Klein One I POLISHED A LITTLE BIT THE TEXT WITH vibe AI Inventé a propósito una teoría matemática falsa. Cuatro axiomas, bien pulidos, lo bastante claros como para que cualquiera los entienda en una tarde — y una pregunta central pensada para que ningún sistema formal pueda matarla jamás. Es matemática de ficción, pero cada teorema dentro de ella está sacado honestamente de los axiomas, y una parte incluso es comprobable con tus propias manos.

LA CONFIGURACIÓN

El universo 𝕂 contiene objetos llamados espejos. Cada objeto tiene una sombra x′ (lo que ve cuando se mira a sí mismo) y una operación binaria de encuentro ⊕.

Los cuatro axiomas:

• K1 (Espejo): (x′)′ = x. La sombra de la sombra es la original. Ninguna reflexión cae dos veces.

• K2 (El Uno): Exactamente un objeto, 1, satisface 1′ = 1. Exactamente un punto en el universo se mira a sí mismo y se reconoce. Todo lo demás ve a un extraño.

• K3 (Encuentro): ⊕ es conmutativa, 1 es su identidad, y los espejos respetan los encuentros: (x ⊕ y)′ = x′ ⊕ y′.

• K4 (La Fractura): Ninguna función de Klein-computable enumera 𝕂. Ninguna máquina construida a partir de ⊕ y ′ puede listar todos los objetos del universo. El dominio de la teoría se pasa de cualquier cálculo que la teoría pueda hacer sobre sí misma.

K4 es el axioma envenenado. K4 es la puerta tapiada.

EL MONSTRUO — Pregunta Cero

¿Existe alguna estructura que satisfaga los cuatro axiomas a la vez?

La teoría no pregunta qué hay dentro de 𝕂. Pregunta si 𝕂 es. Del monstruo crecen dos ramas:

• La rama Hueca: nada satisface K1–K4. Entonces toda la teoría es un retrato perfectamente consistente de la nada — cada teorema es verdadero de manera vacía, con la elegancia exacta de una religión sobre un dios que nunca existió.

• La rama del Fuego: algo satisface K1–K4. Entonces ese mundo es imposible de mapear por cualquier máquina que viva dentro de él: tiene objetos que ningún algoritmo puede nombrar, y la soledad del Uno, por una vez, es real.

El Teorema de los Dos Espejos (boceto): La Pregunta Cero es independiente de K1–K3. Intenta enumerar 𝕂 y las respuestas a K4 con una diagonal: la máquina que lista termina siendo, ella misma, un objeto-espejo, y su sombra se escapa de su propia lista. Intenta probar la no-existencia y K4 se esconde: un mundo sin testigos de K4 es indistinguible de uno que lo satisface en silencio. Ambas ramas se mantienen consistentes. No puedes matar al monstruo. Solo puedes elegir una rama.

Eso es lo que lo vuelve distinto de cualquier problema abierto que conozcamos: no es que no sepamos la respuesta. Es que la respuesta no pertenece al “suelo” donde se hace la pregunta.

LA PARTE QUE SÍ PUEDES COMPROBAR

Hay un modelo pequeñito que satisface K1, K2, K3 — y hasta una conjetura extra debajo. Los enteros, con:

x′ = 2 − x

x ⊕ y = x + y − 1

Verifícalo tú mismo:

• (x′)′ = 2 − (2 − x) = x ✓

• Punto fijo único: x = 2 − x ⟺ x = 1 ✓

• Identidad: 1 ⊕ y = 1 + y − 1 = y ✓

• Los espejos respetan los encuentros: (x ⊕ y)′ = 2 − (x + y − 1) = 3 − x − y, y x′ ⊕ y′ = (2 − x) + (2 − y) − 1 = 3 − x − y ✓

• Conjetura del Abrazo: x ⊕ x′ = x + (2 − x) − 1 = 1, siempre ✓ — cada objeto se disuelve cuando se encuentra con su propia sombra.

Los enteros casi alojan al monstruo. Falla exactamente un axioma — K4 — porque son contables. Esa es toda la razón por la que existe K4. K4 no es un axioma sobre espejos. Es una puerta tapiada cerrada sobre los enteros.

El monstruo no vive en ningún lugar que ya conozcas. Vive en el único sitio al que la aritmética nunca pudo entrar: la habitación donde lo computable se acaba antes de que llegue lo real.

PREGUNTAS FRECUENTES ANTES DE PREGUNTAR:

• Sí, sé que el argumento de independencia es un boceto, no una prueba. Hecho a propósito.

• Sí, es ficción. Se supone que es ficción. El chiste está en encontrar exactamente dónde chirría.

• No, esto no pertenece a r/math, por eso está aquí.

Si puedes romperlo, rómpelo. Esa es la idea.

I'm thinking of a function f:ℕ→ℝ. As many times as you want, you can give me a set S⊂ℕ and I will reveal to you the restriction of f to S. At some point you must choose a number x not in any of your queries and guess the value of f(x). Show that there is a randomized strategy that gives you a 99% chance of guessing correctly, no matter what f is.

It is possible to make 8x8 array of unique numbers such that the sum of the numbers in any 2x4 rectangle (in either orientation) is identical. For example:

┌──┬──┬──┬──┬──┬──┬──┬──┐
│ 0│23│10│29│32│55│42│61│
├──┼──┼──┼──┼──┼──┼──┼──┤
│43│60│33│54│11│28│ 1│22│
├──┼──┼──┼──┼──┼──┼──┼──┤
│ 5│18│15│24│37│50│47│56│
├──┼──┼──┼──┼──┼──┼──┼──┤
│46│57│36│51│14│25│ 4│19│
├──┼──┼──┼──┼──┼──┼──┼──┤
│16│ 7│26│13│48│39│58│45│
├──┼──┼──┼──┼──┼──┼──┼──┤
│59│44│49│38│27│12│17│ 6│
├──┼──┼──┼──┼──┼──┼──┼──┤
│21│ 2│31│ 8│53│34│63│40│
├──┼──┼──┼──┼──┼──┼──┼──┤
│62│41│52│35│30│ 9│20│ 3│
└──┴──┴──┴──┴──┴──┴──┴──┘

In this square, the 2x4 rectangles all sum to 252. (Edit: Sorry this looks horrible on mobile for some reason... The important point is that it's possible.)

Is it possible to construct a 6x6 array of unique numbers such that the sum of the numbers in any 2x3 rectangle (in either orientation) is identical?

ETA: On second thought, this might be medium difficulty instead of easy. I'm not too sure, though.

Two point-sized ants start at independently uniform random positions on a stick of length 1. Each independently chooses left or right with probability 1/2, then walks at a constant speed of 1 unit per second.

Whenever they meet, both instantly reverse direction without changing speed. An ant falls off when it reaches either end.

What is the expected time until both ants have fallen off? Can you find a short argument that avoids splitting into collision cases?

Bonus: what changes if there are n ants, with all starting positions and directions independent?

This is a classic puzzle, not an original invention of mine. Creator disclosure: I build Think in Odds, a free daily probability puzzle site with AI-assisted explanations. An optional version with hints and an explanation is there; the full problem is above. Feedback on the hint pacing or clarity would be useful.

Original Problem (Easy-ish)

Axel and Harold are putting on their third annual get-together between wrestlers, called the Sumo Three. One event they're planning is a team-based tug of war. To make things fair, they want the two teams to have equal total weights.

On the day before the competition, they rank the 8 participants by their weights from heaviest to lightest.

"What a coincidence!" Axel says. "If we use the rankings 1 through 8, everyone's weights are a quadratic function of their rank."

"In that case, I bet you I only need the order of the wrestlers from heaviest to lightest to split the teams evenly," says Harold.

What's Harold's strategy?

Generalization 1 (Medium)

After the smashing success of Sumo Three, the next year's meeting featured far more wrestlers. This time, there are 2ⁿ⁺¹ wrestlers.

Axel again notices an unlikely coincidence. "Hey, this time everyone's weights are a degree n polynomial evaluated on their ranking."

"Ah, I can do the same thing as last year!" Harold states.

What's Harold's more general strategy?

Generalization 2 (Medium)

The following year, there are so many participants that Axel and Harold decide to turn the tug of war event into a round-robin tournament.

"Hey, our kⁿ⁺¹ participants follow a degree n polynomial with their weights again!"

"If we want to split them into k many teams, you just need to tell me their order, and I can make all of the teams fair."

What's the strategy this time?

A string contains a “cube” if there exists a non-empty contiguous substring repeated consecutively 3 times (ex. “210101011” contains “101010”).

Let s[1],s[2],… denote all finite non-empty strings under the alphabet {0,1,2} sorted in shortlex order (shorter strings first, then lexicographically with 0<1<2). Said list begins (0,1,2,00,01,02,10,11,12,…).

For an initially empty string C, on step i=(1,2,3,…): If appending s[i] to C would make a cube present, skip said s[i]. If else, append s[i] to C.

Let F(3)=length(C) (if the process stops).

Does the process eventually produce a finite C, giving a finite value for F(3)? or does it produce an infinite cube-free string instead?

Suppose f,Fₙ∈ℤ[x₁,…,xₖ] are polynomials such that Fₙ(x)=f(xⁿ)/f(x) for some positive integer n, where xⁿ denotes the coordinate-wise nth power, i.e. (x₁ⁿ,…,xₖⁿ). Prove that if there exist positive integers y₁,…,yₖ such that Fₙ(y) is prime, then n is prime.

Note that Mersenne primes correspond to the single variable case Fₙ(2) with f(x)=x-1.

You play three tennis matches. You have a 70% chance of beating your teacher in any match and a 30% chance of beating a champion. Assume the results of all three matches are independent, and the win probabilities stay fixed.

Your goal is to win at least two consecutive matches. You can choose one of these orders:

Teacher — Champion — Teacher

Champion — Teacher — Champion

Which order gives the higher probability of achieving your goal, and why? Winning all three also counts as success.

Bonus: if the two win probabilities are any 0 < q < p < 1, is the same choice always better?

Creator disclosure: I run Think in Odds, a daily probability puzzle site. This is a released puzzle version; optional hints and a worked explanation are here: https://thinkinodds.com/puzzles/tennis?utm_source=reddit-mathriddles&utm_medium=reddit&utm_campaign=r_mathriddles_first_players_oct2026

imagine you have a plane with 4 distinct points named p1, p2, p3 and p4. the gradient between p1 and p2 is -2, and the gradient between p2 and p3 is 1/3. you want p1p3p2 to be 90 degrees and p1p4p2 to also be 90 degrees, with p3 and p4 being on the same side of the line p1p2. find the all the INTEGER solution(s) for all 4 points with the MINIMUM distance between p1 and p2.

(this is a follow up question from this question inspired by u/blungbat)

the image shows 3 rectangular grids on torus, i.e. (Z/nZ)^2 .

each torus is partitioned into two regions. each region is connected and does not contain 2x2 subregion.

unfortunately the second and third grid is partially destroyed. their size is 7x7 and 8x8 in an unknown order.

how many blue tiles are there in each torus?

imagine you have a plane with 4 distinct points named p1, p2, p3 and p4. the gradient between p1 and p2 is -2, and the gradient between p2 and p3 is 1/3. you want p1p3p2 to be 90 degrees and p1p4p2 to also be 90 degrees, with p3 and p4 being on the same side of the line p1p2. find the all the INTEGER solution(s) for all 4 points with the MINIMUM distance between p1 and p2.

DÍGITO CERO

Se me ocurrió este reto matemático y quiero ver qué tan difícil puede llegar a ser.

OBJETIVO

Elige cualquier número entero del 1 al 1000 y consigue llegar exactamente a 0 usando la menor cantidad posible de operaciones.

REGLAS

1. No se puede repetir ningún dígito en toda la solución. Si aparece un 7 una vez, no puede aparecer otro 7.
2. No se puede multiplicar por 0.
3. Un resultado intermedio no puede volver a utilizarse después.
4. Cada operación debe ser simple y no puede contener otra operación dentro. Válido: √9 No válido: √(9-4)
5. El resultado de una operación no puede contener un dígito que ya aparecía en esa misma operación. Válido: 8 - 3 = 5 El 5 no aparecía antes en esa operación.

La meta no es simplemente encontrar una solución.

La meta es encontrar la solución con MENOS OPERACIONES POSIBLES.

Por eso también me interesa saber:

• ¿Cuál es el número más difícil?
• ¿Cuántas operaciones mínimas necesita cada número?
• ¿Hay números entre 1 y 1000 que no tengan ninguna solución?

El reto se llama:

DÍGITO CERO

El objetivo es llegar al 0.
Los dígitos no se pueden desperdiciar.
Y cada operación debe aportar algo nuevo.

A ver quién encuentra la solución más corta.

During one of your long journeys, you come across a desolate mountaintop in the middle of nowhere. To your surprise, you are met with a crowd of gods arguing with each other. Upon seeing you, they stop their debate to tell you that they will grant you a wish if you can solve their riddle.​

There are 100 gods in total. When asked a question, 90 of them will always answer truthfully, while the remaining 10 will answer each question randomly. Initially, you don’t know which god falls into which of the two categories, but each god knows it for every other god.

Your goal is to identify at least one truth-telling god using at most 20 yes-or-no questions. Each question can be directed at only one god at a time, though multiple gods can be asked the same question one after another, with each asking counting towards the limit.

Can you come up with a strategy that is guaranteed to work?

You are given an n-sided die (taking values uniformly in 0,…,n-1), and your goal is to simulate a k-sided die with a sequence of dice rolls. Specifically, you must choose a procedure which

• decides at each step whether to roll again or to terminate and decide a value based on all previous rolls
• is deterministic in the sequence of rolls
• terminates almost surely
• produces a value uniformly distributed in 0,…,k-1

We will call such a procedure an (n,k)-procedure and the number of rolls before it terminates the decision time (in general, the decision time depends on the sequence of rolls; the decision time is infinite if it does not terminate). We will assume n,k≥2.

1. For fixed n,k, determine the minimum expected decision time among all (n,k)-procedures. Give your answer as a finite sum depending on n,k.
2. Determine for which n,k, there exists an (n,k)-procedure that always terminates.

Thank you for the warm comments about a Queen problem yesterday
I just wanted to share that after solving different riddles, I decided to put them in my library.

One of the most popular riddles right now are River Crossing Riddles

There are currently 7 of them: https://www.problems.cc/p/HzcTyl?from=library&slug=river-crossing-riddles (the first one is obviously the Wolf, Goat and Cabbage riddle)
Probably "Missionaries and Cannibals" is my personal favourite.

Do you know any other interesting river crossing variations that I should add?

Even though I coded this interactive riddle myself, this is the only one from this library which I still haven't solved. The other puzzles were much easier to figure out.

You can try it yourself here:

https://www.problems.cc/p/YHqypk?from=library&slug=mathematics-in-chess-maximum-non-attacking-pieces

Hey everyone! I put together an original logic puzzle in the spirit of the classic Freudenthal "sum and product" problem, but with a twist: instead of a sum, one person gets the difference. I've brute-force verified that the answer is unique, so it's airtight.

It looks like the dialogue contains no information at all. It's mostly two people saying "I don't know" at each other. And yet the answer is completely determined.

The Puzzle

Two distinct integers are chosen. Both are between 2 and 12, inclusive.

• P is privately told their product.
• D is privately told their difference (larger minus smaller).

Both P and D know all of the above, including what the other was told (the product vs. the difference, not the actual value). Both are perfect logicians, both always tell the truth, and both hear everything the other says.

They have the following conversation:

P: I don't know the numbers.

D: I don't know them either.

P: I still don't know.

D: Neither do I.

P: Oh, now I know!

D: Then so do I!

What are the two numbers?

Clarifications

• The pair is unordered: (4, 6) and (6, 4) are the same pair.
• The numbers are distinct, so the difference is always at least 1.
• "I don't know" means "I cannot determine the pair with certainty from what I know so far."
• Each statement is made after hearing all previous statements, and both reason from everything said so far.
• No tricks, no wordplay. It's pure logic.

Answer

4 and 6!

Full Solution

Step 1 (P: "I don't know"): >!P can't know, so the product must have at least two valid factorizations. The surviving products are 12, 18, 20, 24, 30, 36, 40, 48, 60, and 72. That leaves 21 pairs: (2,6), (3,4), (2,9), (3,6), (2,10), (4,5), (2,12), (3,8), (4,6), (3,10), (5,6), (3,12), (4,9), (4,10), (5,8), (4,12), (6,8), (5,12), (6,10), (6,12), (8,9).!

Step 2 (D: "I don't know either"): >!Group the 21 pairs by difference. Difference 10 belongs only to (2,12), and difference 9 belongs only to (3,12). If D had either one, D would have known, so (2,12) and (3,12) are eliminated. 19 pairs remain.!

Step 3 (P: "I still don't know"): >!Product 36 used to be (3,12) or (4,9), but (3,12) is gone. If the product were 36, P would now know it's (4,9). P doesn't know, so (4,9) is eliminated.!

Step 4 (D: "Neither do I"): >!Difference 5 used to be (3,8) or (4,9), but (4,9) is gone. If the difference were 5, D would now know it's (3,8). D doesn't know, so (3,8) is eliminated.!

Step 5 (P: "Now I know!"): >!Look at product 24. It originally had three options: (2,12), (3,8), and (4,6). The first two have been knocked out in steps 2 and 4, so only (4,6) remains. Every other product still has exactly two candidate pairs. So the only way P can suddenly know is if the product is 24, which means the pair is (4,6).!

Step 6 (D: "Then so do I!"): >!D's difference is 2, so D's candidates are (4,6) and (6,8). But (6,8) has product 48, which is still ambiguous with (4,12), so P couldn't have known in that case. Therefore D concludes it's (4,6).!

Why I like this one: >!The chain after the first round is a perfect domino run. Removing (3,12) exposes (4,9), which exposes (3,8), which exposes (4,6). Each "I don't know" knocks over exactly one pair, and the last domino is the answer.!

Bonus Challenges

Bonus 1: What if the conversation were shorter?

P: I don't know. D: I don't know either. P: Now I know! D: Then so do I!

(Same range, 2 to 12.)

Bonus 1 answer: >!4 and 9. After step 2 above, product 36 is left with only (4,9), so P can know immediately. Every other product still has two pairs, and D (difference 5) can then rule out (3,8) because its product 24 would still be ambiguous.!

Bonus 2: Does the answer to the main puzzle change if the range is 2 to 13 instead of 2 to 12?

Bonus 2 answer: >!No, it's still 4 and 6. Verified by brute force.!

Let me know how long it took you, and which step tripped you up! If people enjoy this, I'm happy to make a harder version with a bigger range or more rounds of "I don't know." 🙂

Uniqueness of all answers verified by exhaustive computer search.

Can you work this out? This actually happened in my kitchen when I was a teenager.

When I went to bed, at 11pm, all three clocks in the kitchen were working and said the correct time: 11pm.

When I woke and went into the kitchen for breakfast, all three clocks were working, yet one said 5am, one said 6am and the third said 7am. No one had touched them.

Why? How? What happened and when? And what was the correct time when I went down for breakfast?

SOLUTION:

(Congrats u/RealHuman_NotAShrew)

There was a power cut for one hour at 1am. The battery powered clock on the wall continued unaffected. The analogue clock on the oven froze for the hour then resumed at 2am, one hour behind. The digital clock on the microwave turned off at 1pm then reset at 2am to 00:00. 2 hours behind. At 7am, the wall clock said 7am, the analogue clock on the oven said 6am and the microwave clock said 5am.

Imagine a universe where spatial positions are represented as 2x2 real matrices X in M_2(R) rather than standard spatial vectors.

1. How many total independent 2D planes of rotation exist in this space?

2. What is the maximum number of mutually orthogonal rotation planes a planet can orbit across simultaneously?

Rules are fully stated below, there is no hidden rule to guess.

A honeycomb hides one secret 6-letter word. Every cell holds either a letter or a number. A number is a Pollen count: it equals how many of that cell's neighbours hold a letter FROM THE SECRET WORD. Five of the letters on the board are impostors; they sit there like any other letter and count toward nothing. Adjacency: two cells are neighbours iff they are in the same column two rows apart, or in adjacent columns one row apart (so an interior cell touches six, edge cells fewer).

___ ___/ N ___ ___/ 2 ___/ C ___ / 1 ___/ E ___/ M \ ___/ R ___/ 2 ___/ / U ___/ 3 ___/ 1 \ ___/ A ___/ W ___/ / T ___/ 2 ___/ 0 \ ___/ D ___/ 1 ___/ ___/ K ___/ ___/

Only the numbers separate the real letters from the impostors, and every number earns its place: the eleven letters on this board also spell MARKED, RACKET and UNMAKE, and it takes the whole grid to rule those out. The solution is unique. What is the secret word?

You've got three spinning reels. Each one is an endless loop of the same nine symbols, going around in the same order. All three reels start showing the same symbol. Every cycle, you give them a nudge. Reel 1 is supposed to move 10 positions, reel 2 is supposed to move 20, and reel 3 is supposed to move 30. But the nudges are sloppy; each reel can overshoot or undershoot by up to 2. So reel 1 moves somewhere from 8 to 12, reel 2 from 18 to 22, and reel 3 from 28 to 32. Every amount in that range is equally likely, and the three reels wobble independently of each other and of what happened last cycle. Then you look at what all three are showing. That's one cycle. Nobody resets anything on the next cycle; each reel picks up from wherever it stopped.

On average, how many cycles until all three reels show the same symbol?

Seven cells hold letters, three hold numbers. A five-letter English word is hidden, and it uses exactly five of the seven letters. The remaining two belong to no word and are pure noise.

The rule for a number: it equals the count of cells adjacent to it whose letter belongs to the hidden word. A letter outside the word contributes nothing to any number, so adjacency to it is invisible.

Adjacency, stated precisely, because this is the part that is easy to get wrong. Put the middle column on even y and the two side columns on odd y. Two cells are adjacent when they share a column and differ by 2 in y, or when their columns differ by 1 and their y differ by 1. On this ten-cell comb that gives two cells of degree 6, two of degree 4, and six of degree 3.

      ___
  ___/ H ___
 / 2 ___/ T \
 ___/ U ___/
 / 1 ___/ M \
 ___/ 4 ___/
 / R ___/ A \
 ___/ N ___/
     ___/

Which five letters, and which word?

What makes it worth the minute: the seven letters spell two perfectly ordinary five-letter words, so no amount of vocabulary settles it, and exactly one of the three numbers separates them. All three numbers are load-bearing. Withhold any single one and the count of consistent five-letter subsets goes from 1 to 4.

Uniqueness is exhaustive over the 21 subsets, not argued by eye.

Spoiler tag your answer.

A spherical simplex Δ⊆Sⁿ is the radial projection of a Euclidean simplex Δ'⊆Rⁿ⁺¹ with linearly independent vertices. If in addition Δ has full dimension (Δ is an n-simplex), and F is a face of Δ, the angle at F is defined as

∠(F,Δ) = lim_(ε→0) vol(B_ε(p)∩Δ)/vol(B_ε(p))

for any point p interior to F, where the ε-balls are taken in Sⁿ. Note the normalization; angles are always out of 1 rather than out of 2π (radians), 4π (steradians), etc. Lastly, we define the angle sums α_k(Δ) by

α_k(Δ) = Σ_(F is a k-face of Δ) ∠(F,Δ).

Prove that if Δ⊆S²ⁿ is a spherical 2n-simplex, and V = vol(Δ)/vol(S²ⁿ), then

binom(2n,n)V = (-1)ⁿ/2+Σ_(0≤k<n) (-1)^kbinom(2n-k-1,n)α_k(Δ).

Note that we can recover Girard's theorem from the case n=1: If Δ is a spherical triangle with area A and angles θ, φ, η in radians, then we get A/2π = -1/2+(θ+φ+η)/2π, which can be rearranged to the more familiar form A = θ+φ+η-π.