Arithmetic mean
On a blackboard, there are 2016 integers, from 1 to 2016 (including 1 and 2016). You are allowed to remove two numbers you like, but you have to add the arithmetic mean as a new number. (If you delete for example 10 and 11, you have to write 10.5 as the new number). Therefore, the number of integers on the board decrease in every turn, and at the end of the game, only one integer will remain. What is the lowest value of this number?
The answer is 2.
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The lowest integer cannot be lower than 1, because 1 is the lowest number. Also the lowest integer cannot be equal to 1, because the arithmetic mean of 1 and a integer greater than 1 is greater than 1 too. Now we have to show that 2 is possible: At the beginning there are the 2016 numbers: 1, 2, 3, ..., 2014, 2015, 2016 You remove 2014 and 2016 and add the arithmetic mean (2015). 1, 2, 3, ..., 2013, 2015, 2015 Then remove 2015 twice and add 2015. 1, 2, 3, ..., 2013, 2015 In the next turns, remove the two highest numbers. As their difference is 2, the new number will always be an integer. In the end you get: 1, 2, 3, 4, 6 1, 2, 3, 5 1, 2, 4 1, 3 2 Therefore, 2 is possible to be the last integer.