let S(0) = x

then

S(1) = 5x + 4

S(2) = 25x + 4*5 + 4

S(3) = 125x + 4*5^2 + 4*5 + 4

S(n) = x*5^n + c(n), where c(n) = 4(5^(n-1) + 5^(n-2) + ... 1)

Because 5^n is relatively prime to 2016, there are infinitely many values of x that make x*5^n = any chosen value (mod 2016). In particular, there are infinitely many values of x that make x*5^n = -c(n) (mod 2016).

Therefore, there are infinitely many values of x that make S(n) = 0 (mod 2016) for a specific n.

Therefore, there are infinitely many values of x that make S(54) = 0 (mod 2016).

*Edited on ***February 25, 2016, 3:17 pm**