# Tracking Down the Meaning of a Problem

#### (Archive question of the week)

Sometimes, we get a question that seems familiar, and can look back at past questions to see if we have already answered it, or to get ideas about what it means. This is the story of a problem that seems to float around as folklore, in varied forms. At first it seemed unanswerable, because each version seemed to lack information; but with time we figured out how to make sense of it. Its meaning may depend on knowledge of the cultural context in which it was first asked, so that in other cultures it can become an entirely different kind of puzzle.

## A puzzle and its variations

Here is how I first met the problem, as sent by Mark in 2002 (bold indicates features I’ll be pointing out later):

The Case of the 90 Apples

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.

Clearly, it involves some sort of trick. I remembered seeing similar problems, so I searched, hoping for clarification. Just a month earlier, Kumar had asked this less detailed version, which went unanswered:

A Father has three children.  He gave each of them some mangoes as follows:
He gave 50 mangoes to his first son,
He gave 30 mangoes to his second son, and lastly
He gave 10 mangoes to his third son.

They have to sell the mangoes with same price and they should bring home same amount of money.

For example, if first son sells 5 mangoes for $10.00, 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 are sharing mangoes with each other.

How is this possible?

The previous year, Cam had sent the following version, which sounds more ancient, but was his Problem of the Week:

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.


Doctor Budrow answered this one, suggesting they sold apples to one another; I suggested that the only possible answer was that they all gave away all their apples (not just “some”). These sorts of trick answers are not satisfying; but I supported my idea by pointing out that zero was discovered/invented in a setting that matches the story.

In 1997 Benno had sent us the following version, which he said came from a book, The Man Who Counted: A Collection of Mathematical Adventures, by a Brazilian math teacher writing as “Malba Tahan”, which is presumably the original source:

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.


(I later found the English translation of the book online; it turns out that Cam’s version above is a more or less direct quotation from the book, while Benno’s is a paraphrase with different spelling, possibly because it was translated from the original version?)

Notice that the details of the rules vary, but all of them have a feature that I missed: they leave open the possibility that the prices could be changed.

A version sent by Scott in 2000, however, doesn’t seem to allow for changing prices:

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, andgoes home.

All three farmers make the same amount of money. What was the selling price of the eggs?

The same is true of this very simplified version from Greg in 2002 , which has no clues as to the trick:

Three Vendors are to sell their apple 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.

All vendors after selling their apples will end up with the same amount of monies.

## A solution

In 2004, Paul, having read the discussion above, wrote to tell me that allowing prices to change from time to time made a solution possible:

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.

Finally we have a more or less satisfying answer; some of the versions we’ve seen seem to disallow this, so I have to suspect that their authors, if they claimed to have an answer, probably assumed it was the sort of trick I had guessed.

But we still have the question, how would you solve this? I discovered that if the mere possibility of prices varying from time to time is the trick, then there are many solutions. I showed that there are two sets of amounts that could be sold at the prices specified in Paul’s version, and that there are other pairs of prices that work; for example, if they are first sold at $1 per apple (selling 49, 25, and 1 apple, respectively), and then at$6 per apple (selling 1, 5, and 9), then everyone earns $55. A problem with no genuine solution, and one with many possible solutions, are both disappointing to a mathematician (unless we were asked for all possible solutions and the rules were clear – that would be a nice challenge). But there’s more! In 2013, another Paul wrote, suggesting a slightly different interpretation: that prices don’t change, but rather are sold at a combination of prices, like$2 per dozen and $1 each for individual apples. My answer:$2 per dozen and $1 ea for the remainder. 50 = 4 doz at$2 a doz ( $8 ) & the remainder at$1 ea ( $2 ) =$10
30 = 2 doz at $2 a doz ($4 ) & the remainder at $1 ea ($6 ) = $10 10 = 0 doz ($0 ) & the remainder at $1 ea ($10 ) = $10 This doesn’t fit most versions, but does work in some that don’t allow for changing prices – including, if you read it right, the version I started with: “a sign that states how much each apple costs, or how much as many apples as he prefers can be sold for”. I then found that Paul’s solution was in fact one of those I had found for the varying price model; and interestingly, his was based on dozens, while Paul 2004’s solution was based on sevens, which probably was the cultural equivalent of dozens in the original problem. This suggests that people in different cultures might make their own assumptions about how fruit would be sold, and that assumption provides the missing key to the problem. I proceeded to try out different possible assumptions and found that not all package sizes yield solutions, but more than one do. ## A new variation And still there’s more! In 2016 Iqbal sent us this similar problem: The Case of the 90 Apples, Redux Four men have 70, 80, 90, and 100 oranges, respectively. Each got$20 selling all the oranges he had using the same logical sentence.

He wrote back 7 months later to show the solution he had found:

I am contacting you after a long time to announce that my puzzle has been solved.

|      |Price|        |         |Price|        |
Oranges|Dozens| per |Subtotal|Remainder| per |Subtotal|Grand
|      |Dozen|(Dozens)|         |Piece|(Pieces)|Total
------------------------------------------------------------
70      5     2      10        10      1      10     20
80      6     2      12         8      1       8     20
90      7     2      14         6      1       6     20
100      8     2      16         4      1       4     20

I recognized then that his interpretation of the cryptic problem makes it equivalent to the qadi problem. In response, I referred him to The Case of the 90 Apples, and commented,

I did not figure out how the question was meant to be taken because neither Mark nor Paul stated the problem clearly (especially in some of its versions). Even after subsequent submissions, I'm not sure I fully understood how the problem was supposed to be interpreted. My response was to try to analytically find all possible solutions within a general class of rules, because it was clear to me that the solution given was not unique, but appeared to be the result of guessing, perhaps based on cultural assumptions that fruit would be sold in sevens or in dozens. In fact, I showed several solutions using different groupings.

Your version was extremely unclear (because of the vague term "logical sentence"), but your answer has the same form. A clearer restatement might read like so:

Four men have 70, 80, 90, and 100 oranges respectively.
They sold all their oranges using the same pricing scheme,
with set prices for different size packages of oranges,
and got 20 dollars each. What was that scheme?

The scheme has the form, "$X for each ___, and$Y for each additional orange," and your solution is that if the scheme was "$2 for each dozen, and$1 for each additional orange," then each of the four would make the required \$20.

I found a way to efficiently arrive at Iqbal’s solution, and that because his version is more restrictive than the others, it may have only one solution. He then created a couple more similar problems by modifying the numbers. (That’s the sign of a mathematician: not stopping with a solution, but extending the problem!)

This last version, in my formulation, cleans up both the meaning of the problem, and the method of solution. Once you know what the problem means, it’s actually not that hard.

I had always wondered why no one ever quoted the answer from the various sources where they got the problem; no one ever told us they had an authoritative answer, even the one who referred to the book. I finally was able to read the next page of the book (which was hidden when I first located it), and found that it does give the answer:

The solution to the problem with which the qadi of Damascus tormented the three sisters is the following:
"Fatima starts selling her apples at a price of 7 apples for 1 dinar. She sells 49 of her apples at this price, but keeps back 1.
"Cunda sells 28 of her apples at this price, but keeps back 2.
"Shia sells 7 of her apples at this price, but keeps back 3.
"Then Fatima sells her 1 remaining apple for 3 dinars. In accordance with the rules of the qadi, Cunda then sells her 2 remaining apples for 3 dinars each. And Cunda then sells her 3 remaining apples for 3 dinars each.
"Thus:
Fatima
First phase:    49 apples bring a profit of 7 dinars
Second phase:    1 apple brings a profit of 3 dinars
------------------------------------
Total:          50 apples bring a profit of 10 dinars
Cunda
First phase:    28 apples bring a profit of 4 dinars
Second phase:    2 apple brings a profit of 6 dinars
------------------------------------
Total:          30 apples bring a profit of 10 dinars
Fatima
First phase:     7 apples bring a profit of 1 dinar
Second phase:    3 apple brings a profit of 9 dinars
------------------------------------
Total:          10 apples bring a profit of 10 dinars

"Therefore, each made a profit of 10 dinars, and thus the problem set by the envious qadi of Damascus was solved."


So, yes, the key is to use the stated prices sequentially; and, no, no explanation is given of how to solve it. But it would seem that we are to take the meaning as, “Using the two prices I suggested, find appropriate quantities so that they each get the same amount of money.” If it had been stated that way, it would have been still a challenge, but less frustrating.

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