document.write( "Question 1044811: 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? \n" ); document.write( "
Algebra.Com's Answer #660273 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
The sequence of the first 13 leaps (or 14 arrangements) are as follows:\r
\n" ); document.write( "\n" ); document.write( "ABC --> BAC ( B leaps to form the next arrangement)
\n" ); document.write( "ABC --> ACB --> CAB --> CBA --> BCA --> BAC (C leaps to form next arrangement)
\n" ); document.write( "ABC --> ACB --> CAB --> CBA --> BCA --> BAC (C leaps to form next arrangement)
\n" ); document.write( "...........................\r
\n" ); document.write( "\n" ); document.write( "(cycle repeats itself after 6 arrangements after the first two arrangements)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "For any k leaps, there are k+1 arrangements.
\n" ); document.write( "For 1991 leaps there are 1992 arrangements. Subtract the first two arrangements to start the count for the other arrangements,
\n" ); document.write( "to get 1992 - 2 = 1990.
\n" ); document.write( "1990/6 = 331 + 4/6 (leaving a remainder of 4), meaning the last leap arrangement is CBA, which is not the initial position ABC.
\n" ); document.write( " \n" ); document.write( "
\n" );