The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. 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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