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What is the sum of all natural numbers $n$ such that the product of the digits of $n$ (in base $10$ ) is equal to $n^2-10 n-36 ?$
$12$
$13$
$124$
$2612$
Solution
(b)
Given, $n^2-10 n-36$
$n$ is a natural number.
$\therefore$ Product of its digits is $\geq 0$
$\therefore n^2-10 n-36 \geq 0$
$n=\frac{10 \pm \sqrt{100+144}}{2}$
$n=5 \pm \sqrt{61}$
$n \in(-\infty, 5-\sqrt{61}) \cup(5+\sqrt{6} 1, \infty)$
But $n$ is positive integer.
$\therefore \quad n \geq 13$
When $n$ is two digits numbers, then maximum product $=9 \times 9=81$
$\therefore \quad n^2-10 n-36 \leq 81$
$n^2-10 n-117 \leq 0$
$n \in[5-\sqrt{142}, 5+\sqrt{142}]$
$n$ is taken two digit number.
$\therefore \quad n \in[13,17)=13,14,15,16$
$\therefore$ Product of digits $=3,4,5,6$
When put $n=13$
$13^2-10 \times 13-36=169-166=3$
$n=13$ satisfies
