The Cannibals of an Island have the habit of eating each other.
One evening, the cannibals threw a dinner party. Six cannibals turned up and they decided to eat each other in turn. So someone was selected for everyone to eat (except the victim!), and when he had been eaten, someone else was selected, and so on.
If it took one cannibal two hours on his own to devour one person, how long was it before just one consumer remained?
When the first course arrived (cannibal 1) there were 5 other cannibals there to eat him. Therefore it will take 2/5 hours to eat him.
When cannibal 2 comes out, ther are only 4 cannibals left to eat him. Therefore taking 2/4 hours.
Three cannibals are left to eat cannibal 3, taking 2/3 hours.
2 cannibals eat cannibal 4 in 2/2 hours.
And the last cannibal eats cannibal 5 in 2/1 hours.
Thus taking a total of:
2/5 + 2/4 + 2/3 + 2/2 + 2/1 Hours
= 4 hours and 34 minutes.
This solution assumes that the rate of eating remains constant regardless of the amout of eating already done.
It also ignores the fact that if one canibal eats another, he now contains the mass of two cannibals and would theoretically take twice as long to eat.
Solution
