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4.
If a number x = a * 11 + b (where b is the remainder of the division), then
x^2016 = A * 11 + b^2016, where A is some integer expression, a polynomial where the argument is the number 11.
In fact, if x = 15, then b = 4.
x^2016 = A * 11 + 2^4032
We notice: if the remainder were equal to 1, then 1 to any power would remain 1.
For powers of two:
2^10 is divisible by 11 with a remainder of 1. Then
2^4032 = (2^10)^403 * 2^2 = (c * 11 + 1)^403 * 2^2 = (C * 11 + 1) * 2^2
It is clear that the final remainder is four.