My students (Matvey Borodin, Eric Chen, Aidan Duncan, Boyan Litchev, Jiahe Liu, Veronika Moroz, Matthew Qian, Rohith Raghavan, Garima Rastogi, Michael Voigt) and I recently wrote a paper connecting the stable marriage problem and Sudoku. I just blogged about it. By the way, my students are in grades 7-9.
On the way, we invented a new type of Sudoku, which we call joint-groups Sudoku. This type is in contrast to a famous type of Sudoku, called disjoint-groups Sudoku. In a disjoint-groups Sudoku, in a particular place in a 3 by 3 box, all the digits are distinct across all the boxes. For example, the top-left corners of nine boxes have all the digits 1 thought 9. This creates nine additional disconnected regions (depending on the placements inside a 3 by 3 box) to add to columns, rows, and boxes that have to contain distinct digits.
For our new type, we wanted the digits in a particular place in each box, instead of being different, to be the same as much as possible. How much of sameness is possible? The first row contains three top-left corners. Thus, by Sudoku rules, these top-left corners have to be distinct. Thus, the top-left corners in all nine boxes have to contain at least three distinct digits. So here is the rule for the joint-groups Sudoku: the nine digits in a particular place in a 3 by 3 box contain not more than three distinct digits. It is easy to see that it means they contain exactly 3 distinct digits, each of them three times.
Here are two Sudoku puzzles from our paper. Each puzzle, when completed, forms a joint-groups Sudoku.
I recently posted my puzzle designed for the MoMath meet-up.
What’s in the Name?
4, 6, X, 9, 10, 12, 14, 15, 16
1, 2, 6, 24, 120, X, 5040, 40320, 362880
2, X, 3, 4, 7
1, 2, 3, 4, 5, 6, X, 8, 9, 153, 370, 371
X, 2, 3, 4, 5, 6, 7
6, 28, 496, 8128, X, 8589869056, 137438691328
0, 1, 1, X, 4, 7, 13, 24, 44, 81
Now it is time for the solution.
The solvers might recognize some sequences and numbers. For example, numbers 6, 28, and 496 are famous perfect numbers. Otherwise, the solvers are expected to Google the numbers and the pieces of the sequences with or without X. The best resource for finding the sequences is the Online Encyclopedia of Integer Sequence.
The first “AHA!” happens when the solvers notice that the sequences’ names are in alphabetical order. The order serves as a confirmation of the correctness of the names. It also helps in figuring out the rest of the sequences’ names. The alphabetical order in such types of puzzles hints that the real order is hidden somewhere else. It also emphasizes that the names might be important. The sequences names in order are:
Composite
Factorial
Lucas
Narcissistic
Natural
Perfect
Tribonacci
The second “AHA!” moment happens when the solvers realize that the Xs all have different indices. The indices serve as the final order, which in this case is the following:
Natural
Lucas
Composite
Tribonacci
Perfect
Factorial
Narcissistic
The third “AHA!” moment happens when the solvers realize that the number of terms is different in different sequences. It would have been easy to make the number of terms the same. This means that the number of terms has some significance. In fact, the number of terms in each sequence matches the length of the name of the sequence. The solvers then can pick the letter from each of the names corresponding to X. When placed in order, the answer reads: NUMBERS.
The answer is related to the puzzle in two ways:
The puzzle is about numbers.
The sequences’ names do actually need the second word: Lucas numbers, composite numbers, and so on.
The advantage of this puzzle for zoomed group events is that the big part of the job — figuring out the sequences — is parallelizable. Additionally, it has three “AHA!” moments, which means different people can contribute to a breakthrough. The puzzle also has some redundancy in it:
Due to the redundancy of the English language, it is possible to solve this puzzle without figuring out the names of all the sequences.
If the solvers can’t figure out the order, they can anagram the letters to get to the answer.
If the solvers do not realize that they have to use the letter indexed by the X, there is another way to see the answer: read the diagonal when the sequences’ names are in order.
This is the puzzle I designed for yesterday’s event at the Museum of Mathematics. This puzzle is without instructions — figuring out what needs to be done is part of the fun. Solvers are allowed to use the Internet and any available tools. The answer to this puzzle is a word.
This is the sequence of numbers n such that 3 times the reversal of n plus 1 is the number itself. In other words, n = 3*reversal(n)+1. For example, 742 = 3*247+1. In fact, 742 is the smallest number with this property. How does this sequence continue, and why?
I recently published Sergei Bernstein’s awesome Star Battle called Swiss Cheese. Another lovely Star Battle from him is called Hooks. You can play it online at puzz.link.
Star Battle is one of my favorite puzzle types. The rules are simple: put two stars in each row, column, and bold region (one star per cell). In addition, stars cannot be neighbors, even diagonally.
My son, Sergei Bernstein, recently designed a Star Battle with a beautiful solve path. This is my favorite Star Battle so far. I like its title too: Swiss Cheese.
A problem from the 2021 Moscow Math Olympiad went viral on Russian math channels. The author is Dmitry Krekov.
Problem. Does there exist a number A so that for any natural number n, there exists a square of a natural number that differs from the ceiling of An by 2?
Konstantin Knop, the world’s top authority on coin-weighing puzzles, suggested the following problem for the 2019 Russian Math Olympiad.
Puzzle. Eight out of sixteen coins are heavier than the rest and weigh 11 grams each. The other eight coins weigh 10 grams each. We do not know which coin is which, but one coin is conspicuously marked as an “Anniversary” coin. Can you figure out whether the Anniversary coin is heavier or lighter using a balance scale at most three times?
This is one of my favorite problems given at the 2017 Moscow Olympiad to grades 6 and 7. It was suggested by one of my favorite problem writers: Alexander Shapovalov.
Problem. We are given eight unit cubes. The third of the total number of their faces are blue, and the rest are red. We build a large cube out of these cubes so that exactly the third of the unit cube’s visible faces are red. Prove that you can use these cubes to build a large cube whose faces are entirely red.
