Imagine you are the best math-logic puzzle creator in the world. You are to make one single puzzle that will revolutionize the universe of puzzles by using math and logic. The puzzle will be unique, like no other ever existed, and it shall be the hardest puzzle ever created and almost impossible to solve, even for the best thinkers in the world and there will be only one concrete answer, without any paradoxes. https://discord.gg/wCxJ6ueC

Generalized version of my old post

There are n users on a social network called Mathbook, and some of them are Mathbook-friends. (On Mathbook, friendship is always mutual and permanent.)

Starting now, Mathbook will only allow a new friendship to be formed between two users if they have at least two friends in common. What is the minimum number of friendships that must already exist so that every user could eventually become friends with every other user?

Prove that there exists a positive constant c such that the following statement is true: Consider an integer n > 1, and a set S of n points in the plane such that the distance between any two distinct points in S is at least 1. It follows that there is a line l separating S such that the distance from any point of S to l is at least c * n^(-1/3).

(A line l separates a set of points S if some segment joining two points in S crosses l.)

Note: Weaker results with c * n^(-1/3) replaced by c * n^(-alpha) may be awarded points depending on the value of the constant alpha > 1/3.

Let alpha ≥ 1 be a real number. Hephaestus and Poseidon play a turn-based game on an infinite grid of unit squares. Before the game starts, Poseidon chooses a finite number of cells to be flooded. Hephaestus is building a levee, which is a subset of unit edges of the grid (called walls) forming a connected, non-self-intersecting path or loop.

The game begins with Hephaestus moving first. On each of Hephaestus's turns, he adds one or more walls to the levee, as long as the total length of the levee is at most alpha * n after his n-th turn. On each of Poseidon's turns, every cell adjacent to an already flooded cell and with no wall between them becomes flooded.

Hephaestus wins if the levee forms a closed loop such that all flooded cells are contained in the interior of the loop, stopping the flood and saving the world. For which values of alpha can Hephaestus guarantee victory in a finite number of turns, no matter how Poseidon chooses the initial flooded cells?

Note: Formally, the levee must consist of lattice points A0, A1, ..., Ak, which are pairwise distinct except possibly A0 = Ak, such that the set of walls is exactly {A0A1, A1A2, ..., Ak-1Ak}. Once a wall is built, it cannot be destroyed. If the levee is a closed loop (i.e., A0 = Ak), Hephaestus cannot add more walls. Since each wall has length 1, the length of the levee is k.

A snake of length k is an animal that occupies an ordered k-tuple (s1, s2, ..., sk) of cells in an n x n grid of square unit cells. These cells must be pairwise distinct, and si and si+1 must share a side for i = 1, 2, ..., k-1. If the snake is currently occupying (s1, s2, ..., sk) and s is an unoccupied cell sharing a side with s1, the snake can move to occupy (s, s1, ..., sk-1) instead.

The snake has turned around if it occupied (s1, s2, ..., sk) at the beginning, but after a finite number of moves occupies (sk, sk-1, ..., s1) instead.

Determine whether there exists an integer n > 1 such that one can place a snake of length 0.9 * n^2 in an n x n grid that can turn around.

An n times m matrix is nice if it contains every integer from 1 to mn exactly once and 1 is the only entry which is the smallest both in its row and in its column. Prove that the number of n times m nice matrices is (nm)!n!m!/(n+m-1)!.

Prove that for all sufficiently large positive integers n and a positive integer k <= n, there exists a positive integer m having exactly k divisors in the set {1,2, ....., n}

What is the minimum value of

[ |a + b + c| * (|a - b| * |b - c| + |c - a| * |b - c| + |a - b| * |c - a|) ] / [ |a - b| * |c - a| * |b - c| ]

over all triples a, b, c of distinct real numbers such that

a^2 + b^2 + c^2 = 2(ab + bc + ca)?

A Nim-style game is defined as follows: Two positive integers k and n are given, along with a finite set S of k-tuples of integers (not necessarily positive). At the start of the game, the k-tuple (n, 0, 0, ..., 0) is written on the blackboard.

A legal move consists of erasing the tuple (a1, a2, ..., ak) on the blackboard and replacing it with (a1 + b1, a2 + b2, ..., ak + bk), where (b1, b2, ..., bk) is an element of the set S. Two players take turns making legal moves. The first player to write a negative integer loses. If neither player is ever forced to write a negative integer, the game ends in a draw.

Prove that there exists a choice of k and S such that the following holds: the first player has a winning strategy if n is a power of 2, and otherwise the second player has a winning strategy.

A. Two players play a cooperative game. They can discuss a strategy prior to the game, however, they cannot communicate and have no information about the other player during the game. The game master chooses one of the players in each round. The player on turn has to guess the number of the current round. Players keep note of the number of rounds they were chosen, however, they have no information about the other player's rounds. If the player's guess is correct, the players are awarded a point. Player's are not notified whether they've scored or not. The players win the game upon collecting 100 points. Does there exist a strategy with which they can surely win the game in a finite number of rounds?

b)How does this game change, if in each round the player on turn has two guesses instead of one, and they are awarded a point if one of the guesses is correct (while keeping all the other rules of the game the same)?

It is known and not too hard to prove that any 5 points in the plane define a unique conic section.

My riddle for you is:

Given 5 points in the plane, how would you construct the tangents to the conic they define at one of the points?

A clock has 6 hands instead of 3, each moving at a different speed. Here are the speed values for each hand:
1: Moves forward by x/12 degrees each minute
2: Moves forward by x^2 degrees each minute
3: Moves backward by x degrees each minute
4: Moves forward by x/2 degrees each first minute and 2x degrees each second minute
5: Moves forward by x degrees each minute
6: Moves backward by sqrt(x+y) degrees each five minutes
We know that two of these hands are the real minutes and hours hands, but that there is no seconds hand.
y is a prime number that is a possible value for minutes in a clock, e.g.: 59 works, but not 61.
At the start, the clock shows midnight, which is the actual time. After a certain amount of time, 4 hands meet in one one spot, while the other 2 meet in another spot.

Question: What is the time?

I got inspired to think about this problem from the movie 13 https://www.youtube.com/watch?v=D4Kdbx3q_Rs

Imagine a scenario where you have a loaded gun and there is 1/6 chance that each chamber has a bullet. All participants fire only once. All participants only get one bullet.

You're in a circle of X number of men that all point at the head of the person infront. When would it be prudent to hold your fire and fire last and when to fire first?

If there are three men in the circle (or rather triangle here) then its obviously better to hold your fire, as the person infront of you might shoot the person that will shoot you and if he has a bullet you're dead.

But if there are four people it's obviously the other way around, you want to shoot the person infront of you to save the person who might shoot the guy that will shoot you!

So what about 5 or more people? How would you calculate the probalistic aspects of whether you should wait to fire or not? Is it associated with the number of bullets in the chamber or not? Is it irrelevant at that point?

My gut feeling tells me that alternating between shooting first an shooting last (5-6-7 etc number of men), in game where all the other participants actions are randomized is the "best" solution. My second best guess is that its irrelevant. But Id love to see the math on this for 5+ players

A gangster, hunter and hitman are rivals and are having a quarrel in the streets of Manchester. In a given turn order, each one will fire their gun until one remains alive. The gangster misses two of three shots on average, the hunter misses one of three shots on average and the hitman never misses his shot. The order the three shooters will fire their gun is given by these 3 statements, which are all useful and each will individually contribute to figuring out in which order the rivals will go. We ignore the possibility that a missed shot will hit a shooter who wasn't targeted by that shot.

• A shooter who has already eaten a spiced beef tartar in Poland cannot shoot before the gangster.
• If the hitman did not get second place at the snooker tournament in 1992, then the first one to shoot has never seen a deer on the highway.
• If the hitman or the hunter is second to shoot, then the hunter will shoot before the one who read Cinderella first.

Assuming that each of the three shooters use the most optimal strategy to survive, what are the Gangster's chances of survival?

Let {p > 2} be a prime, and let {a_1, a_2, …, a_{2n}} be integers not divisible by {p}, such that

{{ (ra_1) / p } + { (ra_2) / p } + … + { (ra_{2n}) / p } = n}

for any integer {r} not divisible by {p}. Prove that there exists {2 ≤ j ≤ 2n} such that {a_1 + a_j} is divisible by {p}.

Let X ~ Geo(1/2), Y ~ Geo(1/4), not necessarily independent.

How large can P(X=Y) be?

There are 2022 users on a social network called Mathbook, and some of them are Mathbook-friends. (On Mathbook, friendship is always mutual and permanent.)

Starting now, Mathbook will only allow a new friendship to be formed between two users if they have at least two friends in common. What is the minimum number of friendships that must already exist so that every user could eventually become friends with every other user?

Consider an infinite grid G of unit square cells. A chessboard polygon is a simple polygon (i.e. not self-intersecting) whose sides lie along the gridlines of G

Nikolai chooses a chessboard polygon F and challenges you to paint some cells of G green, such that any chessboard polygon congruent to F has at least 1 green cell but at most 2020 green cells. Can Nikolai choose F to make your job impossible?

The card is a 2x2 square with either 0 or 1 written in each grid cell.

There is the following operation: 1) take two cards. then for each of the 4 squares,
take the numbers from these two cards at the same coordinates, and write them into the draft card.
2) then we take a draft card and some third card. we look at the contents of the draft card at the (x, y) coordinate, let's say it is (a, b), and write the number from the (a, b) coordinate of the third card and write it on the (x, y) coordinate of the new card.

Initially there are сards:
[0 0] and [0 1]
[1 1] [0 1]

If at the beginning we have these 2 initial cards and some third card and start performing operation with these 3 cards (and the also with new cards we get from operation), what numbers should be on the third card, so that after performing operations few times, its possible to get cards with every existing number combination?

bonus: what if instead of being 2x2 and holding 2values (0 and 1), the cards are 3x3 and can hold 3 values? (the initial ones are [[0 1 2] [0 1 2] [0 1 2]] and [[0 0 0] [1 1 1] [2 2 2]])

https://preview.redd.it/nqrt5pi1ip2e1.png?width=1280&format=png&auto=webp&s=1aa50ace8f19cee52d1ca37b8cf5deecb6f77c8b

Definitions:
Even integers N and M are given such that 6 ≤ N ≤ M.

A singly even number is an integer that leaves a remainder of 2 when divided by 4 (e.g., 6, 10).
A doubly even number is an integer that is divisible by 4 without a remainder (e.g., 4, 8).

When N is a singly even number:
Let S = N + 2.
Let T = ((NM) − 3S)/4.

When N is a doubly even number:
Let S = N.
Let T = ((NM) − 3S)/4.

Problem:
Prove that it is possible to place S L-trominoes and T Z-tetrominoes on an N × M grid such that: Each polyomino fits exactly within the grid squares. No two polyominoes overlap. Rotation and reflection of the polyominoes are allowed.

There are 99 other prisoners and you isolated from one another in cells (you are also a prisoner). Every prisoner is given a positive integer code (the codes may not be distinct), and no prisoner knows any other prisoner's code. Assume that there is no way to distinguish the other 99 prisoners at the start except possibly from their codes.

Your only form of communication is a room with 2 labelled light bulbs. These bulbs cannot be seen by anyone outside the room. Initially both lights are off. Every day either the warden does nothing, or chooses one prisoner to go to the light bulbs room: there the prisoner can either toggle one or both lights, or leave them alone. The prisoner is then lead back to their cell. The order in which prisoners are chosen or rest days are taken is unkown, but it is known that, for any prisoner, the number of times they visit the light bulbs room is not bounded.

At any point, if you can correctly list the multiset of codes assigned to all 100 prisoners, everyone is set free. If you get it wrong, everyone is executed. Before the game starts, you are allowed to write some rules down that will be shared with the other 99 prisoners. Assume that the prisoners will follow any rules that you write. How do you win?

Harder version: What if the initial position of the lights is also unknown?

Bonus: Is there a way for all 100 prisoners to know the multiset of codes? (I haven't been able to solve this one yet)

To divide a heritage, n brothers turn to an impartial judge (that is, if not bribed, the judge decides correctly, so each brother receives (1/n)th of the heritage). However, in order to make the decision more favorable for himself, each brother wants to influence the judge by offering an amount of money. The heritage of an individual brother will then be described by a continuous function of n variables strictly monotone in the following sense: it is a monotone increasing function of the amount offered by him and a monotone decreasing function of the amount offered by any of the remaining brothers. Prove that if the eldest brother does not offer the judge too much, then the others can choose their bribes so that the decision will be correct.

Let G be a directed acyclic graph. Suppose k is a positive integer, such that the lengths of maximal paths in G do not cover all residue classes modulo k. Prove that chromatic number of G is at most k.

warning: if you do not like algebra crunching, please skip this.

When a spacecraft wants to raise its orbital radius around a celestial body from r to R, it can either do Hohmann transfer or bi-elliptic transfer. (see below for more details)

There exist a constant k such that when R / r > k, bi-elliptic transfer always require less Δv (thus less fuel) than a Hohmann transfer even though it require one more engine burn.

k is a root of a cubic polynomial. Find this cubic polynomial.

For those who do not want to deal with physic stuff, here are some starting assumptions (axiom) that i work from:

!1. Kepler's first law: the spacecraft orbit is an ellipse, where the celestial body is at one of the focus. (engine burn changes the shape, but still an ellipse)!<

!2. Kepler's second law: at apoapsis (furthest) and periapsis (closest), r1 v1 = r2 v2 (unless engine burn is performed)!<

!3. Conservation of energy: at any point, 1/2 v^2 - μ / r is a constant (unless engine burn is performed), where μ is another constant related to the celestial body. wlog you can set μ=1.!<

!4. An engine burn spend fuel to change velocity. A bi-elliptic transfer has 3 engine burns!<(diagram) >!, first burn brings the apoapsis from r to x, where x>R. Then at apoapsis, second burn brings periapsis from r to R, finally when back to periapsis, third burn brings the apoapsis back from x to R, circularizing the orbit. if x=R, then it is reduced to Hohmann transfer!< (diagram) . >!the problem ask for which k, ∀x>R, bi-elliptic is better.!<

note: i discovered this problem when playing ksp , and the solution i found became my new favorite constant. part of the reason for this post is to convince more people: this constant is cool! :)

too easy? try this variant: There exist a constant k2 such that when R / r < k2, bi-elliptic always require more Δv (thus more fuel) . k2 is a root of 6th degree polynomial.

For 5 distinct positive integers a, b, c, d and e, the following statements are true:

1. a is equal to the sum of squares of two distinct integers.
2. e is the second to the smallest among five integers.
3. cd is a perfect number.
4. The sum of all digits of b is equal to 13.
5. d and e are coprimes.
6. Dividing a+b+d by 12, we get 7 as the remainder.
7. d+2 is an abundant number.
8. a<d
9. ae is a multiple of 3.
10. There are at least two squares of integers among a, b, c, d and e.
11. The sum of the maximum and the minimum among the five integers is less than 100.

If there exists a pentagon whose lengths of edges are equal to a, b, c, d and e respectively, what is the minimum perimeter of the pentagon?

For each positive integer n, how many integer pairs (j,k) exist such that j^k = k^j (mod n) and 0 < j < k < n?

my teacher challenged us with this puzzle/problem and no matter how hard i try i can’t seem to solve it or find it online (chatgpt can’t solve it either lol) i’m really curious about the solution so i decided to try my luck here. it goes like this: there are three people, A,B and C. Each of them has a role, they are either a knight, a knave or a joker. The knight always tells the truth, the knave always lies, and the joker tells the truth and lies at random (there is only one of each, there can’t be two knights, for example). Find out who is who by asking only 3 yes or no questions. You can ask person A all three questions or each of them one question, however you wish, but they can ONLY answer with yes or no. :))))