# Practice Paper 2 Question 14

Find all positive integers \(n\) such that \(n+3\) divides \(n^2+27.\)

## Related topics

## Warm-up Questions

- 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 \(n^2 - 3n + 2.\)

## Hints

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

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at oi.[email protected]. Please do not write to this address regarding general admissions or course queries.