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(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.
The humongous number is the potential number of answers, not any particular answer or set of answers.

Posted by Charlie
on 20160321 17:02:49 