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Inverting Triangles (Posted on 2005-08-15) Difficulty: 1 of 5
A piece of paper had the following diagram:
               o              o o o o
      From:   o o         To:  o o o
             o o o              o o
            o o o o              o
Below it, it read "Given the initial formation of ten coins, move exactly # coins to produce the end formation." It was pretty obvious that # stood for a digit, but it was smudged and couldn't be read. What possible numbers could it have been so the problem was solvable?

To allow explaining the solution, number the coins like this:

           0 
          1 2
         3 4 5
        6 7 8 9
Note: This problem was inspired by a forum question by Nicole.

No Solution Yet Submitted by Erik O.    
Rating: 2.1667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution re: # = 3 solutions Plus Three | Comment 4 of 14 |
(In reply to # = 3 solutions by brianjn)

Had 'owl' not posted since my last response, I would have merely edited my comment.

I can further move 9 coins.  Consider that one of the vertices of the triangle be fixed, eg 9, then all other 9 coins need to be moved to the right of the 9 and also below that line.

To 'owl', each of the solutions in my "# = 3 solutions" can be accomplished by ensuring that the moved coin remains in contact with one or two other coins; I actually visualised this in the documentation of my solutions. 

I shall not do this for this additional solution, but it may readily been seen that it is indeed possible.

If I consider the block of:
  1 2
   4       as immovable, an the top left 'vertex' I can move 7 coins to form the desired inversion.

If I consider the group of:
 8  9  as immovable, I can move 8 coins to create the desired array; two to the right and 6 below.

So, in all I have 3, 4, 6, 7, 8 and 9.  Clearly I cannot have 1 or 2.

And 5?  I could do this, but I would consider this to be an illegitimate solution:
If this   3 4 5
             7 8        is my immovable block at the top left of my array, I would have to move the 9 from its position at some time and replace it with another coin - feel that that is not in the spirit of the puzzle,
so,
 I'll stay with 6 solutions; 3, 4, 6, 7, 8 and 9.


  Posted by brianjn on 2005-08-15 05:47:15

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