Question: Here is an old puzzle that people are frequently writing to me about. Seven applewomen, possessing respectively 20, 40, 60, 80, 100, 120, and 140 apples, went to market and sold all the apples at the same price, and each received the same sum of money. What was the price?
Answer: Each woman sold her apples at seven for I¢, and 3¢ each for the odd ones over. Thus, each received the same amount, 20¢. Without questioning the ingenuity of the thing, I have always thought the solution unsatisfactory, because really indeterminate, even if we admit that such an eccentric way of selling may be fairly termed a "price." It would seem just as fair if they sold them at
different rates and afterwards divided the money; or sold at a single rate with different discounts allowed; or sold different kinds of apples at different values; or sold the same rate per basketful; or sold by weight, the apples being of different sizes; or sold by rates diminishing with the age of the apples; and so on. That is why I have never held a high opinion of this old puzzle.
In a general way, we can say that n women, possessing an + (n - 1), n(a + b) + (n - 2), n(a + 2b) + (n - 3), ... ,n[a + b(n - 1) 1 apples respectively, can sell at n for a penny and b pennies for each odd one over and each receive a + b(n - I) pennies. In the case of our puzzle a = 2, b = 3, and n = 7.
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