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# Logic contest (Day 3)

I suggest everybody to read the rules as I have created some different rules.

Now we are continuing day 3 after successful day 2 and day 1 the logic contest with rules as follows

1. Suppose problem $$i$$ is posted. The one who solves the problem and posts the solution becomes able to publish the $$i+1$$ problem.

2. This will continue until the problem posted is not answered within $$6$$ hours. If the solution is not posted, the problem maker himself starts with the next problem, posting a solution of the previous one.

3. The deadline question will be declared later. Meaning, the question with which the contest ends.

4. Whosoever posts a new problem should post on slack in #general that new problem is up!

5. The new problem poster should know the solution of his posted problem.

6. If the new problem is not posted in 15 minutes of answering the previous question, ANYBODY can post a new question.

7. Marking scheme is none, This time the question setter will decide the marks, Out of 5 and the one who answers will be given that credit. The marks will be set acc. To the level of the question as in brilliant!

8. We are starting from Question 37

Sharky: 25 marks

Prince: 18 marks

Kaustubh: 5 marks

Julian: 4marks

AA: 4marks

Alex: 4 marks

Deeparaj: 2 marks

Note by Prince Loomba
8 months, 1 week ago

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Problem 39(4 marks)

Consider five holes in a line. One of them is occupied by a fox. Each night, the fox moves to a neighboring hole, either to the left or to the right. Each morning, you get to inspect a hole of your choice. What strategy would ensure that the fox is eventually caught? · 8 months, 1 week ago

I would check the holes in this order from the left: 2nd, 3rd, 4th, 2nd, 3rd, 4th.

Explanation:

We will split this into 2 cases: The fox is either in an even hole on the first night or an odd hole.

If the fox is in an even hole, it is either the 2nd or 4th hole. If it is in the 2nd hole, we find him in the first night. If he is in the 4th hole, he can either move to 3rd hole or 5th hole the next night. Thus, we check the 3rd hole next and either catch him, or not. However, if he goes to hole 5, he must go back to hole 4, where we catch him.

If the fox started in an odd hole, we haven't found him yet, by parity of the holes and its movement. Thus, we check the 2nd hole on the 4th day since it must be in an even hole by now (because of parity), and the above repeats. · 8 months, 1 week ago

Problem 55: (3 marks)

A king is chucking a party and you are all invited! However, 48 hours before the party, the king has realised that one of the 240 barrels of wine have been poisoned with a substance that kills at exactly on the midnight just after it has been drunk. Fortunately, the king has been given 2 midnights between now and the party to determine which barrel is poisoned.

The king decides to test each barrel on some slaves, but the kingdom is going through a slave recession so the king wants to use as few slaves as possible. What's the minimum number of slaves required to determine the poisoned barrel? · 8 months ago

The answer is 5, so everyone got it wrong. In this case, we have 2 days before the party, but the poison takes 24 hours. Label the barrels in ternary, and give to corresponding prisoners based on the number (First digit for first prisoner, etc.). If there is a 0, do not give to the prisoner. If there is a 1, give to the prisoner before the first midnight. If there is a 2, give to the prisoner before the second midnight but after the first midnight. The deaths of the prisoners will tell us which barrel is poisoned based on when they died and who did.

In general, for $$n$$ number of midnights and $$N$$ barrels, the answer is $$\lfloor \log_{n+1} N \rfloor$$. · 8 months ago

I will give 4 marks to you · 8 months ago

But the problem has 3 marks! · 8 months ago

20? · 8 months ago

Incorrect. Please, try again. · 8 months ago

Answer is 8.http://www.folj.com/puzzles/very-difficult-analytical-puzzles.htm Go to this · 8 months ago

Incorrect. Read the question again. · 8 months ago

8 I suppose. I heard with 1000 bottles. Anyways the number of prisoner is $$log_{2}N$$ · 8 months ago

Incorrect. Read the question again. · 8 months ago

Problem 54 (3 marks)

You and your opponent shall play a game with three dice: First, your opponent chooses one of the three dice. Next, you choose one of the remaining two dice. The player who throws the higher number with their chosen dice wins. Now, each dice has three distinct numbers between 1 and 9, with pairs of opposite faces being identical. Design the three dice such that you always win! In other words, no matter which dice your opponent chooses, one of the two remaining dice throws a number larger than your opponent, on average. · 8 months ago

If the dice are as follows: $$A(1, 6, 8)$$, $$B(2, 3, 9)$$ and $$C(4, 5, 7)$$, then, on average, $$A$$ should beat $$C$$, $$C$$ should beat $$B$$ and $$B$$ should beat $$A$$. Thus, this dice combination should enable a greater chance of success. · 8 months ago

Post next · 8 months ago

Problem 53 (4 marks):

An old treasure map says as follows:

There is an unmarked grave and two tall oak trees. Walk from the grave to the left tree, counting the number of steps. Upon reaching the left tree, turn left by 90 degrees and walk the same number of steps. Mark the point with a flag. Return to the grave. Now, walk towards the right tree, counting the number of steps. Upon reaching the right tree, turn right by 90 degrees and walk the same number of steps. Mark this point with another flag. The treasure lies at the midpoint of the two flags.

A crew of pirates reach the island. They find a pair of tall oak trees merrily swaying in the wind. However, the unmarked grave is nowhere to be found. They are planning to dig up the entire island. It'll take a month. Can they do any better? · 8 months ago

It does not matter much where the grave is! There are only two places where the treasure can be. If you construct a square with the two trees at two diametrically opposite corners, then the treasure can be on either of the other two corners. There is a degenerate case where the grave would be on the same line as the trees. in this case the treasure would be buried exactly between the trees. This degenerate case is, however, excluded by the wording of the instructions, since in that case there would not be a left and a right tree.

Proof with vectors: Let the unmarked grave be the origin. Let A and B denote vectors to the two oak trees. The first flag is at A + iA. The second flag is at B - iB. The mid-point is ½ (A + B) + ½ (A - B)i, which may be rewritten as A + ½(B - A) - ½(B - A)i. Note that -iX denotes counterclockwise rotation of vector X by 90 degrees. Now the last expression has three summands, which can be visualized as follows: "A" denotes moving from the unmarked grave to the left oak tree, "½ (B - A)" denotes moving half the distance between the two oak trees from A to B, and "- ½(B-A)i" denotes turning counterclockwise by 90 degrees and moving the same amount as the previous step. Finally, the unmarked grave could be on either side of the line joining A and B. So there are two points where the treasure might be. · 8 months ago

Problem 52 (4 marks)

We have three Containers the Maximum volume of the first is 10 liters and the second is 7 liters and the third is 3 liters.

We put in the first 10 liters of water (or any liquid).

We need Divide the water to 5 liters in the first and 5 liters in the second without any signs or any measuring tools. · 8 months ago

1. Fill the 10L container.

2. Pour 10L into 7L, leaving 3L in 10L container.

3. Pour 7L into 3L, leaving 4L in 7L container.

4. Pour 3L into 10L, leaving 6L in 10L container.

5. Pour 7L into 3L, leaving 1L in 7L container.

6. Pour 3L into 10L, leaving 9L in 10L container.

7. Pour 7L into 3L, leaving 1L in 3L container.

8. Pour 10L into 7L, leaving 2L in 10L container.

9. Pour 7L into 3L leaving 5L in 7L container.

10. Pour 3L into 10L, leaving 5L in 10L container.

· 8 months ago

Problem 48 (4 marks)

You have 100 green eyed people on an island. They have no way to know that they're green eyed. They can see everybody on the island. Each day, they can leave the island if they go up to the guards and say if they're green eyed or not. If they are wrong, they fall into a pit of lava and die. What could you say to free all of the people in the least number of days?

Note: you can't tell them something that they don't know. · 8 months ago

"At least 99 of you have green eyes."

This takes only one day. · 8 months ago

How is mine violating the rules? Explain · 8 months ago

Look at the note. · 8 months ago

"All of you can go to the guard and save their lives" · 8 months ago

That's violating the rules. · 8 months ago

"At least one of you is green-eyed." · 8 months ago

wrong. · 8 months ago

BTW, the people have to be sure that they don't fall in the pit. · 8 months ago

Problem 44 (3 marks)

$$K$$ is a collection of $$n$$ different sets. Each set has distinct elements. Every set has exactly 1 element in common with every other set.

Given that $$K$$ contains only elements {1, 2, 3, 4, 5}, and

“1” appears in $$K$$ 2 times

“2” appears in $$K$$ 2 times

“3” appears in $$K$$ 3 times

“4” appears in $$K$$ 4 times

“5” appears in $$K$$ 5 times

What is the value of $$n$$. · 8 months ago

This problem indeed has an issue. · 8 months ago

Do you mean by the fact that "each set has distinct elements" that a set can't have repeating elements like 2 2s or that the number of elements between sets is different? · 8 months ago

@A A Yes obviously · 8 months ago

The first one · 8 months ago

But then how is it possible ?

There are 5 5s which have to be in 5 different sets and each of the 5 sets should be different meaning that there can't be more than 4 different numbers of the remaining values that should be given to all of the sets which include 5s differently and as such that you get something like 51 52 54 53 5 which makes impossible a construction.

So if every set has a distinct numbers it seems impossible.

Or I am missing soemthing. · 8 months ago

Problem 43 (4 marks)

You are trapped in a prison and the guard decides to play a game. If you win, you will be allowed to leave. But if you lose, you die. The guard says that he is thinking of Number 1, Number 2, or Number 3. You are allowed to ask one question to find out which of these three numbers the guard has in mind. However, the guard will only answer with a “yes”, “no”, or “I don’t know”. What should you say to the guard? · 8 months ago

Let the numbers be a, b and c.

If a, b and c are not distinct, the solution to this problem will be trivial. I would not go through these cases.

If a, b and c are distinct, without loss of generality, assume $$a<b<c$$.

"If I pick any number from this set: {a,b}, is my number smaller than yours?"

If the guard says yes, then his number must be the greatest number, which is c. If the guard says no, then his number must be the smallest number, which is a. If the guard says idk, then his number will be greater if you pick a, but equal if you pick b, so his number must be b.

On an unrelated note, I noticed many of the comments has a downvote. Is this a glitch or is somebody really disliking every comment? · 8 months ago

Haha Ayush disliked all of mine comments. Downvoted! · 8 months ago

Post the next problem · 8 months ago

Is the sum of the remaining numbers less than four? · 8 months ago

Haha. How can 4 work, the numbers are number 1 number 2 number 3 not 1,2,3 · 8 months ago

i said the SUM · 8 months ago

Is it Number 1 or Number 2 or Number 3? · 8 months ago

To this question he will answer yes only, what benefit? · 8 months ago

Wrong he can answer only yes no or idk · 8 months ago

When does he answer no or idk ? Isn't it true that the number is one of the 3 aayway ? · 8 months ago

@A A AAway · 8 months ago

@A A Yes it is. The question is to ask him one question to get the number he is thinking of · 8 months ago

Problem 42 (3 marks)

You are furniture shopping for your elderly parents in another town, when you come across the perfect piece that would go well in their dining room. You need to inform them of your discovery, but find that the only way to send a message in the small town is by telegraph. Now, the owner of the telegraph is an unscrupulous man who see's an opportunity to make a lot of money of you. He offers you the following deal-

10 cents for the first word

1 dollar for the second word

10 dollars for the third word

100 dollars for the fourth word, etc.

Being clever, you are able to inform your parents that the furniture they want is waiting for them, and that they need to pick it up, while only spending 10 cents

Notes:

You need to send a complete sentence, i.e. with a verb and noun

No, you don't force the owner of the telegraph at gunpoint to send your message for only 10 cents · 8 months ago

The answer is COMFORTABLE. Think it before asking how. · 8 months ago

How does saying comfortable do anything? · 8 months ago

I-found-the-perfect-furniture-for-you-come-pick-it-up , is it right ? Theoretically anyway that would be one long compound word which would be read as a sentence. · 8 months ago

@A A No. 1 word · 8 months ago

Problem 51 3 marks

Rebuses:

NME NME NME NME I am NME NME NME NME

D B 1 N B 2 A B 3 T B 4 S B 5

$${ 1 }^{ 2 }$$ $$\leftarrow$$

give get

give get

give get

give get

2CHECK

M E A N I N G

E

A

N

I

N

G

Poli4cy

     |

|


com $$\rightarrow$$ $$Na^{ +1 }$$

     |

|

· 8 months ago

Please dont post a collection of questions, a single one. I am awarding 2 marks to sharky and 1 to me. · 8 months ago

Back to squared one · 8 months ago

Stand up and be counted · 8 months ago

I am surrounded by enemies.

Stand backwards???? (should be pointing upwards so really, it should be Stand up) and be counted.

No good television? (It would sound better as Nothing good on Television)

(Still figuring out)

Forgive, Forget

Double Check (Chess terminology)???

The beginning of the end

Foreign Policy? · 8 months ago

2 marks to you · 8 months ago

I changed it by the way and you got 5 right.

The nothing good on television I counted as right. · 8 months ago

Problem 50 (4 marks)

You have 8 boxes which each having different weights from 1 to 8 kg , thelightest has 1 kg , the next 2 kg an so on anyway. You also have a scale with 2 pans. What is the minimum number of weights necessary to separate the boxes into 2 groups of 18 kg each anyway ? · 8 months ago

@A A 4? · 8 months ago

Emmmm , you should justify that. Can you show that it can be done in 4 ? · 8 months ago

Problem 49 (5 marks)

You are carrying water across the desert to hydrate your friend. Your camel can only carry 1000 bottles of water at a time. The desert is 1000 meters across. You have 3000 water bottles. Each mile, your camel drinks one bottle of water. When you go back, your camel doesn't drink water. What is the maximum number of water bottles you can get across? · 8 months ago

Its 833.

https://brilliant.org/problems/appleville-to-bananaville-transportation/

This is the exact same problem I posted months ago. This time I wrote 500/3 as 166 so got 834. But its 167 and 833 · 8 months ago

I posted correct answer first! I should get points · 8 months ago

I thinkk 833. · 8 months ago

correct. · 8 months ago

Is the change from "meter" to "mile" intentional, or a typographical error? · 8 months ago

It's a typo. · 8 months ago

That's a good question. I don't know the miles unit but I suppose that it should change the problem's answer anyway. · 8 months ago

Is the answer 832? · 8 months ago

Hey! SCORE in technothlon ? · 8 months ago

9.0000 · 8 months ago

Problem 47 (2 marks)

Rebus puzzle: Decode its meaning

13579 vs. U · 8 months ago

The odds are against you
· 8 months ago

Shouldn't you post the next problem ? · 8 months ago

@A A I'm posting. · 8 months ago

Ok? · 8 months ago

Sure. It's according to the rules. · 8 months ago

Problem 46 (2 marks)

A , B and C received the marks 8 , 7 , 9 at english , philosophy and logic anyway. It's know that B doesn't have any 8 , the 8 wasn't got at logic and the biggest was at philosophy.The problem has more solutions. But because it is an easy problem you should find anyway all of the solutions which meet the constraints. · 8 months ago

@A A Philosophy 9 logic 7 english 8

Philosophy Logic English

A,B,C

B,C,A

C,B,A

B,A,C

4 solutions · 8 months ago

Problem 45 (4marks)

In a town was commited a murder and 8 suspects have been identified.

By separate investigations two detectives , A and B , reduced the number of suspects to anyway 2 and their lists contain 3 names from which only 1 is common , the culprit.

The 2 detectives met at the police station to compare their notes and find the name of the principal suspect but can't communicate between them unless are watch be the police at the station. Try to find a strategy by which the 2 of them find the name of the principal suspect without the police founding out too.

If the two detectives would give the police 2 lists of 2 names in which is to be included the principal suspect but don't want the police to find out who is the principal suspect and therefore that wouldn't be the right strategy. This is a cute problem and so the solution I suppose so I let you soem time to think as it is a little bit a hard problem anyway.

Assume the strategy doesn't imply making sign or gestures or something like that but only talking.

That is the two detective should know just from talking with each other which is the principal suspect but not the policeman. · 8 months ago

Problem 41 (2 marks)

Sherlock Holmes went to Transilvania to solve some cases of vampirism. Getting in the village where he had to inspect the case he found it populated by vampires and humans alike.

Vampires always lie and humans always tell the truth. Moreover half of the citizens are insane (mentally that is anyway) both vampires and humans while the other part is sane.

Crazy people consider true all false statements and false all true statements , while the sane citizens do exactly the contrary considering true true statements and false the other. Therefore sane humans and crazy vampires say just true statements and crazy humans and sane vampries make just false statements.

At a moment of his investigation Sherlock meets 2 sisters out of which by his unsurpassed deductive skills of course he immediately found that one is a humand and the other is a vampire but since all human mental capacities are limited and anyway even if they wouldn't they would still be limited in deductive capacities he didn't knew how sane are them anyway.

Sherlock being of course Sherlock Holmes found out from this discussion which is which so what conclusion did the famous detective with flawless logic made of the sisters anyway ? Tell what are both A and B vampire/human or who is who anyway. The problem can have more solutions. · 8 months ago

@A A Case A is sane human

A is telling lie that both are crazy. Not possible

Case A is sane vampire

A is telling lie. So B should be sane human as crazy human would support him

Case A is crazy human

A is telling lie. So B is sane. B is not supporting means B is sane human. Not possible as 1 is vampire

Case A is crazy vampire

She is telling truth. So other is also crazy. Other disagrees means she is telling lie. So she is crazy human.

So 2 cases only. Sane vampire Sane human and crazy vampire crazy human · 8 months ago

Right. That is the correct solution. · 8 months ago

@A A I knew it. So I posted next anyways · 8 months ago

@A A A can be crazy vampire then B is crazy human Or A can be crazy human then B is human Or A can be vampire then B is crazy vampire or human · 8 months ago

One is human one is vampire and it is already solved so 1 of your cases is wrong · 8 months ago

The person who tells first shouts the sum of all the numbers he can see. Consequently, the second person can shout out his number correctly by calculating the sum of all numbers he can see (except the first person's) and subtracting it from the number the first person screams out and thus be freed. The third person uses the same stratgey to figure out his number and thus get freed, and so on till the last person. So, we can save n - 1 people. · 8 months, 1 week ago

Again the same. They all shout at the same time · 8 months, 1 week ago

Problem 40 (5 marks):

The king has put all his prisoners on a death sentence, but he decides to have a bit of fun with these prisoners before they die. He plays a game with them.

Each prisoner (there are $$n$$ prisoners) is to wear a hat with the numbers in the interval $$[0, n-1]$$. However, duplicates are allowed so 2 or more prisoners can have the same number. Any prisoner can see everyone else's number but not their own. When a bell rings, all prisoners must shout out what they think their number is at the same time. If any one of them get it right, all the prisoners are released.

The prisoners are given an evening to discuss a strategy. How can the prisoners possibly be freed? · 8 months, 1 week ago

We will do a preliminary investigation first with $$n=2$$. Thus, the numbers must be in the interval $$[0, 1]$$. The following strategy works for any of the 4 combinations: the first prisoner shouts out the number on the second prisoner's hat, and the second prisoner shouts out the number that isn't on the first prisoner's hat. If the 1st prisoner and second prisoner hat numbers are $$(0, 0)$$ or $$(1, 1)$$, then the first prisoner has correctly identified his own number, whereas if the numbers are $$(0, 1)$$ or $$(1, 0)$$, the second prisoner correctly identifies his number. Thus, the prisoners escape.

Doing further investigations with $$n=3, 4, 5$$ will help us see the answers of the prisoners to escape must include the sum of the total numbers seen and $$\pmod{n}$$.

In fact, for the general case, this strategy works:

Let the $$i$$-th prisoner say $$i-s \pmod{n}$$ where $$s$$ is the sum of the numbers each prisoner sees and for $$1 \leq i \leq n$$. Then, if the total sum of the numbers is $$m \pmod{n}$$, then the $$m$$th prisoner successfully says his number. The $$m$$th prisoner exists since $$0 \leq m \leq n-1$$.

Let me explain why. Let $$h$$ be the number on $$m$$'s hat. Thus, $$s=m-h$$. Subbing this back in, we get $$m$$ says $$m-(m-h)=h$$. Thus, the $$m$$th number says $$h$$, which is the number on his hat. · 8 months ago

I got the strategy but not the proof of how it works before. · 8 months ago

6 hours have passed · 8 months ago

Sorry, will post solution. Do I get the marks or do they remain unclaimed? · 8 months ago

Look I have already changed your marks from 7 to 12 · 8 months ago

Each prisoner should count the antimode (That is the minimum frequent number) he can see and speak it. If it is not unique, he can speak any of the antimode numbers. · 8 months, 1 week ago

Can you explain how this strategy works? · 8 months, 1 week ago

Yep.Lets suppose after the first person says wrong,then second person knows minimum frequency observation 1st person thought.So he can deduce info from it Now 3rd can do the same and the following conttinues · 8 months, 1 week ago

The trick is that they all shout at same time. So my logic went wrong · 8 months, 1 week ago

Problem 38 (4 marks)

"You are in a room with very many bags of 1oz gold coins, one bag of counterfeit gold coins (which weigh 0.9 oz each), ONE penny, and a "penny scale"--which is the type of scale where you put your object to weigh on the platform, put the penny in the pay slot, and a ticket is ejected with the weight of the object on the scale platform. You are told that if you can determine which bag is counterfeit, then you have can all the bags. You can weigh one set of objects on the platform, one time."

Here the number of coins and number of bags is random · 8 months, 1 week ago

Let there be $$n$$ bags. Label these bags 1 to $$n$$. Take 1 coin from bag 1, 2 coins from bag 2, ..., $$n$$ coins from bag $$n$$. Weigh these altogether. If there were no counterfeit coins, then the weight should be $$\frac {n(n+1)}{2}$$ oz, but since there is, subtract this weight from the non-counterfeit weight. Whatever this difference is, that is the number of the bag which contains the counterfeit coins. · 8 months, 1 week ago

Hmm... It's cool · 8 months, 1 week ago

Thanks. Answer the problem. · 8 months, 1 week ago

Problem 37(3 marks)

Four guys are led into room with four chairs, and forced to sit down. They build a brick wall between one of them and the other three. They blindfold them, place hats on their heads, then remove the blindfolds.

The four guys are given the following information:

1. We put hats on all of their heads.

2. Two of them are blue and the other two are red.

3. If one of them can tell us the color of his hat, they will not be murdered

And they can't look behind them, of course. So how did they survive?

They cant talk among themselves. · 8 months, 1 week ago

WLOG the person who is bricked away has a black hat. If the person at the back of the three people sees 2 white hats in front of him, he knows that his hat is black so they won't be murdered. If there aren't 2 white hats in front of him, then he won't say anything. Because of this silence, the person in the middle of the three knows that the last person can't determine the colour of his hat. From this, he gets that his hat and the hat in front of him are of different colours. Therefore, he can determine the colour of his hat. Thus, they all remain living. · 8 months, 1 week ago

Put the next question and set up the marks · 8 months, 1 week ago