A Western Rancher finding himself well advanced in years, called his boys together and told them that he wished to divide his herds between them while he yet lived. "Now, John," He told the eldest, "you may take as many cows as you think you could conveniently care for, and your wife Nancy may have one ninth of all the cows left." To the second son he said, "Sam, you may have the same number of cows that john took, plus one extra because john had the first pick. To your wife, Sally, i will give one ninth of what will be left." To the third son he made a similar statement. he was to take one cow more than the second son, and his wife was to have one ninth of those left. the same applied to the other sons. Each took one cow more than his next oldest brother, and each son's wife took one ninth of the remainder. After the youngest son had taken his cows, there were none left for his wife. Then the rancher said: "Since horses are worth twice as much as cows, we will divide up my seven horses so that each family will own livestock of equal value." The problem is to tell how many cows the rancher owned and how many sons he had. i am basically completely at a loss on where to even START this problem. i know that the number the eldest takes away is unknown and that everytime a wife takes 1/9 from the leftovers the number changes cause the number of cows goes down. but depending on the number of cows the eldest takes this guy sure does have a crapload of cows 0_0!!