Let's think about this problem.

This operation @ is consisted of two of multiplying or dividing plus one constant a(a=/ 0).

a @ a= a

2 @ 3a = 2/3

2a@ 2 = a^2

1 @ 1 = ?

 

hint

@ : ( ) # ( ) # ( )

where ( ) means constant or variables

where # means multiplication or division.

 

The first operator searching problem.

Edited January 7, 201114 yr by alpha2cen

I'm going to guess that x@y = x*a/y, so 1@1=a, but I wouldn't be able to say if that's a unique solution or not.