The 97 rational numbers 49/1, 49/2, 49/3, ..., 49/97 are written on a blackboard.
Two of the above numbers X and Y are chosen and replaced by X*YXY+1.
The procedure is repeated until a single number Z(say) remains on the board.
Determine the possible values of Z.
(In reply to
re: Two special results by Ady TZIDON)
The first is if you combine 49/1 with 49/2 as X and Y in the stated algorithm; then combine that result with 49/3; and then that result with 49/4, etc. until you get the one number at the end.
The other is the final result when you start at the other end: combine 49/96 with 49/97 as the first X, Y pair, then the result of that with 49/95, etc.
This is as opposed to the random order used in the 500 examples in my first post.

Posted by Charlie
on 20160320 22:49:12 