Long ago, there was a king who had six sons. The king possessed a huge amount of gold, which he hid carefully in a building consisting of a number of rooms. In each room there were a number of chests; this number of chests was equal to the number of rooms in the building. Each chest contained a number of golden coins that equaled the number of chests per room. When the king died, one chest was given to the royal barber. The remainder of the coins had to be divided fairly between his six sons.

Now: Is a fair division possible in all situations?

(In reply to my answer by Shaun)

if the equation is x^3 - x and is factored as (x)(x-1)(x+1) which is 3 consecutive numbers, then every other number is even and every third number divisible by 3. Thereby any selected group of 3 consecutive numbers will have exactly one number divisible by 3 in it and one or two numbers divisible by 2 in it depending on if the first number is even or odd. Since all 3 numbers are multiplied together THis means 2 x 3 is a part of every equation but when there is one coin one chest and one room and the equation is 0*1*2

Posted by Shaun on 2005-10-12 06:01:36