54 2. Let the geometric sequence be $a, a \cdot \frac{1991}{1990}, \cdots, a\left(\frac{1991}{1990}\right)^{1989}$. Then when $a=n, a \cdot\left(\frac{1991}{1990}\right)^{1989} = n+k-1$, $k$ takes the minimum value. Since $k$ must be a natural number, when $a=n=1990^{1989}$, $k$ takes the minimum value, $k=1991^{1989}-1990^{1889}+1$, and its last digit is 2.
54 Let $n, n+1, \cdots, n+k-1$ be $k$ consecutive natural numbers that contain exactly 1990 numbers forming a geometric sequence with a common ratio of $\frac{1991}{1990}$. Then, when $k$ takes the minimum value, the unit digit of $k$ is $\qquad$.