40 children are standing in a circle and one of them (say child-1) has a ring. The ring is passed clockwise. Child-1passes on the 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?

Question

Accepted Answer

The correct answer is option B. The ring is passed with increasing increments: +1, +2, +3, +4, and so on. We need to find the number of passes (changes) until the ring returns to Child-1. Let C_n be the child holding the ring after 'n-1' passes (C_1 is the starting child). The position of the child is calculated modulo 40. The sequence of children holding the ring is: 1, 2 (1+1), 4 (2+2), 7 (4+3), 11 (7+4), 16 (11+5), 22 (16+6), 29 (22+7), 37 (29+8), 6 (37+9 = 46 mod 40), 16 (6+10), 27 (16+11), 39 (27+12), 12 (39+13 = 52 mod 40), 26 (12+14), 1 (26+15 = 41 mod 40). The ring returns to Child-1 after 15 passes. The question asks for 'how many such changes (including child-1)', which refers to the number of passes made to return to child-1, which is 15.

Detailed Solution

More Basic Numeracy Questions

More from UPSC 2023 CSAT

Ace UPSC with AI-powered Practice