I am always clueless in doing this kind of problem
how are you supposed to intelligently guess the numbers?
It helps to simplify the notation a bit, and not much trial and error is needed. (Perhaps none is, but I used some.)
Each of a, b, and c is a positive integer less than 10, and none is equal to the either of the others.
[MATH]x = 100b + 10c + c = 100b + 11c \text { and } y = 40 + c \implies 122 \le x \in \mathbb Z,\ y \in \mathbb Z, \text { and } 41 \le y \le 49.[/MATH]
It is given that [MATH]y = \dfrac{x}{a} \implies ay = x \implies x \le 9 * 49 = 441 \implies 122 \le x \le 433.[/MATH]
[MATH]\text {If } a = 5, \text { integer } y = \dfrac{x}{5} = \dfrac{100b + 11c}{5} = 20b + 2c + \dfrac{c}{5}.[/MATH]
But no permissible value of c will cancel 5 so we have a contradiction. Thus, a is not 5.
Playing around with
[MATH]41a \le ay \le 49a \text { and } ay = x = 100b + 11c \ge 122 \implies[/MATH] we find the following:
[MATH]a \ge 3,[/MATH]
[MATH]b = 1 \iff a = 3 \text { or } 4,[/MATH]
[MATH]a = 6 \implies b = 2,[/MATH]
[MATH]a = 7 \implies b = 2 \text { or } 3,[/MATH]
[MATH]a = 8 \implies b = 3,\text { and}[/MATH]
[MATH]a = 9 \implies b = 3 \text { or } 4,[/MATH]
[MATH]\text {If } a = 4, \text { integer } y = \dfrac{x}{4} = \dfrac{100b + 11c}{4} = 25b + 2c + \dfrac{3c}{4}.[/MATH]
The only permissible value of c that will cancel 4 is 8, which entails that
[MATH]y = 40 + 8 = 48 \text { and } x = 100 * 1 + 11 * 8 = 188, \text { but } 4 * 48 = 192 \ne 18[/MATH][MATH]8[/MATH].
We have a contradiction. Thus, a is not 4. In short, a = 3, 6, 7, 8, or 9.
[MATH]a(40 + c) = ay = x = 100b + 10c + c \implies ac = 10d + c.[/MATH]
Now we run through the multiplication tables and find
[MATH]a = 3 \implies c = 5 \implies x = 155 \text { and } y = 45, \text { but } 3 * 45 = 135 \ne 155,[/MATH]
So a is not 3.
[MATH]a = 7 \implies c = 5 \implies x = 355 \text { and } y = 45, \text { but } 7 * 45 = 315 \ne 355,[/MATH]
So a is not 7. And a = 8 does not work at all.
[MATH]a = 9 \implies c = 5 \implies x = 455 \text { and } y = 45, \text { but } 9 * 45 = 405 \ne 455,[/MATH]
Thus a = 6. So b = 2, and c must equal 2, 4, or 8. But c cannot equal b so c must be 4 or 8.
[MATH]44 * 6 = 264 \ne 244.[/MATH]
So c must equal 8. Let's check.
[MATH]48 * 6 = 288.[/MATH]