate, I would take a prime number and see in what power it belongs to the answer. It’s simplest to consider a prime number p that is slightly below 5000. Then p should appear as a factor in the product of all odd positive integers below 10000 exactly once. Now let us look at the choices. Number p appears in 5000! once and in 10000! twice (as p and 2p). Hence, it appears in (A) zero times, and twice each in (B) and (C). We also can rule out (E) as the product of odd numbers below 10000 must be divisible by primes between 5000 and 10000, but 5000! doesn’t contain such primes. Thus the answer must be (D).

The method I just described won’t produce the formula. But the ideas in this method allow you to eliminate all the choices except the right one. Moreover, this method provides you with a sanity check after you derive the formula. It also helps to build your mathematical intuition.

I hope that you will find my essays about AMC useful. And good luck on February 9!

I teach students to solve math problems by appreciating the big picture, or by noticing the problem’s inner symmetries, or through a deep understanding of the problem. In the long run, one thing leads to another: such training structures their minds so that they are better at understanding mathematics and, as a consequence, they perform well at math competitions.

That is why, when AMC is still far away, I do not give my students a lot of AMC problems; rather, I pick problems that contain useful ideas. When I do give AMC problems, I remove the multiple choices, so they understand the problems completely, instead of looking for shortcuts. For example, this problem from AHSME 1999 is a useful problem with or without choices.

What is the largest number of acute angles that a hexagon can have?

As AMC approaches, we start discussing how to solve problems given multiple choices. Training students for AMC is noticeably different from teaching mathematics. For example, some problems are very specific to AMC. They might not even exist without choices. Consider this problem from the 2001 AMC12:

A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?

Here we really need the choices in order to pick one complex number, such that its real part is an integer or a half integer and, in addition, the product of the number with its conjugate produces an integer.

Sometimes the choices distract from solving the problem. For example, in the following problem from the 2005 AMC12, having choices might tempt students to try to eliminate them one by one:

The sum of four two-digit numbers is 221. None of the eight digits is 0 and no two of them are same. Which of the following is not included among the eight digits?

Without the choices, students might start considering divisibility by 9 right away.

On some occasions, the choices given for the problems at AMC make the problem more interesting. Here is an example from the 2000 AMC10:

Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

The choices are 21, 60, 119, 180 and 231. We can immediately see that the answer must be odd. Because the span of the three remaining choices is so wide, we suspect that we can eliminate the smallest and the largest. Trying for 5 and 7 — the two smallest primes in the range — we can eliminate 21. Similarly, checking the two largest primes in the range, we can eliminate 231. This leaves us with the answer: 119. If the choices were different, we might have lost the interplay between the solution and the list of choices. Then, solving the problem would have been slower and more boring. There are ten pairs of prime numbers to check. And we would need on average to check five of them until we stumbled on the correct choice.

In other cases having multiple choices makes the problem more boring and less educational. Here is another problem from the same competition.

Two non-zero real numbers, a and b, satisfy ab = a – b. Find a possible value of a/b + b/a – ab.

Solving this problem without choices can teach students some clever tricks that people use when playing with expressions. Indeed, when we collect a/b + b/a into one fraction (a² + b²)/ab, we might remember that a² + b² is very close to (a – b)², and see from here that a/b + b/a – 2 is (a – b)²/ab, which, given the initial condition, equals ab. Thus, we can get the answer: 2.

On the other hand, if you look at the multiple choices first: -2, -1/2, 1/3, 1/2, 2, you might correctly assume that the answer is a number. Thus, the fastest way to solve it is to find an example. If a = 1, then b must be 1/2, and the answer must be 2. This solution doesn’t teach us anything new or interesting.

My next example from the 2002 AMC10 is similar to the previous one. The difference is that the solution with multiple choices is even more boring, while the solution without these choices is more interesting and beautiful.

Let a, b, and c be real numbers such that a – 7b + 8c = 4 and 8a + 4b – c = 7. Then a² – b² + c² is: 0, 1, 4, 7, or 8.

Given the choices, we see that the answer is a number. Hence we need to find any solution for the system or equations: a – 7b + 8c = 4 and 8a + 4b – c = 7. For example, if we let c = 0, we have two linear equations and two variables a and b that can be solved by a straightforward computation. Then we plug the solution into a² – b² + c².

Without knowing that the result is a number, we need to look at the symmetries of our two initial equations. We might discover a new rule:

If we have two expressions ax + by and bx – ay, where a and b are switched between variables and there is a change in sign, it is a good idea to square each of them and sum them up, because the result is very simple: (a² + b²)(x² + y²).

Hence, in our initial problem we need to move the term with b in our two linear equations to the right; then square them and sum up the results. This way we may get a very simple expression. And indeed, this trick leads to a solution and this solution provides insight into working with algebraic expressions.

This is the perfect problem to linger over, assuming you’re not in the middle of a timed competition. It might make you wonder for which parameters this problem works. You might discover a new theorem that allows you to create a very similar problem from any number that can be represented as a sum of two non-trivial squares in two different ways.

To prepare my students for AMC I need to teach them tricks that are not useful at USAMO, or in mathematics in general for that matter. Many tricks distract from new ideas or from understanding the problem. All they give us is speed.

This bothers me, but to pacify myself, I keep in mind that most of my students will not become mathematicians and it might be useful in their lives to be able to make split-second decisions among a small number of choices.

However, it seems like Americans have the opposite problem: we make quick decisions without thinking. I’m concerned that training for multiple choice tests and AMC competitions aggravates this problem.

Should you try to guess an answer to a multiple-choice problem during a test? How many problems should you try to guess? I will talk about the art of boosting your guessing accuracy in a later essay. Now I would like to discuss whether it makes sense to pick a random answer for a problem at AMC 10.

Let me remind you that each of the 25 problems on the AMC 10 test provides five choices. A correct answer brings you 6 points, a wrong answer 0 points and not answering at all gives you 1.5 points. So guessing makes the expected average per problem to be 1.2 points. That is, on average you lose 0.3 points when guessing. However, if you are lucky, guessing will gain you 4.5 points per problem, and if you are unlucky, it will lose you 1.5 points per problem.

So we see that on average guessing is unprofitable. But there are situations in which you have nothing to lose if you get a smaller score and a lot to gain if you get a better score. Usually the goal of a competitor at AMC 10 is to get to AIME. For that to happen, you need to get 120 points or be in the highest one percentile of all competitors. This rule complicates my calculations. So I decided to simplify it and say that your goal is to get 120 points and then see what mathematical results I can get out of that simplification.

First, suppose you are so accurate that you never make mistakes. If you have solved 20 problems, then your score without guessing is 127.5. If you start guessing and guess wrongly for all of the last five questions, you still have your desired 120 points. In this instance it doesn’t matter whether you guess or not.

Suppose on the other hand that you are still accurate, but less powerful. You have only solved 15 problems, so your score without guessing is 105. Now you must be strategic. Your only chance to get to your goal of 120 points is to guess. Suppose you randomly guess the answers for the 10 problems you didn’t solve. To make it to 120, you need to guess correctly at least five out of the ten remaining problems. The probability of doing so is 3%. Here is a table of your probability of making 120 points if you solve correctly n problems and guess the other problems.

We can see that if you solved a small number of problems, then the probability of getting 120 points is minuscule; but as the number of problems you solved increases, so does the probability of getting 120 points by guessing.

The interesting part is that if you have solved 19 problems, you are guaranteed to get to AIME without guessing. On the other hand, if you start guessing and all your guesses are wrong, you will not pass the 120 mark. The probability of having all six problems wrong is a not insignificant 26%. In conclusion, if you are an accurate solver and want to have 120 points, it is beneficial to guess the remaining problems if you solved fewer than 19 problems. It doesn’t matter much if you solved fewer than 10 problems or more than 19. But you shouldn’t guess if you solved exactly 19.

If you are not 100% accurate things get more complicated and more interesting. To decide about guessing, it is crucial to have a good estimate of how many mistakes you usually make. Let’s say that you usually have two problems wrong per AMC test. Suppose you gave answers to 20 problems at AMC. What’s next? Let us estimate your score. Out of your 20 answers you are expected to get 18*6 points for them plus 5*1.5 points for the problems you didn’t answer. Your expected score is 115.5. You are almost there. You definitely should guess. But does it matter how many questions you are trying to guess?

The correct answer to one question increases your score by 4.5 with probability 0.2, and the wrong answer decreases your score by 1.5 with probability 0.8. One increase by 4.5 is enough for you goal of 120 points. So if you guess one question, with probability 0.2, you get 120 points. If you guess two questions, then your outcome is as follows: you increase your score by 9 points with probability 0.04; you increase your score by 3 points with probability 0.32; and you decrease your score by 3 points with probability 0.64. As you need at least a 4.5 increase in points, it is not enough to guess one question out of two. You actually need to guess correctly on both of them. The probability of this happening is 0.04. It is interesting, but you have a much greater chance to get to your goal if you guess just one question than if you guess two. Overall, here is the table of probabilities to get to 120 points where m is the number of questions you are guessing.

Your best chances are to guess all the remaining questions.

By the end of the test you know how many questions you answered, but you don’t know how many errors you made. The table below tells you what you need in order to get 120 points. Here is how you read the table: The number of problems you solved is in the first column. If you are sure that the number of mistakes is not more than the number in the second column, you can relax as you made at least 120 points. The last column gives the score.

If the number of mistakes you made is one more than in the corresponding row of the table, you should start guessing in order to try to get 120 points. Keep in mind that there is a risk: if you are not sure how many problems you solved already and start guessing, you might ruin your achievement of 120 points.

In the next table I show how many questions you can guess without the risk of going below 120 points. The word “all” means that it is safe to guess all the remaining questions.

You can see that if your goal is to get 120 points, your dividing line is answering 22 questions. If you solved 22 questions or more, there is no risk in guessing. Namely, if you have already achieved more than 120 points, guessing will not take you below that. But if you made more errors than are in the table, then guessing might be beneficial. Hence, you should always guess in this case — you have nothing to lose.

Now I would like to show you my calculations for a situation in which you are close to 120 points and need to determine the optimum number of questions to guess. The first column is the number of answered questions. The second column is the number of mistakes. The third column is your expected score without guessing. The fourth column is the optimum number of questions you should guess. And the last column lists your chances to get 120 points if you guess the number of questions in the fourth column.

You can see that almost always if you are behind your goal, you should try to guess all of the remaining questions, with one exception: if you answered 19 questions and one of them is wrong. In this case you should guess exactly two questions — not all that remain.

Keep in mind that all these calculations are very interesting, but don’t necessarily apply directly to AMC 10, because I simplified assumptions about your goals. It may not be directly applicable, but I hope I have expanded your perspective about how you can use math to help you understand how better to succeed at math tests and how to design your strategy.

I plan to teach you how to guess more profitably, and this skill will also advance your perspective.

Should you allocate time for checking your answers during important tests? I will use AMC 10/12 as an example, but you can adjust the calculations for any other test.

AMC 10/12 is a math competition that asks 25 multiple-choice questions that you need to solve in 75 minutes. You get 6 points for a correct answer, 1.5 points for an unanswered question and 0 points for a wrong answer.

Whether or not to take time to check your answers depends on you and the situation. If you finished your test and have some time left over, then surely you should use the extra time for checking your answers. If you only have three minutes left and your next problems are too complex to be dealt with in that time, then it is logical to use these moments to check back.

Sometimes, though, it isn’t worth it to check your answers. If you haven’t finished the test, but are a super-accurate person and never make mistakes, then it is better to continue working on the next problem than to waste time checking your correct answers. Also, if you rarely catch your own mistakes anyway, it doesn’t make sense to check.

But things are not usually so clear. By the end of the test, most people need to make a decision: continue working through the problems or use the final moments to check the answers? How can you best decide if you should allocate time for checking and, if so, how much time?

The problems in AMC tests increase in difficulty. I suggest that each time you take the test or practice for AMC, take note of two things. How long did it take you to solve the test’s last problem and what is the level of your accuracy for it. Suppose you know that at the end of the AMC test you can solve a problem in about 10 minutes and it is correct about 90% of the time. That means that investing the last ten minutes in solving the next problem will give you on average 5.4 points. If you remember that a blank answer gives you 1.5 points, you should realize that solving the last problem increases your score by 3.9 points. If you are very accurate, your score can increase more, but not more than 4.5 points.

Try conducting the following experiment. Take an AMC test from a past year. Do it for 65 minutes — the time of the test minus the time you need for that last problem. Then spend the last 10 minutes checking and correcting your answers. Now let us calculate how profitable that would be. Compare the scores you would have gotten without your corrections and with your corrections. If checking increases your score by more than 3.9 points, it is more profitable to check than to solve the next problem. If you do not make errors when you’re trying to make corrections, the rule of thumb is that correcting one mistake is better than solving one problem. Indeed, your score increases by 6 points if you correct a mistake, and by not more than 4.5 points if you solve the next problem.

On all tests that punish wrong answers, correcting an error produces more points than solving a new question.

If you find that checking is profitable, but you can’t check all the problems in ten minutes, you should consider allocating more time. Keep in mind, though, that you should adjust the sample calculation above for the last two problems. Remember, the next to the last problem is generally easier than the last problem. So if it takes you ten minutes to solve the last problem in the test, it most probably will take you less than twenty minutes to solve the last two. Also, since the difficulty increases throughout the test, the accuracy of the second to last problem might be better than the accuracy of the last problem. In addition, the first ten minutes that you check may be more productive than the next ten minutes of checking. So if you wonder if you should forgo the last two problems in order to check your earlier work, you have to redo the experiment anew, measuring both how long it takes you to solve those two problems and the benefits of checking.

This discussion can potentially help you to increase your score. However, there are other strategic considerations to weigh when deciding whether or not to check your work. For example, if the number of mistakes in your tests varies and sometimes you are 100% accurate and you are one problem away from your goal to get to AIME, it is more profitable to go for the last problem and hope for the best. I will discuss the strategic considerations for AMC some other time.

By Tanya Khovanova and Richard Stanley

This essay is written especially for high school and undergrad math lovers who enjoy problem solving and who plan to major in mathematics. One of the authors, Tanya, often received this advice when she was an undergraduate in Russia: “Problem solving is child’s play. You’ll have to change your attitude if you plan to succeed in research.”

Perhaps that’s why some famous problem solvers, even those who won gold medals at IMO, became not-so-famous mathematicians. To help you avoid that fate, we’ll discuss the ways in which research is unlike problem solving.

Is research different from problem solving?

Yes and no. There are many mathematicians who continue problem solving as their form of research. Remember Paul Erdos who used to suggest a lot of problems and even offered money rewards for solutions. Many mathematicians solve problems posed by other people. You might consider Andrew Wiles as the ultimate math problem solver: he proved Fermat’s last theorem, which had been open for 400 years. Though he could not have done it without the many theories that had already been generated in the search to find the elusive proof.

You can become a mathematician and continue to look around you for problems to solve. Even though this is still problem solving, the problems will be very different from competition problems, and you will still need to adjust to this type of research.

Problems you solve during research

So, what is the difference between problems that mathematicians solve during competition and the problems they tackle for their research?

Expected answer. In competition problem solving you know there is a solution. Often you know the answer, but you just need to prove it. In research there is no guarantee. You do not know which way it will go. For this reason finding counter-examples and proving that some ideas are wrong is a positive contribution, for it can eliminate some possibilities. So one adjustment is that you might start valuing negative answers.

Difficulty level. Competition problems are designed to be solved in one hour, so you are expected to generate an idea in just minutes. In research the problem might drag on for years, because it is far more difficult. If you get used to the instant gratification of competition problem solving, you might find the lengthy work of research frustrating. It’s very important to adjust your expectations so that you won’t drop a problem prematurely. You need to measure progress in small intermediate steps and learn to appreciate this different rhythm.

Motivation. Although you miss the euphoria of finding quick solutions, you get a different kind of reward with research. Because no one knows the answer in advance, when you solve the problem, you are the first to do so. You have opened up a new truth.

Time limits. In competitions you have a time limit for every problem. In research you set your time limits yourself. That allows you to put a problem aside and come back later if necessary. In a sense you can think about several problems at the same time.

Your passion. You can choose your problems yourself. Research is much more rewarding if you follow your heart. In competitions you have to spend time on problems you might not like. Here you have an option to choose and pick only the problems that appeal to you. Thus, you become more motivated and as a result more successful.

Finding a problem

After solving problems posed by other people, the next step is to pose math problems yourself. As we mentioned before, in research you do not always have a strictly-defined problem. It is a significant adjustment to move from solving already-defined problems to posing the problems yourself.

Generalizations. Often you can generalize from an existing problem to more general cases. For example, if you see a problem for n=3, you can wonder what happens for any n, or for any prime n.

Being on the lookout. Sometimes a situation puzzles you, but you can’t formulate a specific problem around that situation. For example, why do most of the terms in the sequence end in 9? Is there a reason for that? Or, you might find that a formula from your integrable systems seminar is similar to a formula from your representation theory class. This might lead you to the essential research question: “What is going on?” You always need to be on the lookout for the right questions.

Value. When you create your own research problems it is crucial to always ask yourself: Is the problem I am creating important? What is the value of this problem? There is no a good reason to create random generalizations of random problems. If the problem you found interests you very much, that is the first sign that it might interest other people; nonetheless, you should still ask yourself how this problem will help advance mathematics.

Mathematics is not only problem solving

There are other things to do than solve problems. There are many mathematicians who work differently, who don’t solve problems or don’t only solve problems. Here are some of the many options mathematicians have:

Building structures. You may not be interested in calculating the answer to a question, but rather in building a new structure or a new theory.

Advancing the language. When you invent new definitions and new notations, you will help to simplify a math language so that the new language will allow you to prove your results and other peoples’ results faster and clearer.

Unification. Sometimes you notice two results in two different areas of mathematics with some kind of similarity. Explaining why these results are the same might create a new understanding of things. It is great to unify two different areas of mathematics.

Explaining. Very often proofs are not enough. Why is something true? What’s the reason and what’s the explanation? It is good to ask yourself a “why” question from time to time, such as, “Why is this proof working?” When you find an answer, it might become easier to understand what to do next and how to generalize your proof.

Directions. Many mathematicians are valued not for the problems they solve or suggest, but for ideas and directions they propose. Finding a new direction for research can generate unexpected opportunities and create tons of math problems on the way. It can be valuable to come up with good conjectures, even if you have no hope of solving them yourself. Two example of this are the Weil conjectures (eventually proved by Deligne) and the Langlands program, which is still incomplete but which has generated a huge amount of important research.

Vision. What is the most general thing that can be proved by this technique? What kinds of improvements and refinements are there? It is good to step back from the problem you solved and meta-think about it.

As you can see, problem solving is just the beginning of all that mathematics can offer you. Mathematicians find these other options very rewarding, so it’s worth your while to try these varied aspects of mathematical work to see if you have a taste for other things. If you don’t venture beyond problem solving you might miss the full beauty of mathematics.

I’ve translated two problems from the 2009 Moscow Math Olympiad. In both of them our characters are genetically engineered octopuses. The ones with an even number of arms always tell the truth; the ones with an odd number of arms always lie. In the first problem (for sixth graders) four octopuses had a chat:

• “I have 8 arms,” the green octopus bragged to the blue one. “You have only 6!”
• “It is I who has 8 arms,” countered the blue octopus. “You have only 7!”
• “The blue one really has 8 arms,” the red octopus said, confirming the blue one’s claim. He went on to boast, “I have 9 arms!”
• “None of you have 8 arms,” interjected the striped octopus. “Only I have 8 arms!”

Who has exactly 8 arms?

Not only do octopuses lie or tell the truth according to the parity of the number of their arms, it turns out that the underwater world is so discriminatory that only octopuses with six, seven or eight arms are allowed to serve Neptune. In the next problem (for seventh graders), four octopuses who worked as guards at Neptune’s palace were conversing:

• The blue one said, “All together we have 28 arms.”
• The green one said, “All together we have 27 arms.”
• The yellow one said, “All together we have 26 arms.”
• The red one said, “All together we have 25 arms.”

How many arms does each of them have?

My students enjoyed the octopuses, so I decided to invent some octopus problems of my own. In the first problem, the guards from the night shift at Neptune’s palace were bored, and they started to argue:

• The magenta one said, “All together we have 31 arms.”
• The cyan one said, “No, we do not.”
• The brown one said, “The beige one has six arms.”
• The beige one said, “You, brown, are lying.”

Who is lying and who is telling the truth?

In the next problem the last shift of guards at the palace has nothing better to do than count their arms:

• The pink one said, “Gray and I have 15 arms together.”
• The gray one said, “Lavender and I have 14 arms together.”
• The lavender one said, “Turquoise and I have 14 arms together.”
• The turquoise one said, “Pink and I have 15 arms together.”

What number of arms does each one have?

The book An Introduction to Diophantine Equations by Titu Andreescu and Dorin Andrica is targeted at people preparing for USAMO and IMO. It contains a lot of problems on Diophantine equations from math Olympiads used in various math Olympiads all over the world.

The first chapter discusses several methods for solving Diophantine equations: decomposition, using inequalities, using parameters, modular arithmetic, induction, infinite descent, and other miscellaneous ideas. Each sub-chapter starts with a short description of the method, accompanied by several solutions to sample problems. At the end of each sub-chapter there are a plethora of exercise problems.

The second and the third chapters are more theoretical. The former discusses some classical equations and the latter looks at Pell’s equation. These two chapters also contain problems, but the bulk of the chapters is devoted to basic theory that is essential to an understanding of Diophantine equations.

For those who are training for the Olympiads, this is an important book to own, not only because there are few other books on the subject, but because it provides so many useful problems.

I’ve long complained that most training books for math competitions leave out any discussion of how we choose a method by just looking at a problem. Andreescu and Andrica didn’t fill that gap with this book.

Perhaps in their next book they will point out clues that indicate that a particular problem might be solved by the parametric method. And explain which types of problems are best solved with induction. Let them challenge students to find those clues in a problem that help us to judge which method might be most promising, instead of randomly trying one method after another. Let me give you a sample problem from the book, which originated at the Balkan Mathematical Olympiad:

Prove that the equation x⁵ – y² = 4 has no solutions in integers.

The solution is to take the equation modulo 11, and see that it is impossible.

Is there a reason to start with the modular arithmetic method and not with other methods? If we use modular arithmetic, do we recognize why it’s best to start with 11? I’m convinced that this problem has sufficient clues to suggest starting with checking this equation modulo 11.

I wonder if you, my readers, agree with me. If so, can you explain which hints in the problem lead to taking the equation modulo 11? I believe it should be a part of competition training to learn to identify clues that suggest that one direction might be preferable to the others.

I love Harvard-MIT Math Tournaments. I like the mini-events, especially when I learn a new game. I also like the guts round, where I enjoy the adrenaline rush of watching the progress in real time. I also like the fact that I know many of the kids from different teams: my current students, my former students, the members of my club, my Sergei’s friends.

The problems for the competitions are designed by undergraduate students at MIT and Harvard. Kudos to them. Still, I was somewhat disappointed with the November 2009 problems. Most problems are variations of standard problems with different parameters. It is not easy to design a problem, but I was hoping for something fresh.

My favorite problem from the HMNT 2009 tournament was in the theme round:

There are five guys named Alan, Bob, Casey, Dan, and Eric. Each one either always tells the truth or always lies. You overhear the following discussion between them:

• Alan: “All of us are truth-tellers.”
• Bob: “No, only Alan and I are truth-tellers.”
• Casey: “You are both liars.”
• Dan: “If Casey is a truth-teller, then Eric is too.”
• Eric: “An odd number of us are liars.”

Who are the liars?

My second favorite problem was in the guts round:

Six men and their wives are sitting at a round table with 12 seats. These men and women are very jealous — no man will allow his wife to sit next to any man except for himself, and no woman will allow her husband to sit next to any woman except for herself. In how many distinct ways can these 12 people be seated such that these conditions are satisfied? (Rotations of a valid seating are considered distinct.)

This was the funniest problem:

You are trapped in ancient Japan, and a giant enemy crab is approaching! You must defeat it by cutting off its two claws and six legs and attacking its weak point for massive damage. You cannot cut off any of its claws until you cut off at least three of its legs, and you cannot attack its weak point until you have cut off all of its claws and legs. In how many ways can you defeat the giant enemy crab? (Note that the legs are distinguishable, as are the claws.)

It is difficult to arrange so many problems for four rounds without mistakes. The error in the following problem is not a typo and it bothers me that no one caught it:

Pick a random digit in the decimal expansion of 1/99999. What is the probability that it is 0?

Hey, there is no uniform distribution on an infinite set of integers: picking a random digit is not defined.

In my days of competing in math, I met guys who could solve any geometry problem by using coordinates: first they would assign variables to represent coordinates of different points, then they would write and solve a set of equations. It seemed so boring. Besides, this approach doesn’t provide us with any new insight into geometry.

I find geometric solutions to geometry problems much more interesting than algebraic solutions. The geometric solutions that use geometric transformations are often the shortest and the most beautiful.

I.M. Yaglom wrote a great trilogy called The Geometric Transformations. The first book of this trilogy discusses translations, rotations and reflections. The second one — looks at similarity transformations, and the third one talks about affine and projective transformations. A lot of beautiful problems with their solutions are scattered throughout these books. They include all my favorite problems related to transformations.

I think geometry is the weakest link for the USA math team. So we have to borrow the best geometry books from other countries. This trilogy was translated from Russian and Russians are known for their strong tradition of excellence in teaching geometry.

Below you can find sample problems from Geometric Transformations 1, Geometric Transformations 2 and Geometric Transformations 3 — not necessarily in this order.

I.M. Yaglom wrote a great trilogy called The Geometric Transformations. The first book of this trilogy discusses translations, rotations and reflections. The second one — looks at similarity transformations, and the third one talks about affine and projective transformations. A lot of beautiful problems with their solutions are scattered throughout these books. They include all my favorite problems related to transformations.

I think geometry is the weakest link for the USA math team. So we have to borrow the best geometry books from other countries. This trilogy was translated from Russian and Russians are known for their strong tradition of excellence in teaching geometry.

Below you can find sample problems from Geometric Transformations 1, Geometric Transformations 2 and Geometric Transformations 3 — not necessarily in this order.

Problem 1. Let A be a point outside a circle S. Using only a straightedge, draw the tangents from A to S.

Problem 2. At what point should a bridge be built across a river separating two towns A and B (see figure) in order that the path connecting the towns be as short as possible? The banks of the river are assumed to be parallel straight lines, and the bridge is assumed to be perpendicular to the river.

Problem 3. Suppose you have two lines drawn on a piece of paper. The intersection point A of the two lines is unreachable: it is outside the paper. Using a ruler and a compass, draw a line through a given point M such that, were the paper bigger, point A would belong to the continuation of the line.

Konstantin Knop sent me the following coins puzzle, which was created by Alexander Shapovalov and first appeared at the Regional round of the all-Russian math Olympiad in 2000.

Baron Münchhausen has 8 identical-looking coins weighing 1, 2, …, 8 grams. The Baron knows which coin is which and wants to demonstrate to his guests that he is right. He plans to conduct one weighing on a balance scale, so that the guests will be convinced about the weight of one of the coins. Can the Baron do this?

This being a sequence-lover blog, we want to create a sequence out of this puzzle. The sequence is the following: Let the Baron initially have n identical-looking coins that weigh exactly 1, 2, …, n grams. Then a(n) is the minimum number of weighings on a balance scale that the Baron needs in order to convince his guests about the weight of one of those coins.

The original puzzle can be restated as asking whether a(8) = 1. The sequence is defined starting from index 1 and the first several terms are easy to calculate: 0, 1, 1, 1, 2, 1, 1, 1. Can you continue this sequence?

Let’s look at where ones occur in this sequence:

Theorem. If the weight of a coin can be confirmed with one weighing, then one cup of that weighing must contain all the coins with weights from 1 to some i, and the other cup must contain all the coins with weights from some j to n. Furthermore, either the scale must balance, or the cup containing the 1-gram coin must be lighter.

Proof. What does it mean for the Baron to convince his guests about the weight of some coin with one weighing? From the perspective of the guests, a weighing is a number of coins in one cup, a number of coins in the other cup, and a number of coins not on the scale, together with the result the scale shows (one or the other cup heavier, or both the same weight). For the guests to be convinced of the weight of some particular coin, it must therefore be the case that all possible arrangements of coin weights consistent with that data agree on the weight of the coin in question. Our proof strategy, therefore, is to look for ways to alter a given arrangement of coin weights so as to change the weight given to the coin whose weight is being demonstrated, thus arriving at a contradiction.

First, obviously, the coin whose weight k the Baron is trying to confirm has to be alone in its group: either alone on some cup or the only coin not on the scale. After that we can divide the proof of the theorem into several cases.

Case 1. The coin k is on a cup and the scale is balanced. Then we want to show two things: k = n, and the coins on the other cup weigh 1, 2, …, i grams for some i. For the first part, observe that if k < n, then the coin with weight k+1 must not be on the scale (otherwise it would overbalance coin k). Therefore, we can substitute coin k+1 for coin k, and substitute a coin one gram heavier for the heaviest coin that was on the other cup, and produce thereby a different arrangement with the same observable characteristics but a different weight for the coin the Baron claims has weight k.

To prove the second part, suppose the contrary. Then it is possible to substitute a coin 1 gram lighter for one of the coins on the other cup. Now, if coin k-1 is not on the scale, we can also substitute k-1 for k, and again produce a different arrangement with the same observable characteristics but a different weight for the coin labeled k. On the other hand, if k-1 is on the scale but k-2 is not, then we can substitute k-2 for k-1 and then k-1 for k and the weighing is again unconvincing. Finally, if both k-1 and k-2 are on the scale, and yet they balance k, then k=3 and the theorem holds.

Consequently, k = n = 1 + 2 + … + i is a triangular number.

Case 2. The coin k is on the lighter cup of the scale. Then: first, k = 1, because otherwise we could swap k and the 1-gram coin, making the light cup lighter and the heavy cup heavier or unaffected; second, the 2-gram coin is on the heavy cup and is the only coin on the heavy cup, because otherwise we could swap k with the 2-gram coin and not change the weights by enough to affect the imbalance; and finally n = 2 because otherwise we could change the weighing 1 < 2 into 2 < 3.

Thus the theorem holds, and the only example of this case is k = 1, n = 2.

Case 3. The coin k is on the heavier cup of the scale. Then k = n and the lighter cup consists of some collection of the lightest available coins, by the same argument as Case 1 (but even easier, because there is no need to maintain the balance). Furthermore, k must weigh exactly 1 gram more than the lighter cup, because otherwise, k-1 is not on the lighter cup and can be substituted for k, making the weighing unconvincing.

Consequently, k = n = (1 + 2 + … + i) + 1 is one more than a triangluar number.

Case 4. The coin k is not on a cup and the scale is not balanced. Then, since k must be off the scale by itself, all the other coins must be on one cup or the other. Furthermore, all coins heavier than k must be on the heavier cup, because otherwise we could make the lighter cup even lighter by substituting k for one of those coins. Likewise, all coins lighter than k must be on the lighter cup, because otherwise we could make the heavier cup even heavier by substituting k for one of those coins. So the theorem holds; and furthermore, the cups must again differ in weight by exactly 1 gram, because otherwise we could swap k with either k-1 or k+1 without changing the weights enough to affect the result on the scale.

Consequently, the weight of the lighter cup is k(k-1)/2, the weight of the heavier cup is 1 + k(k-1)/2. Thus the total weight of all the coins is n(n+1)/2 = k²+1. In other words, case 4 is possible iff n is the index of a triangular number that is one greater than a square.

Case 5. The coin k is not on a cup and the scale is balanced. This case is hairier than all the others combined, so we will take it slowly (noting first that all the coins besides k must be on some cup).

Lemma 1. The two coins k-1 and k-2 must be on the same cup, if they exist (that is, if k > 2). Likewise k-2 and k-4; k+1 and k+2; and k+2 and k+4.

Proof. Suppose they’re not. Then we can rotate k, k-1, and k-2, that is, put k on the cup with k-1, put k-1 on the cup with k-2, and take k-2 off the scale. This makes both cups heavier by one gram, producing a weighing with the same outward characteristics as the one we started with, but a different coin off the scale. The same argument applies to the other three pairs of coins we are interested in, mutatis mutandis.

Lemma 2. The four coins k-1, k-2, k-3 and k-4 must be on the same cup if they exist (that is, if k ≥ 5).

Proof. By Lemma 1, the three coins k-1, k-2, and k-4 must be on the same cup. Suppose coin k-3 is on the other cup. Then we can swap k-1 with k-3 and k with k-4. Each cup becomes heavier by 2 grams without changing the number of coins present, the balance is maintained, and the Baron’s guests are not convinced.

Lemma 3. If coin k-4 exists, that is if k ≥ 5, all coins lighter than k must be on the same cup.

Proof. By Lemma 2, the four coins k-1, k-2, k-3 and k-4 must be on the same cup. Suppose some lighter coin is on the other cup. Call the heaviest such coin c. Then, by choice of c, the coin with weight c+1 is on the same cup as the cluster k-1, …, k-4, and is distinct from coin k-2 (because c is on a different cup from k-3). We can therefore swap c with c+1 and swap k with k-2. This increases the weight on both cups by 1 gram without changing how many coins are on each, but moves k onto the scale. The Baron’s guests are again unconvinced.

Lemma 4. The theorem is true for k ≥ 5.

Proof. By Lemma 3, all coins lighter than k must be on the same cup. Further, if a coin with weight k+4 exists, then by the symmetric version of Lemma 3, all coins heavier than k must also be on the same cup. They must be on the other cup from the coins lighter than k because otherwise the scale wouldn’t balance, and the theorem is true.

If no coin with weight k+4 exists, that is, if n ≤ k+3, how can the theorem be false? All the coins lighter than k must be on one cup, and their total weight is k(k-1)/2. Further, in order to falsify the theorem, at least one of the coins heavier than k must also be on that same cup. So the minimum weight of that cup is now k(k-1)/2 + k+1. But we only have at most two coins for the other cup, whose total weight is at most k+2 + k+3 = 2k + 5. For the scale to even have a chance of balancing, we must have

k(k-1)/2 + k+1 ≤ 2k + 5 ⇔ k(k-1)/2 ≤ k + 4 ⇔ k(k-1) ≤ 2k + 8 ⇔ k² – 3k – 8 ≤ 0.

Finding the largest root of that quadratic we see that k < 5.

So for k ≥ 5, the collection of all coins lighter than k is heavy enough that either one needs all the coins heavier than k to balance them, or there are enough coins heavier than k that the theorem is true by symmetric application of Lemma 3.

Completion of Case 5. It remains to check the case for k < 5. If n > k+3, then coin k+4 exists. If so, all the coins heavier than k must be on the same cup. Furthermore, since k is so small, they will together weigh more than half the available weight, so the scale will be unbalanceable. So k < 5 and n ≤ k+3 ≤ 7.

For lack of any better creativity, we will tackle the remaining portion of the problem by complete enumeration of the possible cases, except for the one observation that, to balance the scale with just the coin k off it, the total weight of the remaining coins, that is, n(n+1)/2 – k must be even. This observation cuts our remaining work in half. Now to it.

Case 5. Seven Coins. n = 7. Then 5 > k ≥ n – 3 = 4, so k = 4. Then the weight on each cup must be 12. One of the cups must contain the 7 coin, and no cup can contain the 4 coin, so the only two weighings the Baron could try are 7 + 5 = 1 + 2 + 3 + 6, and 7 + 3 + 2 = 1 + 5 + 6. But the first of those is unconvincing because k+1 = 5 is not on the same cup as k+2 = 6, and the second because it has the same shape as 7 + 3 + 1 = 2 + 4 + 5 (leaving out the 6-gram coin instead of the asserted 4-gram coin).

Case 5. Six Coins. n = 6. Then 5 > k ≥ n – 3 = 3, and n(n+1)/2 = 21 is odd, so k must also be odd. Therefore k=3, and the weight on each cup must be 9. The 6-gram coin has to be on a cup and the 3-gram coin is by presumption out, so the Baron’s only chance is the weighing 6 + 2 + 1 = 4 + 5, but that doesn’t convince his skeptical guests because it looks too much like the weighing 1 + 3 + 4 = 6 + 2.

Case 5. Five Coins. n = 5. Then 5 > k ≥ n – 3 = 2, and n(n+1)/2 = 15 is odd, so k must also be odd. Therefore k=3, and the weight on each cup must be 6. The only way to do that is the weighing 5 + 1 = 2 + 4, which does not convince the Baron’s guests because it looks too much like 1 + 4 = 2 + 3.

Case 5. Four Coins. n = 4. Then the only way to balance a scale using all but one coin is to put two coins on one cup and one on the other. The only two such weighings that balance are 1 + 2 = 3 and 1 + 3 = 4, but they leave different coins off the scale.

The remaining cases, n < 4, are even easier. That concludes the proof of Case 5.

Consequently, by the argument similar to the one in case 4 we can show that the number of coins in case 5 must be the index of a square triangular number.

This concludes the proof of the theorem.

Now we can describe all possible numbers of coins that allow the Baron to confirm a coin in one weighing, or, in other words, the indices of ones in the sequence a(n). The following list corresponds to the five cases above:

1. n is a triangular number. For example, for six coins the weighing is 1+2+3 = 6.
2. n = 2. The weighing is 1 < 2.
3. n is a triangular number plus one. For example, for seven coins the weighing is 1+2+3 < 7.
4. n is the index of a triangular number that is a square plus one. For example, the forth triangular number, which is equal to ten, is one greater than a square. Hence the weighing 1+2 < 4 can identify the coin that is not on the cup. The next number like this is 25. And the corresponding weighing is 1+2+…+17 < 19+20 +…+25.
5. n is the index of a square triangular number. For example, we know that the 8th triangular number is 36, which is a square: our original problem corresponds to this case.

If we have four coins, then the same weighing 1+2 < 4 identifies two coins: the coin that weighs three grams and is not in a cup and the coin weighing four grams that is in a cup. The other case like this is for two coins. Comparing them to each other we can identify each of them. It is clear that there are no other cases like this. Indeed, for the same weighing to identify two different coins, it must be the n-gram on the cup, and the n-1 coin off the scale. From here we can see that n can’t be big.

As usual we want to give something to think about to our readers. We have given you the list of sequences describing all the numbers for which the Baron can prove the weight of one coin in one weighing. Does there exist a number greater than four that belongs to two of these sequences? In other words, does there exist a total number of coins such that the Baron can have two different one-weighing proofs for two different coins?

To conclude this essay we would like to note that the puzzle we are discussing is related to the puzzle in one of Tanya’s previous posts:

You have 6 coins weighing 1, 2, 3, 4, 5 and 6 grams that look the same, except for their labels. 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?

The latter puzzle appeared at the last round of Moscow math Olympiad in 1991. The author of this problem was Sergey Tokarev.