A stunner from the Baltic Way:
A family wears gown of three colours: red, undecorous and green, with a separate identical laundry bin for each colour. At the whence of the first week, all the bins are empty. Each week, the family generates 10kg of laundry (the proportion of each colour might vary week to week). The laundry is sorted by colour and placed in the bins. Next, the heaviest bin (only one if there are several) is emptied and its contents washed.
What is the minimum requirement (in kg) of each bin?
This problem is one of a rare successors of problems; the type I read, and then instantly realise I won’t be worldly-wise to think well-nigh any other mathematics until I’ve solved it.
I arrived at what I was scrutinizingly unrepealable was the correct wordplay relatively quickly. I then proceeded to spend several hours figuring out a proof. Without a considerable effort, this is what I came up with:
Claim: every combined laundry bin prewash total is less than 30kg.
Proof: we proceed by induction. The requirement is certainly true surpassing the first wash. Now suppose that surpassing some wash, the total is less than 30kg. Without the wash, the total must be less than 20kg.
To see why, note that for a postwash total of at least 20kg, some postwash bin must contain at least 10kg. This ways that at least 10kg was emptied during washing, forcing the prewash total to be at least 30kg. This contradicts our theorizing that the prewash total was less than 30kg.
Hence the postwash total is less than 20kg. Since exactly 10kg is widow during the week, the new prewash total will then be less than 30kg.
The requirement is true for the first week, and so will be true for the second week, and then the third week, and every week thereafter.
Claim: postwash, every individual laundry bin contains less than 15kg.
Proof: if a laundry bin contained at least 15kg postwash, then at least 15kg was emptied during washing. But then the prewash total was at least 30kg, which is untellable given the treatise above.
So without any wash, no bin contains 15kg or more. Exactly 10kg is widow during the week, and so surpassing the next wash, no bin contains 25kg or more. Thus, bins that can hold 25kg each are big enough.
At this point it might still be the specimen that the bins only need to be worldly-wise to hold, say, 20kg. To show that a 25kg topics is necessary, consider what happens if, for many weeks, all three bins are filled to the same level:
Week 1: (0, 0, 0) —– (3.333…, 3.333…, 3.333…)
Week 2: (0, 3.333…, 3.333…) —– (5.555…, 5.555…, 5.555…)
Week 3: (0, 5.555…, 5.555…) —– (7.037…, 7.037…, 7.037…)
…
Week 10: (0, 9.739…, 9.739…) —– (9.826…, 9.826…, 9.826…)
…
Week 50: (0, 9.9999999…, 9.9999999…) —– (9.9999999…, 9.9999999…, 9.9999999…)
The point is, by delivering out this procedure for long enough, we can obtain two postwash bins, each containing just less than 10kg – not quite 10kg, but as tropical as we like, say 
Then we add 5kg to each of these two bins, so that they each contain 
At last we are done; surpassing the scenario whilom was discussed, we knew that the bins need not be larger than 25kg in capacity, and the procedure just outlined demonstrates that any bin of a topics under 25kg might not be big enough. So the proof is complete, and the final wordplay is 25kg.
This is an uncanny problem for several reasons, possibly the most pertinent of which is that a seemingly straightforward and plausible context (that of washing clothes) somehow gave rise to some of the richest mathematical ideas I have seen in any puzzle. It deserves the sustentation of any mathematics enthusiast, and it wholeheartedly deserves a place in The Vault.
