$491^{1}+2^{2}+3^{3}+\cdots+1900^{1900}$ The last digit is
A. 1
B. 3
C. 7
D. 9
49 B. For positive integers $a, a^{n}(n=1,2,3, \cdots)$, the last digit of the number cycles with a period of 4 (or 2 or 1). The last digits of $1^{1}, 2^{2}, \cdots, 20^{20}$ are:
$1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0$. The sum of the last digits of the first ten numbers is 7, and the sum of the last digits of the next ten numbers is also 7. Therefore, the sequence of the last digits of all $n^{n}$ forms a periodic sequence with a period of 20. Thus, starting from 1, the sum of the last digits of every ten consecutive $n, n^{n}$ is 7, so the last digit of the original expression is the last digit of $7 \times 199$, which is 3.