The condition is equivalent to: $100a+10b+c=abc(a+b+c)$ <=> $9(11a+b)=(a+b+c)(abc-1)$ Now the cases: i) if $a+b+c=9k$ with $k\leq 3$ .If $a+b+c=9$ the solutions are $a=1,b=3,c=5$ and another solution $a=1,b=c=4$. If $a+b+c=18$ there are no solutions. If $a+b+c=27$ then again there exist no solutions. ii) if $abc-1=9m$, then again there are no solutions iii) if $a+b+c=3n$, $n\leq 8$ and the same time $abc-1=3l$. After all the tests and tries we have one more solution for $a=1,b=c=4$. So there exist two numbers. The numbers $135$ and $144$. This is a problem very boring why you always try and test cases

silouan wrote: The condition is equivalent to: $100a+10b+c=abc(a+b+c)$ <=> $9(11a+b)=(a+b+c)(abc-1)$ Now the cases: i) if $a+b+c=9k$ with $k\leq 3$ .If $a+b+c=9$ the solutions are $a=1,b=3,c=5$ and another solution $a=1,b=c=4$. If $a+b+c=18$ there are no solutions. If $a+b+c=27$ then again there exist no solutions. ii) if $abc-1=9m$, then again there are no solutions iii) if $a+b+c=3n$, $n\leq 8$ and the same time $abc-1=3l$. After all the tests and tries we have one more solution for $a=1,b=c=4$. So there exist two numbers. The numbers $135$ and $144$. This is a problem very boring why you always try and test cases anyone have a better solution without trial and error casework tons of times?