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 seemjust 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,
andn = 7.