Made this one up over the weekend, thought you lot might enjoy it.

Five members turned up to a stargazing night at a country park. The sign-in sheet shows exactly one arrival every five minutes: 8:00, 8:05, 8:10, 8:15 and 8:20. Unfortunately the names next to the times got smudged.

Two of the five were winding everyone up all night and lied in every sentence they said. The other three always told the truth.

At 8:12 the club's best eyepiece disappeared from its case. Whoever took it was one of the two liars, and had already arrived by then.

Here's what each of them said:

Priya: "I got here before Femi."

Dafydd: "Both liars got here before 8:10."

Keturah: "Priya and Lorcan arrived five minutes apart."

Lorcan: "Dafydd arrived exactly 15 minutes before Priya."

Femi: "Priya arrived exactly 10 minutes before Lorcan."

Who arrived when, who are the two liars, and who took the eyepiece?

Please hide answers in spoiler tags, like this

Start with a total of 0. Draw independent random real numbers, each uniformly distributed on (0, 1), and add them to the total. Stop at the first draw that makes the total strictly greater than 1. Count every draw, including the one that crosses the threshold.

What is the exact expected number of draws? Give a proof, rather than a simulation.

A related variation: replace 1 by a fixed threshold t with 0 < t <= 1. How does the expectation depend on t?

This is a classic stopping-time puzzle, not my invention. Creator disclosure: I’m David Chen, building Think in Odds; development, explanations and this post are AI-assisted. The complete puzzle is above. An optional playable version of the threshold-1 case, with hints and a worked explanation, is here: https://thinkinodds.com/puzzles/sum-over-one?utm_source=reddit-mathriddles&utm_medium=reddit&utm_campaign=r_mathriddles_sum_nov2026

If you try the hints, I’d welcome honest feedback on whether the first one leaves room to discover the key step. No account is needed.

A suspect is spotted on a flat desert highway exactly 40 miles due East of a police station. The suspect suddenly speeds off due North at exactly 60 mph.

The police cruiser immediately departs the station, traveling in a straight diagonal line across the open desert floor at a constant 120 mph to intercept.

Question: At what exact angle (in degrees North of due East) must the police cruiser travel to perfectly intercept the suspect?

There are 100 lockers in a long corridor numbered #1...#100, all open. A person walks across the corridor and slams shut every other open locker they encounter, leaving first open, shutting second, leaving third open and so on.

At the end of the corridor the person turns around and walks back in the other direction, shutting every other open locker they encounter starting again from the second one.

The person continues walking back and forth until only one locker remains which they choose. What is the number of this locker?

Now, what if there are 1000 lockers? Can you find some general method to determine the last locker for any N without working it through by hand?

Hint: This may be handy

This type of problem is known as the Josephus problem - a linear variation of it.

Hey guys, I want your help to help me find a mental trick which uses maths. You all might have heard or done the math mental trick in which you tell your friend to do a set of operations on a number he chooses from 1-10. Then your friend arrives at a number which will be the same every time you use the trick on him. But there is something that has been itching my brain since some time. 2 years back a friend of mine used a similar trick on me. The thing was that every time he used it I got another answer and he was able to predict it successfully without doing any calculations. I don’t remember much about the trick but I’ll tell what I remember. So if I’m sure, he used to somehow force the final answer to make it same as the number I thought in my mind. I never realised how he was able to predict the number I was thinking of until I started thinking about the trick he used on me that day. If I’m also sure, he also wrote his prediction on a piece of paper before showing it to me and his prediction was always right. I also think that he asked me the result of one of the operations I did and then wrote his prediction on a piece of paper. I am pretty sure that he asked me atleast 1 result of the operations he told me to do in order to predict my answer. I did find out the trick that day itself. I got to know that the trick works by forcing me to choose the number that he has written on paper without me knowing. In the last step of the trick he somehow forces me into getting the number he has predicted. I also think that in the last step I offer him some numbers to choose from in which my number is there and all the other numbers I have given him are related to it. Please help me in finding this trick guys. Unlike the trick in which we normally arrive at the same answer, this one is different and is really amazing but sadly I don’t remember how to do it. I appreciate your help and support.

3 men walked into a hotel. the desk clerk said, its $30. so the men spilt the money and payed 10 each. later that day the desk cler told to the waiter, those men where kind give the fiive buck back. so the waiter was in the elevator trying to figure out how to split the 5 bucks between 3 people so he decided to give each of the men 1 dollar and keep 2 for himself. this means the men each payer 9 dollars. if you add the money they got together you get 27. if you add 27 and the ammount of money the waiter took thats 29. where did the extra dollar go?

Imagine two rows of 7 unique items (A, B, C, D, E, F, G) placed on a track, sliding toward each other in a continuous, infinite loop.

A fixed viewport creates an asymmetric 3x2 window over the tracks. It views exactly 3 items from the top row and 2 items from the bottom row (aligned directly underneath the first two items of the top row).

As the rows shift step-by-step, the window looks like this for the first 3 steps:

Step 1:
          | A B C | D E F G
* * * * * | * * |

Step 2:
        A | B C D | E F G
  * * * * | * * | *

Step 3:
      A B | C D E | F G
    * * * | * * | * *

The Challenge:

Is there a way to rearrange the sequence of the second row so that all 5 items visible inside the window are completely unique (no duplicate items) at every single step of the infinite loop?

Q: What is the correct arrangement for the second row?

Three friends checked into a hotel room

The room originally cost $30, so each friend paid $10

Later, the hotel manager realized that the actual room price was only $25. He gave the hotel employee $5 and told him to return it to the three friends

The employee decided to keep $2 for himself and returned $1 to each friend

Now, each friend had originally paid $10 but got $1 back, meaning each actually paid $9

So:

$9 × 3 = $27

The employee kept $2.

$27 + $2 = $29

But they originally paid $30!

Where did the missing $1 go?

Take your time before answering. Don't look for the solution too quickly! 😁

alice and bob plays a game:

1. let set S = set of divisors of 441, set A=set B={}

2. each turn: remove x∈S from S and add x into A (for alice) or B (for bob)

3. win condition: player's set contains subset of geometric sequence of length 3

alice goes first, turns alternate. who has the winning strategy?

(another game)

same as above, with the following changes:

1. set S = set of divisors of 200000

2. x∈S can be removed and add to player's set iff x/5 ∉ S

3. win condition: player's set contains subset of geometric sequence of length 4

who wins?

edit: the second game does not work as intended

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?