How many positive integral values of $n$, less than $100$, are there such that $n^3+72$ is completely divisible by $n+7$? How many positive integral values of $n$, less than $100$, are there such that $n^3+72$ is completely divisible by $n+7$ ?
MY WORK :-
Well we know that $a+b$ divides $a^3+b^3$ thus $n+7$ divides $n^3+7^3$ thus it divides $n^3+343$. Now if $n+7$ divides $n^3+72$ then it divides their difference. Thus $n+7$ divides $343-72=271$. Now the factor of $271$ are $1$ and $271$ only. So if $n\lt100$ then there are no values possible; hence answer should be $0$. However in answer key it's said $1$. I want to know why it's $1$ ??

Do the polynomial division with remainder:
$${n^3+72\over n+7}=n^2-7n+49-{271\over n+7}\ .$$
Now $n^2-7n+49$ is an integer for all $n$. Therefore it remains to check, for which $n\in[100]$ the number $271$ is divisible by $n+7$.
