Truth Tellers and Liars in Circle

On a certain island there live only knights, who always tell the truth, and knaves, who always lie.

One day you find 10 island natives standing in a circle. Each one states: "Both people next to me are knaves!"

Of the 10 in the circle, what is the minimum possible number of knights?

Do you think there can be 5?

Identifying Number of Truth Tellers in Circle

What was the task given?

Recalling the given situation. 

On a certain island there live only knights, who always tell the truth, and knaves, who always lie.

One day you find 10 island natives standing in a circle. Each one states: "Both people next to me are knaves!"

 
Every Knight must be surrounded by 2 Knaves and every Knave has to be surrounded by at least one knight to satisfy the given condition. So there must be Knave-Knight-Knave groups must be standing in circle. 

Now if Knave of previous group is counted for the next group, then there will be 5 knights in the circle as shown below.

But the question asks minimum possible number of Knights in the circle.

So after forming 3 groups of Knave-Knight-Knave separately (total 9 in circle), the last person will be obviously surrounded by 2 knaves. Hence, he must be Knight. See below.

This way there can be only 4 knights standing in the circle without violating the given condition.
 

An Island Of Puzzles

There is an Island of puzzles where numbers 1 - 9 want to cross the river.

There is a single boat that can take numbers from one side to the other.

However, maximum 3 numbers can go at a time and of course, the boat cannot sail on its own so one number must come back after reaching to another side.

Also, the sum of numbers crossing at a time must be a square number.

You need to plan trips such that minimum trips are needed.

This should be that minimum number! 

Numbers On An Island Of Puzzles

What was the challenge?

We need only 7 trips to send all digits across the river.

1. Send 2, 5, 9 (sum is 16).

2. Bring back the 9.

3. Send 3,4, 9 (sum is 16).

4. Bring back the 9.

5. Now send 1,7,8 (sum is 16).

6. Bring back the 1.

7. And finally send 1,6,9 (sum is 16).

Distinguish The Fake Coin

You have twelve coins. You know that one is fake. The only thing that distinguishes the fake coin from the real coins is that its weight is imperceptibly different. You have a perfectly balanced scale. The scale only tells you which side weighs more than the other side.

What is the smallest number of times you must use the scale in order to always find the fake coin?
 
Use only the twelve coins themselves and no others, no other weights, no cutting coins, no pencil marks on the scale. etc.

These are modern coins, so the fake coin is not necessarily lighter.

Presume the worst case scenario and don't hope that you will pick the right coin on the first attempt.

Process to identify the fake one! 

Source 

Process To Identify Fake Coin

What was the task given? 

If we knew, the fake coin is lighter or heavier than original one then the process would have been pretty simple like this! But we don't know.

Let's number the coins from 1 to 12. We'll make 3 groups of these coins as 1,2,3,4 in one group, 5,6,7,8 in other group and 9,10,11,12 in one more group.

First of all weigh 1,2,3,4 against 5,6,7,8.

CASE 1 : 1,2,3,4 = 5,6,7,8

 That means coin among 9,10,11,12 is fake one. So weigh 9,10 against 11,8.

   CASE 1.1 : If 9,10 = 11,8 then 12 is fake coin.

   CASE 1.2 : If 9,10 > 11,8 then either 9 or 10 is heavier (hence fake) or 11 is lighter (hence fake). Weigh 9 against 10. If they balance then 11 is fake one. If they don't then heavier of 9 & 10 is fake. 

   CASE 1.3 :  If 9,10 < 11,8 then either 9 or 10 is lighter (hence fake) or 11 is heavier (hence fake). Weigh 9 against 10. If they balance then 11 is fake one. If they don't then lighter of 9 & 10 is fake.

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.