KVPY SA Exam 2018 Question Paper

Product of digits of natural number will be a non negative integer
so, ${\mathit{n}}^2-10\mathit{n}-36\ge 0$
⇒ $\mathit{n}\in \left(-\mathit{\infty },5-\sqrt{61}\right]\cup \left(5+\sqrt{61},\mathit{\infty }\right)$
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 ${\mathit{n}}^2-10\mathit{n}-36\le 81$
⇒ $\mathit{n}\in \left[5-\sqrt{142},5+\sqrt{142}\right]$
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 ${13}^2-10\times 13-36=3$
so 13 satisfies the given condition
Hence it is a solution
chcking for n = 14
product = 1 × 4 = 4
$142-10\times 14-36$ $=196-140-36=20>6$
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
$\mathrm{f}\left(\mathit{n}\right)={\mathit{n}}^2-10\mathit{n}-36;$ $\mathrm{f}\left(100\right)={100}^2-10\times 100-36$ $=8964>729$
and f(n) is increasing Hence no 3 digit Integers and similary any higher integer will not satisfy
⇒ n = 13 is the only answer.