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0|0|0|||||Josephus Flavius Simple Solution|Modular Math|1|23:53:36|08/17/2010|There is a much easier way to solve the Josephus Flavius puzzle than any of the webpages.%0D%0A%0D%0ANote%3A%0D%0AThis solution references the starting position of the circle as 0%21%0D%0A%0D%0AIf x is the increment and n is the total number of people%2C then the position on the circle is given by this formula%3A%0D%0A%0D%0A-%28x%5E%28m%2B1%29%29 %28mod n%29%2C where m is the largest integer st x%5Em %3C%3D n%0D%0Aor alternatively%0D%0A-%28x%5Em%29 %28mod n%29%2C where m is the smallest integer st x%5Em %3E%3D n%0D%0A%0D%0A%0D%0AExplanation%3A%0D%0A%0D%0AThe position is always 0 whenever n is a power of x. This is because in the first pass around the circle all y%2Ax positions are eliminated%2C second pass all y%2A%28x-1%29 positions are eliminated%2C etc until y%2A%28x-x%29 %3D 0 is the only position left.%0D%0A%0D%0AIf n is not a power of x%2C then using x%5Em as a starting point%2C where m is the largest integer st x%5Em %3C n%2C the position increments by x for however much n is in excess of x%5Em. A direct computation results in %28n - x%5Em%29%2Ax. To prevent wraparound references %28ex%3A the 18th position of a 13 person circle%29 the answer must be computed mod n. This results in%3A%0D%0A%0D%0A%28n - x%5Em%29%2Ax %28mod n%29 %3D %28n%2Ax - x%5E%28m%2B1%29%29 %28mod n%29 %3D -%28x%5E%28m%2B1%29%29 %28mod n%29