Let the speed of A, B and C be 5v, v and v respectively. Let the circumference of the track be 12vt. Instance when A meets: B - 3t, 6t, 9t, 12t, 15t, 18t .... C - 2t, 4t, 6t, 8t, lOt, 12t .... In every 6t units of time A meets B twice while A meets C thrice. So in 36t A would give: B - 2 x 6 = 12 cards C - 3 x 6 = 18 cards, (total 30) Now, at 38t A gives C a card (his 31st) at 39t A gives B a card (his 32nd) and at 40t A gives C a card (his last) So the required difference = (18 + 2) - (12 + 1) = 7.