40 children are standing in a circle and one of them (say child-1) has a ring. The ring is passed clockwise.…
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?
14
15
16
17
Solution
The positions are 1, 2, 4, 7, 11, ... where the increments are 1, 2, 3, .... After $k$ passes, the position is $1 + \dfrac{k(k+1)}{2}$. The ring returns to child-1 when $\dfrac{k(k+1)}{2}$ is a multiple of 40, i.e. $k(k+1) \equiv 0 \pmod{80}$. The smallest such $k$ is 15: $15 \times 16 = 240 = 3 \times 80$. So after 15 changes the ring is with child-1 again.