4. Let $\overline{a b}$ denote a two-digit number with tens digit $a$ and unit digit $b$. Find a twodigit number $\overline{x y}$ satisfying $\overline{x y}=(x-y)!(\overline{y x}-3)$.
(1 mark)
設 $\overline{a b}$ 表示十位爲 $a$ 及個位爲 $b$ 的兩位數。求滿足 $\overline{x y}=(x-y)!(\overline{y x}-3)$ 的兩位數 $\overline{x y}$ 。
4. 42
4. Note that the difference between $\overline{x y}$ and $\overline{y x}$ is $(10 x+y)-(10 y+x)=9(x-y)$, which is divisible by 9 . It follows that $\overline{x y}$ is not equal to $\overline{y x}-3$, so from the equation we see that $x-y$ is at least 2 . On the other hand, $x-y$ is at most 4 , for otherwise $(x-y)$ ! would exceed 100 . Rewrite the equation as $10 x+y=(x-y)!(10 y+x-3)$. If $x-y=2$, then eliminating $x$ gives $11 y+20=2(11 y-1)$, which yields $y=2$ and hence an answer 42 . On the other hand, setting $x-y=3$ and $x-y=4$ yield no integral solution, so that the answer is unique.