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# Practice Paper 2 Question 14

Find all positive integers $$n$$ such that $$n+3$$ divides $$n^2+27.$$

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## Warm-up Questions

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 $$n^2 - 3n + 2.$$

## Hints

• Hint 1
Can you simplify $$\frac{n^2+27}{n+3}?$$
• Hint 2
Have you simplified the fraction sufficiently such that there are only irreducible terms?
• Hint 3
Since the original fraction has to reduce to an integer for it to be a solution, what condition should the irreducible term meet?
• Hint 4
What are the possible values of $$n$$ to match the above condition?

## Solution

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.

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