Hardest Puzzles

Toughest Game Show Question

2012-01-09T00:36:00.000-08:00

Toughest Game Show Puzzle / Question

You are on a game show and there are three doors. The presenter tells you that behind one of doors there is a car and behind the other two are goats, if you pick the car you win it. After you have picked a door the presenter opens a different door with a goat behind it, he then gives you the chance to change what door you open, what should you do?

Solution

You want to change your decision. The initial decision has a 1 in 3 shot of being correct. Changing to the alternate door has a 2/3 chance of being correct. Think of it this way, if your initial pick is not the car (which has a 2/3 probability) then it must be that the alternate unopened door IS the car.

This is also known as the Monty Hall Problem.

Toughest Math Statement Question

2011-12-25T20:39:00.000-08:00

Toughest Math Statement Puzzle / Question

I am thinking of a 6-digit number. The sum of the digits is 43.

And only two of the following three statements about the number are true:

(1) it's a square number,
(2) it's a cube number, and
(3) the number is under 500000.

Solution
Discuss Below

Toughest Sherlock Holmes Question

2011-12-19T21:31:00.001-08:00

Toughest Sherlock Holmes Question

Sherlock, A detective who was mere days from cracking an international smuggling ring has suddenly gone missing. While inspecting his last-known location, you find a note:
710 57735 34 5508 51 7718
Currently there are 3 suspects: Bill, John, and Todd. Can you break the detective's code and find the criminal's name?

Solution
Bill. If you read the message upside down,
you'll notice that the numbers resemble letters and that those letters form legible sentences.
The message is 'Bill is boss. He sells oil.'

Solution By

Greentulip05 , Noone , Michael Beck

http://dailybrainteaser.blogspot.com/2011/10/sherlock-holmes-cipher-
puzzle.html

Toughest Probability Question

2011-12-12T03:22:00.000-08:00

Toughest Probability Puzzle / Question

How many people must be gathered together in a room, before you can be certain that there is a greater than 50/50 chance that at least two of them have the same birthday?

Solution
By Himanshi-Negi
&
http://dailybrainteaser.blogspot.com/2011/07/probability-of-having-same-birthday.html

Only twenty-three people need be in the room, a surprisingly small number. The probability that there will not be two matching birthdays is then, ignoring leap years, 365x364x363x...x343/365 over 23 which is approximately 0.493. this is less than half, and therefore the probability that a pair occurs is greater than 50-50. With as few as fourteen people in the room the chances are better than 50-50 that a pair will have birthdays on the same day or on consecutive days.

Toughest Popular Question

2011-12-04T20:59:00.001-08:00

Toughest Popular Puzzle / Question

Outside a room there are three light switches. One of switch is connected to a light bulb inside the room.
Each of the three switches can be either 'ON' or 'OFF'.

You are allowed to set each switch the way you want it and then enter the room(note: you can enter the room only once)

Your task is to then determine which switch controls the bulb ??

Solution

Set the first switches on for abt 10min, and then switch on the second switch and then enter the room.
Three cases are possible
1.Bulb is on => second switch is the ans
2.Bulb is off and on touching bulb , you will find bulb to be warm
=>1st switch is the ans.
3.Bulb is off and on touching second bulb , you will find bulb to be normal(not warm)
=>3rd bulb is the ans.

By Dumanshu Goyal &
http://dailybrainteaser.blogspot.com/2011/08/challenging-puzzle.html

Toughest Microsoft Interview Question

2011-11-28T09:27:00.000-08:00

Toughest Microsoft Interview Puzzle / Question - 29th November

The puzzle question is : On Bagshot Island, there is an airport. The airport is the homebase of an unlimited number of identical airplanes. Each airplane has a fuel capacity to allow it to fly exactly 1/2 way around the world, along a great circle. The planes have the ability to refuel in flight without loss of speed or spillage of fuel. Though the fuel is unlimited, the island is the only source of fuel.
What is the fewest number of aircraft necessary to get one plane all the way around the world assuming that all of the aircraft must return safely to the airport? How did you get to your answer?
Notes:
(a) Each airplane must depart and return to the same airport, and that is the only airport they can land and refuel on ground.
(b) Each airplane must have enough fuel to return to airport.
(c) The time and fuel consumption of refueling can be ignored. (so we can also assume that one airplane can refuel more than one airplanes in air at the same time.)
(d) The amount of fuel airplanes carrying can be zero as long as the other airplane is refueling these airplanes. What is the fewest number of airplanes and number of tanks of fuel needed to accomplish this work? (we only need airplane to go around the world)

Solution
By DaveDodson & WEBSITE

http://dailybrainteaser.blogspot.com/2011/07/aeroplane-hardest-quiz.html

As per the puzzle given above The fewest number of aircraft is 3!

Imagine 3 aircraft (A, B and C). A is going to fly round the world. All three aircraft start at the same time in the same direction. After 1/6 of the circumference, B passes 1/3 of its fuel to C and returns home, where it is refueled and starts immediately again to follow A and C.

C continues to fly alongside A until they are 1/4 of the distance around the world. At this point C completely fills the tank of A which is now able to fly to a point 3/4 of the way around the world. C has now only 1/3 of its full fuel capacity left, not enough to get back to the home base. But the first 'auxiliary' aircraft reaches it in time in order to refuel it, and both 'auxiliary' aircraft are the able to return safely to the home base.

Now in the same manner as before both B and C fully refuelled fly towards A. Again B refuels C and returns home to be refuelled. C reaches A at the point where it has flown 3/4 around the world. All 3 aircraft can safely return to the home base, if the refuelling process is applied analogously as for the first phase of the flight.

Toughest Picture Question

2011-11-21T20:42:00.000-08:00

Toughest Picture Puzzle(Question)

Below the four parts have been reorganized. The four partitions are exactly the same in both arrangements. Why is there a hole?

Solution

By Author
The gradient of the teal hypotenuse is different than the gradient of the red hypotenuse.

Toughest Brain Question

2011-11-14T20:36:00.000-08:00

Toughest Brain Twister Question - 14 November

2+3=8,
3+7=27,
4+5=32,
5+8=60,
6+7=72,
7+8=??
Solve it ?

Solution
Fastest (Amit Singh)
By Tomasz Budzeń (Best Explained)

n A B C=A-B C(n) - C(n-1) = 2

0 2
1 2 + 3 = 8 2*3 = 6 2 4
2 3 + 7 = 27 3*7 = 21 6 6
3 4 + 5 = 32 4*5 = 20 12 8
4 5 + 8 = 60 5*8 = 40 20 10
5 6 + 7 = 72 6*7 = 42 30 12
6 7 + 8 = ?? 7*8 = 56 (42)

(Pascal's difference machine)

A(n) =
B(n) + C(n) = a(n) * b(n) + C(n) = a(n) * b(n) + (n-1) * (n-2) = (n+1) * b(n) + n * (n+1)

So:

A(n) = (n + 1) * (b(n) + n) [it doesn't depend on a]

And:

A(6) = 7 * (8 + 6) = 7 * 14 = 98

Toughest Age Question

2011-11-06T05:27:00.000-08:00

Toughest Age Question - 6 November

Two old friends, Jack and Bill, meet after a long time.
Three kids
Jack: Hey, how are you man?
Bill: Not bad, got married and I have three kids now.
Jack: That’s awesome. How old are they?
Bill: The product of their ages is 72 and the sum of their ages is the same as your birth date.
Jack: Cool… But I still don’t know.
Bill: My eldest kid just started taking piano lessons.
Jack: Oh now I get it.

How old are Bill’s kids?

Solution

By Anirban Mukherji

3,3,8

Lets break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 sum(2, 2, 18) = 22
2, 4, 9 sum(2, 4, 9) = 15
2, 6, 6 sum(2, 6, 6) = 14
2, 3, 12 sum(2, 3, 12) = 17
3, 4, 6 sum(3, 4, 6) = 13
3, 3, 8 sum(3, 3, 8 ) = 14
1, 8, 9 sum(1,8,9) = 18
1, 3, 24 sum(1, 3, 24) = 28
1, 4, 18 sum(1, 4, 18) = 23
1, 2, 36 sum(1, 2, 36) = 39
1, 6, 12 sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Jack was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 sum(2, 6, 6) = 14
3, 3, 8 sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.

Toughest Logic Puzzle

2011-10-30T00:01:00.000-07:00

Toughest Logic Puzzle - 30 October

You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned.

The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.

You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned.

You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.

What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours?

Solution
Fastest Ans : Clive tooth

Best Ans : I <3 puzlzes & Clive Tooth

10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.

Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.

Here is how you would find one poisoned bottle out of eight total bottles of wine.

Bottle 1 Bottle 2 Bottle 3 Bottle 4 Bottle 5 Bottle 6 Bottle 7 Bottle 8
Prisoner A X X X X
Prisoner B X X X X
Prisoner C X X X X
In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.

With ten people there are 1024 unique combinations so you could test up to 1024 bottles of wine.

Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a millilitre from each bottle, they will only consume the equivalent of about one bottle of wine each.

Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combinations where all but one prisoner must sip from the wine. By avoiding these two types of combinations you can ensure no more than 8 prisoners die.

Google Interview Question

2011-10-22T07:25:00.000-07:00

Google Interview Question - 23 October

-> 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

Solution

Fastest Ans : Gerald Edgar

Best Explained Ans : Anmol

Toughest Cipher

2011-10-15T03:53:00.000-07:00

Toughest Cipher - 15 October
What does this message say?

G T Y O R J O T E O U I A B G T

Hint : Count the letters and try splitting the letters up into groups.

Solution
This type of code is known as a Caesar Box (Julius Caesar was the first to write codes this way.) To decipher the message, simply divide the code into four groups of four (you can also divide them into groups such as 5 groups of 5 or 6 groups of 6 depending on the number of letters in the phrase), and rearrange them vertically like this...G T Y O
R J O T
E O U I
A B G T

Fastest Ans : Manish Sehjpal

Best Ans : Sam

Toughest Maths Series

2011-10-10T10:52:00.000-07:00

Toughest Maths Series - 10 October

What are the next three numbers in this series?

4, 6, 12, 18, 30, 42, 60, 72, 102, 108, ?, ?, ?

Solution

First Correct Solution : Clive Tooth
Best Explained Solution : Lavesh Rawat

138 (137 and 139 are prime), 150 (149 and 151 are prime), 180 (179 and 181 are prime)

The series lists numbers that are flanked by two prime numbers.

4 (3 and 5 are prime)
6 (5 and 7 are prime)
12 (11 and 13 are prime)
18 (17 and 19 are prime)
30 (29 and 31 are prime)
42 (41 and 43 are prime)
60 (59 and 61 are prime)
72 (71 and 73 are prime)
102 (101 and 103 are prime)
108 (107 and 109 are prime)

thus

138 (137 and 139 are prime), 150 (149 and 151 are prime), 180 (179 and 181 are prime)