Can $X \times Y = \overline{XY}$? Is it possible that $XY = \overline{XY}$, where $\overline{XY} = 10X+Y$, $X \in \{1,\dots,9\}$ and $Y = \{0,\dots,9\}$?
I am trying to show there is no way to do this using algebra, but I got stuck:

No, there is no such way, because $\overline{XY} = 10X + Y$ ,
$XY - Y = 10X$ ,
$Y(X - 1) = 10X$ ,
$Y = \dfrac{10X}{X - 1}$

I realise there are two unknowns in there, so it is a formula. I understand that I  am required to say something about the characteristics of the formula, but I don't know what.

In human words:

Prove that there is no two digit number which is the product of its digits.

 A: It is because $\overline{XY}=10X+Y\ge10X\gt XY$ (as $Y<10$ and $X>0$).
A: Because "$xy$" is a two digit number, you have an unstated assumption that $x,y$ are both integers between $0$ and $9$ and that $x \ne = 0$
And $\frac {10x}{x-1} > \frac {10x}{x} = 10$ so that is impossible.
Of course, if you have no such restrictions. then of course $y = \frac {10x}{x-1}$ has solutions.  Just let $x \ne 1$ and $y = {10x}{x-1}$.  Sux as $x = 2$ and $y = 20$ but then "$2\_120$" is not a meaningful symbol.  Although it is certainly true that $10*2 + 20 = 2*20$.
A: This is not the shortest solution but may be of interest.
Complete set of solutions in integers:
$$XY=10X+Y\quad\Leftrightarrow\quad
  XY-10X-Y=0\quad\Leftrightarrow\quad
  (X-1)(Y-10)=10\ .$$
So $X-1$ and $Y-10$ are complementary factors of $10$ and each is $\pm1,\pm2,\pm5$ or $\pm10$.  Solutions are
$$\matrix{X={}&2&3&6&11&0&-1&-4&-9\cr Y={}&20&15&12&11&0&5&8&9\cr}$$
There is no case in which $Y$ is a digit and $X$ is a non-zero (positive) digit, so your problem has no solution.
