Answers 227A purely arithmetical solution is not difficult by a method of averages, the
average cost per animal being the same as the cost of a sheep.
By algebra we proceed as follows, working in dollars: Since x + y + z =
100, then fu + ~y + 1hz = 50.
50x + lOy + ~z = 1,000
~x + ~y + 1hz = 50
49~x + 9~y 950by subtraction, or 99x + 19y = 1900. We have therefore to solve this in-
determinate equation. The only answer is x = 19, Y = l. Then, to make
up the 100 animals, z must equal 80.
- THE SEVEN APPLEWOMEN
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 in-
genuity 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 re-
spectively, 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.- A LEGACY PUZZLE
The legacy to the first son was $55.00, to the second son $275.00, to the
third son $385.00, and to the hospital $605.00, making $1,320.00 in all.