**I SUGGEST EVERYBODY TO READ RULES AS I HAVE CREATED SOME DIFFERENT RULES.**

Now we are starting a logic contest with rules as follows

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

This will continue until the problem posted is not answered within 48 hours. If the solution is not posted, the problem maker himself starts with the next problem, posting a solution of the previous one.

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

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

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

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

Marking scheme is that marks for the \(i^{th}\) question is the simple interest on 10 at 10% per annum for \(i\) years

**Prince (me): 29 marks(Q7+Q10+Q12)**

Aman Rajput: 28 marks(Q17+Q11)

Mehul: 27 marks(Q14+Q13)

AA: 20 marks(Q15+Q5)

Archit: 19 marks(Q 9+Q6+Q4 )

Agnishom: 8 marks(Q8)

Angela: 4 marks(Q4)

Svatejas: 3 marks(Q3)

Ivan: 2 marks(Q2)

Deeparaj: 1 mark (Q1)

## Comments

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TopNewestProblem 13.You have a switch that can be turned on and off. In 1 minute you turn it on, in 30 seconds you turn it off, in 15 seconds you turn it on, and so on. After 2 minutes is it on or off? – Alex Wang · 2 months, 1 week ago

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– Prince Loomba · 2 months, 1 week ago

Cannot be determined, its an infinite geometric series.Log in to reply

– Mehul Arora · 2 months, 1 week ago

Cannot be determined.Log in to reply

– Prince Loomba · 2 months, 1 week ago

Are you posting a new problem or shall I?Log in to reply

– Mehul Arora · 2 months, 1 week ago

Please do the honours.Log in to reply

– A A · 2 months, 1 week ago

Justify why to make a complete answer even if it obvious. If it's obvious and improtant you still have to say it anyway.Log in to reply

– Alex Wang · 2 months, 1 week ago

After each off there is an on, and after each on there is an off. This is Grandi's series. 1-1+1-1+1-1+1.........=1/2. The light can't be half on or half off.Log in to reply

This is assuming that the switching can grow smaller and smaller. Anyway. – A A · 2 months, 1 week ago

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– Prince Loomba · 2 months, 1 week ago

I mentioned geometric series, was it wrong?Log in to reply

– Alex Wang · 2 months, 1 week ago

you are right too.Log in to reply

– Alex Wang · 2 months, 1 week ago

Correct.Log in to reply

– Prince Loomba · 2 months, 1 week ago

Awarded to mehulLog in to reply

– Prince Loomba · 2 months, 1 week ago

This question carried 13 marks!Log in to reply

Problem 11You have two equal sized buckets. One contains water and the other contains the same amount of wine. You transfer a cup of wine to the water bucket and mix it in. Next you transfer a cup of the mixture back to the wine bucket. Is there more wine in the water, or water in the wine? – Prince Loomba · 2 months, 1 week ago

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– Mehul Arora · 2 months, 1 week ago

We have to take into consideration whether the buckets are filled to the brim and will overflow on putting the cup inside them.Log in to reply

– Prince Loomba · 2 months, 1 week ago

Dont consider itLog in to reply

Transferring again x amount to y-x. Take some numerical value say x=1 , y=5 , we will have 1/6

After again transferring we have. 5/6/5 =1/6 which is same. – Aman Rajput · 2 months, 1 week ago

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– Mehul Arora · 2 months, 1 week ago

I think Aman is correct.Log in to reply

– Prince Loomba · 2 months, 1 week ago

WrongLog in to reply

– Aman Rajput · 2 months, 1 week ago

Oh why ?Log in to reply

– Prince Loomba · 2 months, 1 week ago

Answering this will be posting the solution to the problem!Log in to reply

– Aman Rajput · 2 months, 1 week ago

okay bu i still think it should be equal.let me try one more timeLog in to reply

– Prince Loomba · 2 months, 1 week ago

Else come to slack.Log in to reply

Jack is looking at Anne, but Anne is looking at George. Jack is married, but George is not. Is a married person looking at an unmarried person? – Archit Agrawal · 2 months, 1 week ago

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– Prince Loomba · 2 months, 1 week ago

Suppose Anne to be married. Anne lookjng at George. Condition satisfied. Anne unmarried. Jack looking at Anne. So condition satisfied. Yes is the answerLog in to reply

– Hung Woei Neoh · 2 months, 1 week ago

This is copied from a Brilliant questionLog in to reply

– Prince Loomba · 2 months, 1 week ago

Actually its quite famous. I knew it beforehandLog in to reply

– A A · 2 months, 1 week ago

A lot of problems posted here are quite famous.Log in to reply

– Prince Loomba · 2 months, 1 week ago

He didnt. It is a pure coincidenceLog in to reply

Problem 9In a palace, there are some finite number of rooms (but you do not know how many) arranged in a circular fashion. Each room is connected to two other rooms and the last room is connected to the first room.

In each of the room there is a lightbulb, either switched on or off. Some of the rooms are lit, some of them are not and there is no pattern that the illumination sequence follows.

You start off at an arbitrary room and are required to turn of the bulbs in all of the rooms.

What is your strategy and how do you ensure that you've actually switched off all the bulbs? – Agnishom Chattopadhyay · 2 months, 1 week ago

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– Archit Agrawal · 2 months, 1 week ago

Switch off all the bulbs which are on and touch the bulbs which are off. When you find a bulb which is off and is warm you will know that all the bulbs are switched off.Log in to reply

– Agnishom Chattopadhyay · 2 months, 1 week ago

This was not a lateral thinking puzzle. The bulbs do not heat up or are touchable. They can carry just one bit of information.Log in to reply

What if the room is very cold or the palace is extremely lengthy or something anyway in that manner ? – A A · 2 months, 1 week ago

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Problem 5Alphonse and his \( n-1 \) friends are sitting in a circular fashion when Edgar meets them (sitting of n people). Each of them owns coin. First person passes 1 coin to the person on his left side. The second person in turn passes 2 coins to the person siting on the left. Third person passes 1 coin to the left, 4th passes 2 coins to his left and so on, so each person receiving coin from the right has to pass 2 coins to the left. Similarly, a person receiving 2 coins from the right has to pass 1 coin to the left.

If at any point of time, a person runs out of coin he is thrown out of the game. The game will terminate if at the end only 1 person is with coins in his possession. There might be values of n where the game may not terminate (e. g. 3 persons left with 4 4 4 coins respectively) For how many different values of n from 4 to 100 does the game terminate? – Angela Fajardo · 2 months, 1 week ago

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Solution for Question 5In first round the players have 1 coin each. Move to second round. 3rd,5th,7th... people have 2 coins each. Move to 3rd round. 7th,11th,15th,... people have 4 coins each. We know that if atleast 2 people have 4 coins, the game wont end. So we require the values from 4 to 10, i.e. 7 values, since from 11th, the 7th and 11th will both have 4 coins and no termination.

Answer is 7 – Prince Loomba · 2 months, 1 week ago

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– A A · 2 months, 1 week ago

Haha , this is the technotlon 2016 4th question. I just think the answer is 6 but in the paper there is no option which has " 6 " as an answer anyway. but btw please modify a little the statement and say rather something like "each of them owns 1 coin" to be more clear and avoid confusion so to say.Log in to reply

– Prince Loomba · 2 months, 1 week ago

Now you understood.Log in to reply

– A A · 2 months, 1 week ago

Hmmmm , actually.... ... yea, well I verified my reasoning and is it seems correct. Anyway , according to that reasoning of mine now I get the 7th value too which anyway i missed at first. I'm pretty sure that for odd numbers after each persons passes a coin once it doesn't end and I've calculated again attentively this time. They each lose and get 1 coin alternating at every round of passing depending on their position , thatis the ones which are in even positions get +1 coins and the other lose at some round while at the next round the odd positions loses and the even gain.Log in to reply

– A A · 2 months, 1 week ago

yep , so I think. You should post the solution to this problem.Log in to reply

– Prince Loomba · 2 months, 1 week ago

LaterLog in to reply

I'm still not that sure it's right. – A A · 2 months, 1 week ago

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Problem 3:Construct a chess position that is illegal (cannot be achieved in any number of moves from the normal position), but is legal if any one unit (other than the kings) is removed. – Ivan Koswara · 2 months, 1 week ago

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Since in chess it is illegal for a pawn on the 8th rank (or in other words if when you place the pawn on the 8th rank and didn't change it to a queen/knight/etc)

since you cant remove the kings, you can only remove the 8th rank pawn, making the game a draw and also legal – Angela Fajardo · 2 months, 1 week ago

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– Ivan Koswara · 2 months, 1 week ago

Whoops, that was a loophole. Nice for spotting that.Log in to reply

– Prince Loomba · 2 months, 1 week ago

Thats why contest has moved onLog in to reply

– Prince Loomba · 2 months, 1 week ago

Its rightLog in to reply

– Prince Loomba · 2 months, 1 week ago

You can post problem 4Log in to reply

– Angela Fajardo · 2 months, 1 week ago

Nope. You do the honors :)Log in to reply

– Prince Loomba · 2 months, 1 week ago

Posted. Try the nextLog in to reply

– Svatejas Shivakumar · 2 months, 1 week ago

Seven pawns on the a file(a1,a2,...a7) with the two kings anywhere.Log in to reply

– Prince Loomba · 2 months, 1 week ago

You should post as soon as you give the answerLog in to reply

– Prince Loomba · 2 months, 1 week ago

Post the next problemLog in to reply

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– Ivan Koswara · 2 months, 1 week ago

You cannot tell that the bishop on the black square is supposed to be a bishop on a white square.Log in to reply

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anypiece is removed, not only one of the bishops.Log in to reply

Btw , sorry for the rushed proof. Hopefully it's still clear enough to be understood. – A A · 2 months, 1 week ago

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So , I should post a solution for that question since I said I will anyway and since Prince awarded me points for it anyway.

The problem is actually pretty beautiful as it implies a synthetic and simple reasoning but quite elegant and cute anyway and more importantly it so as can be solved just by involving any quite of technicality or , if some technicality would be involved , it would be so at the point at which it is immediately conceived in the form of a direct concept.

To understand better the problem , let's define a "profit" as being the gain/lose of coins a player has after he both receives and gives a coin , the act of receiving a giving being a turn. Also define a round to be a continuous sequence of moves in which every player receives such a profit once but observe that this definition of round is incomplete since it is not reduced to the set of turns since if a round would be the set of moves when each player has made a turn then the first player will always gain more profit at the end of each round.

Now , having those stuff defined let's see how the game would go in the conceiving of this so defined terms or of the "language"/interpretation provided to understand it and used anyway. What will be done will anyway be to analyze how the profit of every player varies over the passing of rounds which is simply exactly the only stuff to which the problem itself reduces.

Observe that for each number of players , n , in the first round the first 2 players will anyway be eliminated (the 1st gives his only coin , the 2nd gives the 2 coins) and then after this depending on their positions , each player will alternate receiving 2 coins and giving 1 coin having a profit of +1 coin after a turn and receiving 1 coin and giving 2 coins having a profit of -1 coin each this alternating sequence starting with the 3nd player which receives 2 and gives 1 coin therefore with the 3nd having a +1 profit. Then it can be concluded that after the first round you have that each player in the odd position except the first two , which are in an odd position have a +1 and each player which is in a even position has a -1 profit meaning that at the profit-round representation you have +1 -1 +1 -1 +1-1 +1-1 etc anyway.

That means that after the first round all odd and one even position (player 1) are off the game so the remaining number of players will be n-1/2 which would mean in a properly said interpretation of stuff that you lose a pair of 2 already and from each other pair of 2 players remaining you lose one player anyway. The distribution of this profit can be quickly checked for the parity cases as for even number of players cases at some round after the first the distribution for a configuration after the 1st round will be stable meaning that at each round the same players will have a -1 profit while for the odd the players will alternate losing.

Let's analyze this intuitive and almost immediately seen thinking checking the cases of odd numbers and even number of players after some round after the first.

To check this suppose that at some round after the 1st the number of persons remaining in the game is even. Then at each round the first player has a +1 gain since he receives 2 from the last because the last is on an even place and gives 2 coins and therefore all those on even positions will get +1 , while the odd ones will get -1 until the end of the round where the same thing keeps continuing since at the next round the first also receives 2 coins from the last since the last is on an odd position which runs indefinetely with playerson odd positions loosing that is having -1 profit until they are off the game.

So the idea that a sequence of odd players will have players getting out of the game is right.

Now , check the case of an even number of players. For an even number of players the first player at the second round also gains +1 and also the same alternating lose/win profit stuff happens but because the position of the last person in the sequence is odd after the end of round anyway the person on the first place will not gain but lose coins this same course of alternating happening so at every 2 round you will have that the odds have +1 and the even have -1 at the first , then the odds have -1 and the even +1 at the second. Because every palyer therefore alternates receiving and losing coins their profit remains always the same.

So it's checked that for an odd number of players at a round after the first the game runs in a continuous pattern and never ends as it's unstable every alternating round anyway.

Because of this it means that you need to have just odd sequences at every step of the game and therefore that because at every sequence which has players getting out of the you eventually lose half of the players the number of players which shall remain after the first round should be of the for 2^m anyway.

Then you have that all n which after the first round gives you powers of 2 are ok and all the others , because contain in prime factorization an odd number are not since they will lead eventually to an odd sequence of players which as seen will get to run indefinitely never ever ending anyway so to say!

Expressed mathematically anyway for it to work you need to satisfy , for any number n the condition 2^m= (n-1)/2 with n = 2^m+1 or n = 2^m+2 for the odd and even numbers.

For the numbers between 4 and 100 therefore you have 10 or 11 cases depending whether or not you have to include 4 meaning the powers 1 , 2 , 4 , 3 , 5 , 6 anyway.

Of course there are some details to cover in the proof to make it rigorous such that at every round after the first the players have more than 1 coin for example but such things are pretty much easily understood by anyone and is sufficient to let it so anyway. – A A · 2 months, 1 week ago

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– Prince Loomba · 2 months, 1 week ago

100%Log in to reply

There are at least 2 socks of the same color is all of these cases. Therefore, the probability = 100.

That is my proof. – Alex Wang · 2 months, 1 week ago

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– Alex Wang · 2 months, 1 week ago

Correct. Post problem 19.Log in to reply

– Prince Loomba · 2 months, 1 week ago

Tomorrow. Problem 16 is there todayLog in to reply

Problem 17A knight arrives at a castle carrying a lance that is five feet long. The guard tells him that no-one is allowed in the castle with an object that is over four feet long. The knight is a tad upset, but then has an idea. He goes into town, finds the local carpenter and asks him to make something. The knight then returns to the castle and the guard lets him in.

Due to his clever thinking, the knight finds himself inside the castle with his lance. The lance is fully functioning, and has not been cut in any way.

How is this possible? – Prince Loomba · 2 months, 1 week ago

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– Aman Rajput · 2 months, 1 week ago

Make a cube... Of side 5/sqrt3 . Which is less than 4. Done!Log in to reply

– Prince Loomba · 2 months, 1 week ago

Rather 19 problemLog in to reply

– Prince Loomba · 2 months, 1 week ago

You are eligible to make problem 18Log in to reply

– Prince Loomba · 2 months, 1 week ago

Can it carry 5 m thing?Log in to reply

– Aman Rajput · 2 months, 1 week ago

Yes it can carry diagonally 5 m thing.. I'm correct. Make sure post some correct questions which is not having multiple answers.. I'm correct again.. you verify my resultLog in to reply

– Aman Rajput · 2 months, 1 week ago

It is given that he went to carpenter. My answer is correct.Log in to reply

– Alex Wang · 2 months, 1 week ago

how would you make a cube?Log in to reply

– Alex Wang · 2 months, 1 week ago

did he/she fold it?Log in to reply

– Alex Wang · 2 months, 1 week ago

or did he make something that was four feet long and showed that to the guard instead?Log in to reply

– Prince Loomba · 2 months, 1 week ago

IdkLog in to reply

– Alex Wang · 2 months, 1 week ago

can i make problem 18?Log in to reply

– Prince Loomba · 2 months, 1 week ago

Answer this thenLog in to reply

– Alex Wang · 2 months, 1 week ago

what?Log in to reply

Problem 16

A safe can be opened by inserting strings of an arbitrary length made from the digits from 1 to 9 which don't need to be included just once.

It is known that apart from the combinations which are correct and open the safe there are 2 other types of combinations , some which when entered will have no affect on the safe which are called neuter combinations and some which when inserted will destroy the safe which will be named bad combinations.

It is known that some strings are related to other strings of other by a relation which will be named R by some rules which are about to be presented.

For clarity define the following , for 2 strings of arbitrary length x and y

a) xy is the combination made from x followed by y

b) i(x) is the inverse of the combination x (if x is 21 for example then i(X)=12) anyway

The relation R has the following characteristics :

A) For any combination x , 2x2 is related to x.

B) If x R y then 1x R 2y

C) If x R y then 5x R i(y)

D) If x R y then 9x R yy

E) If x R Y and x is a neuter combination then y is a bad combination

F) If x R y and x is a bad combination then y is a neuter combination.

Find a configuration which will certainly open the safe.

Consider that from a correct string you can't obtain a neuter one or a wrong string.

Also the relation between the string is not reflexive. That is if x R y it doesn't mean that y R x. – A A · 2 months, 1 week ago

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– Prince Loomba · 2 months, 1 week ago

Haha again sameLog in to reply

– Aman Rajput · 2 months, 1 week ago

Latex required.. why are you in so hurry .? Edit again :)Log in to reply

– A A · 2 months, 1 week ago

No need for latex. All notation is i(x) , x, y , xRy anyway.Log in to reply

Problem 15If you were to put a coin into an empty bottle and then insert a cork into the neck, how could you remove the coin without taking out the cork or breaking the bottle? – Mehul Arora · 2 months, 1 week ago

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– Alex Wang · 2 months, 1 week ago

can the coin fall through the hole?Log in to reply

– Prince Loomba · 2 months, 1 week ago

Push the cork into the bottle and shake the coin outLog in to reply

– A A · 2 months, 1 week ago

push the cork ?Log in to reply

@A A – Mehul Arora · 2 months, 1 week ago

Points toLog in to reply

– Prince Loomba · 2 months, 1 week ago

And J posted more if you are comparing time. My words are moreLog in to reply

– Mehul Arora · 2 months, 1 week ago

As I said, if you want the points, you take all of them.Log in to reply

– Prince Loomba · 2 months, 1 week ago

Why, so that I cant achieve more marks?Log in to reply

– Mehul Arora · 2 months, 1 week ago

If you're in it for points, then please take all the points you need.Log in to reply

– Prince Loomba · 2 months, 1 week ago

No no. I was replying to your suggestion. I am not for that. I started this for fun and exercise.Log in to reply

– Prince Loomba · 2 months, 1 week ago

Same solution at same time, both will be awarded if rightLog in to reply

– Mehul Arora · 2 months, 1 week ago

I got A A's comment first.Log in to reply

problem 14Imagine that you are trapped in a prison cell. The only way out is through the door, but the door is locked. There is a combination lock on the door, and the following sequence of numbers appears above the lock:

1

11

21

1211

111221

312211

…to open the door you have to enter the next number in the sequence into the combination lock. What number should you enter? – Prince Loomba · 2 months, 1 week ago

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– Mehul Arora · 2 months, 1 week ago

13112221Log in to reply

– Mehul Arora · 2 months, 1 week ago

It's a look and Say series. Like for 11 there are 2 ones. therefore, the next number is 21. Answer: 13112221Log in to reply

– Prince Loomba · 2 months, 1 week ago

Post the next question. Awarded marksLog in to reply

– Mehul Arora · 2 months, 1 week ago

Posting. Need some time.Log in to reply

– Prince Loomba · 2 months, 1 week ago

How much time bro. New day (night) is coming.Log in to reply

– Prince Loomba · 2 months, 1 week ago

Answer if and only if you have the next problem!Log in to reply

– Prince Loomba · 2 months, 1 week ago

Answer requiredLog in to reply

– Prince Loomba · 2 months, 1 week ago

With solutionLog in to reply

– Mehul Arora · 2 months, 1 week ago

I posted the solutionLog in to reply

Problem 12:On Friday 15,2016 , Aman Rajput went to Cannaught Place. He started talking to strangers and asked some questions. Here is the conversation:

Me: Hello Ma'am my name is aman. I want to ask you a question. I want you to answer in yes or no . Okay ? Stranger : Okay fine! Ask.

Me: Say yes, if you have 2 children including the fact that one should be a girl born on Friday ? Stranger: No ,Sorry,

Proceeding in similar manner. Finally,after a long time I find a stranger who says yes.

What is the probability that this stranger has two girls ?Assume equal probability of giving birth to either sex and equal probability to give brith on any day.

– Aman Rajput · 2 months, 1 week agoLog in to reply

– Prince Loomba · 2 months, 1 week ago

Its 13/27 sureLog in to reply

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– Prince Loomba · 2 months, 1 week ago

Its 7/13 I supposeLog in to reply

– Aman Rajput · 2 months, 1 week ago

wrongLog in to reply

– Prince Loomba · 2 months, 1 week ago

Just change boy to girl and tuesday to fridayLog in to reply

– Prince Loomba · 2 months, 1 week ago

I was confused whether 13 was in numerator or denominator!Log in to reply

One change: My solution to problem 5 is wrong. I have creditted marks to AA. He can post its solution – Prince Loomba · 2 months, 1 week ago

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Problem 10 – Archit Agrawal · 2 months, 1 week ago

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Problem 8The weird and inaccurate soothsayer Nostradamus once announced…..

“On Wednesday 2nd February 2000, an event will take place for the first time in over 1000 years. In fact, the last time this event happened was 28 August 888″. For once, he was right. What was the event? – Prince Loomba · 2 months, 1 week ago

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– Prince Loomba · 2 months, 1 week ago

Wrong. Think moreLog in to reply

– Agnishom Chattopadhyay · 2 months, 1 week ago

I googled the answer. I guess I am not qualified to answer anymore.Log in to reply

– Prince Loomba · 2 months, 1 week ago

Come on. Post solution and the problem nextLog in to reply

– Agnishom Chattopadhyay · 2 months, 1 week ago

The only dates after 888-28-08 that have all even digits are 2000-02-02Log in to reply

Problem 7Chopsticks is a hand game for two players, in which players extend a number of fingers from each hand and transfer those scores by taking turns to tap one hand against another.

Each player uses both hands to play the game, the number of digits extended on a hand showing the number of points that the hand has. Both players start with each hand having one point — one finger extended on each hand. The goal of the game is for a player to force their opponent to extend all of their fingers and thumbs on both hands. A hand with all fingers and its thumb extended is called a "dead hand". Players take turns to tap one of their hands against another hand that is not dead (either their own other hand, or one of their opponent's). The number of points on the tapping hand is added to the number on the tapped hand, and the player with the tapped hand extends their digits to show the new score. The tapping hand remains unchanged.

A player may tap their own hand to transfer points from one hand to the other. For example, if a player had three points on his or her right hand and one on his or her left, the player could rearrange them to have two on each hand. A "dead hand" is treated as having no points, for this purpose, which allows a player to bring a dead hand back into play by transferring points to it. For instance

If you have 5/2 (Dead hand and other hand has 2 points) you can split 1/1. (one hand has 1 point the other has 1 point) If you have 5/3 you can split 2/1 or 1/2. If you have 5/4 you can split 2/2 or 3/1 or 1/3. If you have 2/3 you can switch it and make it 3/2

You cannot prolong the game by not taking your turn.

It is already known that the first player (the player who starts first) has the winning strategy. As the second player, what is the strategy to prolong the game as long as possible? (Assume the first player and second player are both perfect logicians, and that the first player wants to win as fast as possible.)

References – Julian Poon · 2 months, 1 week ago

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– Prince Loomba · 2 months, 1 week ago

The strategy to always winning once you have gone second is to tap the opponents hand as infrequently as possible. For example, when your opponent starts the game and taps your hand, split to three on one hand and zero on the other. At this point your opponent has no choice but to tap your hand again as them splitting to two on one hand would lead to an instant win for you. Once your opponent taps your "three" hand, you have two options to split (2),(2) or (3,1). It is ideal to split (3,1) because the game will be won more rapidly but it is still possible to force a win from (2,2). From (3,1), assume your opponent taps your (3) hand. You are left with (4,1). It is then ideal to split (3,2). If your opponent taps your (3) hand you tap one of their hands with your now (4) hand your opponent will tap your (4) hand and you will be left with (2,0). You split (1,1) and (1,1) against just one finger is an easy win. The way to win in that case will be described later on. If your opponent taps your (2) hand you will be left with (3),(3). Next, tap either hand with a (3). Your opponent will be left with (4,1) If your opponent splits to (3,2), tap their (2) hand. (If you tap their (3) hand your opponent will split and you will wind up with 3,3 against 1,1 again). From this point your opponent has to split their (3). You will then tap their (2), your opponent will tap one of your hands and then you will win. If your opponent taps your hand with a 4, you take out their 4 and your opponent has no choice but to tap your remaining hand with a 1 and you win. If your opponent taps your hand with a 1, you will take out their 4 and your opponent will have no choice but to tap out your 4 hand. You will be left with 3,0 against 1. This scenario will be described later on. If from (3,1) your opponent taps your (1) hand, it is ideal to tap their hand with your (3) hand. Your opponent will be left with (4,1). If your opponent splits to (3,2) it is ideal tap out their (2) hand. Then your opponent will have no choice but to split to (2,1). Then, tap out their (2) hand. Your opponent has no choice but to tap your (2) hand. Then, it is ideal to split to (4,2). Your opponent will then tap your (4) hand, you split (1,1) and (1,1) against just one finger is an easy win. The way to win in that case will be described later on. If your opponent taps your (3) you tap their (4) hand, and you are left with (2,0) against (1,0). Your opponent will be forced to tap your (2), and you will be left with 3,0 against 1. This scenario will be described later on. If your opponent taps your (2) hand, you tap their (4) hand and you are left with (3,0). Your opponent will have to tap you again giving you (4,0) you win by tapping their hand. Image titled Always Win Chopsticks Step 4 4 From (3),(0) against (1,0) and it is your turn, it is ideal to split to (2,1). If your opponent taps your 2 hand, split to 2,2. Your opponent will then tap you leaving you with 3,2. Tap their hand with your 2. Your opponent will have no choice but to split to 1,2. From there tap out their 2 hand and your opponent will have to tap your two hand. Split to 4,2, your opponent will tap your 4 hand and you will be left with 2,0. From there you split (1,1) and (1,1) against just one finger is an easy win. The way to win in that case will be described later on. If your opponent taps your 1 hand, split to 3,1. Your opponent will have to tap your 1 hand. From 3,2, tap their hand with your 2. Your opponent will have no choice but to split to 1,2. From there tap out their 2 hand and your opponent will have to tap your two hand. Split to 4,2, your opponent will tap your 4 hand and you will be left with 2,0. From there you split (1,1) and (1,1) against just one finger is an easy win. The way to win in that case will be described later on. Image titled Always Win Chopsticks Step 5 5 From 1,1 against 1,0, (your opponent's turn) is one of the easiest wins in the game. Your opponent will tap one of your hands leaving you with 2,1. Split to 3,0, your opponent will be forced to tap your 3 leaving you with 4,0. Finally, you tap their hand to win the game. Image titled Always Win Chopsticks Step 6 6 Of course, there is always the person that thinks that they're tricky by starting out splitting to 2,0 as their first move. This situation, however, can be forced into a win as well. Once your opponent splits 2,0 tap their 2 hand (if you try to split 2,0 as well your opponent will just split back to 1,1 and it will go in a never-ending loop. From 3,0 your opponent has to split to 2,1, you then split to 2,0. Their only option is to hit you with their 2 hand to be 4,0. Now you split to 2,2. If your opponent taps one of your 2 hands with their 1, tap out their 2. Your opponent will then have no choice but to tap your 2. Then you will be left with 3,3. Split to 2,4, your opponent will tap out your 4, and then you split (1,1) and (1,1) against just one finger is an easy win. The way to win in that case is described in step 5. If your opponent taps one of your 2 hands with their 2, tap out their 2. Your opponent will then have no choice but to tap out your 4. From there, you split (1,1) and (1,1) against just one finger is an easy win. The way to win in that case is described in step 5.Log in to reply

Copied from – Prince Loomba · 2 months, 1 week ago

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Problem 6Imagine you have a piece of string long enough to stretch around the earth (40,074 km or 4,007,400,000 cm). Then you take an extra meter of string and add it to the string around the Earth. Now you spread this extra string around the Earth, supporting it somehow, so that the string forms a circle off the ground. How high off the ground would the string be? – Prince Loomba · 2 months, 1 week ago

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I am posting the solution myself since archit has told the answer – Prince Loomba · 2 months, 1 week ago

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– Archit Agrawal · 2 months, 1 week ago

1/2π mLog in to reply

– Prince Loomba · 2 months, 1 week ago

Show the solution and post the next problemLog in to reply

– Prince Loomba · 2 months, 1 week ago

Its approx?Log in to reply

Problem 4In the year 1995, A murder happened. Mr. Shalls was found dead on his bed. Famous detective Conan was called. He narrowed the suspects to Charlie, Gotham, Sharky, Hitler and Mrs. Shalls. A letter came to the government of state in which the murder happened. It was written 222---666---66---1---66; 444---7777; 8---44---33; 55---444---555---555---33---777; Can you find out who is the killer, given that The letter is 100% true? – Prince Loomba · 2 months, 1 week ago

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tap the numbers into keypad in phone you get the answer :) – Angela Fajardo · 2 months, 1 week ago

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– Hung Woei Neoh · 2 months, 1 week ago

The second word spells "hs"...Log in to reply

– Prince Loomba · 2 months, 1 week ago

Sorry for misprinting! I made that. Errors can be thereLog in to reply

– Archit Agrawal · 2 months, 1 week ago

I can't post any question now as I have to go, so someone else can post a question.Log in to reply

– Archit Agrawal · 2 months, 1 week ago

ConanLog in to reply

– Prince Loomba · 2 months, 1 week ago

Post the nextLog in to reply

Problem 2b(originally Problem 3, but now just a variant of Problem 2 because too much math):Same as Problem 2, but the coin is not necessarily fair; it gives heads with probability \(q\) where \(q \in (0, 1)\). (Note that \(q \neq 0, 1\).) And now the probability \(p\) can be any number in \([0, 1]\), too. – Ivan Koswara · 2 months, 1 week ago

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– Prince Loomba · 2 months, 1 week ago

This contest is going more towards mathematics!Log in to reply

Problem 2 :You want to make a choice between two options using a coin.

However, due to the nature of the options, you want your choice to be biased, that is you want the probability of choosing the first option to be \(p\) such that \(p \in (0,1)\) and \(2^n \times p \notin \mathbb{N} \quad \forall n \in \mathbb{N}\)

How do you do this? – Deeparaj Bhat · 2 months, 1 week ago

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Start with the interval \([0,1]\). Mark \(p\) in it. Whenever you toss a coin, divide your current interval into two halves; if it comes heads, take the left interval as your new interval, otherwise take the right one. For example, tossing heads, tails, tails, heads will cause your interval to go to \([0, 0.5], [0.25, 0.5], [0.375, 0.5], [0.375, 0.4375]\). If at any time the interval doesn't contain \(p\), stop; if the interval lies to the left of \(p\), choose the first one, otherwise choose the second one.

The expected number of coin tosses being 2 (at each step, there's half chance that you pick the interval not containing \(p\) and thus stop). To see why this works, imagine that we extend throwing the coins until we throw \(n\) times. We now have \(2^n\) possible intervals, and the coin tosses mean we pick one of them randomly. Only one of them contains \(p\), so as \(n \to \infty\), the probability that our interval will always contain \(p\) goes to 0. At any such \(n\), the number of intervals to the left of \(p\) is exactly \(\lfloor 2^n p \rfloor\); as \(n \to \infty\), the probability of picking an interval to the left of \(p\) is \(\frac{\lfloor 2^n p \rfloor}{2^n} \to p\). Together, this means we will eventually stop, and the probability we pick the first choice is \(p\). – Ivan Koswara · 2 months, 1 week ago

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your problem and this solution is more or less the same as what you'd posted in your question.

This question was basedFor Problem 2b, we use the result of the first problem as a sub routine to make the coin give two equally likely outputs and then use problem 2. This may not be optimal with respect to expectations.

I'd like to know your solution to problem 2b as well. – Deeparaj Bhat · 2 months, 1 week ago

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– Ivan Koswara · 2 months, 1 week ago

Problem 2b: The intervals are now open; endpoints are not included. Divide the interval so that the left portion to the right portion has ratio \(q : (1-q)\). If \(p\) falls on the division, then we stop after this toss (because then \(p\) is outside our interval in both cases). The expected number of tosses needed depends on \(p,q\), but it's finite.Log in to reply

@Ivan Koswara – Deeparaj Bhat · 2 months, 1 week ago

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Problem 1:Two men find an old gold coin and want to have a coin toss with it to decide who gets it. The only problem is the coin is heavier on one side so it comes up heads more than tails. What is a fair way for the men to toss the coin and decide who gets the coin? – Prince Loomba · 2 months, 1 week ago

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Solution to Problem 1 :Toss the coin twice.

To see why this is fair, note that the procedure ends only if the result is a permutation of H and T. As both permutations are equally likely, we're done. – Deeparaj Bhat · 2 months, 1 week ago

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– Prince Loomba · 2 months, 1 week ago

Post the secondLog in to reply

– Agnishom Chattopadhyay · 2 months, 1 week ago

Is the probability that a head comes up known?Log in to reply

– Prince Loomba · 2 months, 1 week ago

NoLog in to reply