5. Write down the two-digit numbers from 19 to 96 in sequence to form a natural number \( N, N=19202122 \cdots 949596 \).

If the highest power of 3 in the prime factorization of \( N \) is \( 3^{k} \). Then \( k=(\quad) \).
(A) 0
(B) 1
(C) 2
(D) 3
5. (B).
The remainder when $N$ is divided by 3 is the same as the remainder when the sum $19+20+21+22+\cdots+95+96$ is divided by 3. It is easy to see that the remainder is $0$. The remainder when $N$ is divided by 9 is the same as the remainder when the sum $19+20+21+22+\cdots+95+96$ is divided by 9. It is easy to see that the remainder is 3. That is, $3 \mid N \cdot 3^{2} \times N$. Therefore, $k=$ 1.