Tanya Khovanova's Math Blog

Now that my weight loss is under way, I want to build an exercise program.

I already exercise. I am a member of the MIT ballroom dance team and I go to the gym, where I use the machines, swim, and take gentle yoga and Zumba classes.

Doesn’t sound bad, right? But the reality is that I went to Zumba class a total of three times last year. I skipped half the dance classes I signed up for because I was too tired to attend. I had sincerely planned to go to them, but they’re held in the evenings, when I’m already drooping.

The yoga situation is the worst. I’m scheduled to go to yoga twice a week, but just as it is time to leave for yoga, I get very hungry and cannot resist having a snack. Then I remember that they do not recommend practicing yoga after eating a meal, and voilà—I have found my excuse. I am all set up to exercise five hours a week, but in reality most weeks I do not exercise at all.

How can I motivate myself? I know that with food, if I have gained weight one day, I eat less the next. So if I’ve missed my goal of exercising one week, do I add those hours to the next week? That’s unlikely to actually help, since the problem is that I’m not meeting my exercise goal, period.

Many people suggested that I punish myself. For example, if I do not meet my goal, I should donate money to a cause I do not support. This feels wrong. If I fail, I’ll feel doubly guilty. I’ll go broke paying for psychotherapy.

I can try to reward myself. But what should the reward be? Should I reward myself with a piece of tiramisu? Since I am trying to persuade myself that sugar is bad, I shouldn’t create a situation that makes sugar desirable. So rewarding with food won’t work. Should I buy myself something? If I really want it, I will buy it anyway.

I’ve been thinking about a plan for a long time. Finally I realized that I should find other people to reward me. I do have a lot of friends, and I came up with an idea of how they could help.

The next time I saw my friend Hillary I asked her if she wanted to sponsor my new exercise plan. She said, “I’m in,” without even hearing the plan. Hillary is a true friend. This is what she blindly signed up for.

I decided to push myself to exercise five hours a week. Because of weather and health fluctuations, I pledged to spread 20 hours of exercise over four weeks. I will sent Hillary weekly reports of what I do. This is in itself a huge motivating factor. After the four weeks, we will go to lunch together, which is a great reward for me to look forward to. If I succeed with my plan, she pays for lunch. If I fail, I pay.

Once I saw how enthusiastic Hillary was, I lined up four other friends for the next four-week periods. I hope that after several months of exercise, I will learn to enjoy it. Or at least, I will start feeling the benefits and that itself will be a motivating factor.

Hillary liked my plan so much that she designed a similar exercise plan for herself. Now I am looking forward to two lunches with Hillary.

I was driving on Mass Pike, when the cars in front of me stopped abruptly. I hit the brakes and was lucky to escape the situation without a scratch.

Actually, it wasn’t just luck. First of all, I always keep a safe distance from the other cars. Second, if I see the brake lights of the car in front of me, I automatically remove my foot from the gas pedal and hold it over the brake pedal until I know what the situation is.

On a highway, if the car in front of me has its brake lights on, usually that means that the driver is adjusting their speed a little bit. So, most of the time I don’t have to do anything. Seeing that the car in front of me has its brake lights on is not a good predictor of what will happen next. Only after I see that the distance between me and the car in front of me is decreasing rapidly, do I know to hit my brakes. That means that brake lights alone are not enough information. Differentiating between insignificant speed adjustments and serious braking requires time and can cost lives.

I have a suggestion. Why not create smart brake lights. The car’s computer system can recognize the difference in the strength with which the brakes are hit and the lights themselves can reflect that. They can be brighter or a different color or pulsing, depending on the strength of the pressure.

The drivers behind will notice these things before they will notice the decrease in the distance. This idea could save lives.

I recently posted the following coin weighing puzzle invented by Konstantin Knop:

We have N indistinguishable coins. One of them is fake and it is not known whether it is heavier or lighter than all the genuine coins, which weigh the same. There are two balance scales that can be used in parallel. Each weighing lasts one minute. What is the largest number of coins N for which it is possible to find the fake coin in five minutes?

The author’s solution in Russian is available at his blog. Also, two of my readers, David Reynolds and devjoe, solved it correctly.

Here I want to explain the solution for any number of required weighings.

It is easy to see that for n weighings the information theoretical bound is 5ⁿ. Indeed, each weighing divides coins into five groups: four pans and the leftover pile. To distinguish between coins, there can’t be two coins in the same pile at every weighing.

Suppose we know the faking potential of every coin, that is, each coin is assigned a value: potentially light or potentially heavy. If a potentially light coin is ever determined to be fake, then it must be lighter than a real coin. The same story holds for potentially heavy coins. How many coins with known potential can we process in n weighings?

If all the coins are potentially light then we can find the fake coin out of 5ⁿ coins in n weighings. What if there is a mixture of coins? Can we expect the same answer? How much more complicated could it be? Suppose we have five coins: two of them are potentially light and three are potentially heavy. Then on the first scale we compare one potentially light coin with the other such coin. On the other scale we compare one potentially heavy coin against another potentially heavy coin. The fake coin can be determined in one weighing.

The discussion above shows that there is a hope that any mixture of coins with different potential can be resolved. After each weighing, we want the number of coins that are not determined to be real to be reduced by a factor of 5. If one of the weighings on one scale is unbalanced, the potentially light coins on the lighter pan, plus the potentially heavy coins on the heavier pan would contain the fake coin. We do not want this number to be bigger than one-fifth of the total number of coins we are processing. So we divide coins in pairs with the same potential, and from each pair we put the coins on different pans of the same scale. So in one weighing we can divide the group into five equal groups. If there is an odd number of coins with the same potential, then the extra coin doesn’t go on the scales.

The only thing that we is left to check is what happens if the number of coins is small. Namely, we need to check what happens when the number of potentially light coins is odd and the number of potentially heavy coins is odd, and the total number of coins is not more than five. In this case the algorithm requires us to put aside the extra coin in each group, but the put-aside pile can’t have more than one coin.

After checking small cases, we see that we can’t resolve the problem in one weighing when there are 2 coins of different potential, or when the 4 coins are distributed as 1 and 3.

On the other hand, if we have extra coins that are known to be real, then the above cases can be resolved. Hence, any number of coins with known potential greater than four can be resolved in ⌈log₅n⌉ weighings.

Now let’s go back to the original problem in which we do not know the coins’ potential at the start. After a weighing, if both scales balance, then all the coins on the scale are real and the fake coin is in the leftover pile and we do not know its potential. If a scale doesn’t balance then the fake coin is in one of its two pans: the lighter pan has coins that are potentially light and the heavier pan has coins that are potentially heavy.

Let’s add an additional assumption to the original problem. Suppose we have an unlimited supply of coins that we know to be real. Let u(n) be the maximum number of coins we can process in n weighings if we do not know their potential.

What would be the first weighing? Both scales might be balanced, meaning that the fake coin is in the leftover pile of coins with unknown potential. So we have to leave out not more than u(n−1) coins. On the other hand, exactly one scale might be unbalanced. In this case, all the coins on this scale will get their potential known. The number of these coins can’t be more than 5^n-1. But this is an odd number, so we can use one extra real coin to make this number even, in order to put the same number of coins in each pan on this scale.

So u(n) = 2 · 5^n-1 + u(n−1), and u(1) = 3. This gives the answer of (5ⁿ+1)/2. Now we need to go back and remember that we got this bound using an additional assumption that we have an unlimited supply of real coins. Looking closer, we do not need our additional supply of real coins to be unlimited; we just need not more than two real coins. The good news is that we will have these extra real coins after the first weighing. The bad news is that for the first weighing we do not have extra real coins at all. So in the first weighing we should put unknown coins against unknown coins, not more than 5^n-1 on each scale, and as the number on each scale must be even, the best we can do is put 5^n-1−1 coins on each scale.

Thus the answer is (5ⁿ−3)/2 for n more than 1.

We can generalize this problem to any number of scales used in parallel. Suppose the number of scales is k. Suppose the number of weighings is more than 1, then the following problems can be solved in n weighings:

• If all the coins have known potential, then the maximum number of coins that can be resolved is (2k+1)ⁿ.
• If we do not know the potential of any coin and there is an unlimited supply of real coins, the maximum number of coins that can be solved is defined by a recursion: u(n) = k (2k+1)^n-1 + u(n−1) and u(1)=k + 1. So the answer is: ((2k+1)ⁿ+1)/2.
• If we do not know the potential of any coin and there is no extra real coins, then the answer is u(n) − k = ((2k+1)ⁿ+1)/2 − k.

The methods I described can be used to answer another common question in the same setting: Find the fake coin and say whether it is heavier or lighter. Let us denote by U(n) the number of coins that can be resolved in n weighings when there is an unlimited supply of extra real coins. Then the recurrence for U(n) is the same as the recurrence for u(n): U(n) = 2·5^n-1 + U(n−1). The only difference is in the initial conditions: U(1) = k. This means that U(n) = ((2k+1)ⁿ−1)/2. If we don’t have extra real coins then the answer is: U(n) = ((2k+1)ⁿ−1)/2 − k.

When we don’t need to say whether the fake coin is heavier or lighter, we can add one extra coin to the mix: the coin that doesn’t participate in any weighing and is fake if the scales always balance.

I valued my previous analog scale: I brought it from Europe and it showed my weight in kilograms.

I don’t remember why, a year ago, I decided to buy a digital scale: after all, my old scale is still functioning. But my new digital scale became an important instrument in my weight loss program.

Now I understand how my old scale and my lazy nature played tricks on my mind. For example, let’s say I decide that as soon as I reach one hundred kilograms (220 pounds), I would do something about it. Some time passes, I step on the scale, and it shows 100 kilograms — and maybe more. Is it more or not quite? If I lean to the right, it looks lower. The scale is not precise. So, I step back and adjust the scale at zero a little bit, to make sure it is not more than zero, but it actually could be slightly less. So I have reached my limit, but the scale’s imprecision allows me to pretend that there is a chance I am not over 100 kilograms, and thus do not have to do anything. More time passes, the scale shows 103 kilograms. I realize that I have been deceiving myself, but then it is too late for a fast fix that would lower my weight below 100. So I push the limit, and decide to wait until 105 kilograms.

My digital scale erased any ambiguity. The scale, however, has its own doubts. When I stand on it, the display flips between two numbers for a while. One of the numbers means no dinner tonight. But at the end the scale calls it: it spits out the final number with a beep. I had already decided that the scale is the boss. The flipping is irrelevant; the decision is irrevocable. If it means no dinner, so be it.

My new scale prevents inaction. It also allowed me to design my new plan: my Yellow Road. In this plan, my target weight decreases by 0.1 pounds per day. My behavior depends on my real weight with respect to my target weight. A small difference changes my behavior. So precision is essential.

As you can see in the picture the plan continues to work. I’ve lost 16 pounds since the start of the plan. Actually, I’ve lost 18 pounds, if I weigh myself without the camera.

I recently posted the following problem from the Moscow Olympiad:

There were n people at a meeting. It appears that any two people at the meeting shared exactly two common acquaintances.

• Prove that all the people have exactly the same number of acquaintances at this meeting.
• Show that n can be greater than 4.

Here is the proof for the first bullet. Choose a person X. Take a pair of X‘s acquaintances. These two acquaintances have to share two acquaintances between themselves one of whom is X. In other words, we have described a function from all pairs of X‘s acquaintances to people who are not X. On the other hand, for every person who is not X, s/he and X share a pair of acquaintances. Hence, there is a bijection between people other than X and all pairs of X‘s acquaintances. If the number of X‘s acquaintances is a, and the total number of people is t, then we have shown that (a choose 2) = t−1. As this is true for any X, we see that everyone has the same number of acquaintances. Moreover, this situation can happen only if t−1 is a triangular number.

But wait. There is more work that needs to be done. The smallest triangular number is 1. That means that t might be 2. If there are two people at the meeting, then the condition holds: they have 0 common acquaintances. The next triangular number is 3. So we need to see what would happen if there are four people. In this case, if everyone knows each other, it works. This is why the second bullet asks us to find an example of the situation with more than four people, because four people is too easy.

Let’s look at larger triangular numbers. The situation described in the problem might also happen when there are:

• 7 people total and everyone has 4 acquaintances,
• 11 people total and everyone has 5 acquaintances,
• 16 people total and everyone has 6 acquaintances,
• and so on: a(a−1)/2 + 1 people total and everyone has a acquaintances.

The official Olympiad solution suggests the following example for 16 people total. Suppose we put 16 people in a square formation so that everyone knows people in the same row and column. I leave it to the reader to check that every two people share exactly two acquaintances.

Let me prove that there is no solution for a total of seven people. If there were a solution, then each person would have to know four people. My first claim is that the acquaintance graph can’t contain a four-clique. Suppose there is a four-clique. Then each person in the clique has to have another acquaintance outside of the clique to make it up to four. In addition, this extra acquaintance can’t be shared with anyone in the clique, because the clique contains all the acquaintances that they share. This means we need to have at least four more people.

Next, suppose two people a and b know each other and share an acquaintance c. Any two people in this group of three has to have another shared acquaintance, who is not shared with the third person. That is, there should be another person who is the acquaintance of a and b, a different person who is an acquaintance of a and c, and a third person who is acquainted with b and c. These three extra people are all the acquaintances of a, b, and c. Which means the last person who is not acquainted with a, b, or c, has less than for four acquaintances.

Let’s look at a more difficult problem that I offered at the same posting:

There were n people at a meeting. It appears that any k people at the meeting shared exactly k common acquaintances.

• Prove that all the people have exactly the same number of acquaintances at this meeting.
• Is it possible that n can be greater than 2k?

As in the previous solution, we see that a, the number of acquaintances of a person and t, the total number of people, satisfy the following equation: (a choose k) = (t−1 choose k−1).

For example, if k = 3, the equation becomes (a choose 3) = (t−1 choose 2). This is a question of finding numbers that are both tetrahedral and triangular. They are known and their sequence, A027568, is finite: 0, 1, 10, 120, 1540, 7140. The corresponding number of acquaintances is 3, 5, 10, 22, 36 and the total number of people is 3, 6, 17, 57, 121. The first trivial example involves 3 people who do not know each other. The next example is also simple: it has 6 people and everyone knows everyone else.

What about non-trivial examples? If there are 17 people in the group, then each person has to know 10 people. Does the acquaintance graph exist so that every group of three people share 3 acquaintances?

We see that the problem consists of two different parts. First, we have to solve the equation that equates two binomial coefficients. And second, we need to build the acquaintance graph. Both questions are difficult. We see that for k = 2 we have an infinite number of solutions to the equation with binomial coefficients. For k = 3, that number is finite. What happens with other k? If there are 2k people and they all know each other, then this works. But are there other non-trivial solutions? I am grateful to Henry Cohn for directing me to the works of Singmaster who studied non-trivial repetitions of numbers in Pascal’s triangle. In particular, Singmaster showed that the equation (n+1 choose k+1) = (n choose k+2) has infinitely many solutions given by n = F_2i+2F_2i+3−1 and k = F_2iF_2i+2−1.

This sequence generates the following non-trivial examples (15 choose 5) = (14 choose 6), (104 choose 39) = (103 choose 40), and so on. That means it might be possible that there is a group of 16 people so that every 6 people share 6 acquaintances. In this situation every person must know everyone else except for one other person. That leads us to the structure of the acquaintance graph: it is a complement to the perfect matching graph. I leave it to my readers to check that the corresponding acquaintance graph doesn’t exist. Are there examples of two binomial coefficients that equal each other and that lead to an acquaintance graph that can be built?

Now that I’ve tackled the solution to this Olympiad problem, I see that I generated more questions than I answered.

Many years ago at Gelfand’s seminar in Moscow, USSR, someone pointed out a young girl and told me: “This is Natalia Grinberg. In her year in the math Olympiads, she was the best in the country. She is the next you.”

We were never introduced to each other and our paths never crossed until very recently.

Several years ago I became interested in the fate of the girls of the IMO (International Math Olympiad). So, I remembered Natalia and started looking for her. If she was the best in the USSR in her year, she would have been a gold medalist at the IMO. But I couldn’t find her in the records! The only Grinberg I found was Darij Grinberg from Germany who went to the IMO three times (2004, 2005, and 2006) and won two silver medals and one gold.

That was clearly not Natalia. I started doubting my memory and forgot about the whole story. Later I met Darij at MIT and someone told me that he was Natalia’s son.

I was really excited when I received an email from Natalia commenting on one of my blog posts. We immediately connected, and I asked her about past events.

Natalia participated in the All-Soviet Math Olympiads three times. In 1979 as an 8th grader she won a silver medal, and in 1980 and 1981 she won gold. That indeed was by far the best result in her year. So she was invited to join the IMO team.

That year the IMO was being held in the USA, which made Soviet authorities very nervous. At the very last moment four members of the team did not get permission to travel abroad. Natalia was one of them. The picture below, which Natalia sent to me, was taken during the Soviet training camp before the Olympiad. These four students were not allowed to travel to the IMO: Natalia Grinberg, Taras Malanyuk, Misha Epiktetov, and Lenya Lapshin.

Because of the authorities’ paranoia, the Soviet team wasn’t full-sized. The team originally contained eight people, but as they rejected four, only six traveled to the USA, including two alternates.

I have written before how at that time the only way for a Jewish student to get to study mathematics at Moscow State University was to get to the IMO. I wrote a story about my friend Sasha Reznikov who trained himself to get to the IMO, but because of some official machinations, still was not accepted at MSU.

Natalia’s story surprised me in another way. She didn’t get to the IMO, but she was accepted at MSU. It appears that she was accepted at MSU as a member of the IMO team, because that decision was made before her travel documents were rejected.

Natalia became a rare exception to the rule that the only way for a Jewish person to attend MSU was to participate in the IMO. It was a crack in the system. They had to block visas at the last moment, so that people wouldn’t have time to make a fuss and do something about it. Natalia slipped through the crack and got to study at the best university in the Soviet Union.

Unfortunately, the world lost another gold IMO girl. Three Soviet team members won gold medals that year. Natalia, being better then all of them, would have also won the gold medal.

I recently published the following coin puzzle:

There are four silver coins marked 1, 2, 3, and 5. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is lighter than it should be. Find the fake coin using the balance scale twice.

My readers, David Reynolds and ext_1973756, wrote to me that I am missing a coin of 4 grams. Indeed, the same puzzle with five coins—1, 2, 3, 4, and 5—is a more natural and a better puzzle.

David Reynolds also suggested to go all the way up to 9 coins:

There are nine silver coins marked 1, 2, 3, 4, 5, 6, 7, 8, and 9. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is lighter than it should be. Find the fake coin using the balance scale twice.

It is impossible to resolve this situation with more than nine coins as two weighings provide nine different answers to differentiate coins. But indeed it is possible to solve this problem for nine coins. It is even possible to suggest a non-adaptive algorithm, that is to describe the weighings before knowing the results.

To find such a strategy we need to satisfy two conditions. First, we have to weigh groups of coins of the same supposed weight, otherwise we do not get any useful information. Second, there shouldn’t be any two coins together (in or out of the pan) in both weighings, because it would then be impossible to differentiate between them.

Here is one possible solution of the problem:

• The first weighing: 1, 5, and 9, against 2, 6, and 7
• The second weighing: 1, 6, and 8, against 2, 4, and 9

David Reynolds also suggested a problem in which we do not know whether the fake coin is heavier or lighter:

There are four silver coins marked 1, 2, 3, and 4. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is either lighter or heavier than it should be. Find the fake coin using the balance scale twice.

Again, four coins is the best we can do when in addition to find it, we also want to determine if it is heavier or lighter. Indeed, if there were five coins we would have needed to cover ten different answers, which is too many for two weighings.

Here is the solution for four coins:

The two weighings are 1+3=4, and 1+2=3. If the first weighing balances, then the fake coin is 2 and the second weighing shows if it is heavier or lighter than it should be. Similarly, if the second weighing balances, then the fake coin is four and we can see whether it is heavier or lighter than it should be. If the left pan is lighter/heavier for both weighings, then the fake coin is 1 and is lighter/heavier. But if one pan is heavier on the first of two weighings and the other pan is heavier on the second weighing, then the fake coin is 3. In both cases it is easy to determine whether the fake coin is heavier or lighter.

Now David is missing a coin. If we just want to find the fake coin without determining whether it is heavier of lighter, we can do it with five coins:

There are five silver coins marked 1, 2, 3, 4, and 5. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is either lighter or heavier than it should be. Find the fake coin using the balance scale twice.

We can use the same solution as the previous (four coins) problem. If the scale balances both times, then the fake coin is 5. However, in this case we will not know whether the coin is heavier or lighter.

We can’t extend this problem to beyond five coins. Suppose we have six coins. We can’t use more than three coins in the first weighing. This is because if the scale unbalances, we can’t resolve more than three coins in one remaining weighing. Suppose the first weighing balances; then we have at least three leftover coins we know nothing about and one of them is fake. These three coins should be separated for the next weighing. That means one of the coins needs to be on the left pan and one on the right pan. We can add real coins any way we need. But if the second weighing unbalances we do not know if the fake coin is on the left and lighter or on the right and heavier.

Two interesting research results about male homosexuality are intertwined. The first one shows that the probability of homosexuality in a man increases with the number of older brothers. That is, if a boy is the third son in a family, the probability of him being a homosexual is greater than the probability of a first son in a family being homosexual. The second research result shows that the probability of homosexuality increases with the number of children the mother has. So if a woman is fertile and has many children, the probability that each of her sons is a homosexual is greater than the probability that an only child is a homosexual.

Many people conclude from the first result that a woman undergoes hormonal or other changes while being pregnant with boys that influence the probability of future boys being homosexual. Looking at the second result, researchers conclude that homosexuality has a genetic component. Moreover, that component is tied up with the mother’s fecundity. The same genes are responsible for both the mother having many children and for her sons being homosexual. This assumption explains why homosexuality is not dying out in the evolution process.

In one of my previous essays I showed that the first results influences the second result. If each next son is homosexual with higher probability, then the more children a mother has the more probable it is that her sons are homosexuals. That means that the second result is a mathematical consequence of the first result. Therefore, the conclusion that the second result implies a genetic component might be wrong. The correlation between homosexuality and fecundity could be the consequence of hormonal changes.

Now let’s look at this from the opposite direction. I will show that the first result is the mathematical consequence of the second result: namely, if fertile women are more probable to give birth to homosexuals, then the probability that the second sons are is higher than the probability that the first sons are gay.

For simplicity let’s only consider mothers with one or two boys. Suppose the probability of a son of a one-son mother to be a homosexual is p₁. Suppose the probability of a son of a two-sons mother to be a homosexual is p₂. The data shows that p₂ is greater than p₁. What is the consequence? Suppose the number of mothers with one son is m₁ and the number of mothers with two sons is m₂. Then in the whole population the probability of a boy who is the first son to be gay is (p₁m₁+p₂m₂)/(m₁+m₂) and the probability of a boy who is the second son to be gay is p₂. It is easy to see that the first probability is smaller than the second one.

Let me create an extreme hypothetical example. Suppose mothers of one son always have straight sons, and mothers of two sons always have gay sons. Now consider a random boy in this hypothetical setting. If he’s the second son, he is always gay, while if he is the first son he is not always gay.

We can conclude that if the probability of having homosexual sons depends on fecundity, then the higher numbered children would be gay with higher probability than the first-born. This means that if the genetics argument is true and being a homosexual depends on the mother’s fecundity gene, then it would follow mathematically that the probability of homosexuality increases with birth order. The conclusion that homosexuality depends on hormonal changes might not be valid.

So what is first, chicken or egg? Is homosexuality caused by fecundity, while birth order correlation is just the consequence? Or vice versa? Is homosexuality caused by the birth order, while correlation with fecundity is just the consequence?

What do we do when the research results are so interdependent? To untangle them we need to look at the data more carefully. And that is easy to do.

To show that homosexuality depends on the order of birth independently of the mother’s fertility, we need to take all the families with two boys (or the same number of boys) and show that in such families the second child is more probable to be homosexual than the first child.

To show the dependence on fertility, without the influence of the birth order, we need to take all first-born sons and show that they are more probable to be homosexuals if their mothers have more children.

It would be really interesting to look at this data.

My new favorite hat puzzle was invented by Konstantin Knop and Alexander Shapovalov. It appeared (in a different wording) in March 2013 at the Tournament of the Towns:

A sultan decides to give 100 of his sages a test. The sages will stand in line, one behind the other, so that the last person in the line sees everyone else. The sultan has 101 hats, each of a different color, and the sages know all the colors. The sultan puts all but one of the hats on the sages. The sages can only see the colors of the hats on people in front of them. Then, in any order they want, each sage guesses the color of the hat on his own head. Each hears all previously made guesses, but other than that, the sages cannot speak. They are not allowed to repeat a color that was already announced. Each person who guesses his color wrong will get his head chopped off. The ones who guess correctly go free. The rules of the test are given to them one day before the test, at which point they have a chance to agree on a strategy that will minimize the number of people who die during this test. What should that strategy be?

I loved it so much that I wrote a paper about it. You can find the solution there.

I particularly like these two problems that I gave my AMSA students for homework:

Athos, Porthos, and Aramis were rewarded with six coins: three gold and three silver. Each got two coins. Athos doesn’t know what kind of coins the others got, but he knows his own coins. Ask him one question to which he can answer “Yes,” “No,” or “I do not know,” so that you will be able to figure out his coins.

There are four silver coins marked 1, 2, 3, and 5. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is lighter than it should be. Find the fake coin using the balance scale twice.