IMC 2006 Q20
20. Given that the number 2006 is the correct answer to the calculation
\(1 - 2 + 3 - 4 + 5 - 6 + \ldots + (n - 2) - (n - 1) + n,\)
what is the sum of the digits of \(n\) ?
- A 3
- B 4
- C 5
- D 6
- E 7
Intermediate Mathematical Challenge 2006, Question 20. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D 6
As \(n\) is clearly odd, the series may be written as \(1 + [-2 + 3] + [-4 + 5] + \ldots + [-(n - 1) + n]\). So \(1 - 2 + 3 - 4 + 5 - 6 + \ldots + (n - 2) - (n - 1) + n\) \(= 1 + 1 + \ldots + 1\) \(= \tfrac{1}{2}(n + 1)\). So \(\tfrac{1}{2}(n + 1) = 2006\), that is \(n = 4011\).
Solution © UK Mathematics Trust, from the IMC 2006 solutions.