$30 \cdot 8$ If $a, b, c$ are positive integers, and $a, b$ are odd, then $3^{a}+(b-1)^{2} c$ is
(A) odd.
(B) even.
(C) when $c$ is even, it is odd; when $c$ is odd, it is even.
(D) when $c$ is odd, it is odd; when $c$ is even, it is even.
(E) if $c$ is not a multiple of 3, it is odd; if $c$ is a multiple of 3, it is even.
(43rd American High School Mathematics Examination, 1992)
[Solution] Since $3^{a}$ is an odd number; and $b$ is an odd number, then $b-1$ is even, thus $(b-1)^{2}$ is even, and consequently $(b-1)^{2} c$ is even. Therefore, $3^{a}+(b-1)^{2} c$ is odd. Hence, the answer is $(A)$.