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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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