40 children are standing in a circle and one of them (say child-1) has a ring. The ring is passed clockwise. Child-1 passes on to child-2, child-2 passes on to child-4, child-4 passes on to child-7 and so on. After how many such changes (including child-1) will the ring be in the hands of child-1 again?
Answer & explanation
Answer: (b) 15
The ring moves forward 1, 2, 3, … places on successive passes, so after k passes it has moved 1 + 2 + … + k = k(k + 1)/2 places. It is back with child-1 when this is a multiple of 40, which first happens at k = 15 (120 places, i.e. 3 full rounds).
- Child-1 → 2 → 4 → 7 → …: the jumps are 1, 2, 3, … places.
- After k passes the ring has moved k(k + 1)/2 places round the circle of 40.
- It is back with child-1 when k(k + 1)/2 is a multiple of 40, i.e. when k(k + 1) is a multiple of 80.
- k = 15: 15 × 16 = 240 = 3 × 80 works; no smaller k does (for example 14 × 15 = 210 and 9 × 10 = 90).
- So after 15 passes (120 places, 3 full rounds) the ring returns to child-1.
Remember · Growing jumps 1, 2, 3, … add up to triangular numbers k(k + 1)/2; returning to the start needs a multiple of the circle size.
Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·