$27.1+2^{2}+3^{3}+\cdots+2017^{2017}$ The last digit of this expression is ( )
(A) 1
(B) 3
(C) 5
(D) 6
Answer: (A)
Solution: We are asked to find the last digit, so we only need to consider the last digit of each addend.
The last digits of $1,11^{11}, 21^{21}, \ldots, 2011^{2011}$ are all 1,
The last digits of $2^{2}, 12^{12}, 22^{22}, \ldots, 2012^{2012}$ are:
$4,6,4,6, \ldots$
The last digits of $3^{3}, 13^{13}, 23^{23}, \ldots, 2013^{2013}$ are $7,3,7,3, \ldots$
The last digits of $4^{4}, 14^{14}, 24^{24}, \ldots, 2014^{2014}$ are all 6,
The last digits of $5^{5}, 15^{15}, 25^{25}, \ldots, 2015^{2015}$ are all 5,
The last digits of $6^{6}, 16^{16}, 26^{26}, \ldots, 2016^{2016}$ are all 6,
The last digits of $7^{7}, 17^{17}, 27^{27}, \ldots, 2017^{2017}$ are:
$$
3,7,3,7, \ldots
$$
The last digits of $8^{8}, 18^{18}, 28^{28}, \ldots, 2008^{2008}$ are:
$$
6,4,6,4, \ldots
$$
The last digits of $9^{9}, 19^{19}, 29^{29}, \ldots, 2009^{2009}$ are all 9.
The last digit of $(1+6+5+6) \times 202+(4+6+7+3) \times 101+(6+4) \times 100+6+9 \times 201$ is 1, so the last digit of $1+2^{2}+3^{3}+\cdots+2017^{2017}$ is 1.