Here’s a neat coin puzzle I received by email from my reader s_hskz2 (at twitter.com).
Puzzle. You have 9 coins: 3 gold coins, 3 silver coins, and 3 bronze coins. Within each metal, the coins are indistinguishable. Exactly one gold, one silver, and one bronze coin are counterfeit; the other six are genuine. You are provided with a magic bag that functions as follows: when you place a subset of coins into the bag and cast a spell, the bag glows if and only if the subset contains all three counterfeit coins. Can you identify all three counterfeit coins using at most 5 tests?
I tried to find an easy solution and didn’t. Then I decided to use information theory to guide me to an answer. Unsurprisingly, it worked. The solution wasn’t trivial, but it was a lovely practice in using information theory for such puzzles.
Later, s_hskz2 sent me a more difficult version: There are 10 coins of each kind, and you are allowed to test 10 times, but I was too lazy to try.
Puzzle. Three brothers who are identical triplets live on the seventh, eighth, and ninth floors of the same apartment building. Their apartments are identical and vertically stacked. One day all three step onto their balconies, standing in the same upright posture. The brother on the eighth floor shouts “AAAA!” Which of the other two will hear him first?
I gave this puzzle to my students, and two of them offered the right answer for the wrong reasons. One said, “The seventh-floor brother, because air is warmer closer to the ground and sound travels faster in warmer air.” Another said, “The seventh-floor brother, because the air is denser at lower altitude and sound travels faster in denser air.”
For the last homework assignment, I gave my students the task of finishing a famous Russian joke. Problem. At the height of the Cold War, a U.S. racing car easily beat a Russian car in a two-car race. How did the Russian newspapers truthfully report this in order to make it look as though the Russian car had outdone the American car?
The joke was that the Russian newspapers truthfully reported that the Russian car came in second and the American car second to last.
One of my students, William, got a different idea and wrote a whole article.
AMERICAN CAR STOPS RUNNING LONG BEFORE RUSSIAN CAR FINISHES RACE
A Russian car and an American car were competing in a two-car race. At one point, the American car mysteriously drove off the race course and stopped. Of course, this meant that the race was over for them. All that the Americans could do was watch on the sidelines for the Russian car to reach the end of the course, which it completed successfully. The outcome of the race was in no way uncertain. Congratulations, Russians!
We recently wrote a blog post on how to generalize the game of SET and promised to continue. Here we are. But first, a reminder of what the game of SET is.
In the game of SET, we have 81 cards, each containing one, two, or three of the same object. The object is green, red, or purple, the shape is squiggly, oval, or diamond, and the shading is empty, full, or stripped. Three cards form a set if, for every feature, the attributes are all the same or all different. An example of a set with all features different is shown below. By the way, such sets are usually more difficult to spot. In the game, you need to find sets as fast as you can.
If we assign each attribute value a number 0, 1, or 2, we get an equivalent definition of a set. Three cards form a set if and only if the values for each feature sum to zero modulo 3. Thus, we can see our cards as vectors in the space F34. Three vectors form a set if they sum up to 0.
The generalizations we described in the previous post, were the following. We pick a different group and define a set as a few cards that might need to be in a specific order that multiply to the group’s identity.
However, there is a different way to generalize sets to groups. Three cards that form a set in a classical game of SET, taken in any order, form an arithmetic progression. In other words, if a, b, and c form a set, then vectors b−a and c−b are the same. We can check this. We have c−b = c−(c+b+a)−b =−2b−a = b−a.
Thus, we can generalize the game of SET differently. Suppose cards are vectors in some space. We say three of them, a, b, and c, form a set if and only if b−a = c−b. Now, the order becomes important, similar to our previous generalization. We do not need to use commutative groups like vector spaces. For any group, our condition is equivalent to ba-1 = cb-1. Thus, c = ba-1b.
Interestingly, we do not care much about the identity card, meaning the card deck is a torsor. We introduced the notion of a torsor before, which informally is a group that forgets about its identity. Now, let’s check possible examples.
Suppose the values of one attribute correspond to Z4. This game is not very inspiring as two values, a=0 and b=2, can be completed to a set with the third card, which equals c=(0,0,0) and is already used. The next interesting example is Z5. Here, values a=0 and b=2 can be completed to a third value c=4. We will leave it to the reader to check that for any two cards, a and b, the third card, c, differs from both a and b.
To make it more visual, we can use a pentagon with one marked direction. We fix the pentagon in space. In this case, three cards form a set if and only if the directions of the first and the third card are symmetric with respect to the direction of the second card. If we want to use three pentagons, we can reuse the cards from the game C53T, we described in our previous post. To make this game more visual, we can use three pentagons. We can mark each coordinate with a direction on the corresponding pentagon. The colors are to allow players to visually process the cards faster. For theoretical purposes, the colors can be ignored. Also, the pentagons themselves have different intensities of gray to emphasize that the game is played on each of them separately and also help choose the top of the card.
Let us go back and calculate when it is possible that the element that completes a set is already used. If our initial elements are a and b, then we need ba-1b to complete a set. If this element is equal to b, then a = b, which contradicts the assumption that we start with two different cards. Suppose ba-1b = a, then, equivalently (ba-1)2 = 1. The new element can be the one that is already used if and only if the group contains elements of order 2.
Can we use other decks we described in the previous post? Consider the game ProSet/Socks. The group is commutative, and every element is of order 2, which means ba-1b is always a. We can’t use the deck at all! What about the EvenQuads deck? It can be viewed as Z43, so it is possible that we can’t complete any two cards to a set. However, there is a bigger problem with the deck. To play with it, we need to actually assign values to colors and shapes. We are saying that even if we decide to use the group, we should make different cards! For example, we can style the cards as squares like the pentagons above.
We also mentioned in the previous post the group, which is the wreath product of S2 and S3. Consider the following example of two cards that were screen-printed from the Numberphile video on the variations of the game of SET.
The first card, a, is the inverse of itself, so the third card we are looking for is described as the product bab, which we can visualize as the following. If we ignore the beads, the card that completes the set is a. Luckily, if we do not ignore the beads, it is not a; we need to add two beads to a.
The product of permutations is difficult to visualize, so playing this game with the cards in the Numberphile video might be difficult. The good news is that this group can be visualized in many different ways:
The group of symmetries of a cube, meaning its rotations and reflections;
The group of symmetries of an octahedron, meaning its rotations and reflection;
The wreath product of S2 and S3;
The direct product of S4 and S2.
We want to show you a beautiful deck from the tsetse website that allows you to use any one of the four definitions to play the game. The game is called OCTA Set, as the underlying group is called an octahedral group. An example of a set is in the image below, where the top and bottom shapes represent the same element of the group. Moreover, the deck is a torsor: there is no the identity card.
The top shape represents an octahedron where the opposite faces are the same color. Thus, the top shape shows a unique rotation of an octahedron. All the swirls are the same in the image, but in other cards, the swirl could be white. Two different swirls mean that the octahedron needs to be reflected to get from one shape to the other. In our example above, no reflection is involved, so all the swirls are black.
We can ignore that the top shape represents an octahedron and only look at the choice of colors. The changes of colors represent a permutation in S4 and a swirl represents an element in S2.
The bottom shape represents a cube where the opposite faces are the same color. One of the faces is solid, and the opposite face is hollow. This way, one can reconstruct the complete shape of a cube by the top three faces.
We can ignore that the top shape represents a cube and only look at the choice of colors. The colors form a permutation, and the beads correspond to changing the hollowness of a color. When viewing the cube this way, the permutation acting on the colors is S3, and each color has its hollowness associated with an element of S2. This exactly corresponds to the wreath product of S3 and S2.
Let us prove that this is a set. Consider the bottom cube shape. Comparing the first two cards, the top face doesn’t change. We can see that the symmetry of the cube is the 90-degree clockwise rotation around the line that goes through the centers of green faces. In such a rotation, the left face on the second card keeps the color from the first card, while the right face takes the color from the left face on the first card and swaps hollowness. We see that the third cube completes the set.
For another proof, let us look at the top shape and discuss what happens with the permutation of colors when changing from the first card to the second. The left color moves to the bottom, the bottom color to the right, the right color to the center, and the center color to the left. Not surprisingly, we got a cyclic permutation of order 4, similar to a 90-degree rotation being of order 4. The same thing happens when moving from the left to the right. The swirl stays the same.
As we mentioned, when you play this game and pick two cards then calculate what card completes the set, you might discover that it is one of the cards you picked. The probability that two cards in a specific order can’t be completed into a set is the same as the probability of picking a random element in our group and discovering that it has order 2. The symmetric group S4 has 9 elements of order 2. Thus, the direct product with S2 has 19 elements of order 2, giving a probability of 19/48. For completeness, this group also has 8 elements of order 3, 12 elements of order 4, 8 elements of order 6, not to mention the identity of order 1.
If ba-1 is an element of order three, then the cards a, b, and the card c that completes the set form a set when they are taken in any order. As a tradition in mathematical writing, we leave it to the reader to check that fact. Just a reminder that in the game of SET, the group element ba-1 always has order 3.
Notably, there was nothing special or extraordinary about the group discussed above. It has a pretty visualization as a cube or octahedron, but is not otherwise particularly interesting. The reason why this group allowed for these two platonic solids to be used to visualize it is because the cube is dual to the octahedron. But we could have similarly used any group to play set! One such example might consider using the group of symmetries of the other pair of dual platonic solids, the icosahedron and the dodecahedron. This group is actually equivalent to A5, also known as the alternating group of order 5, which consists of all even permutations of five elements. The tsetse website we mentioned above contains an implementation of such a game called A5SET (pronounced “asset”). The design of the site, games, and cards was done by Andrew Tockman and Della Hendrickson.
The world is full of groups and symmetries. Any group can be turned into a game of SET!
Foams are cool mathematical objects studied by my brother, Mikhail Khovanov. I already wrote about them in my previous blog posts, Foams Made out of Felt and Tesseracts and Foams. Here, I would like to explain why foams are so cool, but first, I need to remind you of their definition. Foams are finite 2-dimensional CW-complexes, such that each point’s neighborhood must be homeomorphic to one of the three objects below.
An open disc. Such points are called regular points.
The product of a tripod and an open interval. Such points are called seam points.
The cone over the 1-skeleton of a tetrahedron. Such points are called singular vertices.
Foams are cool: they are 2-dimensional CW-complexes embedded in 3-space, with singularities only of the most generic kind, which makes them relatively simple. Moreover, they are combinatorially defined, which makes them easier to work with than with many other geometric objects.
My two previous blog posts have some pictures, but now, I just want to discuss a generic planar cross-section of a foam, which is a planar graph. In the cross-section, seams become vertices, and faces (regular points) become edges. The tripod condition above implies that the resulting graph is trivalent: each vertex has degree 3.
The most interesting foams are tricolarble: foams where their faces can be colored in three colors, so that each face has its own color, and, at the seams, three faces of three different colors meet. The cross-section of such a foam makes a tricolorable trivalent graph. This coloring is called Tait coloring. The cool thing is the Tait’s theorem connects the Tait coloring to the 4-color theorem.
Tait’s theorem. The following two statements are equivalent.
Every planar graph is 4-colorable.
The edges of every planar bridgeless trivalent graph are 3-colorable.
I won’t discuss the proof here, but I will explain how to color the edges of a graph in three colors when the faces are colored in four, and vice versa.
Assume that the four colors of the faces form a group of four elements, called the Klein group. Let’s say that gray is the identity, and red, blue, and green are the rest. Then, the product of gray and x is x. The product of any two non-gray colors is the third non-gray color.
Given a trivalent graph G whose edges are colored in three colors, we can color the faces of that graph in the following manner. Color one of the faces a random color. Then, calculate the colors of the other faces so that each edge’s color is the product of the colors of neighboring faces.
Going back, if we have a planar trivalent graph with faces colored in four colors, we can assign an edge a color that is the product of the colors of neighboring faces. As neighboring faces have different colors, the product of those colors is never gray (the identity). Thus, the edges will be colored in three colors. I leave it to the reader to check that three edges incident to a vertex must be colored in different colors.
Kronheimer-Mrowka homology theory of graphs states that the Kronheimer-Mrowka homology of a trivalent graph is non-zero if and only if the graph has no bridge. If one can prove that the rank of the homology group is the number of 3-colorings of the edges (or at least that the non-zero homology implies the existence of the tricoloring of that graph), then the Four-color theorem would follow from Tait’s theorem.
Foams are cool by themselves, but there is hope that they might provide a conceptual proof of the Four-color theorem, making them awesome!
Do you know that 2025 is a composite, deficient, evil, odd, square, and powerful number? I collect properties of numbers at my Number Gossip website, where you can also find detailed definitions of these terms. Provocatively, 2025 is also an apocalyptic power, meaning that 2 to the power of 2025 contains 666 as a substring.
Recently, Tamas Fleischer sent me an email discussing additional fascinating properties of 2025. While I am slowly deciding whether to add them to my database, there is some urgency in posting these properties in anticipation of the coming year. Here’s the material from Tamas, retold in my own words.
Out of the properties mentioned earlier, the square property is the only rare one. On my website, I define a property as rare if fewer than 100 numbers below 10,000 possess it. Square numbers barely make the cut. But 2025 is not just a square number—it is the square of a triangular number. If you remember the formula for the sum of cubes of the first n natural numbers, the result is (n(n+1)/2)2, which is the square of the nth triangular number. Thus, 2025 is the sum of the cubes of all one-digit numbers.
Additionally, 2025 is the product of 25 and 81. My website notes an intriguing property shared by 25 and 2025: both remain square numbers when all their digits are incremented by 1. For example, 25 becomes 36, and 2025 becomes 3136, both of which are squares. Moreover, 25 is the smallest such number, and 2025 is the second smallest. What my website does not mention is that their square roots exhibit a similar pattern. The square roots of 25 and 2025 are 5 and 45, respectively. When their digits are incremented by 1, the results are 6 and 56, the square roots of 36 and 3136, respectively. The original and incremented squares and their square roots are tied together in a surprising way.
2025 also shares an interesting property with 81. Both are square numbers with an even number of digits and if you split the digits in half and sum the halves, the result is the square root of the original number. For 81, splitting into 8 and 1 gives 8 + 1 = 9, which is the square root of 81. Similarly, for 2025, splitting into 20 and 25 gives 20 + 25 = 45, the square root of 2025. Intriguingly, 81 is the smallest number with this property, and 2025 is the second smallest.
Once, I gave a puzzle to my students as homework just to check their level of attention.
Puzzle. A family has two identical twins. One of them is a boy, what is the probability that the other one is a boy?
What can I say? Some of the students didn’t pay attention and gave weird answers like 1/2, 1/3, and 2/3.
The twins are IDENTICAL. The other one has to be a boy! Duh!
One of the students was well-educated and mentioned that it is theoretically possible for different twins to be different genders, though this is extremely rare. When one fertilized egg splits into two, producing two embryos, the genetic material of both eggs is the same, almost. Some errors during splitting are possible, and it seems that some very particular errors can lead to the identical twins being identified as a boy and a girl. I never knew that before!
However, one student thought outside the box. In his vision, a family adopted two identical twins who aren’t each other’s twins, just happen to be identical twins with someone else.
Here is a puzzle for middle school students from the Möbius tournament.
Puzzle. For which natural numbers n greater than 1 it is possible to arrange n numbers 1 through n in a circle so that the difference between two neighbors always divides n?
A new hat puzzle from Gribalko, reminding me of traffic lights.
Puzzle. You and six of your mathematician friends each have a hat placed on your head. Each of you can see the hats of all the others but cannot see your own. You were all told that there were three red, three yellow, and three green hats in total, but two of them were hidden. Your friends began to say the following phrases in sequence:
First: “I don’t know what color my hat is.”
Second: “I also don’t know what color my hat is.”
Third: “I also don’t know what color my hat is.”
Fourth: “I know that my hat is red.”
Fifth: “But I still don’t know what color my hat is.”
Sixth: “And I am sure that my hat is yellow.”
Can you determine what color hat you have on your head?
I once saw a TikTok video featuring a puzzle: four matches were arranged to form a plus sign, and the challenge was to move one match to create a square. The solution shown differed from what first came to my mind, so I decided to share the puzzle with my students. Instead of drawing a diagram, I described it to them in words.
Puzzle. Arrange four matches to form a plus sign. Move one match to form a square.
Most of my students gave the same solution shown in the video: moving one match slightly away from the center to create a square in the center, with the inner edges of the matches forming its borders. Not all the students were as lazy as I was; some drew pictures to illustrate. One example is shown below, where the resulting square is in green.
My solution, also discovered by a couple of students, was to move one match to form the number 4, which is a square.
I am glad I didn’t provide a picture because it led to two unexpected solutions. For the first one, imagine a 3D shape out of four matches where one projection forms three sides of a square, and the other one is a plus sign. We can take the match from the plus sign that is not a side of a square and use it to complete the square. The second solution is shown below. The arrangement already contains a small square, so you can take a match and put it back. Being lazy brings benefits!
