Which unlucky Cowboy will have to shoot himself?
This is spin-off (probably harder) of another Cowboy Puzzle which I read a few days ago.
One hundred drunken cowboys stand in a circle. Let’s say they have been assigned numbers from 1 to 100.
We start with number 1. Number 1 spares (or skips) number 2 (first neighbor in sequence) and shoots number 3 dead. Then he gives the gun to number 4. Number 4 spares/skips TWO of his neighbors( 5 and 6), and shoots number 7. Then number 4 gives the gun to the next guy in sequence, number 8. Number 8 now spares/skips THREE of his neighbors in sequence (9,10 and 11) , kills number 12 and gives the gun to number 13. Then number thirteen spares FOUR cowboys in progression( 14,15,16 and 17) and kills number 18. And so on. So each time someone gets a gun he spares/skips a number of cowboys (in progression) equal to 1 more number than the previous killer spared.
Since they are standing in a circle the killing game continues and they go round and round. Of course only those alive and standing are counted in the deadly game. Also assume that the gun can be reloaded as required.
One cowboy is very unlucky. He will have to shoot himself! So which number is the first Cowboy in sequence that will have to shoot himself?
As a side question (if you want to try) if they continue playing the game after he shoots himself, will there be more cowboys who will have to shoot themselves?
And in your own time if you want to have fun with this then which Cowboy kills the most number of men?
BTW the orginal puzzle I read included cowboy killing his immediate neighbour (1 person) only and asked for the last man standing.
No programming please.