My son, Alexey Radul, designed a puzzle for my students.
Puzzle. Simplify the expression: ln ln a · ln ln b · ⋯ · ln ln z.
Mathematics, applications of mathematics to life in general, and my life as a mathematician.
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My son, Alexey Radul, designed a puzzle for my students.
Puzzle. Simplify the expression: ln ln a · ln ln b · ⋯ · ln ln z.
Every year, I used to blog about math-related puzzles from that year’s MIT mystery hunt. However, I want to skip this year. I participated in the hunt, and a few of the puzzles I saw were not good enough. Moreover, the puzzles had boring steps or were inelegant. I am not motivated to sift through every math puzzle in the hunt and check its quality.
This year, instead of cataloging all the math puzzles, I will present a short list of puzzles from the hunt that were recommended to me by others. Most of these puzzles are unrelated to math, but I checked them out, and they seem cool.
After the hunt, I asked two people about their favorite puzzles. Both of them mentioned Showcase. Looking at it, I know why. I’ll give you a hint: this puzzle will appeal to people doing competitive programming.
Here are some other recommendations.
Puzzle. If you can move exactly two matches, what is the largest possible number you can get?
Many people assume that given that 508 has three digits, the answer also has three digits. They get the number 999 and stop there.
Some students realized they can make an extra 1 out of two matches and gave answers with four digits: 1503, 5051, and 5103.
Another great idea is to take out the two horizontal matches from 0, turning the 0 into 11. For example, one might use the two matches to turn 5 into 8 and get 8118. However, we can combine this idea with the previous one and use the two matches to make a new digit 1 to get 51181.
When I first saw this puzzle several years ago, 51181 was the official answer. But my students went further, much, much further.
Another elegant idea is to rotate the picture and get 81151.
Some students decided that having all the digits be the same height is not very important. One vertical match reads as 1, even if it is half the height. The largest number they got this way was 511811. Combining this idea with flipping the number allows us to get 811511.
However, small-font digits are often used in powers. This train of thought led to a brilliant answer, 511811. This number is way larger than everything we saw before: it has 41 digits. If we combine this idea with rotating the number, we can get 811511, a 43-digit number.
I recently decided to check the Internet again and discovered a chain of videos showing solutions to this puzzle by the same author. He started with a simple solution, 51181, then people left comments in the video with better solutions, so he remade the video. This happened several times. I probably should send him a link to this essay for yet another video.
Here is an interesting solution by one of my students that I haven’t seen anywhere else. The idea is to use scientific notation. The student took two matches from the right side of 0. He used one of them to convert the first digit from 5 to 9 and the second digit from 0 into the letter E. He got 9E8 = 900000000. If we combine this idea with using a small 1, we can get 5E81, an 82-digit number.
Another brilliant idea is using two matches to create a caret symbol representing an exponent. This way, we can get 5^118, which is an 83-digit number. If we combine this idea with the rotation idea, we can get 8^115, a 104-digit number.
If we ignore the relative sizes of the digits, we can have the small digits as the base of the exponent and the larger digits as the power. One of the students suggested 115118, a 5330-digit number. And if we rotate the number, the answer we can get is 118115: the number with 8451 digits. This is a humongous number compared to the mere 3-digit one we started with, to the pathetic 5-digit original solution, and to the pitiful 104-digit previous record. But my students didn’t stop there.
One student suggested snapping two matches and creating a double factorial 505!!, with 575 digits. Combining this with other ideas, we can get 8115!!, a 14102-digit number.
One of my students decided to use two matches from the last digit 8 to create a factorial sign. She glued one of the matches perpendicular to the plane to get 505! in the projection. She even made a picture that you can see below. This number has 1148 digits, and combining this with previous ideas, we can get 5118!, which has 16763 digits. Or even 8115!, which has 28202 digits.
For this puzzle, a factorial works better than a double factorial. Here is my own suggestion: use one of the matches to create a small digit one, and split the other match into a factorial. This way, we can get 81151!, a number with a whooping 363154 digits.
This is an excellent example of a thinking-outside-the-box puzzle. I love this puzzle because it has not one, but many boxes one can think outside off.
Sometimes I mess with my students. Recently, I gave them the following problem for homework.
Puzzle. Don’t read this sentence.
This problem is a paradox: students can’t know not to read it unless they read it! I expected my students to explain the paradox, and a couple of them did. But, most of them provided me with an infinite stream of entertainment. Here are some of their answers divided into categories, starting with the apologizing ones, and there were a lot of those:
I wasn’t surprised by the apologies, as this problem is intended to be intimidating. However, other students tried to wiggle out of it:
Yet other students decided to rebel:
Karma is a boomerang, and this problem got me into a pickle. One of the most brilliant and funny solutions possible was to leave the answer box blank, implying that the sentence wasn’t read. However, how would I grade this? What if they just skipped the problem? I decided to err on the students’ side and gave full credit for an empty answer box.
A couple of students made a point of pretending they didn’t read the sentence:
But I saved the best for last. My favorite answer was the following:
Puzzle. Alice and Bob play the following game on a regular chessboard, where all the pieces move according to the standard rules of chess. Alice has a king, and Bob has a knight. They would place their pieces on the chessboard without attacking each other. Alice starts, and they alternate moves. Alice loses if the knight checks or all available moves place her under the knight’s attack. For which initial positions of the pieces is Bob guaranteed a win?
Puzzle. “I guarantee,” said the pet-shop clerk, “that this parrot will repeat every word it hears.” A customer bought the parrot but found it wouldn’t speak a single word. Nevertheless, the clerk told the truth. Explain.
The official answer:
Indeed, in this case, it is not a lie that the parrot will repeat every word it hears. My students had some other ideas. The following answer differs from the official one by one letter, but the spirit of the solution is the same.
Another idea my students had was to introduce a time component.
And a couple of outside-of-the-box answers.
Puzzle. Alice and Bob divide a pie. Alice cuts the pie into two pieces. Then Bob cuts one of those pieces into two more pieces. Then Alice cuts one of the three pieces into two pieces. In the end, Alice gets the smallest and the largest piece, while Bob gets the two middle pieces. Given that both want to get the biggest share of the pie, what is Alice’s strategy? How much can she get?
I got immediately attracted to the puzzle Oleg Polubasov recently posted on Facebook.
Puzzle. A rectangular clearing in a forest is an N-by-M grid, and some of the cells contain a tower. There are no towers in the cells that neighbor the forest. A tower protects its own cell completely and parts of the eight neighboring cells at a depth of half of a cell. Here by neighbors, we mean the cells horizontally, vertically, and diagonally adjacent to the given cell. In particular, if each cell is one square unit, a tower protects four square units. The protected area forms a square with borders that lie in between the grid lines. A tsar knows the towers’ locations and wants to calculate the protected area. Prove that the following formula gives the answer: the number of 2-by-2 subgrids that contain at least 1 tower.
I like this puzzle because it has an elegant solution. But there is more. The puzzle reminds me of one of my favorite novels by Arkady and Boris Strugatsky: The Inhabited Island, also known as Prisoners of Power. This is a science fiction novel where Max Kammerer, a space explorer, ends up on a planet with desolate people who, twice daily, experience sudden bouts of enthusiasm and allegiance to the government. Later, it becomes clear that the love for the rulers comes from towers that broadcast mind-control signals suppressing critical thinking and making people prone to believe propaganda.
The novel was written in 1969 but accurately describes the modern Russian propaganda machine. It appears that there is no need for a secret signal. Just synchronized propaganda on government-controlled TV turns off people’s brains.Share:
Here is one of my all-time favorite problems.
Puzzle. Four glasses are placed on the four corners of a square rotating table; each glass is either right-side up or upside down. You need to turn them all in the same direction, either all facing up or all down. You may do so by grasping any two glasses and turning either none, one of them, or both over.
There are two catches: 1) You are blindfolded, and 2) the table is spun after each round. Assuming a bell rings when you have them all facing the same way, how do you do it?
When I first heard this problem, the person who tortured me with it forgot to mention the bell. The problem was impossible: there was no feedback. Then, the bell was mentioned. It was an aha moment: you get information, but only a confirmation that the problem is solved. I tried to think about the puzzle backwards. What could be my last step? Assume that I know that the glasses alternate around the table: up, down, up, down. Then, as my last step, I can reach for the diagonal glasses and turn them over.
This is how you might start thinking about this puzzle. You can find the rest of the solution on the puzzle’s Wikipedia page.
Here is a wonderful variation, proposed by Michael Hotiner from Ukraine, that appeared ten years ago at a puzzle competition. This variation is not cheap; the players must pay with their lives, albeit virtual ones.
Puzzle setup. Consider a computer game where at level M, there is an M-sided rotating table with glasses at each corner. The setup is similar to the four-glasses puzzle above. The player is blindfolded, and the table rotates between rounds. As soon as level M is over, if all the glasses face one direction, the bell rings, and the player moves to the next level, M+1.
At each round, the player can decide on the number N of glasses to touch. Upon deciding, they have to pay N! lives for that round. Then, the player can touch the glasses one by one, choosing the next glass depending on how the previous glasses are oriented. After all the glasses are touched, the player can decide which ones to turn upside down.
Let’s see what happens at the very beginning of this game. The game starts at level 1, with only one glass. The round is solved before it starts. The cost is 0. The bell rings, and the game immediately goes to level 2.
Consider level 2. If the bell doesn’t ring immediately, the glasses face different directions. The cheapest way to proceed is to choose N = 1 and turn any glass. The cost is 1.
Consider level 3. A player can solve it in one round by touching all three glasses and turning them the same way. The cost is 3! = 6. There is a cheaper method that costs 4 lives in two rounds. In the first round, the player can touch any two glasses and turn both of them up. If the bell doesn’t ring, then the third glass is upside down. In the next round, the player can touch two glasses. If one is upside down, that glass should be turned up. If both glasses are right side up, then the player should turn both of them over to finish the round.
- Find a way to pass level 3 using only three lives.
- Find the cheapest way to pass level 4.
- Find the cheapest way for level 6.
- Find a way to pass level 5 using only 30 lives.