7. A two-digit number is three times the product of its tens and units digits. What is this two-digit number?



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7. Solution: Let the tens digit be $x$, and the units digit be $y$. According to the problem, we have
$$10 x+y=3 x y$$

Dividing both sides of the equation by $x$ gives
$$10+\frac{y}{x}=3 y$$

Since 10 and $3 y$ are both integers, $\frac{y}{x}$ must also be an integer. Let $z=\frac{y}{x}$, then the original equation becomes
$$3 y-z=10$$

It is easy to see that $y=4, z=2$ is one solution. All integer solutions are
$$\left\{\begin{array}{l}
y=4+t, \\
z=2+3 t,
\end{array} \quad t=0, \pm 1, \pm 2, \cdots\right.$$

Now let's discuss the conditions that $x$ and $y$ should satisfy. A two-digit number must satisfy $0 \leqslant y \leqslant 9, 1 \leqslant x \leqslant 9$. Therefore,
$$z=\frac{y}{x} \geqslant 0$$

That is,
$$2+3 t \geqslant 0, \quad t \geqslant-\frac{2}{3}$$

Also, $y=x z \geqslant z$. Therefore,
$$4+t \geqslant 2+3 t, \text { i.e., } t \leqslant 1 \text {. }$$

Thus, $-\frac{2}{3} \leqslant t \leqslant 1$, so $t$ can only take the values 0 and 1. Substituting $t=0,1$ into the solutions gives
$t=0$ when $y=4, z=2, x=\frac{y}{z}=2$.
$t=1$ when $y=5, z=5, x=\frac{y}{z}=1$.
The two-digit numbers sought are 24 and 15.