Determine the sum of all possible positive integers n , the product of whose digits equals n^2-15 n- 27.

Step-by-step solution

n^2-15 n-27 is always odd number for all n I n must be maximum of two digit number. because maximum product of three digit number is 729 & minimum value of n^2-15 n-27 for 3 digits number is 10000-1500-27 which is greater than 729 . n^2-15 n-27 is increasing function for all n 8,9,10, . at n=17, n^2-15 n-27 is equal to 17 at n=19, n^2-15 n-27 is equal to 49 at n=21, n^2-15 n-27 is equal to 99 And maximum product of digits of two digit number is 81 So n must be less than 21 Between 1 to 15, n^2-15 n-27 is negative S