For a positive integer $n$, $p_n$ denotes the product of the digits of $n$, and $s_n$ denotes the sum of the digits of $n$. What is the number of integers between 10 and 1000 for which $p_n + s_n = n$ ?
Let $n = xy$ be the two digit number satisfying the given condition.
Given, the product of the digits of $n + \text{sum of the digits of $n$} = n$,
\begin{align}
p_n + s_n &= n\\
xy + x + y &= 10x + y\\
9x – xy &= 0\\
x (9 – y) &= 0.
\end{align}
But $x$ is not zero, because $xy$ is a two digit number.
So, $9 – y = 0 ⇒ y = 9$.
So, $xy$ can be 19, 29, 39, 49, 59, 69, 79, 89, and 99. i.e, 9 numbers.
Let $n = xyz$ be the three digit number satisfying the given condition.
Since $p_n + s_n = n$
\begin{align}
xyz + x + y + z &= 100x + 10y + z.\\
99x + 9y – xyz &= 0.\\
xyz &= 99x + 9y.
\end{align}
Best Answer
Given the conditions, your last equation doesn't have any solution. Let's see.
As $x \neq 0$ let's divide the equation by $x$: $$yz=99+ \frac{y}{x}$$ Therefore $x |y$.
But for $y=0$ there's no solution, and for $y \neq 0$:
$$yz >99 \Rightarrow y>10 \, \mathrm{or} \, z>10.$$ Which is impossible since $y$ and $z$ are digits.