PK 2018 Q10
10. A sequence of positive integers starts with one 1, followed by two 2s, three 3s, and so on. (Each positive integer \(n\) occurs \(n\) times.) How many of the first 105 numbers in this sequence are divisible by 3?
- A 4
- B 12
- C 21
- D 30
- E 45
Pink Kangaroo 2018, Question 10. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D 30
The number of terms in the sequence up to and including the \(n\) occurrences of \(n\) is \(1 + 2 + 3 + \ldots + n = \tfrac{1}{2}n(n + 1)\), which equals 105 when \(n = 14\). Hence the numbers divisible by 3 are 3 (three times), 6 (six times), 9 (nine times) and 12 (twelve times), giving a total of \(3 + 6 + 9 + 12 = 30\) numbers.
Solution © UK Mathematics Trust, from the PK 2018 solutions.