Product of digits of natural number will be a non negative integer so, n2−10n−36≥0 ⇒ n∈(−∞,5−√61]∪(5+√61,∞) but n ∈ IN so n ≥ 13; where n ∈ N case-1 for all 2 digit natural numbers max value of product of digits = 9 × 9 = 81 so n2−10n−36≤81 ⇒ n∈[5−√142,5+√142] but n is taken as a 2 digit natural no.; so 13 ≤ n < 17; ⇒ product of digits = 3, 4, 5 or 6 for 13, 14, 15 and 16 respectively checking n = 12 product of digits = 1 × 3 = 3 and 132−10×13−36=3 so 13 satisfies the given condition Hence it is a solution chcking for n = 14 product = 1 × 4 = 4 142−10×14−36=196−140−36=20>6 and n2 – 10n – 36 is increasing function for n > 5; rest of the 2 digit integers won't satisfy the given condition case-2 for all 3-digit integers max product = 9 × 9 × 9 = 729 The smallest 3 digit no. is 100 f(n)=n2−10n−36;f(100)=1002−10×100−36=8964>729 and f(n) is increasing Hence no 3 digit Integers and similary any higher integer will not satisfy ⇒ n = 13 is the only answer.