Tanya Khovanova's Math Blog

Alexander Gribalko designed a new coin problem, two versions of which appeared at the 43rd Tournament of the Towns. I will start with the easier version.

Problem. Alice and Bob found eleven identical-looking coins; four of them are fake. All real coins weigh the same. All fake coins weigh the same but are lighter than the real ones. Alice offered Bob four specific coins. Before agreeing to take them, Bob wants to ensure they are all real. Can he do so by only using the balance scale twice? He can use any of the eleven coins for weighing purposes.

Surprisingly, the more difficult version has fewer coins.

Problem. Alice and Bob found eight identical-looking coins; three of them are fake. All real coins weigh the same. All fake coins weigh the same but are lighter than the real ones. Alice offered Bob three specific coins. Before agreeing to take them, Bob wants to ensure they are all real. Can he do so by only using the balance scale twice? He can use any of the eight coins for weighing purposes.

I recently posted a cute Shapovalov’s puzzle about gnomes. Here is another easier gnome puzzle, also by Alexander Shapovalov.

Puzzle. All gnomes are knights or knaves: knights always tell the truth, and knaves always lie. There is a gnome on every cell of a 4 by 4 chessboard. It is known that both knights and knaves are present in this group. Every gnome states, “Among my neighbors, the number of knaves is the same as the number of knights”. How many knaves are there, if by neighbors they mean orthogonally adjacent gnomes?

The next gnome puzzle has a different author, Alexander Gribalko. Gnomes in this puzzle are not knights or knaves but rather friendly and polite beings.

Puzzle. Nine gnomes repeated the following procedure three times. They arranged themselves on a 3 by 3 chessboard with one gnome per cell and greeted all their orthogonal neighbors. Prove that not all pairs of gnomes greeted each other.

Here is another lovely problem from a prolific problem writer, Alexander Shapovalov.

Problem. Every cell of a 7 by 7 chessboard has a gnome standing on it. For every pair of gnomes whose cells share an edge, their beards’ lengths differ by no more than 1 inch. Now, we take these gnomes and sit them around a table. Prove that we can do so in a way that any two gnomes sitting next to each other have their beards’ lengths differ by no more than 1 inch.

A standard chessboard is 8 by 8. Why would this problem have a 7 by 7 board? Let’s see.

For even-sized boards, the problem is easy. I will explain why, but first, let me construct a graph related to a board, in this case, any board.

Each cell is a vertex, and two vertices are connected by an edge if the corresponding cells are orthogonal neighbors (share an edge) on the board. A cycle that goes through a graph and visits every vertex exactly once is called a Hamiltonian cycle. A Hamiltonian cycle is a potential way to sit the gnomes around the table and solve the problem. When we sit the gnomes in a circle following a Hamiltonian cycle, two neighbors at the table are also neighbors on the board, and so they have their beards’ lengths differ by no more than 1 inch.

The problem is easy for even-sized boards because it is easy to draw a Hamiltonian cycle on them. An odd-sized chessboard can’t have a Hamiltonian cycle. To prove this, let me color the board in a checkerboard manner. Then, cells that share an edge are different colors. And you can’t make a cycle through the board, where you switch colors at each step, but the total number of steps is odd.

It follows that for odd-sized boards, it is impossible to solve the problem by just connecting neighboring cells on the board. There should be another way. Can you find it?

I want to discuss a problem from a test I gave recently.

Problem. My dog, Fudge, likes books. He brought two books to his corner in the morning and three more books in the evening. How many books will he read tonight?

As I expected, many students didn’t pay attention and just summed up the two numbers in the problem and gave five as the answer. Here are three answers that I especially liked.

Answer 1. Zero, because most dogs can’t read.

This cautious student added most to be on the safe side.

Answer 2. You cannot tell how many books he will read. Just because he brings books to his corner doesn’t mean he will read them.

The second answer demonstrates great logical thinking, were Fudge a human. But the third answer made me laugh.

Answer 3. Three. If he brought three more books to his corner in the evening, it means he finished the two from that morning, so there are three books left for him to read.

Here is an old goodie.

Puzzle. A criminal is sentenced to death. He is offered the last word. He is allowed to make one statement. If the statement is true, the criminal will be electrocuted. If the statement is false, he will be hanged. Can you find a good piece of advice for this man?

The standard answer to this puzzle is for the criminal to say, “I will be hanged.” This creates a paradox. If he is hanged, the statement is true, and he should be electrocuted. If he is electrocuted, the statement is false, and he should be hanged.

Thinking about it, any paradox works. If the man says, “This statement is false,” then the type of punishment he gets can’t be determined.

Thinking about it some more, asking a question instead of making a statement will confuse his executioners all the same.

One of my students advised the criminal to stay silent. This was not my favorite solution, as staying silent requires some concentration. My favorite solution was the statement, “It will rain exactly a thousand years from now.” This way, the criminal can relax, at least for a thousand years.

Today I discuss ideas for solving the puzzle I posted previously in my blog: A Scooter Riddle.

Riddle. Alice, Bob, and Charlie are at Alice’s house. They are going to Bob’s house, which is 33 miles away. They have a 2-seat scooter that goes 25 miles per hour with 1 rider, or 20 miles per hour with 2 riders. Each of the 3 friends has a walking pace of 5 miles per hour. What is the fastest time that all three of them can end up at Bob’s house?

Let’s start.

• The scooter and the three people should always be on the move. Moreover, whenever the scooter moves towards Bob’s house, it should carry two people, and only the driver should be on the scooter whenever it is going back.
• We do not care about each individual’s path: whenever people meet at the same place at the same time, we can swap them. Therefore, we can assume that the scooter’s driver is the same person at all times.
• Thinking about it, we can ignore the scooter’s driver and assume that the scooter is self-driving, and there are only two people, Alice and Bob, who need to get to Bob’s house.
• If Alice and Bob arrive at Bob’s house at different times, there is room for improvement: the fastest person should have used the scooter less and walked more, allowing the other person more use of the scooter. Hence, we can assume that they arrive at Bob’s house simultaneously.
• Without loss of generality, we can assume that Alice starts on the scooter first.
• So the scooter should carry Alice for some time, drop her off, come back for Bob, then drop him off, come back for Alice, and so on. Without loss of generality, we can combine all scooter trips by the same person into one ride.
• Thus, the plan is the following. The scooter drives Alice for x hours, drops her off, and she finishes on foot. Meanwhile, Bob starts on foot. The scooter goes back after dropping off Alice, picks up Bob, and drives him x hours the rest of the way.

Now this is just algebra. Each person covers 20x miles on the scooter and 33 − 20x miles on foot. The walking takes 33/5 − 4x hours. Thus the total trip for each person takes 33/5 − 3x hours.

Now we calculate what the scooter does. The scooter covers 20x miles with Alice and the same number of miles with Bob. It covers 40x − 33 miles alone. Thus the scooter spends 2x + (40x − 33)/25 hours in transit. The two times must be the same. So we have an equation: 6.6 − 3x = 2x + 1.6x − 1.32. Thus, x = 1.2, and the answer to the puzzle is 3 hours.

If you can figure out my number without the Internet, call me.

• The number of letters in the first name of Anna Karenina’s love.
• The number of times the word nevermore appears in the famous poem, subtracted from the month when the events took place.
• The only number in the title of a Bergman movie.
• The number of vowels in the original and more historically meaningful name for the sorcerer’s stone in Harry Potter books.
• The number of vertices of the other geometric shape in “Girl on a Ball”.
• The cube root of the age mentioned in one of the earliest Beatles songs.
• The number corresponding to the lexicographically first string representing an integer in English.
• The first digit of the age at which Pushkin and Mozart died.
• The smallest of two primes such that their sum and difference are also prime.
• The only triangular number that is also prime.

(Just in case: this is a joke and not my actual phone number.)

I run Number Gossip, where you can input a number and get some of its cute properties. For example, the number 63 is composite, deficient, evil, lucky, and odd. In addition, it has a unique property: 63 is the smallest number out of two (the other being 69), such that the common alphabetical value of its Roman representation is equal to itself. Indeed, the Roman representation of 63 is LXIII, where L is the 12th digit, X is the 24th, and I is the 9th. Summing them up, we get 12 + 24 + 9 + 9 + 9 = 63 — the number itself.

I have a list of about 50 properties of numbers that my program checks. Each number greater than one gets at least four properties. This is because I have four groups of properties that cover all the numbers. Every number is either odd or even. Every number is either deficient, perfect, or abundant. Every number greater than one is either prime or composite. Every number is either evil or odious.

In addition, I collect unique number properties. During the website’s conception, I decided not to list all possible unique properties that I could imagine but to limit the list to interesting and unusual properties. My least favorite properties of numbers are the ones that contain many parameters. For example, 138 is the smallest number whose base 4 representation (2022) contains 1 zero and 3 twos. If you are submitting a number property to me, keep this in mind.

Some parameters are more forgiving than others. For example, 361 is the smallest number which is not a multiple of 9, whose digital sum coincides with the digital sum of its largest proper divisor. In more detail, the digital sum of 361 is 3 + 6 + 1 = 10, while 19, the largest divisor of 361, has the same digital sum of 10. In this case, the parameter 9 is special: for a multiple of 9, it is too easy to find examples that work, such as 18, 27, and so on. Sequence A345309 lists numbers whose digital sum coincides with the digital sum of their largest proper divisor. The first 15 terms of the sequence are divisible by 9, and 361 is the smallest term that is not divisible by 9.

By the way, another number that is buried deep in a sequence is 945, which is the smallest odd abundant number. There are 231 abundant numbers smaller than 945; all of them are even.

A more recent addition to my collection is related to the sequence of distended numbers (A051772). Distended numbers are positive integers n for which each divisor of n is greater than the sum of all smaller divisors. It is easy to see that for distended numbers, all sums of subsets of divisors are distinct. The opposite is not true: 175 is the smallest number, where all sums of subsets of its divisors are distinct, but the number itself is not distended.

It is difficult to find special properties for larger numbers, so I am less picky with them. For example, 3841 is the number of intersections of diagonals inside a regular icosagon. The word icosagon hides a parameter, but I still like the property.

I invite people to submit number properties to me. And I received many interesting submissions. For example, the following oxymoronic property was submitted by Alexey Radul: 1 is the only square-free square.

The numbers below 200 that still lack unique properties are 116, 147, 155, 162, 166, 182, and 194. The earliest century that doesn’t have unique number properties ranges between 7000 and 7100. The next ones are 8000–8100 and 9100–9200. By the way, my site goes up to ten thousand.

I also have a lot of properties in my internal database that I haven’t checked yet. I am most interested in the proof of the following property: 26 is the only number to be sandwiched between any two non-trivial powers.

Every year, after the MIT Mystery Hunt was over, I would go through all the puzzles and pick out the ones related to mathematics. This year, I didn’t feel like doing it. Besides, I think it is more important to collect quality puzzles rather than all the puzzles by topic. So my new collection is the puzzles recommended to me, which I might like.

I start with math and logic puzzles.

• Crooked Crossings: A lovely mixture of language and logic.
• Albumistanumerophobia: You are given datasets.
• Lists of Large Integers: You are given integers that are products of two primes. Some of the integers are ridiculously large. I love the puzzle.
• The Colour Out of Space: You are given long division cryptarithms.
• Dancing Triangles: You are given Nikoli-type puzzles on triangular grids.

I continue with computer science.

• The Day You Begin: You are given an interactive input box requesting students’ names.
• Red Herring: You are given a weird image.
• Sorcery for Dummies: You are given an interactive box.

I carry on with some non-math fun.

• Oxford Children’s Dictionary: You are given incomplete dictionary definitions.
• The Times They Had: You are given colorful scored epitaphs.
• La Ronde: You are given a crossword with weird numbering. I quite enjoyed working on it.
• Love, Actually: During the hunt, this puzzle had an intermediate interaction during which an extra clue was sent separately.

I conclude with the plot device round. All the puzzles in this round are relatively easy. But our team got stuck on them until we realized that we already had the answer, which was not a single word. Here are some of the puzzles that were specifically recommended.

• Calculated Whisk: A fun word puzzle.
• A Crying Shamus: You are given crossword clues.
• First Contact Lens: You are given crossword clues and letters.
• Heavenly Bodice: A variation on Star Battle.
• THE NONSENSE SHOW: You are given poems. (This puzzle is number 4 on the Fruit Around page.)

I wrote a lot about how during entrance tests for Moscow State University, the examiners were giving Jewish and other undesirable students special (e.g. more difficult) questions during the oral exams. (See, for example, our paper Jewish Problems with Alexey Radul.) Not all examiners agreed to do this. So the administration made sure that there were different exam rooms: brutal rooms with compliant examiners torturing students with difficult questions, and normal rooms with normal examiners testing preapproved students. The administration also had other methods. One of them is the topic of this essay.

The math department of Moscow State University had four entrance exams. The first was a written math test consisting of three trivial problems, a very difficult one, and a brutally challenging one. At the end, I will show you a sample: a trivial problem and a very difficult one from 1976, my entrance year.

What was the point of such vast variation in difficulty, you may ask? There were two reasons.

But first, let me explain some entrance rules. The exam was scored according to the number of solved problems. A score of two or less was a failing score. People with such scores would be disqualified from the next exam. Any applicant with a smidge of mathematical intelligence would be able to solve all three trivial problems. Almost all applicants who qualified for the next test would have the same score of three on the first test, as they wouldn’t be able to solve the last two problems. Thus, mathematical geniuses and people who barely made the cut got the same score.

There was another rule. Officially, people with a gold medal from their high school (roughly equivalent to a valedictorian) could be accepted immediately if they scored 5 on the first exam. So one of the administrative goals was to prevent anyone getting a 5, thus, blocking Jewish applicants from sneaking in after the first exam.

Another goal was to have all vaguely qualified people get the same score. The same goal applied to other exams. After the four admission exams, the passing score, say X, was announced. A few people with a score higher than X were immediately accepted. Then there were hundreds of applicants with a score of X, way more than the quota of people the department was planning to accept. An official rule allowed the math department to pick and choose whoever they wanted from everyone who scored X.

I heard a speech by the famous Russian mathematician, Vladimir Arnold, directed at decent examiners who tested “approved” students at the oral math exam, which was the second admissions exam. His suggestion was brilliant and simple. If the students are good and belong in the department, give them an excellent grade of 5. If not, give them a failing grade of 2. Arnold’s plan boosted the chances of good students doing better than the cutoff passing score X and removed mediocre students from the competition. His idea was not only brilliant and simple but also courageous: he was risking his career by trying to fight the system.

I never experienced the entrance exams firsthand. By ministry order, as a member of the USSR IMO team, I was accepted without taking any exams. I already wrote an essay, A Hole for Jews, about how getting on the IMO team was the only way for Jewish students to get into the Moscow State University, and how the University tried to block them.

But I still looked at the entrance exam problems I would have had to solve to get in. The last two problems scared me. Now I found them again online (in Russian) at: the 1976 entrance test. The trivial problem below is standard and mechanical, while the other problem still looks scary.

Trivial problem. Solve for x:

Solution. We were drilled in school to solve these types of problems, so this one was trivial. First, make a substitution: y = 3^x. This leads to an equation: (2y – 1)(y – 3)/(y² – 2)(y – 1) ≤ 0. From this we get ranges for y: (-∞, -√2], [1/2,1], [√2, 3]. The last step is to take a logarithm.

Very difficult problem. Three spheres are tangent to plane P and to each other. Two of the spheres are the same size. The apex of a circular cone is on P, and the cone’s axis is perpendicular to the plane P. All three spheres are outside the cone and tangent to it. Find the cosine of the angle between the cone’s generatrix and the plane P, if one of the angles of the triangle formed by the intersection points of the spheres and the cone is 150 degrees.