Flipping Items Simultaneously II

The applet below displays a number (N) of small circles in a cyclic arrangement. All the circles are red, but one, which is blue. When you click on a circle then M successive circles starting with this one clockwise change their color: from red to blue and vice versa. The question is, Is it possible after a number of such moves to arrive at the situation where all the circles are red, except the one pointed to by a small arrow? This one must be blue.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

The original puzzle [Mathematical Circles, pp. 127-128] stated for N = 12 and M = 3, 4, 6 has negative solution. The solution is outlined below.

But the problem begs for a generalization: what can be said about other pairs of M and N?

References

  1. D. Fomin, S. Genkin, I. Itenberg, Mathematical Circles (Russian Experience), AMS, 1996

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Copyright © 1996-2017 Alexander Bogomolny


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Assume M divides N. Then the answer to the question is in negative. To see why, check the "Solution helper" box. Some of the circles will get a green "kernel" at their center. The group of the marked circles has an interesting property: among any M successive circles there are exactly 2 marked circles. Note also that the original blue circle is marked, while the target circle pointed to by the arrow is not. This is a crucial observation.

Imagine that the red circles have been replaced with the number one 1 and the blue circles with -1. At the outset, the product of all the numbers in the marked group is -1. Since on every step exactly 2 marked circles are affected, the product of the numbers in the marked circles always remain one. This is an invariant of the problem.

If the problem had a positive solution, we would have all the marked circles red, so that the product of the numbers in the marked circles would be 1. Which is, as we just seen, is not possible.


  1. Merlin's Magic Squares
  2. Merlin Magic Square Game
  3. Lights Out: an interactive gizmo
  4. Mini Lights Out
  5. Flipping Items Simultaneously
  6. Flipping Items Simultaneously II

|Activities| |Contact| |Front page| |Contents| |Algebra| |Eye opener| |Store|

Copyright © 1996-2017 Alexander Bogomolny

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