IMC 2017 Q7
7. Four different positive integers are to be chosen so that they have a mean of 2017.
What is the smallest possible range of the chosen integers?
- A 2
- B 3
- C 4
- D 5
- E 6
Intermediate Mathematical Challenge 2017, Question 7. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 4
The smallest possible range of four positive integers is 3. Let these integers be \(n\), \(n + 1\), \(n + 2\), \(n + 3\). For the mean of these four integers to equal 2017, we have \(\frac{n + n + 1 + n + 2 + n + 3}{4} = 2017\), that is \(n = 2015\frac{1}{2}\). So the smallest possible range is not 3. However, note that the integers 2015, 2016, 2018 and 2019 have mean 2017 and range 4. So the smallest possible range of the integers is 4.
Solution © UK Mathematics Trust, from the IMC 2017 solutions. Longer versions with investigations: Solutions and investigations (PDF).