IMC 2015 Q11
11. Three different positive integers have a mean of 7. What is the largest positive integer that could be one of them?
- A 15
- B 16
- C 17
- D 18
- E 19
Intermediate Mathematical Challenge 2015, Question 11. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D 18
As the 3 numbers have mean 7, their sum equals 3 × 7 = 21. For one of the numbers to be as large as possible the other two numbers must be as small as possible. They must also be different and so must be 1 and 2. Hence the largest possible of the three numbers equals 21 − (1 + 2) = 18.
Solution © UK Mathematics Trust, from the IMC 2015 solutions. Longer versions with investigations: Solutions and investigations (PDF).