The Last Bean in the Pot

A pot contains 75 white beans and 150 black ones. Next to the pot is a large pile of black beans. 

A somewhat demented cook removes the beans from the pot, one at a time, according to the following strange rule: 

He removes two beans from the pot at random. If at least one of the beans is black, he places it on the bean-pile and drops the other bean, no matter what color, back in the pot. If both beans are white, on the other hand, he discards both of them and removes one black bean from the pile and drops it in the pot.
 
At each turn of this procedure, the pot has one less bean in it. Eventually, just one bean is left in the pot. What color is it?

And the color of last bean is...... 

The Color of Last Bean in the Pot

Little story behind the title!

There are 75 WHITE beans in the pot i.e. they are odd in numbers. Since, they are always taken out in pair, in the end there will be 1 WHITE bean left out.

At some point, when there are 3 bean are left in the pot then there has to be 2 BLACK and 1 WHITE beans in the pot. They can't be 1 BLACK and 2 WHITE beans as for that 73 WHITE beans need to be taken out which is not possible since WHITE are always taken in pair.

So if cook pick 2 BLACK beans (or BLACK & WHITE) at this point then anyhow BLACK and WHITE will be left in the pot. Now, when he pick this pair of BLACK and WHITE then he puts BLACK on the pile and drop WHITE back to the pot as per his rule.

Eventually, WHITE bean will be left in the pot.

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.

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.