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?

As long as the sons are not splitting up nothing the answer will always be divisible by 6.

Assuming that the number of rooms chests and coins is not one, wherein the total number of coins is one and the 6 sons evenly split nothing: the equation would be x rooms by x chests by x coins, take away one chest. So x^3 - x, which is factored as (x-1)(x)(x+1)which are 3 consecutive numbers. Each subsequent answer can be written for n linearly. when n is 2, 1*2*3 = 6 which is of course divisible by 6. From there on out the solution for n=x is to take the previous n value(the n of (x-1)) + 6*(summation of the numbers 1 to n-1). Since n = 2 is divisible by 6 and each subsequent number is an addition of 6*(the summation 1 to n-1) which also must be divisible by 6, all answers must be divisible by 6....since 6*a + 6*b = 6(a+b). Therefore the amount of chests can always be divied up evenly to the 6 brothers.

Posted by Shaun on 2005-10-12 05:35:35