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*Y-X-Y+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(2): cardinality of solution set by Brian Smith)

I was referring to that A001147 as being the sequence of cardinalities for the case of 1, 2, 3, 4, etc. members to be considered to be operated upon by a commutative but nonassociative operation. There's only 1 possible outcome when there is only one member or two members; 3 members have 3 possible outcomes and 4 have 15, etc.

Posted by Charlie on 2016-03-21 17:02:49