G10.4 $A$ number $X$ consists of 2 digits whose product is 24 . By reversing the digits, the new number formed is 18 greater than the original one. What is $X$ ? (Reference: 1991 FG6.1-2)
Let the tens digit of $X$ be $a$ and the units digit be $b$.
$$
\begin{array}{l}
X=10 a+b, \text { reversed number }=10 b+a \\
a b=24 \Rightarrow b=\frac{24}{a} \ldots \ldots \text { (1) } \\
10 b+a-(10 a+b)=18 \Rightarrow b-a=2 \ldots \ldots
\end{array}
$$

Sub. (1) into (2): $\frac{24}{a}-a=2$
$$
\begin{array}{l}
24-a^{2}=2 a \\
a^{2}+2 a-24=0 \\
(a-4)(a+6)=0 \\
a=4 \text { or }-6 \text { (rejected) } \\
b=6 \\
X=46
\end{array}
$$