What is the sum of all natural numbers n such that the product of the digits of n (in base 10 ) is equal to n 2 - 10 n - 36 ?

Step-by-step solution

Given, n 2 - 10 n - 36 n is a natural number. ∴ Product of its digits is ≥ 0 ∴ n 2 - 10 n - 36 ≥ 0 n = 10 ± 100 + 144 2 n = 5 ± 61 ∴    n ∈ ( - ∞ , 5 - 61 ) ∪ ( 5 + 61 , ∞ ) But n is positive integer. ∴    n ≥ 13 When n is two digits numbers, then maximum product = 9 × 9 = 81 ∴    n 2 - 10 n - 36 ≤ 81 n 2 - 10 n - 117 ≤ 0 ∴    n ∈ [ 5 - 142 , 5 + 142 ] n is taken two