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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.
A stark raving mad king tells his 100 wisest men he is about to line
them up and that he will place either a red or blue hat on each of their
heads.
Once lined up, they must not communicate among themselves. Nor
may they attempt to look behind them or remove their own hat.The king tells the wise men that they will be able to see all
the hats in front of them. They will not be able to see the color of
their own hat or the hats behind them, although they will be able to
hear the answers from all those behind them.
The king will then start with the wise man in the back and
ask "what color is your hat?" The wise man will only be allowed to
answer "red" or "blue," nothing more. If the answer is incorrect then
the wise man will be silently killed. If the answer is correct then the
wise man may live but must remain absolutely silent.The king will then move on to the next wise man and repeat the question.
The king makes it clear that if anyone breaks the rules then
all the wise men will die, then allows the wise men to consult before
lining them up. The king listens in while the wise men consult each
other to make sure they don't devise a plan to cheat. To communicate
anything more than their guess of red or blue by coughing or shuffling
would be breaking the rules.
What is the maximum number of men they can be guaranteed to save?
Almost all can survive! Click here to know!
Source
Why this master plan needed?
99 can be guaranteed to save! How?
Even if the person behind calls out the color of the hat that next person is wearing both would be survived only if they are wearing same color of hat.
So how 99 can be saved?
For a simplicity, let's assume there are only 10 wise men & (only) assume we are among them. Now, we need to make a master plan to survive from this game of death.
One of us need to agree to sacrifice his life to save 9 of us & this person would be the first one in line. He will be survived of he has good luck.
The first person in line should shout RED if he founds number of RED hats even otherwise he should shout BLUE. Now if he has good luck then the hat color of his own hat would match & he would be survived.
The clue given by the first person is very important. Right from second person everyone need to count number of RED hats in front of him. Additionally, the next person need to keep track of number of RED hats that people behind him are wearing.
One day three Greek philosophers settled under the shade of an olive
tree, opened a bottle of Retsina, and began a lengthy discussion of the
Fundamental Ontological Question: Why does anything exist?
After a while, they began to ramble. Then, one by one, they fell asleep.
While the men slept, three owls, one above each philosopher,
completed their digestive process, dropped a present on each
philosopher's forehead, the flew off with a noisy "hoot." Perhaps the hoot awakened the philosophers.
As soon as they
looked at each other, all three began, simultaneously, to laugh.
Then,
one of them abruptly stopped laughing. Why?
Interesting reason behind it!
Source
What's the story behind?
The one who stopped laughing was the smartest one among! Read how he was the smartest.
We need to think from the smartest Philosopher's point of view. Let's name 2 other Philosophers as A & B.
Now here is what the smartest Philosopher would think.
"If I had nothing on my head then A & B must have been laughing after looking each other's head. And at least one of them is smart enough to realize that the other is laughing only after looking at him. That means A (or B) would have realized that the some thing is on his head too as B (or A) is laughing after looking him (not me if I had nothing on my head). Hence one of them would have stopped laughing. Since they are not stopping to laugh, I too must have something on my head."
Hence the smartest Philosopher stopped laughing after realizing that the fact.
A brilliant puzzle based on the similar logic is here!
A 120 wire cable has been laid firmly underground between two telephone exchanges located 10km apart.Unfortunately after the cable was laid it was discovered to
be the wrong type, the problem is the individual wires are not labeled.
There is no visual way of knowing which wire is which and thus
connections at either end is not immediately possible.
You are a trainee technician and your boss has
asked you to identify and label the wires at both ends without ripping
it all up. You have no transport and only a battery and light bulb to
test continuity. You do have tape and pen for labeling the wires.
What is the shortest distance in kilometers you will need to walk to correctly identify and label each wire?
Know here the efficient way!
Source
What was the task to test the skill?
The shortest distance is 20 km! Surprised? Read further.
Let's name the two exchanges as a 1 & 2 respectively. Now at end 1, let's make a groups of wires having 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 number of wires. Now somebody might ask why not 15 groups having 8 wires in each. After reading the entire process here, we'll get the answer of it.So we 15 groups have total 1 + 2 + 3....+ 15 = 120 wires. Let's name these groups as A, B, C, D ...... O. That means group A has 1, B has 2, C has 3 wires & so on.
Now join together all the wires of the particular group. For example, 2 wires of group B should be joined together, 7 wires of G tied together & so on. The sole wire of A is left as it is.
We will take the battery & bulb to other end traveling 10 km. We will say a wire is paired with the other if the bulb gets illuminated if battery & bulb connected in between.
Now let's take any wire at the other end & find the number of wires that are pairing with that particular wire under test. We will group such wires & label with those exactly how we labeled at end 1.
For example, if we find 2 wires pairing with particular wire then that wire & 2 paired wires together to be grouped in 3 wires & labeled as C.The sole wire not getting paired with any will be labeled as A. And group with wire pairing with 7 other wires together should be labeled as H.
In this way, we will have the exact group structure that we have at end 1. By now, we have identified & labeled correctly wires in groups of 1,2,3,....15 wires at both ends.
Now, we are going to label each wire of group by it's group & count number. For example, the only wire in group A labeled as A1, 2 wires in B are labeled as B1,B2, wires in G are labeled as G1,G2,G3,G4,G5,G6,G7 and so on.
Now, at end 2 itself, what we are going to do is connecting first wire of each group to A1. Second wire of each group to be connected to B2, third of each to be connected to C3 and so on. (refer the diagram above, where labels of wires that are to be connected together are written in same color).
