$n$
such that $n+3$
divides $n^2+27.$

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

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$\frac{(n3)(n+3) + 36}{n+3} = (n3) + \frac{36}{n+3} = k.$
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.