Numbers –23 (CAT - 2005)

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

(1)81               (2)16               (3)18               (4)9

Solution follows here:

Solution:                                                   

The integers should be between 10 and 1000 => the integers shall contain 2 digits are 3-digits. Let us check case by case.

Case-1: 2-digit numbers:

Let the integer be n = xy => n can be written as “10x+y”

Sum of digits = S_n= x+y

Product of digits = P_n= x*y

Given that, P_n+S_n = n => (x*y)+(x+y) = 10x+y

=> x*y = 9x => x(y-9)=0 => x=0 or y=9

As x is the tens digit in the two digit number, it should not be zero => y=9

For y=9, x may take values from 1 to 9 making the possible value of integer to be 19,29,….99 => resulting in 9 values

Case-II: 3-digit numbers:

Let the integer be n = xyz=> n can be written as “100x+10y+z”

Sum of digits = S_n= x+y+z

Product of digits = P_n= x*y*z

Given that, P_n+S_n = n => (x*y*z)+(x+y+z) = 100x+10y+z

=> x*y*z = 99x+9y => z = (99x+9y)/xy = 99/y + 9/x

Observe carefully. The range of values y can take is 1 to 9, for which 99/y results in the range from 99 to 11 respectively. This is quite impossible as z should be equal to a single digit number only.  Hence no possible values arise in this case.

Hence, in total there are 9 possible integers are available.  

Answer (4)