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?

Friedlinguini got the answer by working through the problem forward. I found the answer working backwards:

In the end, each pile had 16 matches. So the final state is:
16 16 16

The last statement says "Take as many from the 3rd group as there are in the 1st group and add them to the 1st group." Well, this means that the quantity in the 1st group is going to get doubled, and we know the result will be 16 matches, so before we follow the last instruction, the 1st group had 8 matches. The 8 matches that double group 1 came from group 3 so our second to last state is:
8 16 24

Next we deal with "Take as many from the 2nd group as there are in the 3rd group and add them to the 3rd." So now the quantity in the 3rd group is going to get doubled, and we know the result will be 24 matches. So before we follow the second to last instruction, the 3rd group had 12 matches. The 12 matches that double group 3 came from group 2 so our third to last state is:
8 28 12

And finally (well, firstly) the first statement says "Take as many matches from the 1st group as there are in the 2nd group and add them to the 2nd." This time the quantity in the 2nd pile is going to be doubled and we know the result will be 28, so before we follow this instruction the 2nd group had 14 matches. These 14 matches came from group 1, so our initial state is:
22 14 12

Tada! Later!

Posted by nikki on 2003-07-29 12:56:40