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?

You already know that $c=5$.
Hence $100a+10b+5=25ab$.
Hence $20a+2b+1=5ab$.
$20a+2b+1$ is odd.
Hence $5ab$ is odd.
Hence $5ab$ ends with $5$.
Hence $20a+2b+1$ ends with $5$.
Hence $2b+1$ ends with $5$.
Hence $2b$ ends with $4$.
Hence $b=2$ or $b=7$.
Since $5ab$ is odd, $b$ must be odd.
Hence $b=7$.
Hence $20a+2\cdot7+1=5a\cdot7$.
Hence $a=1$.
