10. In the set $\{1,6,7,9\}$, which of the numbers appear as the last digit of $n^{n}$ for infinitely many positive integers $n$ ?
(A) $1,6,7$ only;
(B) $1,6,9$ only;
(C) $1,7,9$ only;
(D) 6,7,9 only;
(E) $1,6,7,9$.
10. Answer: (E).
Note that the power of any positive integer $n$ with last digit 1 or 6 is 1 or 6 respectively.
If the last digit of $n$ is 9 , then $n^{2}=1(\bmod 10)$, and $n^{n}=n^{10 k+9}=-1=9(\bmod 10)$.
If the last digit of $n$ is 7 , then $n^{4} \equiv 1(\bmod 10)$. Suppose the second last digit of $n$ is odd. Then $n^{n} \equiv n^{20 k+17} \equiv 7(\bmod 10)$.