$n\gt 2$ paint cans of identical shape and volume are filled with different type of paints. Each of them is filled with $\frac {n-1}{n}$ of their volume. It's allowed to fill the paint of one can to the other can.
Prove that one can fill each can with paint such that in each can there is the same mixture.
Can somebody help me?
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If we are allowed a measuring implement, then this is the method I thought of. The difference form Hagen's is that he empties the last can, then fills it up, then takes care of all the other cans, while I first fill up the last can, then take care of all the others, then empty the last can. It might be interesting to look into why one can swap the order around like this and still get it correctly.
If there is only one can, there is no paint in it to mix, so in this case, mission accomplished, you are done. Otherwise take $\frac{1}{n(n-1)}$ from each of the other cans into the last one. The last one is now full, while all the others are $\frac{n-2}{n-1}$ full. Mix the others by the method described in this very paragraph, but for $n$ one smaller, so that they are all the same. Stir the last bucket, then fill all the other buckets with that paint, so that they get the same amount of paint each from the last bucket, and stir all the other buckets.
Pour from can $n$ into each other cans until they are full (and can $n$ is empty) and stir thoroughly. Now each can has the correct proportion of colour $n$, and this remains true no matter what we do from now on.
Next fill $\frac1n$ of each can into the $n$th can (apparently, this is an allowed step). Now we have the correct final mix in the $n$th can. Each of the other $n-1$ cans is $\frac{n-2}{n-1}=1-\frac1{n-1}$ full so that we have reduced the problem to the case $n-1$.
Ultimately, the case $n=1$ is trivial.