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 sequence of children holding the ring follows the pattern where the position increases by 1, then 2, then 3, and so on (arithmetic progression of differences).
• The position of the child after $n$ changes is given by the sum of the first $n$ natural numbers: $S_n = \frac{n(n+1)}{2}$.
• Since the children are in a circle of 40, the ring returns to child-1 when $S_n$ is a multiple of 40 (i.e., $S_n \equiv 0 \pmod{40}$).
• We test values of $n$: For $n=14$, $S_{14} = \frac{14 \times 15}{2} = 105$, and $105 \pmod{40} = 25$ (Child 25).
• For $n=15$, $S_{15} = \frac{15 \times 16}{2} = 120$, and $120 \pmod{40} = 0$ (Child 1).
• Thus, after 15 changes, the ring is back with child-1, making option (b) correct.
Why the other options are wrong
- (a) 14
- After 14 changes, the ring is with child 25, not child 1.
- (c) 16
- After 16 changes, the ring is with child 26, not child 1.
- (d) 17
- After 17 changes, the ring is with child 28, not child 1.
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.