$13 \cdot 47$ Let $x, y$ be positive integers, and $x<y$. Also, the characteristic of $\lg x$ is $a$, and the mantissa is $\alpha$; the characteristic of $\lg y$ is $b$, and the mantissa is $\beta$, satisfying the conditions $a^{2}+b^{2}=5, \alpha+\beta=1$. Then the maximum value of $x$ is 
(A) 50.
(B) 80.
(C) 100.
(D) 200.
(Anhui Province Hefei City Junior High School Mathematics Competition, 1990)
[Solution 1] $\because a, b$ are integers, and $a^{2}+b^{2}=5$, from $x<y$ we can deduce $a \leqslant b$. Therefore, $a=1, \quad b=2$. Then $\square$
$$
x \cdot y=10^{a+\alpha} \cdot 10^{b+\beta}=10^{a+\alpha+b+\beta}=10^{4} \text {. }
$$

Since $x<y$, it is easy to get $x_{\text {max }}=80, y_{\text {min }}=125$.
Therefore, the answer is (B).
[Solution 2] $\because \lg x=a+\alpha, \lg y=b+\beta$,
$$
\therefore \quad \lg x y=(a+b)+(\alpha+\beta)=(a+b)+1 \text {, }
$$

Since $\because x, y \in N, \therefore a \geqslant 0, \quad b \geqslant 0, \quad$ and $\quad x<y$.
$\therefore a=1, b=2$. Thus, $x$ is a two-digit number.
Since $x y=10^{4}$, then $x$ should be the largest two-digit divisor of $10^{4}$, i.e., $x_{\text {max }}=$
80. Therefore, $y_{\min }=125$.
Therefore, the answer is $(B)$.