# Is there a clever solution to Arnold's "merchant problem"?

There is a problem that appears in an interview$$^\color{red}{\star}$$ with Vladimir Arnol'd.

You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?

This problem is also quoted here.

Here's my solution:

The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let $$s$$ be the volume of a spoonful and $$c$$ be the volume of a cup. The quantity of wine in this second spoonful is $$\frac{s}{s+c}\cdot s$$ and the quantity of tea in this spoonful is $$\frac{c}{s+c}\cdot s$$. Then the quantity of wine left in the cup is $$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$ and the quantity of tea in the barrel now is also $$\frac{cs}{s+c}.$$ So the quantities that we are asked to compare are the same.

However, Arnol'd also says

Children five to six years old like them very much and are able to solve them, but they may be too difficult for university graduates, who are spoiled by formal mathematical training.

Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.

$$\color{red}{\star}\quad$$ S. H. Lui, An interview with Vladimir Arnol′d, Notices of the AMS, April 1997.

• If you like this sort of problem, there's a nice book compiling them: The Chicken from Minsk Commented Nov 17, 2020 at 14:27
• Given that there's only 3 conceivable solutions (by trichotomy) -- (a) greater wine, (b) greater tea, or (c) they're equal -- it's likely that random guessing by a large group of 6-year-olds will result in about one-third being right by chance, and these cases could be held out as anecdotally successful. Commented Nov 17, 2020 at 15:32

At the end the tea cup is as full as at the start. This implies that the added wine is exactly outweighed by the tea that has disappeared.

• I think this is what Arnol'd had in mind in writing, "Children five to six years old ... are able to solve them". This answer strips every bit of unnecessary complexity away. Commented Nov 17, 2020 at 17:42
• Dammit! I tried the puzzle myself after reading the puzzle, but before reading his proposed solution. Got to exactly the same point; even noticed as I was going along, that calculating the total volume in each at the end should zero out. But still missed this :D :facepalm: Commented Nov 18, 2020 at 7:14

The volume of spoon, $$s$$, is the conserved quantity. It is also the amount of wine in the cup.
When you then take some mixture $$\mathit{tea}+\mathit{wine} = s$$ into the spoon,
$$s-\mathit{wine}$$ is the amount of wine left in the cup and the amount of tea poured into the wine barrel.

• Note that it does not require a uniform mixture, if you somehow took all the wine back or just took tea it is still true. Commented Nov 17, 2020 at 3:37
• What is you have a prime number of tea particles? Commented Nov 17, 2020 at 20:21
• @marshalcraft: Then the price of tea in China must also be prime. Commented Nov 18, 2020 at 1:07

To a first approximation, there is a spoonful of wine in the cup and a spoonful of tea in the barrel. How much are each of these approximations off by? Well, there is a bit less than a spoonful of wine in the cup, since a bit of the wine was removed in the second step. And, there is a bit less than a spoonful of tea in the barrel, since there was a little wine mixed into the spoonful that was put into it. But these errors are exactly the same: both are the amount of wine that was in the second spoonful. So the two quantities are the same: both are one spoonful minus the amount of wine that was in the second spoonful.

Or, here's an even slicker way. Notice that the total volumes of liquid of the cup and barrel have not changed, since the two spoonfuls they exchanged cancelled out. So, the overall change must be that the barrel exchanged some volume of wine for the same volume of tea from the cup.

Note that your solution is actually wrong--when you compute the amounts of wine and tea in the second spoonful, you are assuming the cup was mixed uniformly after the first spoonful, which the problem tells you not to assume (that's what the "(nonuniform!)" is all about).

• Hmm, I thought the "nonuniform!" meant that there is both tea and wine in that mixture, so it's not only tea anymore that is in the cup. Commented Nov 17, 2020 at 3:24

After step 1, there is always one spoonful of wine in your cup.

If, on step two:

You place one spoonful of wine back into your barrel, then there is 0 tea in your wine and 0 wine in your barrel.

You place one spoonful of tea back into in your barrel, then there is 1 tea in your wine and 1 tea in your barrel.

You place half a spoonful of tea and half of wine back into in your barrel, then there is 0.5 tea in your wine and 0.5 wine in your tea.

It seems that the quantity (not ratio) of tea in your wine and wine in your tea remains equal regardless.

• I think that this provides a reasonable intuition, but it falls short as a proof since you've really only described 3 points and not offered a description as to whether the relationship between tea and wine in each vessel is linear or otherwise Commented Nov 17, 2020 at 14:56
• True, but the question was "How would a six year old solve this problem?" I doubt that many six years old will be making proper proofs either. Commented Nov 17, 2020 at 15:53

Argument by symmetry

One way to approach the problem is to recognize the importance of the fact that you're expected to find a solution under the assumption that the tea-wine mixture in the teacup is nonuniform. In other words, it's impossible to know whether you're transferring a spoonful of tea back into the barrel, or a spoonful of wine, or some mixture of the two. What this implies is that the relative sizes of the teacup and wine barrel, as well as the proportion of wine that you scoop out of the teacup, are completely irrelevant.

With this knowledge, we can see that we should get the same answer whether we scoop a spoon of wine into the teacup and then go back, or if we scoop a spoon of tea in the wine barrel and then go back. Without knowing (or needing to know) anything about the relative sizes of the containers, we can just fill the barrel with tea and the cup with wine to do the "reverse" experiment. Transferring the wine into the tea will be no different if we have a teacup-sized wine barrel and a barrel-sized teacup - after all, there's nothing to indicate that this isn't the situation being described!

By symmetry, the only logical conclusion is that there is exactly as much tea in the wine barrel as there is wine in the teacup, regardless of whether the teacup or barrel is filled with tea or with wine to begin with. Otherwise, we would arrive at contradictory results when doing both experiments - we can't find more tea in the wine barrel when going one way, and more wine in the teacup when going the other, which is particularly obvious when we just switch the vessels that the liquids are in to begin with.

1. First We have a $$B_{wine}$$ and a $$C_{tea}$$ and a $$S$$poon
2. Now we have $$B_{wine}-S_{wine}$$ and $$C_{tea}+S_{wine}$$
3. Then we have $$B_{wine}-S_{wine}+(\frac{k}{100}S_{wine}+\frac{100-k}{100}S_{tea})$$ and $$C_{tea}+S_{wine}-(\frac{k}{100}S_{wine}+\frac{100-k}{100}S_{tea})$$

Which shows that in cup of tea we have $$\frac{100-k}{100}S_{wine}$$ and in wine barrel we have $$\frac{100-k}{100}S_{tea}$$. Of course $$S_{tea}=S_{wine}$$. (Both are one spoon)!

Imagine the tea in the wine in the form of a tiny ball within the wine. Then that ball must be exactly the amount of wine that is missing from the wine. Ergo, it is the amount of wine that is in the tea. So, the two amounts are the same, with exactly as much tea in the wine as wine in the tea.

Huh. I must be a 5 or 6 year old since I thought this was completely trivial. I have noticed though that I tend to reason visually far more often than algebraically.

Before:

After:

I mean, it has to be! No matter whether it's a spoonful, or a pinch, or whether you moved stuff back and forth 3 or 4 times, or what have you ... in the end, whatever amount of wine gets replaced by tea must have ended up in the cup.

The way I see it intuitively as a venn diagram. Two spheres represent the arbitrary amount moved around, I. This case a tea spoon amount. So when they overlap, you ask which area is greatest of the two spheres which aren't overlapping. But you see any area taken from one must be taken from the other and the area is the same.

It helps to discretise the problem. E.g red and blue marbles. Suppose there are $$M$$ red marbles in container $$1$$ and $$N$$ blue marbles in container $$2$$. We can assume $$M>>N$$ but it is unnecessary. So we start with $$\begin{array}{c|c|c} & \text{Container }1 & \text{Container }2 \\ \hline \text{Red} &M &0 \\ \text{Blue} &0 &N \end{array}$$

Pick $$a\leqslant M$$ red marbles and toss them in with the blue marbles. Thus, $$\begin{array}{c|c|c} & \text{Container }1 & \text{Container }2 \\ \hline \text{Red} &M-a &a \\ \text{Blue} &0 &N \end{array}$$ Finally, pick $$a$$ marbles from container $$2$$ and toss them in container $$1$$. Suppose $$x\leqslant a$$ of them are blue so $$\begin{array}{c|c|c} & \text{Container }1 & \text{Container }2 \\ \hline \text{Red} &M-a+(a-x) = M-x &a-(a-x)=x \\ \text{Blue} &x &N-x \end{array}$$ Even discretisation is unnecessary, but there are some philosophical debates bound to take place due to some intuition about liquids.

• Thanks to @ITA for spotting a mistake. Corrected. Commented Mar 21, 2023 at 13:10