Let's practice

$\begin{aligned} (100a+10b+c)(a+b+c) & = 2005 \\ (99a+9b + {\color{#3D99F6}a + b +c})({\color{#3D99F6}a + b +c}) & = 2005 & \small \color{#3D99F6} \text{Let }d = a+b+c \\ {\color{#D61F06}(99a+9b + {\color{#3D99F6}d})}{\color{#3D99F6}d} & = {\color{#3D99F6}5}\cdot\color{#D61F06} 401 & \small \color{#3D99F6} \text{Both factors of 2005 are primes.} \\ {\color{#D61F06}(99a+9b + {\color{#3D99F6}5})}{\color{#3D99F6}5} & = {\color{#3D99F6}5}\cdot\color{#D61F06} 401 & \small \color{#3D99F6} \text{We can assume }d=5 \text{ and...} \\ 99a+9b + 5 & = 401& \small \color{#3D99F6} \text{Substract both sides by 5.} \\ 99a+9b & = 396 & \small \color{#3D99F6} \text{Divide both sides by 9.} \\ 11a + b & = 44 \\ \implies a & = \boxed{4} & \small \color{#3D99F6} \text{The only possible solution.} \end{aligned}$

We can observe that both the quantities are integers. By prime factorisation of, 2005=5.401.(a,b,c)=(4,0,1) so it hold good. So, the answer is 4