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?

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.

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.

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.

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.

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?

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.

A number is a perfect square,or a square number, if it is the square of positive integer. Among the first 143 thousand square numbers, what is the sum of all the odd squares?

Help me to solve the problem

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.

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.

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?

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.

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/

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 !

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?

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.

I already know the official answer.

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

This is not homework.

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