Home > Just Math
| Gold (Posted on 2005-10-11) |
 |
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?
|
|
Submitted by Hugo
|
|
Rating: 3.1667 (6 votes)
|
|
|
Solution:
|
(Hide)
|
A fair division of the coins is indeed possible.
Let the number of rooms be N. This means that per room there are N chests with N coins each. In total there are N×N×N = N^3 coins. One chest with N coins goes to the barber. For the six brothers, N^3 - N coins remain. We can write this as: N(N^2 - l), or N(N - 1)(N + l). This last expression is divisible by 6 in all cases, since a number is divisible by 6 when it is divisible by 3 and is even. This is indeed so: whatever N may be, the expression N(N - 1)(N + l) always contains three successive numbers. One of those is always divisible by 3, and at least one of the others is even. This even holds when N=1; in that case all the brothers get nothing, which is also a fair division! |
Comments: (
You must be logged in to post comments.)
|
|
Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|
