Tanya Khovanova's Math Blog

Here is a famous math problem I never before wrote about:

Puzzle. Five pirates discovered a treasure of 100 gold coins. They decide to split the coins using the following scheme. The most senior pirate proposes how to share the coins, and all the pirates vote for or against it. If 50% or more of the pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the next most senior pirate making a proposal.

As pirates tend to be a bloodthirsty bunch, if a pirate would get the same number of coins whether he votes for or against a proposal, he will vote against so that the pirate who proposed the plan will be thrown overboard. Assuming that all five pirates are intelligent, rational, greedy, and do not wish to die, how will the coins be distributed?

You can find the solution in many places including Wikipedia’s Pirate game. The answer is surprising: the most senior pirate gets 98 coins, and the third and the fifth pirates by seniority get one coin each. I always hated this puzzle, but never bothered to think through and figure out why. Now I know.

This puzzle emphasizes the flaws of majority voting. The procedure is purely democratic, but it results in extreme inequality.

That means a democracy needs to have a mechanism to prohibit the president from blatantly benefiting himself. With our current president these mechanisms stopped working. Given that Trump does everything to enrich himself, the pirates puzzle tells us what to expect in the near future.

We, Americans, will lose everything: money, clean air and water, national parks, future climate, health, social security, and so on, while Trump will make money.

I was on the writing team of this year’s hunt, which was based on the movie “Inside Out.” One of our goals was to create an easy first round to allow small teams to have a full hunt experience. Our first round consisted of 34 puzzles related to five basic emotions: joy, sadness, disgust, fear, and anger. Each emotion had its own meta puzzle. And the round had a meta-meta puzzle and a runaround. As I tend to write easy puzzles, I contributed three puzzles to this emotions round. The puzzles had references to corresponding emotions that were not needed for the solve path. They were inserted there for flavor.

• A Learning Path (jointly with Xavid) is a Nikoli-type logic puzzle that was targeted for new hunters. It contained self-referencing hints and solving techniques. It was in both disgust and fear.
• Word Search is a word search with a twist. The words are related to fear and sadness.
• Games Club (jointly with Sergei Bernstein) is a puzzle on the topic of combinatorial two-player games. It is pure sadness.

I also wrote another easy puzzle called A Tribute: 2010-2017 (jointly with Justin Melvin, Wesley Graybill, and Robin Diets ). Though the puzzle is easy, it is useful in solving it to be familiar with the MIT mystery hunt. This is why the puzzle didn’t fit the first emotions round.

I also wrote a very difficult puzzle called Murder at the Asylum. This is a monstrosity about liars and truth-tellers.

In mathematics one of the most important questions is why. Let us consider a problem:

Problem. A number has three hundred ones and three hundred zeroes. Can it be a square?

The solution goes like this. Consider divisibility of this number by 9. The sum of the digits is 300. That means the number is divisible by 3, but not by 9. Therefore, it can’t be a square.

Why do we consider divisibility by 9? The divisibility by 9 is a very powerful tool, but why was it the first thing that came to my mind? The divisibility by 9 doesn’t depend on the order of the digits. Whenever I see a problem where the question talks about digits that can be in any order, the first tool to use is the divisibility by 9.

The why question, is very important in mathematics. But it is also very important in life. It took me many years to start asking why people did this or that. I remember my mom was visiting me in the US. Every time I came back from work, she complained that she was tired. Why? Because she did the laundry in the bath tub. She wouldn’t use my washing machine, because she didn’t have such a thing in Russia. I promised her that I’d do the laundry myself when there was a sufficient pile. However, she insisted that the dirty clothes annoyed her. I would point that my water bill went up. And so on.

We argued like this every day. We were both frustrated. Then I asked myself why. Why does she do the laundry? The answer was there. She wanted to be helpful. I calmed down and stopped arguing with her. I sucked it up and paid the water bills. Her time with me turned into the most harmonious visit we ever had. Unfortunately, it was the last.

Puzzle. 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 which rides at 25 miles per hour with 1 rider on it; or, at 20 miles per hour with 2 riders. Each of the 3 friends walks at 5 miles per hour. How fast can all three of them make it to Bob’s house?

This famous trick puzzle is very old:

Puzzle. The professor is watching across a field how the son of the professor’s father is fighting with the father of the professor’s son. How is this possible?

This puzzle is tricky only because of gender-bias. Most people assume that the professor is male and miss the obvious intended solution, in which a female professor is watching her brother fighting with her husband.

I just gave this problem on a test. Here are other answers that I received.

• The professor is gay and is watching his brother fighting with his husband.
• The professor is watching his brother fighting with the father of the professor’s step-son.
• The father of the professor’s son is himself. So he is watching a video of himself fighting with his brother.

Years ago people couldn’t figure out this puzzle at all. So there has been progress. I was glad that my students suggested so many ideas that work. Nonetheless, many of them revealed their gender-bias by initially assuming that the professor is a man.

I can’t wait until this puzzle stops being tricky.

Puzzle. There are five houses of different colors next to each other equally spaced on the same road. In each house lives a man of a different profession.

• The blue house is adjacent to the mathematician’s and con-man’s houses.
• The first house on the left is green.
• The nurse lives immediately to the right of the mathematician.
• The teacher lives halfway between the plumber’s house and the yellow house.
• The nurse’s house is immediately to the right of the red house.

Who lives in the white house?

Correction Nov 11, 2017. Replaced “the same distance from” with “halfway between” to eliminate the possibility of the plumber living in the yellow house. Thank you to my readers for catching this mistake and to Smylers for suggesting a correction.

I do not remember where I saw this problem.

Problem. Invent a connected shape made out of squares on the square grid that cannot be cut into dominoes (rectangles with sides 1 and 2), but if you add a domino to the shape then you can cut the new bigger shape.

This problem reminds me of another famous and beautiful domino-covering problem.

Problem. Two opposite corner squares are cut out from the 8 by 8 square board. Can you cover the remaining shape with dominoes?

The solution to the second problem is to color the shape as a chess board and check that the number of black and white squares is not the same.

What is interesting about the first problem is that it passes the color test. It made me wonder: Is there a way to characterize the shapes on a square grid that pass the color test, but still can’t be covered in dominoes?

I already wrote about two puzzles that Derek Kisman made for the 2013 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. The second puzzle, 50/50, is considered one of the most difficult hunt puzzles ever. Unfortunately, the puzzle is not available, but my description of it is.

Today let’s look at the third puzzle Derek made for the 2013 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.

A while ago I posted my second favorite problem from the 2015 All-Russian Math Olympiad:

Problem. A coin collector has 100 coins that look identical. He knows that 30 of the coins are genuine and 70 fake. He also knows that all the genuine coins weigh the same and all the fake coins have different weights, and every fake coin is heavier than a genuine coin. He doesn’t know the exact weights though. He has a balance scale without weights that he can use to compare the weights of two groups with the same number of coins. What is the smallest number of weighings the collector needs to guarantee finding at least one genuine coin?

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 100ⁱ 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, 100⁷⁰ to one of the coins on the scale and claim that the pan with this coin is heavier. We continue recursively. 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 100 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 2014. The problems are for 7 graders.

Problem. Inside a 5-by-8 rectangle, Bart draws closed paths that follow diagonals of 1-by-2 rectangles. Find the longest possible path.

This problem is really very difficult. The competition organizers offered an extra point for every diagonal on top of 16. 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.

Here is another problem:

Problem. Alice and Bob are playing a game. They start with two numbers: 2014 and 2015. In one move a player can do one of two things:

1. subtract one of the numbers by one of the non-zero digits in any of the two numbers or,
2. divide one number by two if the number is even.

The winner is the person who is the first to get a one-digit number. Assuming that Alice starts, who wins?