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?

I confirm the solution of Popstar David, although I will explain it differently:

The weight of each cannibal increases after each meal, and cannibals remain cannibals, so they devour the stomach contents as well to complete the meal. Assuming that they have infinite appetite, meaning they do not slow down after a meal, then:

He ends up having everybody else in his stomach, so it takes simply 5x2=10 hours non-stop.

If you calculate the weight of each surviving cannibal after a devouring, you end up with:
1.0, 1.2,1.5,2 and 3 times the original weight.
Since there are 5,4,3,2,1 cannibals that participate, the total time is:
2 hrs x (1/5+1.2/4+1.5/3+2/2+3/1)=10 hours also.

Bon appétit!

Posted by P C on 2004-02-04 12:56:10