$33 \cdot 7 \quad 2^{2^{1989}}+1$ the last digit is
(A) 3 .
(B) 5 .
(C) 7 .
(D) 9 .
(Hubei Huanggang Region High School Mathematics Competition, 1989)
[Solution] Since $2^{1989}=2^{3} \cdot 2^{1986}=8 k=4 k_{1}+4$, and the last digit of $2^{n}$ cycles through $2,4,8,6$, $\therefore \quad 2^{2^{1989}}+1$ has a last digit of 7.
Therefore, the answer is $(C)$.