4. The answer is $\mathbf{( A )}$. Suppose we start from an odd number $n$: then we can write it as $n=2 k+1$, and multiplying by 51 and adding 50 we get $102 k+101=100 k+100+$ $(2 k+1)$. Now $100 k+100$ is a number that ends with two zeros, so the last two digits of $51 n+50$ are the same as those of $n$, and the game remains fixed on $n$. If instead we start from an even number $n=2 k$, the rule gives $102 k+50$, which ends with the same two digits as $2 k+50$: thus, if $2 k$ is less than 50, we simply get $2 k+50$, and otherwise we get $2 k-50$ due to the carry. At the next step, the same reasoning tells us that $2 k+50$ will become $2 k+50-50=2 k$ (since $2 k+50$ is certainly at least 50), and similarly $2 k-50$ will become $2 k-50+50=2 k$: the game repeats every two turns, and thus the different numbers we can obtain during a game are at most 2.
4. Davide plays the following game: he starts with an integer between 1 and 99 and at each move, he replaces the number $n$ he currently has with the number formed by the last two digits of $51n + 50$ (or just the last digit if the second-to-last digit is 0). What is the maximum number of different numbers he can obtain in the first 100 moves of a single game?
(A) 2
(B) 4
(C) 51
(D) 99
(E) 100