ToughNut

The Camel & Bananas

noreply@blogger.com (Freddo) — Tue, 20 Mar 2012 11:00:00 +0000

You want to transport 3,000 bananas across 1,000 kilometers. You have a camel that can carry 1,000 bananas at most. However, the camel must eat 1 banana for each kilometer that it walks. What is the largest number of bananas that can be transported?

Can you solve it?

For extra credit, try deriving a general formula if you have B bananas, need to travel D distance, and the camel has a carrying capacity of C.

Source: mindyourdecisions.com

Card Cash

noreply@blogger.com (Freddo) — Wed, 22 Sep 2010 18:33:00 +0000

Someone offers you the following deal:

There is a deck of 100 initially blank cards. The dealer is allowed to write ANY positive integer, one per card, leaving none blank. You are then asked to turn over as many cards as you wish. If the last card you turn over is the highest in the deck, you win; otherwise, you lose.

Winning grants you $50, and losing costs you only the $10 you paid to play.

Would you accept this challenge?

29 hostages and a terrorist

noreply@blogger.com (Freddo) — Tue, 21 Sep 2010 17:45:00 +0000

29 hostages are captured by a terrorist. They are told, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another."

"There is an isolated switch room here, which contains two light switches labelled A and B, each of which can be in either the 'on' or the 'off' position. I am not telling you their present positions. The switches are not connected to anything."

"After today from time to time whenever I feel so inclined, I will select one hostage at random and escort him to the switch room. This hostage will select one of the two switches and reverse its position. He must move one, and only one of the switches. He can't move both and he can't move none either. Then he'll be led back to his cell."

“I will then take the next hostage there, and he'll be instructed to do the same thing. I am going to choose hostages at random. I may choose the same guy three times in a row, or I may jump around and come back."

"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. "No one else will enter the switch room until I lead them."

"At any time anyone of you may declare to me, 'We have all visited the switch room.' and be 100% sure. "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."

What is the strategy they come up with so that they can be free?

Source: www.allinterview.com

Heads I win, Tails I lose.

noreply@blogger.com (Aaditya) — Tue, 19 Jan 2010 13:45:00 +0000

Akshit took out a coin of his pocket and said to me, 'Heads I win, Tails I lose. I bet half the money in my pocket.'
He tossed and lost. And the game continued for number of times, each time betting half the money in his pocket. We don't remember how many times the coin was tossed or how long the game went, but he lost exactly the same no. of times as he won the bet.
What do you think, did he, on the whole, gain or lose?

Number of Triangles.

noreply@blogger.com (Aaditya) — Sun, 17 Jan 2010 11:34:00 +0000

Take a good look at the figure below:

Count the no. of triangles in this figure.

The Cereal Box Surprise

noreply@blogger.com (Aaditya) — Sat, 21 Nov 2009 21:25:00 +0000

Suppose a box of cereal costs 5$, and each box has a toy in it. There are 5 different toys for you to collect; by collecting all of them you can assemble them together and create a giant robot. If the toys have equal probabilities of turning up - that is, each toy is 1/5 likely to appear in a randomly chosen cereal box - how much will you have to spend, on average, before you can assemble the giant robot of your dreams?

Splitting Problem

noreply@blogger.com (Aaditya) — Sat, 21 Nov 2009 21:04:00 +0000

How do you cut a rectangular cake into two equal pieces with one straight cut when someone has already removed a rectangular piece from it? (The removed piece can be of any size or any orientation.)

consider these images:

An Ant and a Cube

noreply@blogger.com (Freddo) — Sat, 21 Nov 2009 02:47:00 +0000

An ant starts eating a 3*3 rubik's cube made up of cheese at a corner(vertex). What is the probability that the last cube it eats is the body-center cube?

The ant can only travel from a cube to the adjacent cubes (i.e. having common faces)

Courtesy: Nitin Basant

Monty Hall problem a.k.a. The 3 door problem

noreply@blogger.com (Freddo) — Wed, 18 Nov 2009 14:11:00 +0000

Although I'm sure most of the ToughNut readers are familiar with this problem but I've met a lot of people with great aptitude who seem to have all sorts of confusion and disagree with the solution. Lets discuss and debate about the conflicting opinions that we all have. Here it goes...

Monty Hall problem

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Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

5 Pirates Puzzle

noreply@blogger.com (Freddo) — Wed, 18 Nov 2009 05:42:00 +0000

Perhaps the most common of all math/logic puzzles being discussed in forums on the internet, yet an interesting one. Here it goes...

There are five rational pirates, A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them.

The Pirates have a strict order of seniority: A is superior to B, who is superior to C, who is superior to D, who is superior to E.

The Pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins. The pirates, including the proposer, then vote on whether to accept this distribution. If the proposed allocation is approved by a majority or a tie vote, it happens. If not, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again.

Pirates base their decisions on three factors. First of all, each pirate wants to survive. Secondly, each pirate wants to maximize the number of gold coins he receives. Thirdly, each pirate would prefer to throw another overboard, if all other results would otherwise be equal

Source: Stewart, Ian (1999-05), "A Puzzle for Pirates", Scientific American: 98–99

Information for puzzle solvers!!

noreply@blogger.com (Freddo) — Mon, 16 Nov 2009 03:16:00 +0000

Hi puzzlers,

The very idea of ToughNut is to bring in as many challenging and brainteasing puzzles as possible and share them with the followers of this blog. To take the idea a step ahead I invite you all to participate in collaborative publishing. I encourage you to email puzzles directly to itsfreddo.toughnut@blogger.com and have them published on this blog. I would strongly encourage all contributors to mention their names in the end of the email so that they are fairly credited for their contribution to the blog.

I cannot guarantee that every puzzle sent to this email id shall be published. However, the puzzles fairly suiting the interests of the niche community of readers that this blog has shall definitely be published.

Warm regards,

Ankit

The Scared Guards Problem

noreply@blogger.com (Freddo) — Mon, 16 Nov 2009 02:59:00 +0000

561 security guards are positioned so that no two pairs of guards are the same distance apart. Every guard watches the guard closest to him. Is there an arrangement of the guards so that every guard is being watched?

Source: http://www.cs.rpi.edu/~magdon/miscellaneous/puzzles/puzzles.html

5 cards magic trick

noreply@blogger.com (Freddo) — Mon, 16 Nov 2009 02:48:00 +0000

Two magicians, John and Hull, perform a trick with a shuffled deck of cards, jokers removed. John asks a member of the audience to select five cards at random from the deck. The audience member passes the five cards to john, who examines them, and hands one back. John then arranges the remaining four cards in some way and places them face down, in a neat pile.
Hull, who has not witnessed these proceedings, then enters the room, looks at the four cards, and determines the missing fifth card, held by the audience member. How is this trick done?

Note: The only communication between John and Hull is via the arrangement of the four cards. There is no encoded speech or hand signals or ESP, no bent or marked cards, no clue in the orientation of the pile of four cards...

6 people in a group

noreply@blogger.com (Freddo) — Sat, 14 Nov 2009 23:02:00 +0000

Prove that in a group of six people, there will always be three people that are mutual friends or mutual strangers. (Assume that “friend” is symmetric-if x is a friend of y, then y is a friend of x.)

Cocktail Party

noreply@blogger.com (Freddo) — Sat, 14 Nov 2009 22:58:00 +0000

Prove that in any cocktail party with two or more people, there must be at least two people who have the same number of friends. (Assume that "friend" is symmetric-if x is a friend of y, then y is a friend of x.)

Hint: Use pigeonhole principle

Bhaddo, Tawar and KT (Tough)

noreply@blogger.com (Freddo) — Fri, 13 Nov 2009 22:55:00 +0000

Mr. Bhaddo choses two different numbers greater than N but less than M & tells their sum to Mr. Tawar and their product to Mr. KT. The following conversation ensues:

Mr. Tawar: I cannot determine the two numbers.

Mr. KT: I cannot determine the two numbers either.

Mr. Tawar: I still cannot determine the two numbers.

Mr. KT: Now I can determine the two numbers.

Mr. Tawar: Now I can determine the two numbers also.

Find the greatest value of M for which this puzzle has a unique solution, for N=1, N=2 and N=3.

3 Families

noreply@blogger.com (Freddo) — Thu, 12 Nov 2009 21:18:00 +0000

Three families make a remarkable discovery. The sum of the ages of their members are all the same, the sum of the squares of the ages of their members are all the same, and the sum of the cubes of the ages of their members are all the same. Everyone in all 3 families has a different age, and nobody is more than 100 years old.

What is the smallest possible sum of their ages? Can this be done with 4 families?

Red-eyed monks and brown-eyed monks on an island?

noreply@blogger.com (Freddo) — Thu, 12 Nov 2009 15:59:00 +0000

There are 1000 monks living on an island, some with brown eyes and some with red eyes. Monks who have red eyes are cursed, and are supposed to commit suicide at midnight. However, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each monk can (and does) see the eye colors of all other monks, but has no way of discovering their own (there are no reflective surfaces).

All the monks are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).

Of the 1000 monks, it turns out that 100 of them have red eyes and 900 of them have brown eyes, although the monks are not initially aware of these statistics (each of them can of course only see 999 of the 1000 monks).

Life goes on, with brown-eyed monks and red-eyed monks living happily together in peace, and no one ever committing suicide. Then one day a tourist visits the island monastery, and not knowing that he's not supposed to talk about eyes, he states the observation "At least one of you has red eyes." Having acquired this new information, what effect, if anything does this have?

Extension of 2 eggs problem

noreply@blogger.com (Freddo) — Wed, 11 Nov 2009 16:11:00 +0000

How do you solve the problem at the link below for 3 eggs?

http://tough-nut.blogspot.com/2009/11/2-eggs-problem.html

How do you do it for k eggs??

B'day twins problem

noreply@blogger.com (Freddo) — Wed, 11 Nov 2009 14:02:00 +0000

Sheila and He-Man are twins; Sheila is the OLDER twin. Assume they were born immediately after each other, an infinitesimally small - but nonzero - amount of time apart. During one year in the course of their lives, Sheila celebrates her birthday two days AFTER He-Man does. How is this possible?

Bonus: What is the maximum amount of time by which Sheila and He-Man can be apart in their birthday celebrations during the same year?

Note: For both Sheila and He-Man, these birthday celebrations happen on the actual birthday date -- it cannot be a celebration that occurs at a date earlier or later than the actual birthday date for whatever reasons of convenience. Also, the solution has nothing to do with the theory of relativity or any other over complicated nonsense like that.

Globe Traversal Problem

noreply@blogger.com (Freddo) — Wed, 11 Nov 2009 13:54:00 +0000

how many places are there on the earth that one could walk one mile south, then one mile west, then one mile north and end up in the same spot? to be precise, let's assume the earth is a solid smooth sphere, so oceans and mountains and other such things do not exist. you can start at any point on the sphere. also, the rotation of the earth has nothing to do with the solution; you can assume you're walking on a static sphere if that makes the problem less complicated to you.

Five selfish women, a monkey, and some coconuts (2 star)

noreply@blogger.com (Aaditya) — Tue, 10 Nov 2009 18:19:00 +0000

Five women crash-land their airplane on a deserted island in the South Pacific. On their first day they gather as many coconuts as they can find into one big pile. They decide that, since it is getting dark, they will wait until the next day to divide the coconuts.

That night each woman took a turn watching for rescue searchers while the others slept. The first watcher got bored so she decided to divide the coconuts into five equal piles. When she did this, she found she had one remaining coconut. She gave this coconut to a monkey, took one of the piles, and hid it for herself. Then she jumbled up the four other piles into one big pile again.

To cut a long story short, each of the five selfish women ended up doing exactly the same thing. They each divided the coconuts into five equal piles and had one extra coconut left over, which they gave to the monkey. They each took one of the five piles and hid those coconuts. They each came back and jumbled up the remaining four piles into one big pile.

What is the smallest number of coconuts there could have been in the original pile?

P.S. Introducing stars...according to the difficulty of problem on the scale of 1 to 5.

Bulbs-Switches matching problem

noreply@blogger.com (Freddo) — Tue, 10 Nov 2009 15:09:00 +0000

There are 1000 bulbs in a room & the switches for these are in another room arranged in a random fashion.You have to find an optimum strategy to match the bulbs to corresponding switches so that the number of times u enter into the room to look at the bulbs is minimum.

25 cards in a dark room

noreply@blogger.com (Freddo) — Tue, 10 Nov 2009 11:43:00 +0000

A deck of 25 cards, 14 of which are facing up & 11 are facing down, is lying on a table in a dark room. You are asked to go in that room and split the deck into two such that total number of cards facing up in each deck are equal. How do you do that?

2 eggs problem

noreply@blogger.com (Freddo) — Tue, 10 Nov 2009 11:34:00 +0000

* You are given 2 eggs.
* You have access to a 100-storey building.
* Eggs can be very hard or very fragile means it may break if dropped from the first floor or may not even break if dropped from 100 th floor.Both eggs are identical.
* You need to figure out the highest floor of a 100-storey building an egg can be dropped without breaking.
* Now the question is how many drops you need to make. You are allowed to break 2 eggs in the process