<s>[INST] Find all positive three - digit integers $\overline{abc}$ such that $\overline{abc}=5abc$ Number is divisible by $5$ if the last digit is $0$ or $5$. Since $c\neq 0$, $c=5$.
Using brute force (checking each combination with three digits), I get that the only solution is $175$ :  $a=1,b=7,c=5$.
How to solve a problem efficiently?
 [/INST] Once you have $c=5$, it becomes clear that $100a+10b+5=25ab$. Manipulating this, we have:
$$
0=5ab-20a-2b-1=5a(b-4)-2(b-4)-8-1\implies \color{blue}{\boxed{9=(5a-2)(b-4)}}.
$$
Note that $5a-2$ can only be $-2$, $3$, $8$ or bigger than $9$. We conclude that $5a-2=3=b-4$, yielding the answer you have found.
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