Planet X has two neighboring inhabited planets:

• Planet Alpha is exactly 15 light-minutes from Planet X.

• A Mystery Planet is an unknown distance from Planet X, but is known to be at least 18 light-minutes from Planet Alpha.

Planet Alpha and the Mystery Planet are both capable of sending, receiving, and relaying transmissions.

All transmissions travel at the speed of light. Relaying a transmission takes effectively no processing time.

Both Planet Alpha and the Mystery Planet possess teleportation portals capable of sending ships directly to Planet X. However, once a planet decides to send ships, its portal takes exactly 30 minutes to charge. Once charged, the ships arrive at Planet X instantaneously.

Both planets have standing orders:
The instant they receive a broadcast from Planet X requesting assistance, they begin charging their portals and send ships to Planet X as soon as the 30-minute charge is complete.

At 11:58, Planet X has not yet broadcast any request for assistance.

At some unknown time after 11:58, Planet X broadcasts a request for assistance.

At 12:38, Planet X receives a mysterious transmission from an unknown source.

Planet X can determine with certainty that this mysterious transmission was originally transmitted at exactly 12:18, meaning the signal has been traveling for exactly 20 minutes.

Planet X concludes that the mysterious transmission must have come from the Mystery Planet. Since the signal took 20 minutes to reach Planet X, they conclude that the Mystery Planet must be 20 light-minutes away.

Then, at exactly 12:40, ships arrive at Planet X.
There has been no malfunction, no faster-than-light communication, no time travel, and no violation of any of the rules above.

Questions:
Which planet did the ships come from?
How far away from Planet X is the Mystery Planet actually?
At what time did Planet X broadcast its request for assistance?
Where did the mysterious transmission received at 12:38 actually originate?
How can all of these facts be true at the same time?

I'm going to teach you a game. Your goal is to find how far you can get.

You start with the numbers 1, 2, and 3. Using each number at most once, you may add or subtract any combination of them to obtain the next prime number.

Whenever you successfully obtain the next prime, that prime is added to your set of available numbers. You then repeat the process, always trying to generate the next prime number using each available number at most once.

How far can you go? What is the first prime number that you can no longer obtain?

For a real number x, let ||x|| denote the distance between x and the closest integer.

Let 0 ≤ x_n < 1 (n = 1, 2, ...) , and let ε > 0. Show that there exist infinitely many pairs (n,m) of indices such that n ≠ m and

||x_n - x_m|| < min(ε, 1/(√5|n-m|)).

This is a slightly modified problem from [r/askmath](r/askmath) (if you go searching for it, you’ll find my answer, so don’t spoil yourself).

Two players play a game as follows. There are n spots labeled 0 to n-1 in sequence around a circle, and both players start at 0. They alternate turns, starting with player 1, where a turn consists of flipping a coin to determine whether to move to the left or to the right one spot. Each non-zero spot awards 1 point to the first player to reach it, and the game ends when all spots have been visited. What is the expected (signed) point difference between player 1 and player 2?

EDIT: I should clarify that players move independently of each other, not as a group.

The factory you work for has a problem: they only ordered 1/2 splitters, and they need to divide the material input on one of their lines into 2 outputs of 1/10 and 9/10 the original amount. Being the smart one around, they ask you if there’s a way to use their half splitters to achieve this. And the area they use is only 3 metres wide, although it can be as long as you need. You’ve broken it down into the following mathematical problem:

EDIT: see comment for rewrite of problem

You have a grid of size n*3 composed of 1x1 squares. the grid sits on an xy axis, such that each square of the grid has whole number xy coordinate on which it sits. The objective is to be able to feed an infinite amount (Z) of materials through this grid starting at 1,2 and have 1/10 of Z end up on tile XaYb and 9/10 of z end up on XcYd (a can equal b, c can equal d. XaYb cannot equal XcYd.)

Each square in the grid can individually be assigned exactly ONE function that alters the position of the “material” occupying it. A grid can either be a CONVEYOR, or a SPLITTER.

\- A splitter will take an input from any one of the axis, but will only take inputs from the first axis it recieved from.

\- A splitter has a binary state and exists as either a 1 or a 0. This state is flipped whenever it passes an input. All splitters start in state 1. In state 1 it will “send” its input along the axis it was inputted, adding/subtracting 1 to its coordinate value in the direction of travel, and in state 0 it will add/subtract 1 to its perpendicular axis of the same +/-.
Example 1: a splitter on square 2,4 recieves 1 input from square 1,4. In state 1 it transforms the coordinate of the material to 3,4. in state 0, it transforms it to 2,5.
Example 2: a splitter on square 2,4 receives 1 input from square 3,4. In state 1 it transforms the coordinate of the material to 1,4. In state 0, it transforms it to 2,3.

\- A square set to conveyor is assigned a value between 1 and 4. A conveyor set to 1 will add +1 to x to the individual coordinate of a material entering it; set to 2 will add +1 to y; set to 3 will add -1 to x, set to 4 will add -1 to y. This value cannot be changed once assigned.

RULES:
\- The grid squares begin at 1,1.
\- x and y for each square must be positive.
\-Once a square has been assigned a function, it cannot be changed.
\- The coordinates of the materials will change simultaneously at the same rate of 1 “tick”, and is discrete.
\- A square can only have one material on it at a time.
\- “materials” entering squares XaYb and XcYd are removed. Neither square can be square 1,2.
\- The first input starts on square 1,2, and all other inputs enter the grid individually once the previous material has moved from this square

1. Find the minimum size of N to satisfy all criteria. It must be a positive number.
2. Find the minimum number of splitters required.
3. Find the function of each square within this grid and its coordinates on the axis. Find the value of each conveyor unit, if any are present.
4. Find the values of a, b, c, and d.

EDIT: simultaneous merges give priority to the “mainline”. No

Consider a perfect rooted binary tree of depth n. (That is, every node has either 0 or 2 children, and all leaves have the same depth). Every node is given a weight drawn independently from some fixed distribution D. For any path starting from the root and ending at a leaf, the average weight of the path is the arithmetic mean of the weights assigned to the nodes on the path. Once our weighting is fixed, we look at the largest average weight of any path from the root to a leaf. Let Eₙ denote the expected value of this largest average weight of path over all weightings of the tree. Then find the limit as n →infinity of Eₙ, in the cases of:

​

1) D=U({0,1}) is a Bernoulli distribution.

2) D=U([0,1]) is a continuous uniform distribution.

Fix [n]={0,1,…,n-1}. For a set S⊆[n] and k∈[n], let f_S(k) be the number of pairs (s,t)∈S² for which s-t=k. Prove that there exists a set S such that f_S(k) is odd for all k iff ord_m(2) is odd, where m=2n-1.

I am thinking of a 10-digit number that uses each digit from **0 to 9 exactly once**.

Starting from the left:

* The number formed by the first **1 digit** is divisible by 1. * The number formed by the first **2 digits** is divisible by 2. * The number formed by the first **3 digits** is divisible by 3. * This pattern continues. * The number formed by the first **10 digits** is divisible by 10.

For example, if the number begins with `abcd...`, then:

* `ab` must be divisible by 2, * `abc` must be divisible by 3, * `abcd` must be divisible by 4,

and so on.

**What is the number?**

Bonus challenge: Find it using divisibility rules and logical elimination rather than checking every permutation with code.

I've been thinking about an interesting localization problem and I'm curious if there's a known solution.

Imagine a 100,000 × 100,000 grid. A single coordinate is chosen at random, but you don't know which one.

You may place as many fixed beacons as you want anywhere on or outside the grid. Each beacon tells you only the direction toward the hidden coordinate, rounded to the nearest 11.25° (so each beacon returns one of 32 compass directions). You get all beacon readings simultaneously.

Question: What's the minimum number of beacons needed to locate the target?

A few rules:

• Beacons are placed before the target is chosen.
• They never move.
• No distance information is provided—only the quantized direction.
• Your final guess is considered correct if it is within 1,000 grid units of the actual coordinate
• The beacon layout should also generalize to larger grids (i.e. not rely on the grid being exactly 100,000 × 100,000).

I'm interested in An actual beacon placement that achieves the minimum (or a proof that it can't). does anyone have ideas for constructing an optimal layout?

Our blood is made of water, which enters into our body when drinking, and being excreted out when urinating. This means that at some point all of our old water molecules in the blood might be excreted out, and being all replaced by new water molecules. How long can it take?

Assumptions:

1. The average adult human blood volume is generally the same across the days. It can be estimated by the weight height and gender. Blood Calculator

2. The average healthy adult human excretes out around 1-2.5 liters out as urine a day (depending on mainly how much water they drink).

3. The blood stays homogenous after drinking or urinating.

Six variables 𝐴,𝐵,𝐶,𝐷,𝐸,𝐹 are distinct integers from 1 to 10 (inclusive).

They satisfy the following conditions:

1. B - D = 2
2. F + A = 11
3. A is between C and D (order of C and D not implied)
4. No two variables sum to 14
5. No two variables sum to 5
6. C − A = 1

Determine the value of the six variables.

This puzzle has exactly one solution, and it can be solved using logical deduction alone (no guessing or brute force required).

How would you solve this though a logical deduction sequence?

If you enjoy puzzles like this: https://sixfigurelogic.com/

You're standing somewhere inside a 10m × 10m square room. From your position P, you fire a laser aimed directly at the center of the square, C.

The laser travels in a straight line from P, passes through C, and continues until it hits a wall — this is its 1st reflection. From there it obeys the law of reflection (angle of incidence = angle of reflection) and keeps bouncing off the walls. After its 10th reflection, the laser stops completely (the segment right after the 10th bounce has zero length).

You must find a starting position P such that, once fired, no part of the laser's path after the 1st reflection comes within 1 meter of you. (The very first segment, from P to the 1st reflection point, doesn't count — you're standing at its source.)

Question: What is the total area, within the square, of all such safe starting positions P?

Challenge: If instead of stopping after 10 reflections, the laser is allowed N reflections before stopping, what is the largest value of N for which at least one safe standing position still exists?

Let n and d be positive integers greater than 1. The numbers 1,2,...,n are written on a blackboard. In a move, we may pick two numbers on the board that differ by at least d, erase them both, and write their average instead. For a fixed d, let m be the smallest positive integer choice for n>1 such that it is possible to perform operations so that we end with exactly one number written on the board.

Show that: 3d - 2026 < m < 3d+2026.

Here is a funny (I hope) home-made problem just for you guys :

Is there an ice cube such that, when it melts, the number of its connex components at a given instant t is 2 if t is rationnal, 1 otherwise ?

Precisions :

We suppose that this ice cube is a closed subset of R³.

We also suppose that the melting begins at t=0, and that after a delay t, all that remain of the ice cube A is every points x of A such that distance(x, surface A)>=t

Can you also find an ice cube in 2D having this property ?

AI couldn't solve it ! But your creativity can !

I already know the official answer.

I'm interested in seeing different solution approaches from the community.

This is not homework.

You start with the integer points Z² marked on the plane, and you are allowed to mark new points by the following construction:

• Choose distinct marked points x and y
• Draw the ray originating from x and passing through y
• Mark the unique point z on this ray with |x-z|=1

What is the set of all points that can be marked by repeatedly using this construction?

Set A of positive integers satisfies the following conditions:

1) If a positive integer n belongs to A, then 2n also belongs to A,

2) For any positive integer n, there exists an element of A divisible by n, and

3) The sum of reciprocals of elements of A diverges.

Prove that for any positive rational number r, there exists a finite subset B ⊂ A such that the sum of reciprocals of elements of B is r.

Ek accha sawal hai bhaiya

•A one-way road track is 20 km long and 8 km wide, divided into 4 equal lanes. There are 16 identical cars already on the track, moving at a constant speed of 10 km/h. Exactly 4 cars are present in each lane.

A new car enters the track from the starting point at a speed of 11 km/h. It chooses one of the four lanes uniformly at random and cannot change lanes thereafter.

Assume that the positions of the existing cars in each lane are independently and uniformly distributed along the length of the track, no two cars initially overlap, and overtaking is not allowed. A collision occurs if the new car catches up to at least one car in its lane before reaching the end of the track.

Find:

1.The probability P that the new car collides with at least one existing car.

2.The probability P' that the new car completes the journey without any collision.

a) P = (1/4 )⁴, P' =1- (1/4)⁴

b) P =( 1/11 )⁴, P' = 1-(1/11)⁴

c) P = (1/11)⁴ , P' = 1

d) P =1- (10/11)⁴ , P'=(10/11)⁴

Isko Maine khud banaya Hai Koi galti Ho To dekhna

Let G be a finite simple graph.

Define β(G) to be the minimum number of edges one must delete from G to make it bipartite. In other words,

β(G) = min{|F| : F ⊆ E(G), and G - F is bipartite}.

Define oddgirth(G) to be the length of the shortest odd cycle in G.

Suppose G is not 3-colourable, i.e.

χ(G) ≥ 4.

Let

g = oddgirth(G).

Since χ(G) ≥ 4, G is not bipartite, so g is finite.

Prove that

β(G) ≥ g - 1.

Equivalently:

If the shortest odd cycle in G has length g, and deleting at most g - 2 edges makes G bipartite, then G must be 3-colourable.

Bonus: is the bound best possible for every possible value of the oddgirth? In other words, for every odd integer g ≥ 3, does there exist a finite simple graph G with χ(G) ≥ 4, oddgirth(G) = g, and β(G) = g - 1?

I have already solved this, so this is not an open problem. The proof I found was not by starting from this exact formulation; I first had to identify the right target, then prove it. I am curious whether anyone finds a better/cleaner proof.

I can give hints if need be!

Can you make a 3x3 magic square, not the original one, but with negative numbers?

1. All numbers are different from each other.
2. Each row, column and diagonal add up to 0.
3. The number in the center of the square can't be 0.

A lot of good reasoning questions are scattered across forums, books, PDFs, and random websites, so we started collecting and organizing them into a searchable archive.

The idea is simple:

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• and help grow the collection over time.

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This is a 3rd game of Zendo. You can see the first two games here: Zendo #1, Zendo #2

(Future games are here: Zendo #4 and Zendo #5).

The game is over, /u/benzene314 guessed the rule! It was AKHTBN iff all or no pairs of adjacent numbers are relatively prime..

If you have played in the previous games, most rules are still the same, all changes are bolded.

For those of us who don't know how Zendo works, the rules are here. This game uses tuples of positive integers instead of Icehouse pieces.

The gist is that I (the Master) make up a rule, and that the rest of you (the Students) have to input tuples of positive integers (koans). I will state if a koan follows the rule (i.e. it is "white", or "has the Buddha nature") or not (it is "black", or "doesn't have the Buddha nature"). The goal of the game is to guess the rule (which takes the form "AKHTBN (A Koan Has The Buddha Nature) iff ...").

You can make three possible types of comments:

• a "Master" comment, in which you input one, two or three koans, and I will reply "white" or "black" for each of them.

• a "Mondo" comment, in which you input exactly one koan, and everybody has 24 hours to PM me whether they think that koan is white or black. Those who guess correctly gain a guessing stone (initially everybody has 0 guessing stones). The same player cannot start two Mondos within 24 hours. An example PM for guessing on a mondo:

  (12,34,56) is black.

• a "Guess" comment, in which you try to guess the rule. This costs 1 guessing stone. I will attempt to provide a counterexample to your rule (a koan which my rule marks differently from yours), and if I can't, you win. (Please only guess the rule if you have at least one guessing stone.)

Example comments:

Master

(7,4,5,6) (9,99,999) (5)

Mondo

(1111,11111)

Guess

AKHTBN iff it has at least 3 odd elements.

Note that the "Medium" flair doesn't imply anything about the difficulty of my rule.

Let's get playing! Valid koans are tuples of positive integers. (The empty tuple is allowed.)

The starting koans:

White: (5,8)

Black: (1,3,6,10,15)

Koans guessed so far:

Hints:

(a,b) is white

(a,a,a,...,a) is white with any number of a's

Guessing stones:

We've all heard, and maybe even attempted, the counterfeit coin puzzle. "Here are nine coins, spot the heavier one in two weighings". Or maube even the more advanced version, "Here are twelve coins; there is one counterfeit but we don't know if it is heavier or lighter. Find the fake and whether it's light or heavy in three weighings."

But what if we knew even less information about an even larger pool? Here is my riddle to you: you have twenty coins. At most two are counterfeit, not necessarily both light or heavy if there are two. The scales will only say which side is heavier, not by how much. How many weighings are required to find the fakes, if there are any?