$30 \cdot 6$ If a number is equal to the square of some natural number, it is called a perfect square. It is known that $a$ and $b$ are two perfect squares, the unit digit of $a$ is 1, and the tens digit is $x$; the unit digit of $b$ is 6, and the tens digit is $y$. Then
(A) $x, y$ are both odd.
(B) $x, y$ are both even.
(C) $x$ is odd, $y$ is even.
(D) $x$ is even, $y$ is odd.
（2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990）
[Solution 1] Since $9^{2}=81$, and $6^{2}=36$, if we let $a=81, b=36$, then we know that (A), (B), (C) are not true.
Therefore, the answer is (D).
[Solution 2] Since the unit digit of $a$ is 1, and $a=m^{2}$, the unit digit of $m$ can only be 1 or 9. The hundred's digit of $m$ does not affect the ten's digit of $a$. Let the ten's digit of $m$ be $c$, then the ten's digit of $a$ is the unit digit of $2c$, or the unit digit of $18c+8$, both of which are even numbers.

Since the unit digit of $b$ is 6, and $b=n^{2}$, the unit digit of $n$ is 4 or 6. Let the ten's digit of $n$ be $c$, then the ten's digit of $b$ is the unit digit of $8c+1$, or the unit digit of $12c+3$, both of which are odd numbers.
Therefore, the answer is (D).