The Case of the 90 Apples
Date: 11/19/2002 at 09:53:11 From: Mark Milovanovic Subject: 3 sons have to sell different numbers of apples, but come up with the same amount of money A father takes his 3 sons to a small town nearby. He places the oldest son at the first corner, the second son at the second corner, and the youngest son at the third corner. He then gives the oldest son a basket of 50 apples, the second son a basket of 30 apples, and the youngest son a basket of 10 apples. He tells them that the oldest son has the right to put up a sign that states how much each apple costs, or how much as many apples as he prefers can be sold for. The other sons have to follow that sign and sell that same amount for the same price. The sons cannot exchange or trade apples. They still have to come up with the same amount of money at the end of the day when the father comes to collect. They each have to sell their full stock of apples. The first has to sell 50, the second 30, and the third 10, and they all have to give the father the same amount of money without sharing among themselves. What throws me off is that the 50 apples can cost the same number of dollars. Please help.
Date: 11/19/2002 at 12:17:56 From: Doctor Peterson Subject: Re: 3 sons have to sell different numbers of apples, but come up with the same amount of money Hi, Mark. I thought it would be interesting to go back and see how the puzzle has been phrased in Dr. Math submissions, to see if it sheds any light on what is intended. The result is an interesting study. This recent version seems essentially the same as yours: A Father has three children. He gives each of them some mangoes as follows: He gives 50 mangoes to his first son, He gives 30 mangoes to his second son, and lastly He gives 10 mangoes to his third son. They have to sell the mangoes at the same price and they should bring home the same amount of money. For example, if the first son sells 5 mangoes for $10.00, the second and third sons also have to sell 5 mangoes for $10.00. However, they should bring the equal amount of money when they come back home. None of them can share mangoes with each other. How is this possible? This one appears to be the original, and is considerably more interesting: There once lived in Damascus a peasant who bragged to a 'qadi', a judge, that his daughters were not only very intelligent, but blessed with rare skills of the imagination. The qadi had the three girls brought before him. Then he said to them, "Here are 90 apples for you to sell in the market. Fatima the oldest, you will take 50; Cunda you will take 30; and Shia, the youngest, you will take 10. If Fatima sells her apples at a price of 7 to the dinar, you other two will have to sell yours at the same price. And if Fatima sells her apples for 3 dinar each, you two will have to do the same. But no matter what you do, each of you must end up with the same amount of money from your different numbers of apples." "But can I not give away some of the apples that I have?" asked Fatima. "Under no circumstances." said the qadi. "These are the terms: Fatima must sell 50 apples. Cunda must sell 30 apples. And Shia must sell the 10 apples that remain. And all of you must earn exactly the same profit in the end." The three sisters need directions to resolve the problem of the 90 apples before they get to the market. Here each has to sell all her apples. The solution is simple, but tricky: We want a price per apple, say X, for which 50X = 30X = 10X The only solution to this equation is X = 0 That is, if Fatima gives away all her apples for free, and the others do the same, then they will all bring home the same amount of money: zero. Since she was told she can't give away some of her apples, it will be necessary to use some trick wording to get around this: I didn't sell SOME, but ALL; or, I sold them, but not for money, or something like that. Considering that zero was discovered in the Arabic setting given to the story, this answer seems quite fitting. (I should add that, depending on the rules, it may be possible to make an alternative solution, by having some or all buy from one another. Your version seems to disallow this.) Back in 1997, someone sent us a set of problems from Brazil, including this one: When I was a teenager I read a very enchanting book by a Brazilian writer known by his pseudonym, "Malba Tahan," who was actually a mathematics teacher. The title of this book is _The Man Who Counts_. The book tells the adventures of a Persian mathematician, Beremiz Samir, in the Near East in the Middle Ages (the time of the "1001 Nights"). A lot of interesting mathematical problems are solved intuitively by this mathematical hiker in an ambience of magical mystery. I think that this marvelous book is not known internationally because it is by a writer from an undeveloped country, but I think it has merit. To spread the notice of this book I am sending you some of its problems, which I have tried to translate into English. The case of the 90 apples. To prove that three rustic daughters have great intelligence, they must solve a problem proposed by a jealous Cadi. The Cadi gives 90 apples to the girls and says: These apples you must sell in the market. Fatima, the eldest, takes 50, Cunda 30, and Siha, the youngest, the remainder (10). If Fatima sells 7 apples for 1 dinar (the Arabian money at that time) the others will sell the apples at the same price (7 for 1 dinar). If Fatima sells 1 apple for 3 dinars this is the price at which the others must sell their apples. The whole transaction must be accomplished in such a way that the three girls get the same amount of money. The problem seemed impossible to them because if they must sell the apples at the same price, the one that sells more apples must necessarily earn a greater amount of money. They go to the Ima (a wise religious man) and ask the way to solve the problem. He says that they must proceed exactly as the Cadi instructs, and then they will earn the same amount selling the apples at the same price. Beremiz explains the way that the girls sold the apples and performed the apparently difficult challenge. What happens, dear Dr. Math? The problem has an interesting history, doesn't it? This book is now available in America, and may be the source of the other versions, or the puzzle may be found in earlier sources I haven't traced. Here is another version that seems to be the same as the above, without the colorful details or restrictions: Three Vendors are to sell their apples at the same price. Vendor (A) has 10 apples, Vendor (B) has 30 apples, and Vendor (C) has 50 apples. All apples must be sold at the same price. After selling their apples, all vendors will end up with the same amount of money. And here's another version we received a couple of years ago that seems to be derived from the others: Farmer A has 10 eggs. He takes them to market, and writes a price on the board. He sells all of his eggs and goes home. Farmer B has 30 eggs. He takes them to the same market. He sees the price written by farmer A and decides to use that price. He sells all of his eggs and goes home. Farmer C has 50 eggs, uses the same price, sells the eggs, and goes home. All three farmers make the same amount of money. What was the selling price of the eggs? Again, only one answer seems possible, despite the absurdity of selling eggs for $0 each. Finally, here is one sent to us in 1999 from Asia: There are three people: the first has 10 oranges, the second has 30 oranges, and the third has 50 oranges. They decide they will go to market and sell them. Here is what they decide: they will sell each orange at the same price (if the first sells at $1 for each orange, so will the second and the third). The question is, in what way can they sell the oranges so that when there are no oranges left, the three persons have the same amount of money? The answer has to be zero. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 04/16/2004 at 11:28:47 From: Paul Subject: The Case of the 90 Apples (new solution!!!) In all your answers to the "Case of the 90 Apples" you state that the only possible solution is to sell them for $0. This is only the answer if the price that they are all selling the apples for cannot change over time. The way the question was stated to me was thus: Farmer A has 50 apples Farmer B has 30 apples Farmer C has 10 apples Farmer A sets the price and they all sell their apples for that price. He may change the set price at any time. They all sell all their apples and go home with the same amount of money. The solution is: Farmer A sets the price at 7 apples/$1. Farmer A sells 49 (makes $7), B sells 28 (makes $4) and C sells 7 (makes $1). Then farmer A changes the price to $3/apple, and they all sell what they have left. They all end up with $10. I believe this is the correct solution to the problem. The version you suggest "appears to be the original" allows for varying prices and even mentions the prices in my solution.
Date: 04/16/2004 at 14:30:35 From: Doctor Peterson Subject: Re: The Case of the 90 Apples (new solution!!!) Hi, Paul. You may be right, though I am bothered by the fact that some versions of the problem clearly do not allow the price to be changed, and yet whoever posed those versions must have expected it to have a solution! Also, the "original" version does not explicitly say (as your version does) that the price can be changed; it just isn't explicitly forbidden. It may be intended as a "lateral thinking" problem, with that as the key. However, doing that means that there is no longer a unique solution, and the problem is not so interesting. If we arbitrarily decide that there will be only TWO different prices, A and B (which is not necessarily true, once we've opened the problem up by allowing the price to change at all), and call the numbers sold under the first price Xa, Ya, and Za, and under the second price Xb, Yb, and Zb, we have: AXa + BXb = AYa + BYb = AZa + BZb Xa + Xb = 50 Ya + Yb = 30 Za + Zb = 10 Eliminating the "b" amounts, AXa + B(50 - Xa) = AYa + B(30 - Ya) = AZa + B(10 - Za) or (A - B)Xa + 50B = (A - B)Ya + 30B = (A - B)Za + 10B Subtracting 10B and dividing by (A - B), Xa + 40B/(A - B) = Ya + 20B/(A - B) = Za Za = n, Ya = n + 20B/(B - A), Xa = n + 40B/(B - A) which gives us a solution for any given A, B, and n. Taking A = 1/7 and B = 3 just as a guess from the problem's examples, we get B/(B - A) = 3/(20/7) = 21/20 Xa = n + 42 Ya = n + 21 Za = n Clearly, we can choose any value for n from 0 to 8 and get valid numbers for Xa and Ya; since we need to sell a whole number of apples at the second price, and thus need the differences between the numbers sold at the first price to be a whole number (actually, just a multiple of 1/3), we can choose either 0 or 7 for n. So either A = 1/7, Xa = 42, Ya = 21, Za = 0 B = 3, Xb = 8, Yb = 9, Zb = 10 each earns $30 or A = 1/7, Xa = 49, Ya = 28, Za = 7 B = 3, Xb = 1, Yb = 2, Zb = 3 each earns $10 But these are not the only solutions. On one hand, I can choose A and B to be any pair of numbers whose ratio is small enough (such as A = 1, B = 6); and on the other hand, only the ratio of A:B matters, so once I find a solution, I can multiply A and B by anything I want (giving solutions like A = 1, B = 21, with the same amounts sold as above). It takes a little work to find a solution, but then there turn out to be too many. Here is one of my alternative solutions, just to make it clear that it is real: A = 1, Xa = 49, Ya = 25, Za = 1 B = 6, Xb = 1, Yb = 5, Zb = 9 each earns $55 Is this the intended meaning of the problem? I have no idea. But it's certainly worth adding to our archive! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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