12. Seven apple vendors. Seven vendors had 20, 40, 60, 80, 100, 120, and 140 apples respectively. They went to the market and sold all their apples at the same price, earning the same revenue. At what price did the vendors sell the apples?
12. All seven vendors sold apples at 1 cent for 7 pieces; in cases where fewer than seven apples remained, they were sold at 3 cents per piece. Thus, each vendor earned 20 cents. Without in any way disputing the ingenuity of this problem, I have always found its solution unsatisfactory due to its ambiguity, even if we assume that with such an eccentric method of trading, one can speak of a single "price" for apples. We could just as well assume that the vendors sell apples at one price but with different discounts; sell apples of different varieties at different prices; sell at one price per basket; sell by weight, while the apples are of different sizes; reduce the price for less fresh apples, etc.

In general, we can say that \(n\) vendors, who have \(n a+(n-1), n(a+b)+(n-2), n(a+2 b)+(n-3), \ldots\), \(n[a+b(n-1)]\) apples respectively, can sell them in bunches of \(n\) pieces for 1 cent, and the remaining apples for \(b\) cents each, and each of the vendors will earn \(a+b(n-1)\) cents. In the case of our problem, \(a=2, b=3, n=7\).