A heap of 48 matches are divided into three groups.

If I take as many matches from the first group as there are in the second group and add them to the second, and then take as many from the second group as there are in the third group and add them to the third, and finally take as many from the third group as there are in the first group and add them to the first group, the number of matches in each heap would be equal.

How many matches were in the three groups originally?

(In reply to

answer by K Sengupta)

At the outset, we observe that the each of the respective number of matches in the three piles after the third operation is each equal to 48/3 = 16

Thus, the respetive total of piles (one, two, three) after the second operation was (16/2, 16, 16+16/2) or, (8, 16, 24).

Similarly, the respetive total of piles (one, two, three) after the first operation was (8, 16 + 24/12, 24/12) or, (8, 28, 12).

Finally, he respetive total of piles (one, two, three) before the first operation was (8 + 28/2, 28/2,12) or, (22, 14, 12).

Consequently, the total number of matches in the three piles originally was as follows:

# Pile One contained 22 matches

# Pile Two contained 14 matches, and:

# Pile Three contained 12 matches