The last product is 1, and the original four numbers are 1,2,3,4, all of them in Nat mod 7. ]]>

So, 2路6 = 12 and 3路4 = 12, therefore 5路x = 12. The remaining product is 12/5.

If x is the common factor of 2 and 3, the original numbers are x, 2/x, 3/x and 2x. (They multiply to 12), they give two equations:

2x^2 = 5 and 6x^-2 = 12/5 => x = sqrt(5/2)

or

6x^-2 = 5 and 2x^2 = 12/5 => x = sqrt(6/5)

The original numbers could be {sqrt(5/2), sqrt(8/5), sqrt(18/5), sqrt(10)} or {sqrt(6/5), sqrt(10/3), sqrt(15/2), sqrt(24/5)}

]]>