The Fearsome Logical Challenge

You and your two friends Pip and Blossom are captured by an evil gang of logicians. In order to gain your freedom, the gang’s chief, Kurt, sets you this fearsome challenge.

The three of you are put in adjacent cells. In each cell is a quantity of apples. Each of you can count the number of apples in your own cell, but not in anyone else’s. You are told that each cell has at least one apple, and at most nine apples, and no two cells have the same number of apples.

The rules of the challenge are as follows: 

The three of you will ask Kurt a single question each, which he will answer truthfully ‘Yes’ or ‘No’. Every one hears the questions and the answers. He will free you only if one of you tells him the total number of apples in all the cells.

    Pip: Is the total an even number?

    Kurt: No.

    Blossom: Is the total a prime number?

    Kurt: No

You have five apples in your cell. What question will you ask?

THIS should be the question that you need to ask!

Logical Response to The Fearsome Challenge

What was the challenge?

Remember, all you have to do is that ask one crucial question to logicians and not necessarily deduce the total count of apples.

Since, each cell has 1 to 9 apples and no two cells have same number of apples, the lowest count of apple is 1 + 2 + 3 = 6 and the highest count would be 7 + 8 + 9 = 24.

That is the total number of apples could be between 6 to 24.

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Now, Pip and Blossom already have gathered some information about the total.

1. The total is not an even number - Hence, only numbers  7,9,11,13,14,15,17,19,21,23 can represent the total count.

2. The Total is not a prime number - Out of the number above, only 9, 15, 24 are non-prime number.

Hence, the total count must be among 9,15 or 24.

Now, your task is easier. All you need to ask the Kert below question -

"Is total is 15?"
 
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CASE 1 : Total is really 15 -

Then Kert would reply with YES to your question and all of you know the total now.

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CASE 2 : Total is 9 -

The Kert's answer to your question would be NO.

If the total is 9 and you have 5 apples then rest of 4 apples must be distributed among Pip and blossom as (1,3) or (3,1) but can't be (2,2) since no 2 cells can have same number of apples.

Now, the friend having 1 apple (or 3 apples) can think that the total can't be 21 as in that case other 2 must have total of 20 (or 18) apples. But the maximum that other two can have is 9 + 8 = 17 apples.

So any of them can deduce that the total is 9.

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CASE 3 : Total is 21 -

Since you have 5 apple other 2 must be having total of 16 apples. One of them must be having 7 apples and other having 9 apples.

The friend having 9 apples can easily deduce the count at 21 since 9 as a total count is impossible in the case as for that the other must have 0 apples.

And the friend with 7 apples know that other can't have 1 + 1 or 2 + 0 (as per given data) apples in order to have total count of 9. Hence, he too can deduce that the total must be 21.

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To conclude, depending on the what Kert answers to your question and the count of apples that each of other 2 friends have one of them (or you too if count is 15) can deduce the total number of apples easily. And eventually, logicians have to set you free as promised.

Challenge of Inverted Playing Cards

One fine day, Mr. Puzzle and Mr. Fry were playing cards, but suddenly power went off and they were getting bored. So Mr. Puzzle randomly inverted position of 15 cards out of 52 cards(and shuffled it) and asked Mr. Fry to divide the card in two pile with equal number of inverted cards (number of cards in each pile need not be equal).

It was very dark in the room and Mr. Fry could not see the cards, after thinking a bit Mr. Fry divided the cards in two piles and quite surprisingly on counting number of inverted cards in both the piles were equal.

What do you think Mr. Fry must have done?


This is what he must have done! 

  

Equating Counts of Inverted Cards in Piles

What was the challenge?

Mr.Fry must have taken top 15 cards & inverted positions of all. So he divides deck into 2 piles - one with 15 cards & other 37 cards.

Now suppose if there are 7 cards that were inverted in top 15 & 8 were inverted in remaining 37. When he flips top 15, 7 remains in normal position & 8 remains in inverted position. That is equal to 8 cards in inverted position from pile of 37 cards.

In short, if there are N cards inverted in top 15 then there are  15 - N cards inverted in remaining 37 cards. So on flipping position of top 15, there will be 15 - N cards in inverted position in top 15. That's how both piles would have equal number of inverted cards i.e. 15-N.  

 

Had Mr.Puzzle inverted positions of 20 cards randomly then Mr. Fry would have flipped top 20 cards. He would have made 2 piles with one with 20 cards & other 32 cards to equate the count of inverted cards in piles.

Wish Of Cigarette Smoking

Bruce is an inmate at a large prison, and like most of the other prisoners, he smokes cigarettes. During his time in the prison, Bruce finds that if he has 3 cigarette butts, he can cram them together and turn them into 1 full cigarette. Whenever he smokes a cigarette, it turns into a cigarette butt.

One day, Bruce is in his cell talking to one of his cellmates, Steve.

“I really want to smoke 5 cigarettes today, but all I have are these 10 cigarette butts,” Bruce tells Steve. “I’m not sure that will be enough.”

“Why don’t you borrow some of Tom’s cigarette butts?” asks Steve, pointing over to a small pile of cigarette butts on the bed of their third cellmate, Tom, who is out for the day on a community service project.

“I can’t,” Bruce says. “Tom always counts exactly how many cigarette butts are in his pile, and he’d probably kill me if he noticed that I had taken any.”

However, after thinking for a while, Bruce figures out a way that he can smoke 5 cigarettes without angering Tom. What is his plan?

That's his master plan!

Fulfilling The Wish Of Cigarette Smoking

What was the challenge? 

Bruce takes 9 of his 10 cigarette butts and make 3 cigarettes using those 9 (9/3=3). Now, he smokes all 3 cigarettes. At this point, he has 3 + 1 = 4 cigarette butts.

Using 3 out of 4 cigarette butts, he make one another cigarette and smokes it. Now he has 1 + 1 = 2 cigarette butts & till now has smoked 4 cigarettes.

Now he borrows 1 Tom's cigarette butts making total number of cigarette butts equal to 3. Using these 3 butts he makes one more cigarette and this way he smokes 5th cigarette. 

After smoking this 5th, he puts back the cigarette butt left in Tom's pile so that Tom won't find anything missing.

The Coconut Problem

Ten people land on a deserted island. There they find lots of coconuts and a monkeys. During their first day they gather coconuts and put them all in a community pile. After working all day they decide to sleep and divide them into ten equal piles the next morning.

That night one castaway wakes up hungry and decides to take his share early. After dividing up the coconuts he finds he is one coconut short of ten equal piles. He also notices the monkey holding one more coconut. So he tries to take the monkey's coconut to have a total evenly divisible by 10. However when he tries to take it the monkey conks him on the head with it and kills him.

Later another castaway wakes up hungry and decides to take his share early. On the way to the coconuts he finds the body of the first castaway, which pleases him because he will now be entitled to 1/9 of the total pile. After dividing them up into nine piles he is again one coconut short and tries to take the monkey's slightly bloodied coconut. The monkey conks the second man on the head and kills him.

One by one each of the remaining castaways goes through the same process, until the 10th person to wake up gets the entire pile for himself. What is the smallest number of possible coconuts in the pile, not counting the monkeys?

Here is that smallest number! 

Source 

Number Of Coconuts In The Pile

What was the problem? 

Absolutely no need to overthink on the extra details given there. Just for a moment, we assume the number of coconuts in the community pile is divisible by 10,9,8,7,6,5,4,3,2,1.

Such a number in mathematics is called as LCM. And LCM in this case is 2520. Since each time 1 coconut was falling short of equal distribution there must be 2519 coconut in the pile initially. Let's verify the fact for all 10 distributions tried by 10 people.Each time monkey kills 1 person & number of persons among which coconuts to be distributed decreases by 1 each time.