Hard – Page 4 – Puzzle Prime

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Prisoners and a Bulb

Section: Brain Teasers Category: Insight

Difficulty: Hard Tags: Featured

There are 100 prisoners in solitary cells. There is a central living room with one light bulb in it, which can be either on or off initially. No prisoner can see the light bulb from his or her own cell. Every day, the warden picks a prisoner at random and that prisoner visits the living room. While there, the prisoner can toggle the light bulb if he wishes to do so. Also, at any time, every prisoner has the option of asserting that all 100 prisoners have already been in the living room. If this assertion is false, all 100 prisoners will be executed. If it is correct, all prisoners will be set free.

The prisoners are allowed to get together one night in the courtyard and come up with a plan. What plan should they agree on, so that eventually someone will make a correct assertion and will set everyone free, assuming the warden will bring each of them an infinite number of times to the central living room?

SOLUTION

First, the prisoners should elect one of them to be a leader and the rest – followers. The first two times a follower visits the living room and sees that the light bulb is turned off, he should turn it on; after that he shouldn’t touch it anymore. Every time the leader visits the living room and sees that the light bulb is turned on, he should turn it off. After the leader turns off the lightbulb 198 times, this will mean that all followers have already visited the room. Then he can make the assertion and set everyone free.

Unknown Author March 22, 2022

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Merlin and Hermes: Mysterious Lines

Section: Brain Teasers Category: Insight

Difficulty: Hard

Two adventurers, Merlin and Hermes, approached a large iron door built into a cliff face.”Well…”, said Hermes, “What do we do now?”. Merlin produced an old, large piece of crumpled paper from his pocket. “Hrm…”, Merlin mumbled. “It says here that we must speak the six letter keyword to open the door and enter the secret chamber, but I don’t remember seeing any signs as to what that keyword might be…”

After a bit of searching, Hermes notices something etched into the ground. “Come over here!”, he yelled, pointing frantically. And sure enough, barely visible and obscured by dust, was a series of lines of different colors etched into the ground:

“Ah”, Merlin said, “So that is the keyword.” Hermes was lost and confused. After staring at it for another thirty seconds, he grumbled “What keyword!? All I see is a bunch of lines!”. Merlin simply responded, “You’re just looking at it the wrong way. It’s obvious!”

Isn’t it?

SOLUTION

The signs are engraved letters on the ground and Merlin and Hermes are looking at them from above (the italic “looking at it the wrong way” is a hint). The darker a part from some sign is, the farther from the ground it is. The only letters which could correspond to this description are U – N – L – I – N – K. Therefore the keyword is “UNLINK”.

Source:

Puzzling StackExchange

Tryth March 19, 2022

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Battleship

Section: Brain Teasers Category: Mathematics

Difficulty: Hard

A battleship starts moving at 12 PM from an integer point on the real line with constant speed, landing on every hour again on an integer point. Every day at midnight you can shoot at an arbitrary point on the real plane, trying to destroy the battleship. Can you find a strategy with which you will eventually succeed to do this?

SOLUTION

If we know the starting point of the battleship and its speed, then we can determine its position at any time after 12 PM.

There are countably many combinations (X, Y) of starting point and speed. We can order them in the following way:

(0, 0) – starting point 0, speed 0;
(0, 1) – starting point 0, speed +1;
(1, 0) – starting point 1, speed 0;
(0, -1) – starting point 0, speed -1;
(1, 1) – starting point 1, speed +1;
(-1, 0) – starting point -1, speed 0;
(0, 2) – starting point 0, speed +2;
(1, -1) – starting point 1, speed -1;
(-1, 1) – starting point -1, speed 1;
(2, 0)- starting point 2, speed 0,
and so on. Of course, we can choose the ordering in many different ways.

Now we can start exhausting all possibilities one after another. First we assume the combination is (0, 0), calculate where the battleship would be at midnight during the first day and shoot there. Then we assume the combination is (0, 1), calculate where the battleship would be at midnight during the second day and shoot there. If we continue like this, eventually we will hit the battleship.

Unknown Author March 7, 2022

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Soccer Ball

Section: Brain Teasers Category: Mathematics

Difficulty: Hard Tags: Featured

Almost everyone knows what a soccer ball looks like – there are several black regular pentagons on it and around each of them – five white regular hexagons. Around each hexagon there are three pentagons and three more hexagons. However, can you figure out how many pentagons and hexagons are there in total on the surface of the ball?

SOLUTION

Let the number of pentagons is equal to P and the number of hexagons is equal to H. Then the number of edges is equal to (5P + 6H)/2 – that’s because every pentagon has five edges, every hexagon has 6 edges and every edge belongs to 2 sides. Also, the number of vertices is equal to (5P + 6H)/3 – that’s because every pentagon has five vertices, every hexagon has 6 vertices and every vertex belongs to 3 sides. Now using Euler’s Theorem we get P + H + (5P + 6H)/3 – (5P + 6H)/2 = 2, or equivalently P/6=2 and therefore P = 12. Since around every pentagon there are exactly 5 hexagons and around every hexagon there are exactly 3 pentagons, we get H = 5P/3 = 20. Therefore, there are 12 pentagons and 20 hexagons on a soccer ball.

Unknown Author March 1, 2022

2 Comments

Canaries in Space

Section: Brain Teasers Category: Science

Difficulty: Hard

NASA was considering sending canaries into space to study them under zero gravity. The project was scrapped when someone realized that in spite of having sufficient water supplies, they could die of dehydration within a few hours. Why?

SOLUTION

Unlike humans, birds need gravity in order to swallow. Thus, in space they wouldn’t be able to drink and will die of dehydration.

Unknown Author November 4, 2021

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Movie Conundrums 8

Section: Casual Puzzles Category: Rebus

Difficulty: Hard Tags: Movie Conundrums

Can you recognize which famous movies are depicted by these images?

SOLUTION

Manhattan
Joan of Arc
Open Range
Inception
The Odd Couple
In the Loop
Sin City
The Exorcist
Transformers
Absolute Power

Spiked Math November 1, 2021

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Envelopes with Numbers

Section: Brain Teasers Category: Mathematics

Difficulty: Hard Tags: Probability

You are given 2 sealed envelopes with numbers inside. You are told that one of the numbers is twice as much as the other one. You grab one of the envelopes and right before you open it, you make the following calculation:

“If this envelope contains X inside, then the other envelope contains either X/2 or 2X inside. Since the chance that the other envelope contains a larger number is exactly 50%, the expected money I will get after switching is X/4 + X = 1.25X > X. Therefore, I should switch!”

Clearly, this reasoning is wrong, since you can’t possibly deduce which envelope of the two contains a larger number. Where is the mistake?

SOLUTION

The trick is that conditionally on the fact that your envelope contains X, it is not true that the other envelope has 50% chance of containing either X/2 or 2X. The reason is that it is impossible that all amounts of dollars appear in the envelopes with the same probabilities (densities). Thus, for example, if it is very unlikely that an envelope contains more than 1000, and you open an envelope with 800 inside, you will not think that the other envelope has 50% chance of containing 1600.

Unknown Author October 10, 2021

5 Comments

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Does Not Divide Another

Section: Brain Teasers Category: Mathematics

Difficulty: Hard Tags: Featured

How many numbers between 1 and 100 can you pick at most, so that none of them divide another?

SOLUTION

You can choose fifty numbers at most: 51, 52, 53, … , 100.

In order to see that you cannot choose more than fifty, express each number in the form 2ⁿ×m, where m is an odd number. Since no two numbers can have the same m in their expressions, and there are only fifty odd numbers between 1 and 100, the statement of the problem follows.

Unknown Author September 13, 2021

1 Comment

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Houses on a Farm

Section: Brain Teasers Category: Mathematics

Difficulty: Hard Tags: Featured, Graph Theory

Is it possible to connect each of the houses with the well, the barn, and the mill, so that no two connections intersect each other?

SOLUTION

No, it is impossible. Here is a convincing, albeit a informal proof.

Imagine the problem is solvable. Then you can connect House A to the Well, then the Well to House B, then House B to the Barn, then the Barn to House C, then House C to the Mill, and finally the Mill to House A. Thus, you will create one loop with 6 points on it, such that houses and non-houses are alternating along the loop. Now, you must connect Point 1 with Point 4, Point 2 with Point 5, Point 3 with Point 6, such that the three curves do not intersect each other. However, you can see that you can draw no more than one such curve neither on the inside, nor the outside of the loop. Therefore, the task is indeed impossible.

More rigorous, mathematical proof can be made using Euler’s formula for planar graphs. We have that F + V – E = 2, where F is the number of faces, V is the number of vertices, and E is the number of edges in the planar graph. We have V = 6 and E = 9, and therefore F = 5. Since no 2 houses or 2 non-houses can be connected with each other, every face in this graph must have at least 4 sides (edges). Therefore, the total number of sides of all faces must be at least 20. However, this is impossible, since every edge is counted twice as a side and 20/2 > 9.

Unknown Author July 1, 2021

1 Comment

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Game of Coins

Section: Brain Teasers Category: Mathematics

Difficulty: Hard Tags: Featured

Kuku and Pipi decide to play a game. They arrange 50 coins in a line on the table, with various nominations. Then, alternating, each player takes on their turn one of the two coins at the ends of the line and keeps it. Kuku and Pipi continue doing this, until after the 50th move all coins are taken. Prove that whoever starts first can always collect coins with at least as much value as their opponent.

Remark: On the first turn, Kuku can pick either coin #1 or coin #50. If Kuku picks coin #1, then Pipi can pick on her turn either coin #2 or coin #50. If Kuku picks coin #50, then Pipi can pick on her turn either coin #1 or coin #49.

SOLUTION

Let’s assume Kuku starts first. In the beginning, he calculates the total value of the coins placed on odd positions in the line and compares it with the total value of the coins placed on even positions in the line. If the former has a bigger total value, then on every turn he takes the end coin which was placed on odd position initially. If the latter has bigger value, then on every turn he takes the end coin which was placed on even position initially. It is easy to see that he can always do this because after each of Pipi’s turns there will be one “odd” coin and one “even” coin at the ends of the line.

Unknown Author June 28, 2021

2 Comments