IMC 2017 Q16
16. The product of two positive integers is equal to twice their sum. This product is also equal to six times the difference between the two integers. What is the sum of these two integers?
- A 6
- B 7
- C 8
- D 9
- E 10
Intermediate Mathematical Challenge 2017, Question 16. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D 9
Let the two positive integers be \(m\) and \(n\). Then \(mn = 2(m + n) = 6(m - n)\).
So \(2m + 2n = 6m - 6n\), that is \(8n = 4m\). Therefore \(m = 2n\). Substituting for \(m\) gives: \((2n)n = 2(2n + n)\). So \(2n^2 = 6n\), that is \(2n(n - 3) = 0\).
Therefore \(n = 0\) or 3. However, \(n\) is positive so the only solution is \(n = 3\).
Therefore \(m = 2 \times 3 = 6\) and \(m + n = 6 + 3 = 9\).
Solution © UK Mathematics Trust, from the IMC 2017 solutions. Longer versions with investigations: Solutions and investigations (PDF).