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

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(B) Let $P(n)$ be the product of the digits of $n$. We are given $P(n) = n^2-10n-36$.Since $P(n) \geq 0$,we have $n^2-10n-36 \geq 0$. Solving $n^2-10n

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$13$

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(B) Let $P(n)$ be the product of the digits of $n$. We are given $P(n) = n^2-10n-36$.Since $P(n) \geq 0$,we have $n^2-10n-36 \geq 0$. Solving $n^2-10n-36 = 0$ gives $n = 5 \pm \sqrt{61}$. Since $n$ is a natural number,$n \geq 5 + \sqrt{61} \approx 12.8$,so $n \geq 13$.If $n$ is a $2$-digit number,$n = 10a+b$,then $P(n) = ab \leq 81$. Thus $n^2-10n-36 \leq 81$,which implies $n^2-10n-117 \leq 0$. The roots are $5 \pm \sqrt{142} \approx 5 \pm 11.9$,so $n \leq 16.9$. Thus $n \in \{13, 14, 15, 16\}$.For $n=13$,$P(13) = 1 	imes 3 = 3$ and $13^2-10(13)-36 = 169-130-36 = 3$. This works.For $n=14$,$P(14) = 4$ and $14^2-10(14)-36 = 196-140-36 = 20 
eq 4$.For $n=15$,$P(15) = 5$ and $15^2-10(15)-36 = 225-150-36 = 39 
eq 5$.For $n=16$,$P(16) = 6$ and $16^2-10(16)-36 = 256-160-36 = 60 
eq 6$.If $n$ is a $3$-digit number,$n \geq 100$,then $n^2-10n-36 > 100^2-1000-36 = 8964$,but the maximum product of digits for a $3$-digit number is $9^3 = 729$. Thus,no $3$-digit or higher number works.The only solution is $n=13$.

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

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