Discussion
31 children are standing in a circle and one of them (say child-1) has a ring. The ring is passed clockwise. Child-1 passes it on to child-2, child-2 passes it on to child-4, child-4 passes it on to child-7 and so on. After how many minimum such passes (including the first pass from child-1) will the ring come in the hands of child-1 again?
Answer: C
The ring returns to child-1 after 30 passes.
The k-th pass moves the ring k places (1 to 2 is one place, 2 to 4 is two, 4 to 7 is three). After n passes it has moved \(1 + 2 + \dots + n = n(n+1)/2\) places, and it is back with child-1 when that total is a multiple of 31.
\[ \begin{aligned} \frac{n(n+1)}{2} &\equiv 0 \pmod{31} \\ n = 30 &: \frac{30 \times 31}{2} = 465 = 15 \times 31 \end{aligned} \]
Since 31 is prime, it must divide n or n + 1, so the smallest n is 30. n = 31 also works (31 × 32/2 = 496 = 16 × 31), but 30 comes first, and 28, 29 and 32 do not give a multiple of 31.
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