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.

By Tanya Khovanova and Gregory Bomash

What object is missing?

My recent entry, where I asked you to choose the odd one out among these images

was extremely popular. It was republished all around the world and brought my blog as much traffic in one day as I used to get in a month. Not only did I read the many comments I received, I also followed up on other peoples’ blogs who reprinted my puzzle — at least those that were in either Russian or English. I also got private emails and had many conversations in person about it. The diversity of answers surprised me, so I would like to share them with you.

As I’ve said before, I do not think there is a correct answer to this type of question, but I was disappointed by some of the answers. For example, those who simply said, “The green one is the odd one out,” made me feel that either they hadn’t read the question or hadn’t thought about it very much. It’s a shame that these people spent more time sharing their opinion with the world than thinking about the problem in the first place.

I wouldn’t mind someone arguing that the green one is the odd one out, but in this case an explanation is in order. Many people did offer explanations. Some told me that we perceive the color difference stronger than all other parameters I used, and the green figure pops out of the picture more than anything else. In fact, I personally perceive color difference the strongest among all the parameters, but since there are people who are color blind, I would disregard my feelings for color as being subjective.

You can create a whole research project out of this puzzle. For example, you can run an experiment: Ask the question, but flash the images above very fast, so there is no time for analysis — only time to guess. This allows us to check which figure is the first one that people perceive as different. Or you can vary the width of the frame and see how the perception changes.

Color was not the only parameter among those I chose — shape, color, size and the existence of a frame — that people thought was more prominent. My readers weighed these parameters unequally, so each argued the primary importance of the parameter they most emphasized. For example, one of my friends argued that:

The second figure should be the odd one out as, first, it is the only one without a frame, and, second, it is the only one comprised of one color rather than two. So it differs by two features, as others differ only by one feature.

A figure having one color is the consequence of not having a frame, so this particular friend of mine inflated the importance of not having a frame.

However, I can interpret any feature as two features. For example, I can say that the circle is the odd one out because not only is it a different figure, but it also doesn’t have any angles. Similarly, the last one is the smallest one and the border width is in a different proportion to its diameter.

On a lighter side, there were many funny answers to the puzzle:

• The one that says I am special.
• The right one because it is right.
• The fourth one, because four is the only composite index.
• The one that says I am not special.

For the which-is-the-odd-one-out questions, the designer of the question is usually expecting a particular answer. So here’s the answer I expected:

There is only one green figure. Wait a minute, there is only one circle. Hmm, there is only one without a frame and there’s only one small figure. I see! The first one is the only figure that is not the odd one, that doesn’t have a special property, so the first must be the odd one out. This is cool!

And the majority of the answers were exactly as I expected.

Since this is a philosophical problem, some of the responses took it to a different level. One interesting answer went like this:

All right, the last four figures have special features; the first figure is special because it is normal. Hence, every figure is special and there are no odd ones here.

I like this answer as the author of it equated regular features with a meta-feature, and it is a valid choice. This answer prompted me to write another blog entry with a picture where I purposefully tried to not have an odd one out:

Though I wrote that the purpose of this second set of images is to show an example where there is no odd one out, my commentators still argued about which one was the odd one out here.

Finally, I would like to quote Will’s comment to my first set of images:

The prevailing opinion is that the first is least unique and is therefore the oddest. But it is the mean and the others are one deviation from it. Can the mean be the statistical anomaly?

And Cedric replied to Will:

Yes, I think the mean can be a statistical anomaly. The average person has roughly one testicle and one ovary. But a person with these characteristics would certainly be an anomaly.

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 first episode of Numb3rs: Season Six reminded me of the hangman’s paradox. Here is a one-day version of the hangman’s paradox:

Suppose you are in a prison and the guard says to you, “You will be hanged tomorrow at noon and it will be a surprise.” You presume that you can’t be surprised since they already told you, so there is a contradiction in what they’ve said. Therefore, you conclude that they can’t hang you and you relax. Next day at noon the guard comes for you, to take you to be hanged, and you are utterly surprised. Oops.

What I do not like about this paradox is that it assumes that you do not know about the paradox. I, on the other hand, imagine that you, my reader, are logical and intelligent. So the moment the guard tells you that you will be hanged tomorrow at noon and it will be a surprise, you realize that the situation depends on what you decide to believe in now. If you decide that you won’t be hanged tomorrow, then you will have a relatively relaxing day today and you will be caught by surprise tomorrow and die. If you decide that you will die tomorrow, then you will have a nerve-wracking day today, but the guard may release you, to save his honor, since you won’t be surprised.

The original hangman’s paradox in which the guard tells you that you will be hanged on a weekday the following week and that you will be caught by surprise, also assumes that you are not aware of the paradox. If you are aware of the paradox, you know that usually guards in this paradox come for you on Wednesday, so you can prepare yourself. Actually, to guarantee your survival, if not your feeling of moral superiority, you can daily persuade yourself to belief that you will be hanged at noon the next day. This way, you will never be caught by surprise. If you are a person who can control your own beliefs, you may be able to save your life.

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.

Last month I gave my students a problem from Raymond Smullyan’s book The Riddle of Scheherazade:

A certain sheik named Hassan has eight horses. Four of them are white, three are black, and one is brown. How many of them can each say that it is the same color as another one of Hassan’s horses?

Half of my students failed to notice the trick and gave the wrong answer. Recently I gave them the continuation of the problem from the same book:

A certain sheik named Hassan has eight horses. Four of them are white, three are black, and one is brown. Assuming now that Hassan’s horses can talk, how many of them can each say that it is the same color as another one of Hassan’s horses?

This time the majority of my students didn’t notice the trick. This motivated me to continue playing jokes with them. Unfortunately though, Raymond Smullyan had only two problems about Hassan’s horses, so I have to invent the next one myself. Here is what I plan to give my students next time:

A certain sheik named Hassan has eight horses. Four of them are white, three are black, and one is brown. Assuming now that Hassan’s horses can talk and always tell the truth, how many of them will say that it is the same color as another one of Hassan’s horses?

Feel free to continue the series.

My recent blog puzzle where my readers had to choose the odd one out became extremely popular and was republished in many blogs around the world. Some commentators decided that my posting was a joke and an example where the odd one out didn’t exist. I have to disappoint them: as a protest against find-the-odd-one-out questions and to illustrate that sometimes there is no good choice for the odd one out I would have chosen a different picture:

Can you find the odd one out?