Find all positive integers $n$ such that $n+3$ divides $n^2+27.$
1. How many integers $k$ are there such that $7(k^2+k) < 1337?$
2. List all prime factors of $48$ and then find all factors of $48$ by combining the primes.
3. Factorise $n^2 - 3n + 2.$

Can you simplify $\frac{n^2+27}{n+3}?$

Have you simplified the fraction sufficiently such that there are only irreducible terms?

Since the original fraction has to reduce to an integer for it to be a solution, what condition should the irreducible term meet?

What are the possible values of $n$ to match the above condition?

If $n+3$ divides $n^2 + 27$ then $\frac{n^2+27}{n+3} = k$, where $n, k$ are integers. We can use algebraic manipulation (or polynomial long division) to reduce the fraction to $\frac{(n-3)(n+3) + 36}{n+3} = (n-3) + \frac{36}{n+3} = k.$ For $k$ to be an integer, $\frac{36}{n+3}$ must also be an integer. This means $n+3$ has to be a factor of $36,$ which is only possible if and only if $n \in \{1, 3, 6, 9, 15, 33\}.$

Note: $0$ is not a positive integer.