Product and sum of digits equation

Case 1: $n$ is a 2-digit number, $n = 10a + b$. $p_n = ab$, $s_n = a + b$. $ab + a + b = 10a + b \implies ab = 9a \implies b = 9$ (since $a \ne 0$). So $b$ must be 9, and $a$ can be any digit from 1 to 9. Number of solutions $= 9$ (19, 29, 39, 49, 59, 69, 79, 89, 99).

Case 2: $n$ is a 3-digit number, $n = 100a + 10b + c$. $abc + a + b + c = 100a + 10b + c \implies abc = 99a + 9b$. Since $a, b, c \le 9$, $abc \le 81 \times 9 = 729$, but $99a \ge 99$, which yields no valid 3-digit solutions. Total solutions $= 9$.