15. There are three piles of marbles, A, B, and C, totaling 96 marbles. The first time, marbles equal to the number in pile B are taken from pile A and added to pile B; the second time, marbles equal to the number in pile C are taken from pile B and added to pile C; the third time, marbles equal to the number left in pile A are taken from pile C and added to pile A. At this point, the number of marbles in each pile is equal. How many marbles were originally in each pile?
【Solution】Solution: Three exchanges only changed the numbers of each pile, while the total number remained unchanged. The final result is that the three piles have the same number, so the total number should be a multiple of 3.

Therefore, when the three piles of marbles are equal, each pile has: $96 \div 3=32$ (marbles),
Since the third exchange took out the same number of marbles from pile C as the remaining marbles in pile A and added them to pile A, pile A had $32 \div 2=16$ (marbles) after the second exchange, and pile C had $32+16=48$ (marbles) at that time. Since the second exchange took out the same number of marbles from pile B as pile C and added them to pile C, making pile C have 48 marbles, pile C originally had $48 \div 2=24$ (marbles);

Therefore, pile B had $32+24=56$ (marbles) after the first exchange,
Since the first exchange took out the same number of marbles from pile A as pile B and added them to pile B, pile B originally had $56 \div 2=28$ (marbles);

Therefore, pile A originally had $16+28=44$ (marbles);
Answer: Pile A originally had 44 marbles, pile B originally had 28 marbles, and pile C originally had 24 marbles.