Tanya Khovanova's Math Blog

I’ve been invited to help with the Puzzle Column at the MSRI newsletter Emissary. We prepared six puzzles for the Fall 2018 issue.

I love the puzzles there. Number 2 is a mafia puzzle that I suggested. Number 6 is a fun variation on the hat puzzle I wrote a lot about. Here is puzzle Number 3.

Puzzle. Let A = {1,2,3,4,5} and let P be the set of all nonempty subsets of A. A function f from P to A is a “selector” function if f(B) is in B, and f(B union C) is either equal to f(B) or f(C). How many selector functions are there?

(Photo by Rebecca Frankel.)

When I was in grade school, one of the teachers called me Cave Lioness. She hated my unruly hair, which reminded her of a lion’s mane. This teacher was obviously very uninformed, for female lions do not have manes.

This name calling had the opposite to the desired effect. I became proud of my mane and didn’t ever want to cut it. When I grew older, I opted for convenience and started to cut my hair short—sometimes very short.

Last year I was too busy for barbers, and my hair grew more than I intended. As it turned into a mane, I remembered the story of this nickname.

Once I was giving a math lecture and my phone, which I’ve never quite understood, was on the desk in front of me. Suddenly it rang. I didn’t pick it up, as I proceeded with my lecture. The ringing stopped, while I was explaining a particularly interesting mathematical point. After a minute, my phone said, “I do not understand a word you are saying.”

Detective Radstein is investigating a robbery. He apprehends three suspects: Anne, Bill, and Caroline. The detective knows that no one else could have participated in the robbery. During the interrogation the suspects make the following statements:

• Anne: I didn’t do it. Bill did it alone.
• Bill: I didn’t do it. Caroline did it.
• Caroline: I didn’t do it. Bill did it.

Detective Radstein also discovered that all three suspects are members of a club called The Halfsies. Every time they speak, they make two statements, one of which is a lie and the other one is true. Who committed the robbery?

I already wrote about my first experience crocheting hyperbolic surfaces. In my first surface I added two more stitches per current stitch. It took me hours to crochet the last row: the same hours it took me to crochet the rest.

For my next project, I reduced the ratio. The light blue thingy has ratio 3/2. I continued making my life simpler. The next project, the purple surface on the left, has ratio 4/3. The last project on the right has a ratio of 5/4 and is my favorite. Mostly because I am lazy and it was the fastest to make.

How much time will it take you to answer the following question?

Can the equation 29x + 30y + 31z = 366 be solved in natural numbers?

Happy 2019, the first 4 digit number to appear 6 times in the decimal expansion of Pi.

By the way:

2019 = 1⁴ + 2⁴ + 3⁴ + 5⁴ + 6⁴.

Also, 2019 is the product of two primes 3 and 673. The sum of these two prime factors is a square.

This is not all that is interesting about factors of 2019. Every concatenation of these two prime factors is prime. Even more unusual, 2019 is the largest known composite number such that every concatenation of its prime factors is prime. [Oops, the last statement is wrong, Jan 3,2019]

Happy Happy-go-Lucky year, as 2019 is a Happy-go-Lucky number: the number that is both Happy and Lucky.

In case you are wondering, here is the definition of Happy numbers: One can take the sum of the squares of the digits of a number. Those numbers are Happy for which iterating this operation eventually leads to 1.

In case you are wondering, to build the Lucky number sequence, start with natural numbers. Delete every second number, leaving 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, …. The second number remaining is 3, so delete every third number, leaving 1, 3, 7, 9, 13, 15, 19, 21, …. The next number remaining is 7, so delete every 7th number, leaving 1, 3, 7, 9, 13, 15, 21, …. The next number remaining is 9, so delete every ninth number, etc.

Alex Bellos sent me his new book Puzzle Ninja: Pit Your Wits Against The Japanese Puzzle Masters. What has he done to me? I opened the book and couldn’t close it until I solved all the puzzles.

This is a fantastic book. There are many varieties of puzzles, including some types that I’ve never seen before. Also, the beautifully designed puzzles are great. Often puzzles of the same type target different solving ideas or have varied cool themes.

This book is more than a bunch of puzzles; it also contains poetic stories about puzzle histories and Japanese puzzle designers. Fantastic puzzles together with a human touch: this might be my favorite puzzle book.

I present two puzzles from the book. The puzzle type is called Wolf and Sheep Slitherlink. The Slitherlink is a famous puzzle type with the goal of connecting some of the neighboring dots into a single non-self-intersecting loop. A number inside a small square cell indicates how many sides of the square are part of the loop. Wolf and Sheep Slitherlink is a variation of Slitherlink in which all sheep should be kept inside the fence (loop) and all the wolves outside.

Ignore the numbers in the title as they just indicate the order number of Wolf and Sheep Slitherlink puzzles in the book. The number of ninja heads shows the level of difficulty. (The hardest puzzles in the book have four heads.) The difficulty is followed by the name of the puzzle master who designed the puzzle.

The first puzzle above is slightly easier than the second. I like the themes of these two puzzles. In the first one, only one cell—lonely wolf—marks the relationship to the fence. In the second one, the wolf in the center—who needs to be outside the fence—is surrounded by a circle of sheep who are in turn surrounded by a circle of wolves.

My friend Alex Ryba uses interesting math questions in the CUNY Math Challenge. For the 2016 challenge they had the following problem.

Problem. Eve owns two six-sided dice. They are not necessarily fair dice and not necessarily weighted in the same manner. Eve promises to give Alice and Bob each a fabulous prize if they each roll the same sum with her dice. Eve wishes to design her two dice to minimize the likelihood that she has to buy two fabulous prizes. Can she weight them so that the probability for Alice and Bob to get prizes is less than 1/10?

The best outcome for Eve would be if she can weight the dice so that the sum is uniform. In this case the probability that Alice and Bob get the prizes is 1/11. Unfortunately for Eve, such a distribution of weight for the dice is impossible. There are many ways to prove it.

I found a beautiful argument by Hagen von Eitzen on the stack exchange: Let a_i (correspondingly b_i) be the probabilities that die A (correspondingly B) shows i + 1. It would be very useful later that that i ranges over {0,1,2,3,4,5} for both dice. Let f(z) = ∑ a_izⁱ and g(z) = ∑ b_izⁱ. Then the desired result is that f(z)g(z) = ∑_j=0¹⁰ z^j/11. The roots of the right side are the non-real roots of unity. Therefore both f and g have no real roots. So, they must both have even degree. This implies a₅=b₅=0 and the coefficient of z¹⁰ in their product is also 0, contradiction.

Alex himself has a straightforward argument. The probabilities of 2 and 12 have to be equal to 1/11, therefore, a₀b₀ = a₅b₅ = 1/11. Then the probability of a total 7 is at least a₀b₅ + a₀b₅. The geometric mean of a₀b₅ and a₀b₅ is 1/11 (from above), so their arithmetic mean is at least 1/11 and their sum is at least 2/11. Therefore, the uniform distribution for sums is impossible.

So 1/11 is impossible, but how close to it can you get?

I just got this picture from my friend Victor Gutenmacher, which I never saw before. My 1975 IMO team is posing at our training grounds before the Olympiad trip to Bulgaria.

Left to right: Boris Yusin, Yuri Ionin, Zoya Moiseyeva (front), Gregory Galperin (back), me, Ilya Yunus, Valentin Skvortsov, Aleksandr Kornyushkin, Sergei Finashin, Sergei Fomin (front), Alexander Reznikov (back), Yuri Shmelev (front), Yuri Neretin (back), Victor Gutenmacher.

Our coaches are in the shot as well. Surprisingly, or not surprisingly, all of them moved to the USA. Yuri Ionin, now retired, was a professor at Central Michigan University. Gregory Galperin is a professor at Eastern Illinois University. Sergei Fomin is a professor at the University of Michigan. Victor Gutenmacher worked for BBN Technologies and Siemens PLM Software, and is now retired.

There are two more adults in the picture: Valentin Anatolievich Skvortsov, our leader and Zoya Ivanovna Moiseyeva, our deputy leader. Skvortsov was working at the math department of Moscow State University at that time. The University was angry that he didn’t block some students with Jewish heritage from the team thus allowing them to be accepted to Moscow State University without exams. I wrote a story of how Zoya persuaded Alexander Reznikov not to go to Moscow University to help Valentin. It ruined Alexander’s live, and didn’t even help Valentin. 1975 was Valentin’s last trip as the leader.