Henry Ernest Dudeney/Modern Puzzles/20 - The Seven Applewomen

by : $20$

 * The Seven Applewomen


 * 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 their apples at the same price,
 * and each received the same sum of money.


 * What was the price?

Solution
Each woman sold her apples at:
 * $7$ for $1 \oldpence$
 * $3 \oldpence$ for the odd ones left over.

Thus each received the same amount::
 * $1 \shillings 8 \oldpence$

As put it:
 * 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 different kinds of apples at different values;
 * 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 $a n + \paren {n - 1}$, $\paren {a + b} n + \paren {n - 2}$, $\paren {a + 2 b} n + \paren {n - 3}$, $\ldots$. $\paren {a + \paren {n - 1} b} n$ apples respectively,
 * can sell at $n$ for the penny and $b$ pence for each odd one left over,
 * and each receive $a + b \paren {n - 1}$ pence.


 * In the case of our puzzle $a = 2$, $b = 3$, and $n = 7$.