For a positive integer n, let Pn denotes the product of digits of n, and Sn denotes the sum of the digits of n. The number of integers between 10 and 1000 for which Pn+ Sn = n is?
(1)81 (2)16 (3)18 (4)9
Solution follows here:
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 = Sn= x+y
Product of digits = Pn= x*y
Given that, Pn+Sn = 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 = Sn= x+y+z
Product of digits = Pn= x*y*z
Given that, Pn+Sn = 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)