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

The correct option is D 9
Let n be a two digit number.

$n=10a+b⟹P_n=ab,S_n=a+b$

then, $ab+a+b=10a+b$

$⟹ab=9a⟹b=9$

There are 9 such numbers $19,29,39,....,99.$

Then let n be a three digit number.

$n=100a+10b+c⟹P_n=abc,S_n=a+b+c$

Then $abc+a+b+c=100a+10b+c$

$⟹abc=99a+9b$

$⟹bc=99+\frac{9b}{a}$

But the maximum value for $bc=81$.

And RHS is more than $99.$ Hence, no such number is possible.