8.36 If $|x-\lg y|=x+\lg y$, where $x$ and $\lg y$ are real numbers, then
(A) $x=0$.
(B) $y=1$.
(C) $x=0$ and $y=1$.
(D) $x(y-1)=0$.
(E) None of these.
(23rd American High School Mathematics Examination, 1972)
[Solution] If $x-\lg y \geqslant 0$, then the given equation becomes
$$
x-\lg y=x+\lg y,
$$

so
$$
-\lg y=\lg y=0, \quad y=1 .
$$

On the other hand, if $x-\lg y \leqslant 0$, then
$$
-(x-\lg y)=x+\lg y,
$$

so
$$
2 x=0 \text { . }
$$

Notice that $x(y-1)=0$, which implies $x=0$ or $y=1$. Therefore, the answer is $(D)$.