Question

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 (CAT 2005)

• A. 81
• B. 16
• C. 18
• D. 9

Answer 1

The correct option is A 81

Sol:

$\left(a\right)10<n<1000$

Let n is two digit number

$n=10a+b\hat{a}{†}^{\prime }{P}_n=ab,S_n=a+b$

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

$\hat{a}{†}^{\prime }ab=9a\hat{a}{†}^{\prime }b=9$

There are 9 such numbers $19,29,33,....,99.$

Then let n is three digit number

$\hat{a}{†}^{\prime }n=100a+10b+c\hat{a}{†}^{\prime }{P}_n=abc,S_n=a+b+c$

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

$\hat{a}{†}^{\prime }abc=99a+9b$

$\hat{a}{†}^{\prime }bc=99+9$

But the maximum value for $bc=81$

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