Question 1044811
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Three grasshoppers play leapfrog along a line. At each turn, one grasshopper leaps over another, but not over two others. 
Can the grasshoppers return to their initial positions after 1991 leaps?
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The initial configuration is like this:


-----0--------A------B-------------C----------


Three grasshoppers, named A, B and C, were sitting on the straight line (number line !).

Let's consider the function 


f = {{{((X[B]-X[A])*(X[C]-X[B])*(X[C]-X[A]))/(abs(X[B]-X[A])*abs(X[C]-X[B])*abs(X[C]-X[A]))}}}


where {{{X[A]}}}, {{{X[A]}}}, {{{X[C]}}} are coordinates of the grasshoppers A, B and C at each current turn of the game.


The function f takes the values +1 or -1 depending on the position of A, B and C at each current moment. 

In the initial position the value of the function f is +1.

Notice that f changes the sign to the opposite at each and every turn of the play.

Then after 1991-th turn f will have the opposite (negative) sign to that "+1" it had initially.

It means that the grasshoppers can not return to (can not be at) their initial positions after 1991-th turn.
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Solved.

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I'd like to make a comment regarding a Robertb' solution to this problem.


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Robertb starts his solution saying 
"The sequence of the first eight leaps (or 9 arrangements) are as follows:

ABC --> BAC --> ABC --> ACB --->" . . . 

He assumes that the sequence of leaps is strictly pre-assigned.
It leads directly to cycling of the sequence . . . 

But the condition does not require it. The condition does not require strictly pre-assigned sequence.
The sequence can be 

ABC --> BAC --> BCA --> CBA --->" . . .  

It is not prohibited by the condition. It is allowed.


So, Robertb solves, actually, another problem . . . 

Or, if you want, we solved different problems.

In my solution, much more wide class of leaps is considered.

And the resulting statement was proved for much wider class of leaps - for all the leaps allowed by the condition.

It was proved disregarding if there is OR there is no cycling in the sequence.
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