IMC 2011 Q24
24. Given any positive integer \(n\), Paul adds together the distinct factors of \(n\), other than \(n\) itself. Which of these numbers can never be Paul's answer?
- A 1
- B 3
- C 5
- D 7
- E 9
Intermediate Mathematical Challenge 2011, Question 24. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 5
Firstly we give examples to show that Paul's answer could have been any of A, B, D or E.
A: If \(n\) is prime then the only factor of \(n\) other than itself is 1.
B: Take \(n = 4\). Its factors are 1, 2 and 4, and 1 + 2 = 3.
D: Take \(n = 8\). Its factors are 1, 2, 4 and 8, and 1 + 2 + 4 = 7.
E: Take \(n = 15\). Its factors are 1, 3, 5 and 15, and 1 + 3 + 5 = 9.
We now show that Paul's answer cannot be C. If the sum were 5, then the factors of \(n\) other than itself would have to be 1 and 4, as we are not allowed to repeat any number in the sum. However, if 4 is a factor of \(n\), then 2 is also a factor, which produces a contradiction.
Solution © UK Mathematics Trust, from the IMC 2011 solutions. Longer versions with investigations: Solutions and investigations (PDF).