IMC 2018 Q24
24. A list of 5 positive integers has mean 5, mode 5, median 5 and range 5.
How many such lists of 5 positive integers are there?
- A 1
- B 2
- C 3
- D 4
- E 5
Intermediate Mathematical Challenge 2018, Question 24. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: B 2
As the mode of the list is 5, at least two of the integers in the list are 5. The median of the list is 5, so when the list is written in ascending order the third integer is 5. So, since the range of the list is 5, the two possible lists, in ascending order, are \(x, y, 5, 5, x + 5\) or \(x, 5, 5, y, x + 5\), where \(x\) and \(y\) are positive integers and \(x \leqslant y \leqslant x + 5\). The mean of the integers is 5, so their sum is 5 × 5 = 25, for each possible list. Therefore \(x + y + 5 + 5 + x + 5 = 25\). So \(2x + y = 10\). The positive integer solutions of this Diophantine equation are (1, 8), (2, 6), (3, 4), (4, 2). Note that neither \(x\) nor \(y\) equals 5, so there are exactly two 5s in the list. Of the solutions, neither (1, 8) nor (4, 2) satisfy the condition \(x \leqslant y \leqslant x + 5\).
So the only two possible lists are 2, 5, 5, 6, 7 and 3, 4, 5, 5, 8.
Solution © UK Mathematics Trust, from the IMC 2018 solutions. Longer versions with investigations: Solutions and investigations (PDF).