# Last Edition Puzzle

There are 10 big buckets (each capable of holding up to 10 litres of water). Each day, I pour a total of 1 litre of water in the buckets in an arbitrary way (i.e. I can pour the entire 1l  in a chosen bucket, or distribute it equally or in any other way).ou pick any one bucket at random and empty it. This goes on for 9 days. On the 10th day, after I pour the 1litre water in a similar arbitrary manner I pick one bucket of water for my consumption. If I want to maximize the capacity of the bucket I choose on the 10th day, what should be my strategy each day to distribute the water and, what is the maximum volumey I can get?

For example, if I pour my 1litre in any ONE empty bucket each day, then you might empty that each day. So I am just guaranteed the 1litre of the water I pour in a bucket on the last day.

Alternatively, if I distribute the water equally every day, then since you can empty only 9 buckets, at least one bucket will have 1/10 litre from each of the days (this ratio is slightly less on the second day, third day so on), and so on overall I can get roughly 1 + 9/10 litres, i.e. slightly less than 2 Litres.

Is there a strategy by which you can guarantee more than 2 litres in a bucket on the final day?

**Solution**: 

Here is a strategy that works. 

On day 1, I will pour the 1litre water equally among all buckets, so each bucket has 1/10 Litre. 

On day 2, I will pour the 1litre water equally among the non-empty buckets, so each non-empty bucket gets 1/10+1/9.

On day 3, similarly I will pour the 1litre water equally among the non-empty buckets, so each non- empty bucket gets 1/10+1/9 +1/8 etc ...

Finally, one bucket will have, after day 10, 1+1/2+1/3+... 1/10, this is already > 1+1/2+1/3+1/4 = 25/12 > 2.

**Congratulations to Adya of Marutham Farm School, Thiruvannamalai who wins a** ₹**1000 voucher for solving the puzzle.**