A Warden Killing The Boredom

A warden oversees an empty prison with 100 cells, all closed. Bored one day, he walks through the prison and opens every cell. Then he walks through it again and closes the even-numbered cells. On the third trip he stops at every third cell and closes the door if it’s open or opens it if it’s closed. And so on: On the nth trip he stops at every nth cell, closing an open door or opening a closed one. At the end of the 100th trip, which doors are open?

'These' special cells will have doors open!

Similar Puzzle! 

Open Doors in an Empty Prison!

Read the story behind the title!

Let's take few cells into consideration as representatives.

The warden visits cell no. 5 twice - 1st & 5th trip.1 & 5 are only divisors of 5.

He visits cell no. 10, on 1st, 2nd, 5th, 10th trips i.e. 4 times. Here, divisors of 10 are 1,2,5,10.

He stops cell no. 31 only twice i.e. on 1st and 31st trip.

He visits cell no.25 on 1st, 5th, 25th trips that is thrice.

In short, number of divisors that cell number has, decides the number of visits by warden.

For example, above, cell no.5 has 2 divisors hence warden visits it 2 times while 10 has 4 divisors which is why warden visits it 4 times.

But when cell number is perfect square like 16 (1,2,4,8,16) or 25 (1,5,25) he visits respective cells odd number of times. Otherwise for all other integers like 10 (1,2,5,10) or like 18 (1,2,3,6,9,18)  or like 27 (1,3,9,27) he visits even number of time.

For prime numbers, like 1,3,5,....31,...97 he visits only twice as each of them have only 2 divisors. Again, number of visits is even.

Obviously, the doors of cells to which he visits even number of times will remain closed while those cells to which he visits odd number of time will remain open.

So the cells having numbers that are perfect square would have doors open. That means cells 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 would be having their doors open.

The Little Johnny's Dilemma

Little Johnny is walking home. He has $300 he has to bring home to his mom. While he is walking a man stops him and gives him a chance to double his money. 

The man says -

 "I'll give you $600 if you can roll 1 die and get a 4 or above, you can roll 2 dice and get a 5 or 6 on at least one of them, or you can roll 3 dice and get a 6 on at least on die. If you don't I get your $300."
 
What does Johnny do to have the best chance of getting home with the money?

THIS is the BEST he can do! 

The Loss Making Deal !

What was the deal?


The little Johnny should not accept the deal.

The problem is actually finding the probability of winning in each case dice throw.

---------------------------------------------

CASE 1 :  

Condition - Roll a die and get 4 or above.

The probability in the case is 3/6 = 1/2 (getting 4,5,6 from 6 possible outcomes). 

That is only 50% chances of winning the deal.

---------------------------------------------

CASE 2 :

Condition - Roll 2 dice and a 5 or 6 on at least one of them.

Let's find the probability that none get a 5 or 6. 

Probability that single die doesn't get a 5 or 6 is 4/6 = 2/3 by getting 1,2,3,4 in possible 6 outcomes.

Since, these 2 results are independent, the probability that neither dice gets a 5 or 6 is 2/3 x 2/3 = 4/9.

So, the probability that at least one of them gets a 5 or 6 is 1 - 4/9 = 5/9.

That is, about 56% chances of winning the deal.

By other approach, there are 36 possible combinations in 2 dice throw.

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) 

There are total 20 combinations (last 2 rows and 2 columns) where at least one die gets a 5 or 6.

Hence, the probability in the case is 20/36 = 5/9 i.e. about 56% of winning chances.

----------------------------------------

CASE 3 : 

Condition - Roll three dice and get a 6 on at least on die.  

Again finding probability that none gets a 6.

Probability that single die doesn't get a 6 is 5/6 by getting 1,2,3,4,5 out of 6 possible outcomes.

Hence, the probability that three dices doesn't get a 6 = 5/6 x 5/6 x 5/6 = 125/216.

Then the probability that at least one of them get a 6 = 1 - 125/216 = 91/216 

That is only 42% chances of winning the deal.

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On the other hand Johnny can be 100% sure that he will take $300 to home if he doesn't accept the man's deal. 

The Challenge Ahead of New Manager

In a wood-cutting factory, five large sawing machines stand in a windowless room. Each machine has an on/off switch attached, there being no doubt as to which switch controls which machine.

Outside the door to the room are five back-up on/off switches, one for each machine inside. The power for each machine must first pass through the back-up switch, and then the machine switch before reaching the saw.

The problem is, the new manager cannot decide how these back-up switches match with the machines inside the room. One day, the manager's brother visits. The manager takes him inside the sawing room where all five machines are at work and explains the problem. The brother announces that he intends to leave the room and that when he returns he will be able to match correctly the five switches outside the room to the five machines inside. The brother works alone, cannot see the machines from outside the room and solves the problem purely by operating switches. 

How is it possible?

Read how brother managed to do it! 

Genius Moves by Brother of Manager!

What was the challenge?

Suppose you are the brother of that manager & have accepted the challenge. 

For sake of convenience, we will name switches inside the room operating machines as an 'operating switches' & those which are outside the room as a 'back up switches'.

Though windowless, we assume that operating sound of machine(s) can be heard from outside room but can't see which machine(s) is (are) in operating condition. 

Here is what you should do!

1. Let's label the machines which we are going to keep ON as ACTIVE & others as INACTIVE. 

2. Turn off operating switches of 2 machines so that there are three machine ACTIVE and two are INACTIVE inside the room.

3. Go outside and now your task is to find the 3 back up switches controlling 3 active machines inside the room.

4. You have to switch off three switches in all possible combinations. There are 10 such possible combinations (5C3) of 3 controlling switches out of 5.

5. If 0 represents OFF position and 1 represents ON position then you should try all possible below combination. 

11000
10100
10010
10001
01100
01010
01001
00110
00101
00011

6. There will be exactly one combination in which all three ACTIVE machines  will be OFF & there will be no sound coming out of the room. The three switches having value of 0 are controlling ACTIVE machines while other 2 must be controlling INACTIVE machines.

7. Still we don't know the exact switch operating the each machine. Label 3 switches controlling ACTIVE machines as A, B and C & those controlling INACTIVE machines as D & E. 

Allow some time to cool down all those ACTIVE machines.

8. Now, turn on 2 switches A & B and keep switch E ON. Machines connected to A and B will start working.

9. After a while, turn off the switch B and go inside the room. 

10. The machine which is still operating must be controlled by back up switch A.

11. Touch other 2 which were labelled as ACTIVE & check which has got warmer.

12. The machine which is warmer must be controlled by back up switch B.

13. And since we didn't turn on the switch C (after giving time all to cool down), the machine having normal temperature must be controlled by the switch C.

14. Remember, we had turned off operating switches of INACTIVE machines initially. And before entering into room again we have turned on switch E. 

15. Now, turn on operating switch of one of the INACTIVE labelled machine. 

If the machine starts working it must be connected to the switch E & other to the switch D. And if the machine doesn't start working it must be connected to D & other to E. 

This way, you will find every back up switch located outside the room controlling operating switch of each machine inside the room.

 

How much does each container weigh?

There are five containers of oil of different weight. They are weighed in pairs of two with all possibilities. The weights in kgs are 165, 168, 169.5, 171, 172.5, 174, 175.5, 177, 180, and 181.5 . 

How much does each container weigh?

Weight Calculation of each container is here! 

Weight Calculation of Each Container

What was the given data?

Let's name five containers as C1, C2, C3, C4 and C5 with their increasing order of weights. That is weight of C1 <= C2 <= C3 <= C4 <= C5.


As per given data, weight of these containers when weighed in pairs -

165, 168, 169.5, 171, 172.5, 174, 175.5, 177, 180, and 181.5 . Here, weights too are in the increasing order.

It's quite obvious that the the container having lowest weight and the second lowest weight will weigh lowest in all possible pairs.That is 165 kg must be weight of pair C1-C2.

C1 + C2 = 165  

And the next higher pair weight of 168 kg must be involving lowest weighing container C1 and third lowest weighing container C3.

C1 + C3 = 168

Similarly, C4 and C5 together must weigh highest when weighed in pair.

C4 + C5 = 181.5

And same way, C3 and C5 makes second highest weighing pair.

C3 + C5 = 180.

It's quite clear that each of the container is weighed 4 times as each of them appearing in 4 pairs.

So, if weight all above pairs are added then, 

4 ( C1 + C2 + C3 + C4 + C5 ) = 165+168+169.5+171+172.5+174+175.5+177+180+181.5= 1734 kg


( C1 + C2 + C3 + C4 + C5 ) = 433.5 kg     .......(1)

Putting C1 + C2 = 165 and C4 + C5 = 181.5 in above,

165 + C3 + 181.5 = 433.5 

C3 = 87 kg.

Since, C1 + C3 = 168, 

C1 = 168 - 87

C1 = 81 kg.

Also, we have, C3 + C5 = 180

C5 = 180 - 87

C5 = 93 kg.

Since, C4 + C5 =  181.5 

C4 = 181.5 - 93 = 88.5 kg. 

Again, C1 + C2 =  165

C2 = 165 - 81 = 84 kg.

Hence, the weights of five containers are 81, 84, 87, 88.5 and 93 kg.

What day of the week is it?

A group of campers have been on vacation so long, that they've forgotten the day of the week. 

The following conversation ensues. 

Darryl: "What's the day? I don't think it is Thursday, Friday or Saturday." 

Tracy: "Well that doesn't narrow it down much. Yesterday was Sunday." 

Melissa: "Yesterday wasn't Sunday, tomorrow is Sunday." 

Ben: "The day after tomorrow is Saturday." 

Adrienne: "The day before yesterday was Thursday." 

Susie: "Tomorrow is Saturday." 

David: "I know that the day after tomorrow is not Friday." 

If only one person's statement is true, what day of the week is it?

So it was THIS day of the week!

The Forgotten Day of Week is Wednesday!

What was the puzzle?

Let's see once again the conversation that campers had - 

--------------------------------------------------------


Darryl: "What's the day? I don't think it is Thursday, Friday or Saturday." 

Tracy: "Well that doesn't narrow it down much. Yesterday was Sunday." 

Melissa: "Yesterday wasn't Sunday, tomorrow is Sunday." 

Ben: "The day after tomorrow is Saturday." 

Adrienne: "The day before yesterday was Thursday." 

Susie: "Tomorrow is Saturday." 

David: "I know that the day after tomorrow is not Friday."

-----------------------------------------------------

Let's see what day statement of each is suggesting -

Darryl - Sunday, Monday, Tuesday, Wednesday.

Tracy - Monday.

Melissa - Saturday.

Ben - Thursday.

Adrienne - Saturday.

Susie - Friday.

David - Sunday, Monday, Tuesday, Thursday, Friday or Saturday.  

If we assume David's statement is TRUE then one of statements of Darryl (Sunday, Monday, Tuesday are common) or Tracy (Monday is common) or Melissa & Adrienne (Saturday is common) or Susie (Friday is common) or Ben (Thursday is common) has to be also TRUE. But this is against the given data that only 1 of the statement is TRUE.

Hence, David's statement must be FALSE and the only day that isn't pointed by David is Wednesday.

So the day must be Wednesday as suggested correctly by Darryl and thereby making statements of every other camper including David FALSE. 

Crack The 10-Digits Password

In an attempt to further protect his secret chicken recipe, Colonel Sanders has locked it away in a safe with a 10-digit password.  Unable to resist his crispy fried chicken, you'd like to crack the code. 

Try your luck using the following information:

All digits from 0 to 9 are used exactly once.

No digit is the same as the position which it occupies in the sequence.

The sum of the 5th and 10th digits is a square number other than 9.

The sum of the 9th and 10th digits is a cube.

The 2nd digit is an even number.

Zero is in the 3rd position.

The sum of the 4th, 6th and 8th digits is a single digit number.

The sum of the 2nd and 5th digits is a triangle number.

Number 8 is 2 spaces away from the number 9 (one in between).

The number 3 is next to the 9 but not the zero.

Click here to know how to CRACK it! 

Cracking The 10-Digits Password

What was the challenge? 

Let's suppose the 10-digits password is ABCDEFGHIJ.

---------------------------------

Take a look at the given clues to crack down the password.

1. All digits from 0 to 9 are used exactly once.

2. No digit is the same as the position which it occupies in the sequence.

3. The sum of the 5th and 10th digits is a square number other than 9.
 
4. The sum of the 9th and 10th digits is a cube.
 
5. The 2nd digit is an even number.
 
6. Zero is in the 3rd position.
 
7. The sum of the 4th, 6th and 8th digits is a single digit number.
 
8. The sum of the 2nd and 5th digits is a triangle number.
 
9. Number 8 is 2 spaces away from the number 9 (one in between).
 
10. The number 3 is next to the 9 but not the zero. 

--------------------------------------

STEPS : 

Respective Clues are in ().

1] Since, every digit is used only once (1), the sum of 2 digits can't exceed 
9 + 8 = 17 and minimum sum of 2 digits can be only 0 + 1 = 1.

2] So, the sum of 5th & 10th digits which is square (3) can be - 1, 4, 16.
The sum of 9th & 10th digits which is cube (4) can be - 1, 8.
The sum of 2nd & 5th digits which is triangle number (8) can be - 1, 3, 6, 10, 15.

3] As per (6), 0 is already at third position i.e. C = 0. Hence, 2nd, 5th, 9th or 10th digits can't be 0. Therefore, the sum of 5th & 10th or 9th & 10th or 2nd & 5th can't be 1. Revising list of possible values of those in step 2.

The sum of 5th & 10th digits which is square (3) can be -  4, 16.
The sum of 9th & 10th digits which is cube (4) can be -  8.
The sum of 2nd & 5th digits which is triangle number (8) can be -  3, 6, 10, 15.

4] Possible pairs of 9th & 10th digits - (1,7), (7,1), (2,6), (6,2), (3,5), (5,3)

If IJ = 71, then 5th digit E must be 4 - 1 = 3 & possible values of 2nd digit B are 0,3,7.
But as per (5), B must be even & C = 0 already.

IJ = 26 or 62 or 35 doesn't leave any valid value for 5th digit E.

If IJ = 53, then 5th digit E must be 1 & possible values of 2nd digit B are 2,5,9
That is B = 2 (5). But as per (2), no digit is the same as the position which it occupies in the sequence. So, the number 2 can't be at 2nd position where B is there.

Hence, IJ must be 17 i.e. I = 1, J = 7.

5] This leaves only 9 as valid value for 5th digit E so that (3) is true. 

So, E = 9 and hence B = 6 with sum of 2nd digit B and 5th digit E as 15.

6] So far, we have, B = 6, C = 0, E = 9, I = 1 and J = 7. 

As per (10), the number 3 is next to the 9 but not the zero. Hence, the 6th digit F must be 3. F = 3.

7] As per (9), the 8 is 2 spaces away from 9 i.e. here 5th digit E=9. With 3rd position already occupied by C = 0, the digit 8 must be at 7th position. 
So, G = 8. 

8] As per (7), D + F + H must be single digit. We have, F = 3 already, 
so possible values of D + H are 0, 1, 2, 3, 4, 5, 6.

D + H can't be 0 (0+0), 1 (1+0), 2(1+1/2+0), 3(1+2/3+0), 4(2+2/3+1/4+0), 5(2+3/4+1/5+0) since digits are repeated with C=0, I=1, F=3 already.

Hence, D + H = 6. 

9] Now, D = 5 and H = 1 is not possible. Also, D/H can't be 0 or 6. Both D and H can't be 3 at the same time. D can't be 4 as D is at 4th position. 

Hence, D = 2, H = 4.

10] With B = 6, C = 0, D = 2, E = 9, F = 3, G = 8, H = 4, I = 1, J = 7, the only digit left for the first position A is 5. So A = 5.

CONCLUSION : 

A = 5, B = 6, C = 0, D = 2, E = 9, F = 3, G = 8, H = 4, I = 1, J = 7.

So, the 10-digits password ABCDEFGHIJ is 5602938417.

 

The Candle Light Study

Power went off when Vipul was studying. It was around 2:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last four hours and the thin one one hours less. When he finally went to sleep, the thick candle was twice as long as the thin one.

For how long did Vipul study in candle light?

Well, he studied in candle light for....hours. Know here! 

The Candle Light Studying Hours

What was the question?

It's clear that, the thick candle lasts for 4 hours and thin one lasts only 3 hours.

If L is length of these candles (they are of equal length) then thick candle burns L/4 per hour and thin one burns L/3 per hour.

Let's assume Vipul studied for X hours.

In X hours, amount of thick candle burnt = XL/4

 

In X hours, amount of thin candle burnt = XL/3

 

After X hours, amount of thick candle remaining = L - XL/4

 

After X hours, amount of thin candle remaining = L - XL/3

 

Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep.

 

(L - XL/4) = 2(L - XL/3 )

 

L(1 - X/4) = 2L (1 - X/3)

(1 - X/4) = 2(1 - X/3)

 

(4 - X)/4 = 2(3 - X)/3

3(4 - X) = 8(3 - X)

12 - 3X = 24 - 8X

5X = 12

X=12/5

 

Converting 12/5 hours into minutes as X = 12/5 * 60 = 144 minutes.

 

Hence, Vipul must have studied for 2 hours and 24 minutes.

 

 

  

The Toaster Twist!

Jasmine has a toaster with two slots that toasts one side of each piece of bread at a time, and it takes one minute to do so.

If she wants to make 3 pieces of toast, what is the least amount of time she needs to toast them on both sides?

She needs only....minutes. Click to know. 

Smarter Use of Toaster!

What was the task given?

Jasmine has 3 pieces of bread and she want to toast total 6 sides of toasts. Since, the toaster can toast 2 sides (in two slots) at a time in 1  minute, she needs only 3 minutes to toast all her 3 pieces. All that she need to do use the toaster smartly.

Let A-B, C-D and E-F be the names of sides of toast that Jasmine has.

1. She puts 2 pieces - A-B and C-D in the two slots of toaster so that 2 sides say A and C are toasted but B and D remains untoasted. 1 minute used in the process.

2. For next minute, she takes out 1 piece say C-D and puts E-F in toaster while she flips A-B so that B is toasted. With that, B and E are toasted in this minute.

3. Now, only F and D are remain untoasted. In third minute, using 2 slots of toaster she toasts these 2 sides. 

Who Lives Where?

There are 4 big houses in my home town.

They are made from these materials: red marbles, green marbles, white marbles and blue marbles. 

Mrs Jennifer's house is somewhere to the left of the green marbles one and the third one along is white marbles.

Mrs Sharon owns a red marbles house and Mr Cruz does not live at either end, but lives somewhere to the right of the blue marbles house. 

Mr Danny lives in the fourth house, while the first house is not made from red marbles. 

Who lives where, and what is their house made from ?

HERE is step by step process to know who lives where! 

Location and Owner of Each House

What are the hints given?

Take a look at the given details once again - 

-----------------------------------------------------------------

1. Mrs Jennifer's house is somewhere to the left of the green marbles one.

2. The third house along is white marbles.

3. Mrs Sharon owns a red marbles house.

4. Mr Cruz does not live at either end.

5. Mr Cruz lives somewhere to the right of the blue marbles house. 

6. Mr Danny lives in the fourth house, 

7. The first house is not made from red marbles.

--------------------------------------------

Let's make a table of location, owner and material used for house as below and then fill up as per given details.

STEPS : 

---------------------------------------------

1.As per clue no. 6 and 7, we have,

-----------------------------------------------------------------

2. As per clue no.2, we have,

-----------------------------------------------------------------

3. The clue no.3 suggests, Mrs Sharon owns a red marbles house. The only location available for this combination is second house. Hence,

-----------------------------------------------------------------

4. The clue no.4 indicates that Mr Cruz does not live at either end. 

So, the only place that is not at the either end is house no.3 made up of white marbles. Hence, Mr. Cruz must be living at this house.

----------------------------------------------------------------

5. As per clue no.1, Jennifer needs to be somewhere to the left of the green marbles one. 

That green house has to be the fourth house in row and Jennifer must be living in the first house made up of blue marbles for above clue to be true.

-----------------------------------------------------------------

Hidden Wallet of Question Mark

After recent events, Question Mark is annoyed with his brother, Skid Mark. Skid thought it would be funny to hide Question's wallet. 

He told Question that he would get it back if he finds it. So, first off, Skid laid five colored keys in a row. One of them is a key to a room where Skid is hiding Question's wallet. 

Using the clues, can you determine the order of the keys and which is the right key? 

Red: This key is somewhere to the left of the key to the door. 

Blue: This key is not at one of the ends. 

Green: This key is three spaces away from the key to the door (2 between).

Yellow: This key is next to the key to the door. 

Orange: This key is in the middle.

Logical Steps to know the RIGHT key are here!

Key to Get Question Mark's Wallet

Why to find Question Mark's Wallet?

Let's take a look at the given clues once again - 

1.Red: This key is somewhere to the left of the key to the door. 

2.Blue: This key is not at one of the ends. 

3.Green: This key is three spaces away from the key to the door (2 between). 

4.Yellow: This key is next to the key to the door.

5.Orange: This key is in the middle. 

Now, let's draw a table of locations and color of keys as below - 

From Clue No.5, it's clear that the orange key is in the middle.

As per Clue No.3, the green key is three spaces away from the key to the door (right key) (2 between). So, the possible positions of green and right keys
are -

Since as per Clue No.4, yellow key is next to the key to the door, the key to the door can't be positioned at 2 or 5. Hence, it might be positioned at 1 or 4 with Green key at 4 or 1.

But as per Clue No.1, red key is somewhere to the left of the key to the door. For that, the right key can't be positioned at 1 as in that case, no position would be left for the red key as per clue.

Hence, Green key must be at 1 and right key at 4.

 
And now as per Clue No. 4, yellow key is next to right key.

As per Clue No.1, red key is somewhere left to the right key,hence it must be at position 2.

And as per Clue No.2, blue isn't at either end. Rather blue key is the only key left. So, the position left for it is position 4 which is of Right key. Hence, Blue key itself is RIGHT key to the door of room where Question Mark's wallet is placed.

Bags of Marbles - Puzzle

There are three bags, each containing two marbles. Bag A contains two white marbles, Bag B contains two black marbles, and Bag C contains one white marble and one black marble. 

You pick a random bag and take out one marble, which is white. 

What is the probability that the remaining marble from the same bag is also white?

Here is the solution!