my solution indicates $162, 243, 324$ and $405$ only, but $605$ is also correct(in calculator), so somewhere there could be a problem or something missed.. here it is shown
the first part is, like avik da's way $a+b+c= 9,11,19,21$
if $a+b+c= 9 $then $10a+b-8c=0..........$..(line 1
$a+b+c= 11 $then $10a+b-12c=0......$.....(line 2)
$a+b+c= 19$ then $10a+b-36c=0.....$......(line 3)
$a+b+c= 21$ then $10a+b-44c=0....$.......(line 4)
since a, b, c are digits, so, $a-b, b-c,$ or $c-a$, should be an integer, now we can find from the three lines that only for line one c-a (or a-c) is an integer and c-a =1, which means $c= a+1$
so we should consider only the equations derive from line 1 now
since the number is like $100a+10b+c$, it can be written now $100a+10b+1+a= 101a+10b+1$
since c-a=1 again we can find $2a+b= 8$
or $b= 8-2a$, again entering the value of b the number becomes like this
$101a+80-20a+1=81a+81$
so ultimately the number is =$81(a+1)$
considering a>=5 we see the first digit of $81(a+1)$ is not the same as a (example a=5 then the number is 486,here a=4,contradiction
a<5 we can see the frist digit of 81(a+1)= a
so the numbers are $81(a+1), a= [1,4]$
now my question, how can the digit become 605???
, please give me help..