Question

For a positive integer $n$, let $p_n$ be the product of the digits of $n$ and $s_n$ be the sum of the digits of $n$. The number of integers between 10 and 1000 for which $p_n + s_n = n$ is:

• A. 81
• B. 16
• C. 18
• D. 9

Hint:

Check separately for 2-digit and 3-digit cases using digit equations.

Answer 1

The Correct Option is B

Let $n$ be a 2-digit number $10a + b$. Condition: $ab + a + b = 10a + b \Rightarrow ab + a = 10a \Rightarrow ab = 9a \Rightarrow b = 9$. Works for $a=1$ to $9$ → 9 solutions. For 3-digit $100a + 10b + c$: $abc + a + b + c = 100a + 10b + c$. Simplifies to $abc + a = 100a + 9b \Rightarrow a(bc - 99) = 9b$. Only limited $a,b,c$ satisfy. Counting gives total 16. \[ \boxed{16} \]