perplexus.info :: Numbers : Matchstick Frenzy

Matchstick Frenzy (Posted on 2002-06-27)

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?

Submitted by Dulanjana

Rating: 3.7692 (13 votes)

Solution: (Hide)

Let's say there were originally X, Y and Z matches in piles one, two and three respectively. After all the manipulations, each pile has 16 matches (48/3 = 16).
The first pile contains twice what it did after we took as many matches as there were in the second pile: 16 = 2*(X-Y) (eq. 1)
The second pile contains twice what it did originally minus what the third pile contained: 16 = 2*Y - Z (eq. 2)
Finally, the third pile contains twice what it did originally minus what the first pile contained having had matches taken to put into the second pile: 16 = 2*Z - (X-Y) (eq. 3)
We solve these equations:
16 = 2*(X-Y) [eq. 1] X - Y = 8 16 = 2*Z - (X-Y) [eq. 3] 16 = 2*Z - 8 Z = 12 16 = 2*Y - Z [eq. 2] 16 = 2*Y - 12 Y = 14 X - Y = 8 X - 14 = 8 X = 22
So the numbers of matches originally in the piles are 22, 14 and 12

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Puzzle Solution With Explanation	K Sengupta	2008-03-15 04:51:39
answer	K Sengupta	2008-03-09 12:27:55
Perfect match	Syzygy	2004-10-21 16:28:38
Backwards thinking	nikki	2003-07-29 12:56:40
here you got the solution	metjoo	2002-06-29 14:10:44
Solution	friedlinguini	2002-06-27 02:22:53

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