Logic Puzzles

## The Seven Applewomen

Difficulty:

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 theu apples at the same price, and each
received the same sum of money. What was the price?

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.

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.

See also:

• A Poultry Poser (Difficulty: 2)

• Apple Transactions (Difficulty: 2)

• Around The Equator (Difficulty: 2)

• A Sidecar Problem (Difficulty: 3)

• Bicycle Thieves (Difficulty: 4)