1. ## Re: Was he fair

> Each chest contained the same number of coins that were in the room.

Something missing here?

2. ## Was he fair

<P ID="edit" class=small>(Edited by Jezza on 20-Jul-05 14:49. Bad typing error. Now changed in red)</P>My father has 5 brothers and no sisters. Now my Grandfather always tried to be a fair man and in his will he wanted to share out his large stash of gold coins to his sons. What I have not told you so far is my father never got on with my uncles and they mutually distrusted each other especially about Grandads gold. Grandad knowing this decided that we would hide the gold coins in this way. He hid the coins in a secret building consisting of a number of rooms. In each of these rooms was a number of chests, the number of chests in each room was equal to the number of rooms in the building. Each chest contained the same number of coins <font color=red>as there were chests in the room</font color=red>. When Grandad died one of the chests and its content was given to his best friend, Albert. The rest of the coins were divided between my father and uncles.

Question:
a) Is a fair division possible in all situations?
[img]/forums/images/smilies/cool.gif[/img] Why

Hint:<span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">no...not yet</span hide>

3. ## Re: Was he fair

<span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">a) Yes
[img]/forums/images/smilies/cool.gif[/img] Say that the house contains n rooms. There are n chests in each room, and n coins in each chest, so n * n * n = n^3 coins in all. One chest with n coins is given to Albert, so there remain n^3 - n = n * (n^2 - 1) = n * (n - 1) * (n + 1) coins. Among n, n-1 and n+1 there is always at least one number divisible by 2, and one number divisible by 3, so the product is always divisible by 6.</span hide>

4. ## Re: Was he fair

Perfect

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