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Say you have two containers, each with the capacity of 1,1 gallons. In one of the containers you've got 1 gallon of water and the other contains 1 gallon of 100% concentrated alcohol. You are only allowed to transfer 0,1 gallon from one container to the other back an forth as many times as you like, not more than 0,1 gallon and no less. How many times do you have to transfer fluids back and forth to reduce the concentration to 50% alcohol in one of the containers?

Ofcourse you will also mix the solutions after each transfer.

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  • $\begingroup$ in theory it looks like it's infinite, the concentrations will never drop to 50% in either of the initially 100% containers. In reality, there are finite number of molecules in each, and statistics could be used to find a number of transfers that would give effective dilutions of 50% - mathematically though, I think it can be shown to take infinite switches. $\endgroup$
    – Cato
    Apr 18, 2017 at 9:20
  • $\begingroup$ How can you show this mathematically? Can it be done by setting up a recursive formula or representing it with some infinite series? $\endgroup$
    – Parseval
    Apr 18, 2017 at 9:22
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    $\begingroup$ I'd work out formulas for the amount of alcohol in each container, starts at 1 and zero - switch 1 gives you .1 gallon of alcohol in B, switch 2 brings back .1 * .1/1.1 of alcohol, giving you .9 + .1 * .1/1.1 in A - remember that the total volume of liquid in each container is alternating - yes you need to derive an infinite series, or recursive formula - don't worry about atoms etc, it's out of scope $\endgroup$
    – Cato
    Apr 18, 2017 at 9:29
  • $\begingroup$ Yes I did about 50 iterations in excel but it was really tedious, took me 3 hours. But for the life of me I could not figure out any simpler way. I just don't see how to derive this recursive formula, eventhough my intuition says it is possible. $\endgroup$
    – Parseval
    Apr 18, 2017 at 9:31

2 Answers 2

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I am going to assume that you have to repeat a cycle of moving 0.1 gallon from container A to container B, mixing, then moving 0.1 gallon back from container B to container A, and mixing again.

If you start with a alcohol concentration of x (between 0 and 1) in one container then the concentration in the other container must be 1-x (because taken over both containers together we always have equal amounts of water and alcohol).

If you do the maths you will find the that over one cycle the concentration of alcohol changes from x to $\frac{10x+1}{12}$ (assuming prefect mixing in between transfers). Interestingly, you get the same result no matter whether your first transfer is from A to B or from B to A.

So the concentration after n cycles is given by $x_n$ where $$x_n = \frac{10x_{n-1}+1}{12}$$ and $x_0$ = 1.

A more useful way of expressing this is to let y be the excess concentration above 0.5 - so $x_n = 0.5+y_n$ and we have $$y_n = x_n - 0.5 = \frac{10x_{n-1}-5}{12} = \frac{10(0.5 + y_{n-1})-5}{12}= \frac{5y_{n-1}}{6}$$

From this you can see that $y_n$ (which starts at 0.5) gradually approaches 0, but is always positive - it never actually reaches 0. So, as Cato said above, no finite number of cycles can reduce the concentration of alcohol to exactly 50% - although you can get as close to 50% as you like.

(P.S. I misread the question and thought each container started with 1.1 gallons of liquid. If each container starts with 1 gallon of liquid then the numbers in the above change - see comments - but the general conclusion is the same)

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  • $\begingroup$ how did you get $\frac{10x+1}{12}$? $\endgroup$ Apr 18, 2017 at 11:17
  • $\begingroup$ Add 0.1 gallons from A with concentration x to 1.1 gallons in B with concentration 1-x and mix. The concentration in B - call it z - is now the weighted sum of x with weight 1 and 1-x with weight 11. Now take 0.1 gallons from B with concentration z and add it to the 1 gallon left in A with concentration x and mix. Weighted sum of z with weight 1 and x with weight 10 gives you the result. $\endgroup$
    – gandalf61
    Apr 18, 2017 at 12:11
  • $\begingroup$ B has quantity 1-x in it, and we add 0.1x from A, giving 1 - 0.9x in 1.1 gallons. what we send back of x is (1 - 0.9x) / 11 = 1/11 - (9x/110). So A then has (9/10)x + 1/11 - (9x/110) = (9x + 1) / 11 $\endgroup$
    – Cato
    Apr 18, 2017 at 15:00
  • $\begingroup$ if you started with 1 gallon, then you send away 0.1 gallons, but bring 1/11 of it back, giving you 1 - 1/10 + 1 / 110 = 110 - 11/110 +1/ 110 = 100/110 = 10 / 11 after one cycle $\endgroup$
    – Cato
    Apr 18, 2017 at 15:07
  • $\begingroup$ my bad - I misread and though each container started with 1.1 gallons. If each container starts with 1 gallon then I agree the new concentration after 1 cycle is (9x+1)/11 $\endgroup$
    – gandalf61
    Apr 18, 2017 at 15:15
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Follow up: there is a simple argument to show that if we start with equal amounts of alcohol and water and the initial concentrations in the two containers are not both 50% then no method of transfers between the containers can reach 50% concentration unless we are allowed to empty one container completely (which obviously creates a 50% concentration in the other container).

The argument is as follows. Suppose we can reach equal concentrations of 50% in both containers. If the final transfer is from container A to container B then the concentration in container A is not changed by this final transfer. Therefore the concentration in container A was 50% before the final transfer. But we have equal amounts of alcohol and water across both containers. Therefore the concentration in container B must also have been 50% before the final transfer unless container B was empty. So starting from initial concentrations that are different from 50%, we cannot reach a 50% concentration unless we are allowed to empty one container into the other.

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  • $\begingroup$ Wonderful, I really appreciate your help! May I ask what you work with/study and your age? $\endgroup$
    – Parseval
    Apr 19, 2017 at 10:12
  • $\begingroup$ @Parseval - I work in IT but I did a mathematics degree (many years ago) and I have an amateur interest in recreational mathematics and number theory. $\endgroup$
    – gandalf61
    Apr 19, 2017 at 11:19

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