$32 \cdot 2$ The digits of a certain three-digit number from left to right are $h, t$ and $u$, and $h>u$. When the number formed by reversing the digits is subtracted from the original number, if the difference in the units digit is 4, then the other two digits from right to left are
(A) 5 and 9.
(B) 9 and 5.
(C) Cannot be determined.
(D) 5 and 4.
(E) 4 and 5.
(6th American High School Mathematics Examination, 1955)
[Solution] Let the original number be $100h + 10t + u$, and the reversed number be $100u + 10t + h$. Their difference is (note that $u < h$):
\begin{tabular}{|c|c|c|}
\hline $100h$ & $+10t$ & $+u$ \\
\hline -) $100u$ & $+10t$ & $+h$ \\
\hline
\end{tabular}

From $10 + u - h = 4$, which means $h - u = 6$, we know that the number is 594.
Therefore, the answer is $(B)$.