$n$ such that $n+3$ divides $n^2+27.$
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- 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
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Can you simplify
$\frac{n^2+27}{n+3}?$ -
Have you simplified the fraction sufficiently such that there are only irreducible terms?
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Since the original fraction has to reduce to an integer for it to be a solution, what condition should the irreducible term meet?
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What are the possible values of
$n$to match the above condition?
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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.