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 ?
- (a) 14
- (b) 15 ✓ UPSC's answer
- (c) 16
- (d) 17
Why the answer is (b)
• The ring moves 1, then 2, then 3 places, and so on, so after n passes it has moved 1 + 2 + … + n = n(n + 1)/2 places.
• It is back with child-1 when this total is a multiple of 40.
• n = 14 gives 105 (child 26 — not a multiple of 40); n = 15 gives 120 = 3 × 40.
• So the ring returns to child-1 after 15 changes, option (b).
Why the other options are wrong
- (a) 14
- After 14 passes the ring has moved 105 places, not a multiple of 40.
- (c) 16
- The ring is already back after 15 passes; after 16 it has moved 136 places.
- (d) 17
- The ring is already back after 15 passes; after 17 it has moved 153 places.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2023, held on 28 May 2023. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.