Forty poor thieves Forty thieves have 4000 gold coins to split between them. A group of five thieves is $poor$ if together they have less than or equal to 500 gold coins. Let N be the minimum number of poor groups of five thieves among all groups of five thieves. Find N. I know the answer can be calculated using Subtraction rule(total groups-the groups which do not satisfy the equation). But how do I apply it?
 A: Suppose that we give $101$ coins to each of $39$ thieves; that’s $3939$ coins altogether, so Leifr hinn óheppni1, the $40$-th thief, gets only $61$ coins. Let $F$ be a group of five thieves. 


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*Is $F$ poor if Leifr is in $F$? 

*What if Leifr is not in $F$?

*How many poor groups of five thieves are there if we use this distribution of coins?


A harder question is whether we can do any better than this.
1 Leif the Unlucky
A: You have two problems: Calculate the number of poor groups given a particular coin distribution - relatively straightforward - and find (and prove) the best coin distribution. 
The first thought - after checking that an equal distribution will, in fact, mean all groups are poor - is to plunder one thief's wealth by distributing an extra coin from his hoard to each of the other thieves. Let's call that thief Dopey. Then only groups containing Dopey are poor - a total  of ${{39}\choose{4}}$ groups. The challenge then is to find out whether this is in fact the best solution. 
