Binary options for mironov reviews

This blog has a modern version at http: This page contains the copies of recent blog entries. The older copies are here:. How to restore justice: Create a folder named Justice.

Go to the trash bin and click restore. When you sign up, you are friends with everyone. Then you send un-friend requests. I already wrote about two puzzles that Derek Kisman made for the MIT Mystery Hunt. The first puzzle is now called the Fractal Word Search. It is available on the Hunt website under its name In the Details. I posted one essay about the puzzle and another one describing its solution. Unfortunately, the puzzle is not available, but my description of it is.

Today let's look at the third puzzle Derek made for the Hunt, building on an idea from Tom Yue. This is a non-mathematical crossword puzzle. Derek tends to write multi-layered puzzles: You think you've got the answer, but the answer you've got is actually a hint for the next step.

Often multi-layered puzzles get solvers frustrated, but the previous paragraph is a hint in itself. If you expect the difficulty, you might appreciate the fantastic beauty of this puzzle. Welcome to Ex Post Facto. Every time I visit Princeton, or otherwise am in the same city as my friend John Conway, I invite him for lunch or dinner.

I have this rule for myself: I invite, I pay. If we are in the same place for several meals we alternate paying. Once John Conway complained that our tradition is not fair to me. From time to time we have an odd number of meals per visit and I end up paying more. I do not trust my memory, so I prefer simplicity. I resisted any change to our tradition. We broke the tradition only once, but that is a story for another day. Let's discuss the mathematical way of paying for meals.

Many people suggest using the Thue-Morse sequence instead of the alternating sequence of taking turns. When you alternate, you use the sequence ABABAB…. If this is the order of paying for things, the sequence gives advantage to the second person. So the suggestion is to take turns taking turns: If you are a nerd like me, you wouldn't stop here. This new rule can also give a potential advantage to one person, so we should take turns taking turns taking turns.

Continuing this to infinity we get the Thue-Morse sequence: Some even call this sequence a fair-share sequence. Should I go ahead and implement this sequence each time I cross paths with John Conway? Actually, the fairness of this sequence is overrated. I probably have 2 or 3 meals with John per trip. If I pay first every time, this sequence will give me an advantage. It only makes sense to use it if there is a very long stretch of meals. This could happen, for example, if we end up living in the same city.

But in this case, the alternating sequence is not so bad either, and is much simpler. Many people suggest another use for this sequence. Suppose you are divorcing and dividing a huge pile of your possessions. A wrong way to do it is to take turns. First Alice choses a piece she wants, then Bob, then Alice, and so on. Alice has the advantage as the first person to choose. An alternative suggestion I hear in different places, for example from standupmathsis to use the Thue-Morse sequence. I don't like this suggestion either.

If Alice and Bob value their stuff differently, there is a better algorithm, called the Knaster inheritance procedurethat allows each of them to think they are getting more than a half. If both of them have the same value for each piece, then the Thue-Morse sequence might not be good either. Suppose one of the pieces they are dividing is worth more than everything else put together. Then the only reasonable way to take turns is ABBBB….

The beauty of the Thue-Morse sequence is that it works very well if there are a lot of items and their consecutive prices form a power function of a small degree ksuch as a square or a cube function.

You might think that if the sequence of prices doesn't grow very fast, then using the Thue-Morse sequence is okay. Here is the sequence of prices that I specifically constructed for this purpose: Alice gets an extra 1 every time she is in the odd position.

This is exactly half of her turns. That is every four turns, she gets an extra 1. If the prices grow faster than a power, then the sequence doesn't work either. Suppose your pieces have values that form a Fibonacci sequence. Take a look at what happens after seven turns. We see that Alice gets more by F n This value is bigger than the value of all the leftovers together.

I suggest a different way to divide the Fibonacci-priced possessions. If Alice takes the first piece, then Bob should take two next pieces to tie with Alice. So the sequence might be ABBABBABB…. I can combine this idea with flipping turns. So we start with a triple ABB, then switch to BAA. After that we can continue and flip the whole thing: Then we flip the whole thing again.

And again and again. At the end we get a sequence that I decided to call the Fibonacci fair-share sequence. A while ago I posted my second favorite problem from the All-Russian Math Olympiad:. Now it's solution time.

First we show that we can do this in 70 weighings. The strategy is to compare one coin against one coin. If the scale balances, we are lucky and can stop, because that means we have found two real coins. If the scale is unbalanced, the heavier coin is definitely fake and we can remove it from consideration. In the worst case, we will do 70 unbalanced weighings that allow us to remove all the fake coins, and we will find all the real coins. The more difficult part is to show that 69 weighings do not guarantee finding the real coin.

We do it by contradiction. Suppose the weights are such that the real coin weighs 1 gram and the i -th fake coin weighs i grams. That means whatever coins we put on the scale, the heaviest pan is the pan that has the fake coin with the largest index among the fake coins on the scale. Suppose there is a strategy to find a real coin in 69 weighings. Given this strategy, we produce an example designed for this strategy, so that the weighings are consistent, but the collector cannot find a real coin.

For the first weighing we assign the heaviest weight, 70 to one of the coins on the scale and claim that the pan with this coin is heavier.

If a weighing has the coins with assigned weights, we pick the heaviest coin on the pans and claim that the corresponding pan is heavier. If there are no coins with assigned weights on pans, we pick any coin on the pans, assigned the largest available weight to it and claim that the corresponding pan is heavier.

After 69 weighings, not more than 69 coins have assigned weights, while all the weighings are consistent. The rest of the coins can have any of the leftover weights. For example, any of the rest of the coins can weigh grams. That means that there is no coin that is guaranteed to be real. I stumbled upon a couple of problems that I like while scanning the Russian website of Math Festival in Moscow The problems are for 7 graders.

This problem is really very difficult. The competition organizers offered an extra point for every diagonal on top of The official solution has 24 diagonals, but no proof that it's the longest.

I'm not sure anyone knows if it is the longest. The year is The man on the left is my first husband, Alexander Goncharov.

Although we were out of touch for a decade, when I married my third husband, Joseph Bernstein on the rightGoncharov started visiting us. It wasn't me he was interested in: I found this situation hilarious, so I took this photo. But that's not all. My second husband, Andrey Radul, is not in the picture. But all four of us were students of Israel Gelfand. In short, my three ex-husbands and I are mathematical siblings — that is, we are all one big happy mathematical family.

The Best Writing on Mathematics is out. I am happy that my paper The Pioneering Role of the Sierpinski Gasket is included. The paper is written jointly with my high-school students Eric Nie and Alok Puranik as our PRIMES project. At the end of the book there is a short list of notable writings that were considered but didn't make it. The "short" list is actually a dozen pages long. And it includes two more papers of mine: Cookie Monster Plays Games —another PRIMES project written jointly with my high-school student Joshua Xiong.

Mathematical Research in High School: The PRIMES Experiencewritten jointly with PRIMES Chief Research Advisor Pavel Etingof and PRIMES Program Director Slava Gerovitch. To continue bragging, I want to mention that my paper A Line of Sages was on the short list for volume.

And my paper Conway's Wizards was included in the volume. I like Odd-One-Out puzzles that are ambiguous. That is why I bought the book Which One Doesn't Belong? Look at the cover: The book doesn't include answers, but it has nine more examples in each of which there are several possible odd-one-outs. I married an American citizen and moved to the US in At the time I was a very patriotic Russian. It took me a year of pain to realize that some of my ideas had been influenced by Soviet propaganda.

After I washed away the brainwashing, I fell in love with the US. For 25 years I thought that America was great. For the last several months I've been worried as never before in my life. I feel paralyzed and sick. To help myself I decided to put my feelings in words. My mom was 15 when World War II started.

The war affected her entire life, as well as the lives of everyone in the USSR. Every now and then my mom would tell me, "You are lucky that you are already 20 and you haven't witnessed a world war. From time to time I tell myself something like, "I am lucky that halfway through my expected lifetime, I haven't had to live through a world war. To maintain the peace is a difficult job.

Everything needs to be in balance. Trump is disrupting this balance. I am worried sick that my children or grandchildren will have to witness a major war. I've noticed that, as a true showman, Trump likes misdirecting attention from things that worry him to fantastic plot twists that he invents.

What's the best way to make people forget about his tax returns? It's the nuclear button. Dropping a nuclear bomb some place will divert people from thinking about his tax returns. As his plot twists are escalating, is he crazy enough to push the button? The year was the warmest on record. The year was even warmer.

And last year,was even warmer than that. I remember Vladimir Arnold's class on differential equations. He talked about a painting that had been hanging on a wall for 20 years.

Then it unexpectedly fell off. Mathematics can explain how such catastrophic events can happen. I keep thinking about our Earth: My grandchildren might not be able to enjoy beaches and forests the way I did.

What if, like the fallen painting, the Earth can spiral out of control and completely deteriorate? But Trump is ignoring the climate issues. Does he care about our grandchildren? I am horrified that Trump's policies will push climate catastrophe beyond the point of no return. Wiretapping is not wiretapping. Phony jobs numbers stopped being phony as soon as Trump decided that he deserved the credit.

The news is fake when Trump doesn't like it. Trump is a pathological liar; he assaults the truth. Being a scientist I am in search of truth, and Trump diminishes it.

I do not understand why people ignore his lies. Two plus two is four whether you are a democrat, or a republican, or whomever. Facts are facts, alternative facts are lies. I am scared that lies have become acceptable and no one cares about the truth any more.

I lived in Russia for the first unhappy half of my life, and in the US for the second happy half. I do not want to go back. There is something fishy between Putin and Trump. Whether it is blackmail or money, or both, I do not know the details yet, But Trump is under Putin's influence.

Trump didn't win the elections: I do not want to go back to being under Russian rule. I grew up in a country where the idea of a good husband was a man who wasn't a drunkard. That wasn't enough for me. I dreamed of a relationship in which there would be an equal division of work, both outside and inside the home. I could not achieve that because in Russian culture both people work full-time and the wife is solely responsible for all the house chores.

Moreover, Russia was much poorer than the US: The life in USSR was really unfair to women. Most women had a full-time job and several hours of home chores every day.

When I moved to the US, I thought I was in paradise. Not only did I have diapers and a washing machine, I was spending a fraction of the time shopping, not to mention that my husband was open to helping me, and didn't mind us paying for the occasional babysitter or cleaner. For some time I was blind to gender issues in the US because it was so much better. Then I slowly opened my eyes and became aware of the bias.

For some years it has felt like gender equity was improving. Now, with a misogynistic president, I feel that the situation might revert to the dark ages. When women are not happy, their children are not happy, and they grow up to be not happy. If the pursuit of happiness is the goal, the life has to be fair to all groups.

But Trump insults not only women but also immigrants, Muslims, members of the LGBTQ community, as well as the poor and the sick. The list is so long, that almost everyone is marginalized.

This is not a path towards a happy society. Trump attacks the press and attempts to exclude them. Trump has insulted the intelligence community and the courts. He seems to be trying to take more power to the presidency at the expense of the other branches of government. He ignores his conflicts of interest. Trump disregards every rule of democracy and gets away with it.

I am horrified that our democracy is dying. Trump's tax returns could either exonerate him or prove that he is Putin's puppet. The fact that he is hiding the returns makes me believe that the latter is more probable. Why the Republicans refuse to demand to see his returns is beyond my understanding. Trump does so many unethical things. Most of his decisions as president seem to be governed by Trump trying to get richer. Let us consider his hotel in Azerbaijan—a highly corrupt country.

Having lived in a highly corrupt country myself, I know how it works. For example, an Azerbaijani government official who has access to their country's money can make a deal that involves a personal kickback. This means that their government is paying more than necessary for a service or product in order to cover that kickback. This is how national money makes its way into individual pockets. Since all the deals in Azerbaijan are reputed to be like that, I imagine that when Trump built his hotel there, the Trump organization was overpaid in order to cover the bribe to local officials.

Will our country become as corrupt as Azerbaijan? The biggest shock of the election was that so many people were so gullible and actually voted for Trump. They didn't see that his agenda is focused on his own profit, and that he lies and makes promises he doesn't plan to deliver. It really terrifies me that there are some people who are not gullible but still voted for Trump. I recently wrote about my way of playing Nim against a player who doesn't know how to play.

If my move starts in an N-position, then I obviously win. If my move starts in a P-position, I would remove one token hoping that more tokens for my opponent means more opportunity for them to make a mistake. But which token to remove?

Does it make a difference from which pile I choose? Consider the position 2,4,6. If I take one token, my opponent has 11 different moves. If I choose one token from the first or the last pile, my opponent needs to get to 1,4,5 not to lose. If I choose one token from the middle pile, my opponent needs to get to 1,3,2 not to lose. But the first possibility is better, because there are more tokens left, which gives me a better chance to have a longer game in case my opponent guesses correctly.

That is the strategy I actually use: I take one token so that the only way for the opponent to win is to take one token too.

This is a good heuristic idea, but to make such a strategy precise we need to know the probability distribution of the moves of my opponent. At least one of them goes to a P-position. If there are 2 or 3 heaps, then the best strategy is to go for the longest game. Something interesting happens if there are more than three heaps.

In this case it is possible to have more than one winning move from a N-position. It is not obvious that I should play the longest game. If I remove one token, then my opponent has three winning moves to a position with 14 tokens. On the other hand, if I remove 2 tokens from the second or the fourth pile, then my opponent has one good move, though to a position with only 12 tokens.

What should I do? I leave it to my readers to calculate the optimal strategy against a random player starting from position 1,3,5,7. Alex Bellos wrote a puzzle book Can You Solve My Problems?

Ingenious, Perplexing, and Totally Satisfying Math and Logic Puzzles The book contains a mixture of famous puzzles and their solutions. Some of the puzzles are not mathematical in the strictest sense, but still have an appeal for mathematicians. For example, which integer comes up first when you alphabetize all the integers up to a quadrillion? Recognize the puzzle on that book cover? That's my Odd One Out puzzle.

Doesn't it look great in lights on that billboard in London? Mine isn't the only terrific puzzle in the book. In fact, one of the puzzles got my special attention as it is related to our current PRIMES polymath project. The beautiful Pascal triangle has been around for many years.

Can you say something new about it? Of course you can. Mathematicians always find new way to look at things. In RSI student, Kevin Garbe, did some new and cool research related to the triangle. Consider Pascal's triangle modulo 2, see picture which was copied from a stackexchange discussion. A consecutive block of m digits in one row of the triangle modulo 2 is called an m -block.

If you search the triangle you will find that all possible binary strings of length 2 are m -blocks. Will this trend continue? Yes, you can find any possible string of length 3, but it stops there.

The blocks you can find are called accessible blocks. So, which blocks of length 4 are not accessible? There are only two strings that are not accessible: It is not surprising that they are reflections of each other.

Pascal's triangle respects mirror symmetry and the answer should be symmetric with respect to reflection. You can't find these blocks on the picture, but how do we prove that they are not accessible, that is, that you can't ever find them? The following amazing property of the triangle can help.

Every odd row has every digit doubled. Moreover, if we take odd rows and replace every double digit with its single self we get back Pascal's triangle.

Obviously the two strings and can't be parts of odd rows. What about even rows? The even rows have a similar property: If you remove these zeros you get back Pascal's triangle.

The two strings and can't be part of even rows. Therefore, they are not accessible. The next question is to count the number of inaccessible blocks of a given length: This and much more was done by Kevin Garbe for his RSI project. I was the head mentor of the math projects. His paper is published on the arxiv.

I wonder if there exists a direct proof of this formula without considering odd and even rows separately. This RSI result was so pretty that it became a question at our entrance PRIMES test for the year In the test we changed the word accessible to admissible, so that it would be more difficult for applicants to find the research.

Besides, Garbe's paper wasn't arxived yet. The pretty picture above is from the stackexchange, where one of our PRIMES applicants tried to solicit help in solving the test question. Ten people are sitting around a round table.

Some of them are knights who always tell the truth, and some of them are knaves who always lie. Two people said, "Both neighbors of mine are knaves. Today at least three members of the English club came to the club. Following the tradition, each member brought their favorite juice in the amount they plan to drink tonight. By the rules of the club, at any moment any three members of the club can sit at a table and drink from their juice bottles on the condition that they drink the same amount of juice.

Prove that all the members can finish their juice bottles tonight if and only if no one brings more than the third of the total juice brought to the club. Three piles of nuts together contain an even number of nuts. One move consists of moving half of the nuts from a pile with an even number of nuts to one of the other two piles.

Prove that no matter what the initial position of nuts, it is possible to collect exactly half of all the nuts in one pile. N people crossed the river starting from the left bank and using one boat. Each time two people rowed a boat to the right bank and one person returned the boat back to the left bank. Before the crossing each person knew one joke that was different from all the other persons' jokes.

While there were two people in the boat, each told the other person all the jokes they knew at the time. For any integer k find the smallest N such that it is possible that after the crossing each person knows at least k more jokes in addition to the one they knew at the start.

Spoiler for Problem 2. I want to mention a beautiful solution to problem 2. Let's divide a circle into n arcs proportionate to the amount of juice members have. Let us inscribe an equilateral triangle into the circle. In a general position the vertices of the triangle point to three distinct people.

These are the people who should start drinking juices with the same speed. We rotate the triangle to match the drinking speed, and as soon as the triangle switches the arcs, we switch drinking people correspondingly. After degree rotation all the juices will be finished.

I already posted a funny true story that Smullyan told me when I last visited him. Raymond Smullyan died recently at the age of 97 and my mind keeps coming back to this last visit. The year was and I was about to drive back to Boston after my talk at Penn State.

Smullyan's place in the Catskills was on the way—sort of. I wanted to call him, but I was apprehensive. Raymond Smullyan had a webpage on which his email was invisible.

You could find his email address by looking at the source file or by highlighting empty space at the bottom of the page. Making your contact information invisible sends a mixed message. While this was a little eccentric, it meant that only people who were smart enough to find it, could access his email address.

I already knew his email because he had given it to me along with his witty reply to my blog post about our meeting at the Gathering for Gardner in In our personal interactions, he always seemed to like me, so I called Raymond and arranged a visit for the next day around lunch time. When I knocked on his door, no one answered, but the door was open, and since Smullyan was expecting me, I walked right in. So I sat down in his library and picked up a book.

When he woke up, he was happy to see me, and he was hungry. He told me that he didn't eat at home, so we should go out together for lunch. I was hungry too, so I happily agreed. Then he said that he wanted to drive. I do not have a poker face, so he saw the fear in me.

My only other trip with a nonagenarian driver flashed in front of my eyes. The driver had been Roman Totenberg and it had been the scariest drive I have ever experienced. I said that I wanted to drive myself. Annoyed, Raymond asked me if I was afraid of him taking the wheel. I told him that I have severe motion sickness and always prefer to drive myself. Raymond could see that I was telling the truth.

I got the impression that he was actually relieved when he agreed to go in my car. We went to Selena's Diner. He took out playing cards with which he showed me magic tricks. I showed him some tricks too. This was probably a bad move as he abandoned me to go to the neighboring table to show his magic tricks to a couple of young girls. They were horrified at first"his unruly hair, his over-the-top energy, his ebullient behavior"but between me and the waitress, we quickly reassured them.

The girls enjoyed the tricks, and I enjoyed my visit. Like many people, I was appalled by Trump's immigration ban. On the Internet I found many essays that explained that he did not include in the ban those majority-Muslim countries in which he has business interests. See for example, an article at Forbes with a nice map, and an article at NPR.

Now the countries that are excluded are motivated to continue to support Trump's businesses, and to offer him bribes and good deals in exchange for staying out of the ban.

The countries on the list are also motivated to approach Trump and offer him a sweet business deal. So even if the courts stopped the ban, he has already succeeded in showing every country in the world that to be on his good side requires that they pay up. And China got the hint and granted Trump a trademark he's been seeking for a decade. Here is my son Sergei's solution. Divide the coins into nine groups of equal size and number the groups in ternary: On each scale we put three groups versus three groups.

On the first scale we compare the three groups that start with 1 with the three groups that start learn to be a forex trader in india 2. For the second scale we do the same using the last digit instead of the first one, and for the third scale we use the sum of two digits modulo 3. Any pair of scales, if they are assumed to be normal, pivot point trading tips point to one out of nine groups as the group containing the fake coin.

If all three pairs of scales agree on one group, then this is the group containing the fake coin. Thus in three weighings, we reduce the number of groups of coins by a factor of nine. If the pairs of scales do euro sterling exchange rate february 2010 agree, then the random scale produced a wrong weighing and thus can be found out. How do we do that? We have three out of nine groups of coins each of dnkn stock options might contain the fake coin.

We compare two of the groups on all three scales. This way we know exactly which group contains the fake coin and, consequently, which scale generated a wrong weighing.

If we know the random scale, we can speed up the rest of the process of finding the fake coin. The big idea here is that as soon as the random scale shows a wrong weighing result it can be found out. So in the worst case, the random scale behaves as a normal scale and messes things up at the very end.

Stock brokers north plainfield you do android dialer with video call option The paper contains an even stronger solution that provides a better asymptotics.

The first time I visited the US was in at the invitation of an old nedbank namibia forex, Joseph Bernstein. After my arrival Joseph proposed commonwealth bank foreign exchange rates fees I accepted, but my essay is not about that.

Joseph reintroduced me to his daughter, Mira, who was then in her late teens.

I was struck by Mira's charm. I had never before met teenagers like her. Of course, Joseph got points for that as I was hoping to have a child with him. When I moved to the US I met some other kids who were also incredibly charming. It was too late to take points away from Joseph, but it made me realize what a huge difference there was between Soviet and American teenagers.

American teenagers were happier, more relaxed, better mannered, and less cynical than Soviet ones. My oldest son, Alexey, was born in the USSR and moved to the US when he was eight. One unremarkable day when he was in middle school Baker public school in Brooklinethe principle invited me for a chat. I came to the school very worried. The principal explained to me that there was a kid who was bugging Alexey and Option put spread definition pushed him back with a pencil.

While the principal proceeded to explain the dangers of a pencil, I tuned out. I needed all my energy to conceal my happy smile. This was one of the happiest moments of my life in the US.

What a great country I live in where the biggest worry of a principal in a middle school is the waving of a pencil! I remembered Alexey's prestigious school in Moscow. They had fights every day that resulted in bloody noses and lost teeth. When I complained to his Russian teacher, she told me that it was not her job to supervise children during big breaks. Plus the children needed to learn to be tough. No wonder American children are happier. I was wondering if there were any advantages to a Soviet upbringing.

For one thing, Soviet kids grow up earlier and are less naive. They are more prepared for harsh realities than those American kids who are privileged. Naive children grow up into naive adults. Naive adults become naive presidents. I watched with pain as naive Bush "I looked the man in the eye. I found him to be very straightforward and trustworthy. Putin is and, according to Forbes Magazine, has been for the last four years the most powerful person in the world. Even though the US kept its distance from Russia, he was able to manipulate us from afar.

Now that Forex calendar app iphone wants to be close to Putin, the manipulation will be even easier. Putin is better at this game. He will win and we will lose. How many times during the 20th century the date in the format month day and the last two digits of the year can be written with the same digit? I had a distant relative Alla, who was brought up by a single mother, who died in a car crash when the girl was in her early teens.

Alla was becoming a sweet and pleasant teenager; she was taken in by her aunt after the accident. Very soon the aunt started complaining that Alla was turning into a cheater and a thief. The aunt found a therapist for Alla, who explained that Alla was stealing for a reason. Because the world had unfairly stolen her mother, Alla felt entitled to compensation in the form of jewelry, money, and other luxuries.

I was reminded of Alla's story when I was reading The Honest Truth About Dishonesty: How We Lie to Everyone—Especially Ourselves by Dan Ariely.

Ariely discusses a wide range of reasons why honest people cheat. But to me he neglects to look at the most prominent reason. Often honest people cheat when they feel justified and entitled to do so.

One group was asked to write a text avoiding letters x and z. The other group was asked to write a text avoiding letters a and n. The second task is way more difficult and requires more energy.

After the tasks were completed the participants were given a test in which they had a chance to cheat. For this experiment, the participants were compensated financially according to the number of questions they solved. Not surprisingly, the second group cheated more. The book concludes that when people are tired, their guard goes down and they cheat more. I do not argue with this conclusion, but I think another reason also contributes to cheating.

Have you ever intrinsic value call options to write a text without using the letters a and n? Fun as it is, this is cruel and unusual punishment. The request is more difficult than most people expect at an experiment. It could be that participants cheated not only because their capacity for honesty was depleted, but because they felt entitled to more money because the challenge was so difficult.

In another experiment, the participants received a high-fashion brand of sunglasses before day trading and swing trading the currency market pdf download test. Some of them were told that the sunglasses were a cheap imitation of the luxury brand when they really were not.

This group cheated more than the group asian stock markets allstocks thought that they got a real thing.

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The book concludes that wearing fake sunglasses makes people feel that they themselves are fake and so they care less about their honor. Unfortunately, the for stock options prices on excel template doesn't explain market quote stock 2016 vs 2016.5 detail what was actually promised.

It looks like the participants were promised high-fashion sunglasses. In this case, the fake group would have felt deceived and might have felt more justified to cheat. May I suggest the following experiment. Invite people and promise them some money for a minute task. Pay them the promised minimum and give them a test through which they can earn more. Construct it so that they can earn a lot more if they cheat. Then make the non-control group wait for half an hour.

If I were in this group, I would have felt that I am owed for the total of 45 minutes—three times more than what I was promised. I do not know if I would cheat or just leave, but I wouldn't be surprised that in this group people would cheat more than in the control group. If like me, you fancy Raymond Smullyan and his books, then you've heard about knights and knaves.

Knights always tell the truth and knaves always lie. In addition to knights and knaves, there are normal people who sometimes tell the truth and sometimes lie. Here is a puzzle. We can draw a parallel between people and coins. We can say that knights correspond to real coins, and knaves to fake coins that are lighter than real ones. Inspired by normal people, my coauthor Konstantin Knop invented chameleon coins.

Chameleon coins can change their weight and behave like real or fake coins. I just wrote a post about chameleon coins. Normal people are too unpredictable: So logicians invented a simpler type of person, one who switches from telling the truth in one sentence to a lie in the next and then back to the truth.

Such people are called alternators. Here is another puzzle:. Continuing a parallel between people and coins we can define alternator coins: For the purposes of this essay, we assume that the fake coins are lighter than real ones.

Unlike the chameleon coin, which might never reveal itself by always pretending to be real, nti work at home complaints alternators can always be found.

How do you find a single alternator among many real coins? There is a simple strategy: This strategy allows us to find an alternator among 9 coins in four weighings. Can we do better? I used the alternator coins as a research project for my PRIMES STEP program where we do math research with students in seventh and eighth grade.

The students started the alternator project and immediately discovered the strategy above. The next step is sbi forex rates singapore describe a better strategy. For example, what is the maximum number of coins containing one alternator such that the alternator can always be found in four weighings?

But first we count possible outcomes. Suppose there is a strategy that finds an alternator. In this strategy we can't have two unbalanced weighings in a row. To prove that, let us suppose there was an unbalanced weighing. Then the alternator switches foreign exchange market graph shifts weight to a real coin and whether or not the alternator is on the scale, the next weighing must balance.

The beauty of it is that given a strategy each outcome has to point to a particular coin as an alternator. That means the number of outcomes bounds the total number of coins that can be processed. Counting the number of possible outcomes that do not have two unbalances in row is a matter of solving a recurrence, which I leave to the readers to find.

The result is Jacobshtal numbers: For example, the total number of possible outcomes of four weighings is Since each outcome of a strategy needs to point to a coin, the total number of coins that can be processed in four weighings is not more than But 11 is better than 9 in our previous strategy.

Can we process 11 coins in four weighings? I will describe the first part of the strategy. So we have 11 coins, one of which is an alternator. In the first weighing binary options for mironov reviews compare 5 coins against 5 coins. Average daily range - custom mt4 indicator the weighing unbalances, the alternator is on a lighter pan.

Our problem is reduced to finding the alternator among five coins when we know that it is in the real state. If the weighing balances, then we know that if the alternator is among the coins on the scale it must now be in the light state.

For the second weighting, we pick two sets of three coins out of this ten coins and compare them against each other.

Notice that 3 is a Jacobsthal number, and 5, the number of coins outside the scale, is also a Jacobsthal number. If the second weighing balances, the alternator must be among 5 coins outside the scale. All but one of these coins are in the light state, and I leave it to the readers to finish the strategy.

If the weighing unbalances, we need to find the alternator among 3 coins that are in the real state now. This can be done in two weighings, and again the readers are to the rescue. It appears that Jacobsthal numbers provide the exact lower bound of the number of coins day trading simulator asx can be processed. This is what my middle-schoolers discovered and proved. We wrote a paper on the subject.

The strategy in the paper is adaptive. That means it changes depending on the results of the previous weighings. Can we find an oblivious strategy?

I will tell you in later posts. Suppose we have 3 n identical-looking coins. One of the coins is fake and lighter than the other coins which all are real. We also have a random scale. That is a scale that at each weighing behaves randomly. Find the fake coin in the smallest number of weighings. It is impossible to find the fake coin. The scale can consistently misbehave in such a way as to blame a specific real coin for what is the purpose of backdating stock options fake.

Let's try something else. Suppose we have two scales: Find the fake coin. What am I thinking? The normal scale can point to one coin and the random scale can point to another coin and we are in a "she said, he said" situation which whether on the binary options trading at the weekend can't resolve.

Now, in my final try, I'll make it right. We actually have three scales, one of which is random. So here we go, with thanks to my son Sergei for giving me this puzzle:. Let's start with this jquery get selected option value this At least two of the scales will agree, thus pointing to the true result.

This way we can use a divide-into-three-equal-groups strategy for one scale to find the fake coin. It will require 3 n weighings. Of course, we can. We can repeat every weighing on two scales. If they agree we do not need the third scale.

If they do not agree, one of the scales is random and lying, and we can repeat the weighing on the third scale to "out" the random scale. After we identify one normal scale, the process make extra money during residency faster.

Can we do even better? I will leave it to the readers to find a beautiful solution that is asymptotically better than the previous one. Update on Dec 24, The total number of coins should be 3 2 nnot 3 n. We are looking at the worst case scenario, when the random scale is adversarial.

We all have played with problems in which we had real coins and fake counterfeit coins. For this post I assume that the fake coins are always lighter than the real coins. My coauthor Konstantin Knop invented a new type of a coin: This coin can mimic a fake or a real coin.

It can also choose independently which coin to mimic for each weighing on a balance scale. You cannot find the chameleon coin in a mix with real coins if it does not want to be found, because it can consistently behave as a real coin.

Let's add classic fake coins to the mix, the ones that are lighter. Still the task of identifying the chameleons using a balance scale cannot be achieved: We can't identify the fake coins either, as the chameleons can mess things us up by consistently pretending to be fake. What we can do is to find a small stock market occupations of coins some of which are guaranteed to be fake.

Consider the simplest setup, when we have one fake coin and one chameleon in our mix of N coins. Our task now is to find TWO coins, one of which has to be fake. As usual we want to do it in the smallest number of weighings that guarantees that we'll find the two coins. Let me give you a fun problem to solve:. If you want to learn more, we just wrote a paper titled Chameleon Coins. Why did the easy way to make cash runescape angle go to the beach?

Because it was over 90 degrees. I have always wanted to be an honest person and have followed my honor successful forex trade records. Soviet Russia had two honor codes.

One code was for dealing with people and money to make decision synonym other for dealing with businesses and the government. I remind you that in Soviet Russia, all businesses were owned by the government. The money paid for work didn't have any relation to the work done.

The government paid standard salaries and the businesses did whatever. Generally, that meant that they were doing nothing. Meanwhile, the government got its income from selling oil. All people were being screwed by the government, so they had no motivation to play fair. Just as American workers do, Soviet Russians might use the work copier to make personal copies. The difference was that we didn't feel guilty at ripping off the government.

We wouldn't just make a few necessary copies; we would make copies for our friends, our family, for strangers—as many copies as possible. I moved to the United States in Several of my friends took time to explain to me the difference between Soviet Russia and the US. One of the friends, let's call her Sarah, was working as staff at Harvard University.

She told me the story of a recent visit by a famous Russian professor. After he left, the department received a bill. It appears that the professor, in a short visit, used several months-worth of the department's budget allocation for copying and phone calls. I was impressed, in a good way, by the professor who I assume spent a good deal of his time making a lot of copies of papers unavailable in Russia, presumably not only for himself but also for students and colleagues.

On the other hand, it was clearly wrong. My new friend Sarah told me that in the US money does not come from nowhere, and I should include Harvard University in my honor code for people. Actually, not only Harvard, but also any place of work and the government too.

Sarah also told me that since that budget problem, she was asked to talk to every incoming Russian visitor and explain to them how capitalism works. Most Russian visitors were ready to accept the rules.

I too was delighted with that. It is much easier to follow one honor code than two. I was best penny stocks at the moment very happy for my son, Alexey.

He was eight when we moved to Boston. Before our move I had a dilemma. Should I tell him that Stock broker exams india and Stalin were bad guys and killed millions of people? If I gave him a truthful explanation, I would also have to teach him to lie.

Otherwise, if he mentioned this at school or on the street, we would be at risk of going to prison. If I didn't teach him the truth, he might become brainwashed and grow up believing in communism, which would be very, very bad. This is one of those puzzles that I love-hate. I hate it because it is easy to answer this puzzle by inventing a specific mating pattern that ends with one animal.

It is possible to get the correct answer without seeing the beauty. I love it because there is beauty in the explanation of why, if the mating ends with one animal, it has to be a specific animal. The solution is charming, but being a mathematician this problem makes me wonder if ii is always possible to end with one animal. So I add another puzzle. Today I have two new coin puzzles that were inspired by my son, Alexey Radul:.

If you think about it, this problem is isomorphic to a known problem Young money trouble maker lyrics wrote about before:. The following problem was at a entrance test for the MIT PRIMES STEP program. Most of the students didn't think that the paper might be two-sided, but they came up with other inventive ideas.

Below are some of their pictures, and I leave it to you to explain why they work. All the students who submitted these pictures got a full credit for this problem on the test.

If you are a high-school student who wants to conduct research in mathematics, you should check out the MIT PRIMES program.

If you enjoy solving the problems in our entrance test, that's the first indication that you might want to apply. But to determine if the program is right for you, and you are right for the program, please read the following questions and answers which have been prepared for you by Tanya Khovanova, the PRIMES Head Mentor.

Stock market occupations only addresses applications to PRIMES Math, and only to the research track.

I do not like math competitions. Math competitions are completely separate from research in mathematics. If you enjoy thinking about mathematics for long periods of time and are fascinated by our test questions, you should apply. I am good at math, but I really want to be a doctor.

PRIMES requires a huge time commitment, so math should really be your most significant interest. I want to get into What are vested employee stock options, and PRIMES looks good stock options restricted stocks a resume.

PRIMES does look good on a resume. But if you are more passionate about, say, climate change than math, what would Harvard's admission committee see? Our experience in the program is that if math isn't your top interest, your math student may not be sufficiently impressive to be accepted at Harvard as a math researcher.

At the same time, you will not be accepted as the top climate change student as stock broker define didn't invest your time in that. Math research is a hard way to earn points for college. See also, the essay, Thoughts on research by Simon Rubinstein-Salzedo.

My parents want me to apply. I like to make money get turnt parents will not be accepted to the program. Do not apply if you do not really, really want to. Your website suggests that I should spend ten indian stock brokers salary a week on the PRIMES project. I can only spend five.

But I am a genius and faster than other people. We already assume that you are a genius and faster than anyone else you know. Five hours a week are not enough for a successful project. I looked at the past PRIMES projects and nothing excites me as much as my current interest in Pascal's triangle. I doubt I should apply. When you start working on a project, you will learn a lot about it. You will understand why, for rbc foreign exchange cash rates calculator, Cherednik algebras are cool.

The excitement comes with knowledge and invested time. Not yet being excited about Cherednik algebras is not a good reason not to apply. Besides a lot of exciting mathematics is done between several different fields.

I really want to do nothing else than study Pascal's triangle. We try to match our projects to students' interests as much as we can. But we almost never can fulfill a specific request as above. You might get a project related to Young diagrams, which b binary options managed accounts connected to quantum Pascal's triangle.

If this connection doesn't excite you, you shouldn't apply. I think I will be better positioned for research if I spend five more years studying.

There is nothing wrong with this approach. For many years the standard was to start research in graduate school. Our program is innovative. At PRIMES we are trying a different model. It may sound scary, but you will learn everything you need to know in order to do your project.

If the project is in representation theory, for example, you will only learn what you need—not the whole theory. Our hope is that eventually you will take a course in representation theory and expand your grasp of it and see the bigger picture behind your project. We have a reading track for people like you who reside in Boston area. I love math, but I am not sure that I want to be a mathematician. Many people start loving math early in life and then discover that there are many other things that require a similar kind of brain: We do not require from our students a commitment to become mathematicians.

If you want to try research in math, you should apply. If students decide that they do not want to do research in math after finishing our program, we do not consider that a negative result. One way or another, the experience of PRIMES will help you understand better what you want to do with your life.

I want to get to the International Math Olympiad. I am afraid that the time the research project takes prevents me from preparing for competitions. People who are good at Olympiads often have fantastic brain power that helps in research. On the other hand, research requires a different mind set and the transition might be painful.

It is possible, but not trivial to succeed in both. It is up to you to decide how you want to spend your time. I like number theory, but I do not see past PRIMES projects in number theory.

Doable number theory projects are hard to come by and we have fewer number theory projects than students who want to do number theory. There are many high-school programs that teach number theory including PROMYS and Ross programs.

Binary options sergey mironov model trains - gechabvovir’s blog

Our applicants like number theory because they were exposed to it. During PRIMES you will be exposed to something else and might like it as much. I found a local professor to work with on a research project.

Should I apply to PRIMES? PRIMES requires that you devote 10 hours a week to research for a year. It is unrealistic to do two research projects in parallel. Working with someone in person may be better than by Skype at PRIMES.

Also, usually our mentors are not professors, but rather graduate students. On the other hand, they are MIT grad students and projects are often suggested by professors. Our program is well structured. We guarantee weekly meetings in the Spring, we give extra help with your paper, and we have a conference.

It is up to you to decide. Alexander Shapovalov is a prolific puzzle writer. He has a special webpage of his river-crossing puzzles in Russian. Here is one of these puzzles.

In my paper with Joshua Xiong, Nim Fractalswe produced a bijection between P-positions in the three-pile Nim and a three-branch Ulam-Warburton automaton.

We also defined a parent-child relationship on games that is induced by this bijection. Namely, two consecutive P-positions in a longest optimal game of Nim are the ones that correspond to a parent-child pair in the automaton.

A cell in the Ulam-Warburton automaton has exactly one parent. See our paper for more details. Now I want to explicitly write out the rules of an automaton which will generate the Nim P-positions in 3D. Let me restrict the evolution of the automaton to the non-negative octant. That is, we consider points abc in 3D, where each coordinate is a non-negative integer. We define the neighbors of the point abc to be the points that differ from abc in two coordinates exactly by 1. So each point strictly inside the octant has 12 neighbors.

There are three ways to choose two coordinates, and after that four ways to choose plus or minus 1 in each of them. There is a geometric interpretation to this notion of neighborhood. Let us correspond a unit cube to a point with integer coordinates. The center of the cube is located at the given point and the sides are parallel to the axes.

Then two points are neighbors if and only if the corresponding cubes share one edge. Now it becomes more visual that a cube has 12 neighbors, as it has 12 edges. Here is the rule of the automaton. We start with the patriarch, 0,0,0one point being alive. The non-alive point is born inside the non-negative octant if it has exactly 1 alive neighbor that is closer to the patriarch. It follows that the points that are born at the n -th step has a coordinate sum 2 n. Consider for example the starting growth.

At the first step the points 0,1,11,0,1 and 1,1,0 are born. At the next step the points 0,2,2 and 2,0,2 and 2,2,0 are born. In the resulting automaton, the points that are born at step n are exactly the P-positions of Nim with the total of 2 n tokens. Only the points with an even total can be born. Now we proceed by induction on the total number of tokens. The base case is obvious. Suppose we proved that at step n exactly P-positions with 2 n tokens are born.

Consider a P-position of Nim: Remember, that bitwise XOR of aband c is zero. Consider the 2-adic values of aband c aka the smallest powers of 2 dividing aband c. There should be exactly two out of these three integers that have the smallest 2-adic value. Suppose these are a and b. That means by the inductive hypothesis abc has exactly one alive neighbor. Now we need to proof that nothing else is born. For the sake of contradiction suppose that abc is the earliest N-position to be born.

That means it has a live neighbor that is a P-position closer to the patriarch. Then their binary representations can't have the same number of ones at the end. Otherwise, abc is a P-position. That is a and b have different 2-adic values. Thus we can always find a second neighbor with the same number of tokens. That is, both neighbors are alive at the same time; and the N-position abc is never born.

One might wonder what happens if we relax the automaton rule by removing the constraint on the distance to the patriarch. Suppose a new point is born if it has exactly one neighbor alive. This will be a different automaton.

Let us look at the starting growth, up to a permutation of coordinates. At step one, positions 0,1,1 are born. At the next step positions 0,2,2 are born. At the next step positions 0,1,31,2,3 and 0,3,3 are born. We see that 0,1,3 is not a P-positions.

What will happen later? Will this N-position mess up the future positions that are born? Actually, this automaton will still contain all the P-positions of Nim.

In the new automaton, the points that are born at step n and have total of 2 n tokens are exactly the P-positions of Nim with the total of 2 n tokens.

So the points have total of 2 n tokens and are born at step n match exactly the first automaton described above. In the game of Nim you have several piles with tokens. Players take turns taking several tokens from one pile.

The person who takes the last token wins. The strategy of this game is well-known. You win if after your move the bitwise XOR of all the tokens in all the piles is 0. Such positions that you want to finish your move with are called P-positions.

I play this game with my students where the initial position has four piles with 1, 3, 5, and 7 tokens each. I invite my students to start the game, and I always win as this is a P-position.

Very soon my students start complaining that I go second and want to switch with me. My idea is to make the game last long to have many turns before ending to increase the chances of my students making a mistake.

So what is the longest game of Nim given that it starts in a P-position? Clearly you can't play slower then taking one token at a time. The beauty of Nim is that such an optimal game starting from a P-position is always possible. I made this claim in several papers of mine, but I can't find where this is proven. One of my papers with Joshua Xiong contains an indirect proof by building a bijection to the Ulam-Warburton automaton.

But this claim is simple enough, so I want to present a direct proof here. Actually, I will prove a stronger statement. In an optimal game of Nim that starts at a P-position the first player can take one token at each turn so that the second player is forced to take one token too.

- Fout:

Consider a P-position in a game of Nim. Then find a pile with the lowest 2-adic value. That is the pile such that the power of two in its factorization is the smallest. Suppose this power is k. Notice that there should be at least two piles with this 2-adic value.

The first player should take a token from one of those piles. That means the second player has to take tokens from another pile with 2-adic value k. Moreover, the second player is forced to take exactly one token to match the heap-sum. In the position 1,3,5,7 all numbers are odd, so I can take one token from any pile for my first move, then the correct move is to take one token from any other pile. My students do not know that; and I usually win even as the first player.

Plus, there are four different ways I can start as the first player. This way my students do not get to try several different options with the same move I make. After I win several times as the first player, I convince my students that I win anyway and persuade them to go back to me being the second player.

After that I relax and never lose. This looks like a simple linear algebra question with three variables and three equations. But it has a pretty geometrical solution. Again we have six sides and four faces. There should be many solutions. Can you find a geometric one? Do you know how to cut a cake?

There is a whole area of mathematics that studies cake cutting. Mathematicians usually assume that each person has their own idea of what is the best part of the cake.

Suppose three sisters are celebrating the New Year by having a cake. Anna likes only icing, Bella likes only chocolate chips, and Carol likes only pieces of walnuts mixed into the cake. Mathematicians want to cut cakes fairly. But what is fair? Fair is fair, right? There are several different notions of fair cake division. There is a proportionate division.

In such a division every sister gets at least one-third of her value of the cake. Let's see an example. Anna gets one third of the icing, Bella gets one third of the chocolate chips, and Carol gets everything else.

This is a fair proportionate division. Each of the sisters believes that she got at least one-third of the cake, in their own value. But it doesn't seem quite fair. There is a stronger notion of fairness. It is called envy-free. In this division each sister gets at least one-third of the cake and, in addition, none of the three sisters would improve their value by swapping pieces.

That means, if Anna wants only icing, not only does she get at least one-third of the icing, but also no one else gets more icing than Anna. The previous example of the proportionate division is not envy-free. Carol got two-thirds of the icing, so Anna would want to switch with her. Let's try a different division. Anna gets one third of the icing, Bella gets the chocolate chips and another third of the icing, and Carol gets all the walnuts and another third of the icing.

Formally, this is envy-free cake cutting. What do you think Anna feels when she sees the smiles of contentment on the faces of her sisters? Whoever invented the name doesn't understand envy. Anna got one-third of the cake by her value, but the other sisters got the whole cake!

Luckily mathematicians understand this conundrum. So they invented another name for a cake division. They call a division equitable if everyone values all the pieces the same. So the division above is envy-free but not equitable. Let's give each sister one-third of all the components of the cake. This division is very good mathematically: By the way, envy-free division is always proportionate.

This division seems fair. But is it a good division? There is another term here: Pareto-efficient division means that it is impossible to make one person feel better, without making another person feel worse. All divisions above are not Pareto-efficient.

Moving some icing from Carol to Anna, doesn't decrease the value for Carol, but increases the value for Anna. There is an even better way to divide the cake. We can give the icing to Anna, the walnuts to Bella, and the chocolate chips to Carol. This division is proportionate, envy-free, and Pareto-efficient. Mathematicians even have a word for it. They call it a perfect division.

Mathematically this division is perfect. Unfortunately, sisters are not. I know an Anna who would still envy Bella. Why are manhole covers round?

The manhole covers are round because the manholes are round. But the cute mathematical answer is that the round shapes are better than many other shapes because a round cover can't fall into a round hole. If we assume that the hole is the same shape as the cover but slightly smaller, then it is true that circular covers can't fall into their holes.

But there are many other shapes with this property. They are called the shapes of constant width. Given the width, the shape with the largest area is not surprisingly a circle. The shape with the smallest area and a given constant width is a Reuleaux Triangle. Here is how to draw a Reuleaux triangle. Draw three points that are equidistant from each other at distance d. Then draw three circles of radius d with the centers at given points.

The Reuleaux triangle is the intersection of these three circles. Can we generalize this to 3d? What would be an analogue of a Reuleaux Triangle in 3d?

Of course, it is a Reuleaux Tetrahedron: Take four points at the vertices of a regular tetrahedron; take a sphere at each vertex with the radius equal to the edge of the tetrahedron; intersect the four spheres.

Is this a shape of the constant width? Many people mistakenly think that this is the case. Indeed, if you squeeze the Reuleaux tetrahedron between two planes, one of which touches a vertex and another touches the opposite face of the curvy tetrahedron, then the distance between them is equal to d: This might give you the impression that this distance is always d.

If you squeeze the Reuleaux tetrahedron between two planes that touch the opposite curvy edges, the distance between these planes will be slightly more than d. To create a shape of constant width you need to shave off the edges a bit.

Theoretically you can shave the same amount off every edge to get to a surface of constant width. But this is not the cool way to do it. The cool way is to shave a bit more but only from one edge of the pair of opposite edges. You can get two different figures this way: These two bodies are called Meissner bodies and they are conjectured to be shapes of the constant width with the smallest volume.

On the picture I have two copies of a pair of Meissner bodies. The two left ones have three edges that share a vertex shaved off. The very left shape gives a top view of this vertex and the solid next to it has its bottom with holes looking forward. The two shapes on the right show the second Meissner body in two different positions.

I recently discovered a TED-Ed video about manhole covers. It falsely claims that the Reuleaux tetrahedron has constant width. I wrote to TED-Ed, to the author, and posted a comment on the discussion page.

There was no reaction. They either should remove the video or have an errata page for it. Knowingly keeping a video with an error that is being viewed by thousands of people is irresponsible.

I am running a PRIMES-STEP program for middle school students, where we try to do research in mathematics. In the fall of we decided to study the following topic in logic. My students and I discovered a lot of interesting things about these four types of people and wrote a paper: This paper contains 11 cute logic puzzles designed by each of my 11 students. I envied my students and decided to create two puzzles of my own. You have already solved the one above, so here is another, more difficult, puzzle:.

The following cute puzzle of unknown origins was sent to me by Martina Balagovic and Vincent van der Noort:. I received a book Really Big Numbers by Richard Schwartz for review. I was supposed to write the review a long time ago, but I've been procrastinating. Usually, if I like a book, I write a review very fast.

If I hate a book, I do not write a review at all. With this book I developed a love-hate relationship. Let me start with love. I enjoyed reading the first 80 pages. The pictures are great, and some explanations are very well thought out. Plus, I haven't thought much about really big numbers, so the book helped me understand them. I was impressed with how this book treats very difficult ideas with simple explanations and illuminating images. I was captivated by it.

The beginning of the book is suitable for small children. Most of the book is suitable for advanced middle-schoolers who like mathematics. The last part is very advanced. Is it a good idea to show children a book that looks like a children's book, but which soon becomes totally out of their reach?

Richard Schwartz understands it and says many times that pieces of this book might be read several years apart. What child is ready to wait several years to finish a book? How would children feel about the book and about numbers when no matter how hard they try, they cannot understand the end of the book? As a reviewer, I can't recommend the full book for kids who are not ready to grasp the notion of the Ackermann function or arrow notation.

Even if the child is capable of understanding these ideas, there are mathematical issues that would prevent me from recommending it. Let me start by explaining the notion of plex. We call an n-plex a number that is equal to 10 n.

For example, 2-plex means 10 2 which isand plex means 10 The fun part starts when we plex plexes. The number n-plexplex means 10 to the power n-plex which is 10 10 We can continue plexing: When you are hunting for really big numbers, it is easier to write the number of plexes rather than writing plexes after plexes.

Richard Schwartz introduces the following notation to help visualize the whole thing. He puts numbers in a square. Number n in a square means 1-plexed n times. For example, 2 inside a square means 10 Ten inside a square is 1-plexplexplexplexplexplexplexplexplexplex. We can start nesting squares. For example, 2 inside a square means 1-plex-plex or 10 Let's add a square around it: To denote 10 nested in n squares Richard Schwartz uses the next symbol: For example, 1 inside a pentagon is 10 inside a square.

I wrote this number in the previous paragraph: Similarly, n inside a hexagon means 10 inside n nested pentagons. We can continue this forever: What bothers me is why a square? Why not a triangle? If we adopt this scheme, what is the meaning of a number in a triangle? Let's try to unravel this. Following Richard Schwartz's notation we get that n inside a square is the same as 10 inside n nested triangles. What do we do n times with 10 to get to 1 plexed n times?

There is a disconnect in notation here. For example, 10 in two nested triangles should mean 2 inside a square that is 10 This doesn't make any sense. I started googling around and discovered the Steinhaus—Moser notation.

In this notation a number n in a triangle means n n. A number n in a square means the number n inside n nested triangles. A number n in a pentagon means the number n inside n nested squares. This makes total sense to me. If we move down the number of sides, we can say that the number n inside a 2-gon means n times n and the number n inside a 1-gon means n. Schwartz changed the existing notation in two places. First he made everything about This might not be such a bad idea except his 10 inside a square doesn't equal 10 inside a square in Steinhaus—Moser notation.

In Steinhaus—Moser notation 10 inside a square means 10 plexed 10 times. The author removed one of the plexes. He made 10 inside the square to mean 1 plexed 10 times, and as a result it stopped working. Even though the first 83 pages are delightful and the pictures are terrific, the notation doesn't work. I have solved too many puzzles in my life. When I see a new one, my solution is always the intended one.

But my students invent other ideas from time to time and teach me to think creatively. For example, I gave them this puzzle:. I gave a talk about thinking inside and outside the box at the Gathering for Gardner conference. I mentioned this puzzle and the inventiveness of my students. After the conference a guy approached me with another answer which is now my favorite:. Consider a triangle with vertices A, B, and C. Let O be its orthocenter. Let's connect O to the vertices.

We get six lines: These six lines are pair-wise orthogonal: It is easy to see that A is the orthocenter of the triangle OBC, and so on: We say that these four points form an orthocentric system. I heard a talk about this structure at the MOVES conference by Richard Guy. What I loved in his talk was his call to equality and against discrimination.

The point O plays the same role as the other three points. It should be counted. Richard Guy suggested calling this system an orthogonal quadrangle. I am all for equality. This is not a triangle, this is a quadrangle! I want the world to be a better place. My contribution is teaching people to think. People who think make better decisions, whether they want to buy a house or vote for a president. When I started my blog, I posted a lot of puzzles. I was passionate about not posting solutions.

I do want people to think, not just consume puzzles. Regrettably, I feel a great push to post solutions. My readers ask about the answers, because they are accustomed to the other websites providing them. I remember I once bought a metal brainteaser that needed untangling.

The solution wasn't included. Instead, there was a postcard that I needed to sign and send to get a solution. The text that needed my signature was, "I am an idiot. I can't solve this puzzle. In a long run I am glad that the brainteaser didn't provide a solution.

That was a long time ago. Now I can just go on YouTube, where people post solutions to all possible puzzles. I am not the only one who tries to encourage people to solve puzzles for themselves. Many Internet puzzle pages do not have solutions on the same page as the puzzle.

They have a link to a solution. Although it is easy to access the solution, this separation between the puzzle and its solution is an encouragement to think first. But the times are changing.

My biggest newest disappointment is TED-Ed. They have videos with all my favorite puzzles, where you do not need to click to get to the solution. You need to click to STOP the solution from being fed to you. Their video Can you solve the prisoner hat riddle?

To my knowledge, this puzzle first appeared at the rd All-Russian Mathematical Olympiad in Here is the standard version that I like:.

The video is beautifully done, but sadly the puzzle is dumbed down in two ways. First, they explicitly say that it is possible for all but one person to guess the color and second, that people should start talking from the back of the line. I remember in the past I would give this puzzle to my students and they would initially estimate that half of the people would die. Their eyes would light up when they realized that it's possible to save way more than half the people.

They have another aha! This way each person sees or hears everyone else before announcing their own color. In the simplified adapted video, there are no longer any discoveries. There is no joy. People consume the solution, without realizing why this puzzle is beautiful and counterintuitive. Nowadays, I come to class and give a puzzle, but everyone has already heard the puzzle with its solution.

How can I train my students to think? I stumbled upon a TED-Ed video with a frog puzzle:. My first thought was that male frogs croak to attract female frogs. That means the second frog in the clearing is probably an already-attracted female. The fact that the stump frog is not moving means it is male. This puzzle didn't assume any knowledge of biology.

The puzzle assumes that each frog's gender is independent from other frogs. Thus this puzzle is similar to two-children puzzles that I wrote so much about. I not only blogged about this, but also wrote a paper: As in two-children puzzles, the solution depends on why the frog croaked. It is easy to make a reasonable model here. Suppose the male frog croaks with probability p.

Now the puzzle can be solved. Consider the two frogs in the clearing before the croaking: The probabilities corresponding to our outcome—a non-croaking frog on the stump and one croaking frog in the clearing—are in bold. It doesn't matter where you go for the antidote.

The TED-Ed's puzzle makes the same mistake that is common in the two-children puzzles. I don't want to repeat their incorrect solution. The TED-Ed's frog puzzle is wrong.

When I receive a bank statement, I review all the transactions. The problem is my failing memory. I do not remember when and where I last took money from an ATM, and how much. So I decided to create a pattern. I used to take cash in multiples of Probably most people do that. A better idea is to always take the amount that has a fixed remainder modulobut not 0. For example, let's say I take one of the following amounts: Sometimes I need more money, sometimes less.

This set of numbers covers all of my potential situations, but my pattern is that all the numbers end in This way if someone else gets access to my account, they will almost surely take a multiple of I will be able to discover a fraud without remembering the details of my last withdrawal. In addition, when I first started doing this, I was hoping I wouldn't need to wait until I review my own statement to discover problems.

My hope was that if a thief tried to take cash from my account, my bank would notice a change in the pattern and notify me immediately. Now I realize that this was wishful thinking. I doubt that banks are as smart as I am.

One day I should try an experiment. I should go to an ATM I never use and withdraw dollars. I wonder if my bank would notice. One of my jobs is giving linear algebra recitations at MIT. The most unpleasant aspect of it is dealing with late homework.