What is the sum of all natural numbers n such | vedclass

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Question

(b)

Given, $n^2-10 n-36$

$n$ is a natural number.

$	herefore$ Product of its digits is $\geq 0$

$	herefore n^2-10 n-36 \geq 0$

$n=\fr

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(b)

Given, $n^2-10 n-36$

$n$ is a natural number.

$	herefore$ Product of its digits is $\geq 0$

$	herefore 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.

$	herefore \quad n \geq 13$

When $n$ is two digits numbers, then maximum product $=9 	imes 9=81$

$	herefore \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.

$	herefore \quad n \in[13,17)=13,14,15,16$

$	herefore$ Product of digits $=3,4,5,6$

When put $n=13$

$13^2-10 	imes 13-36=169-166=3$

$n=13$ satisfies

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

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