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)}

]]>2 + 5 x 2 = 12

3 + 6 x 3 => 21

5 + 8 x 5 => 45 or 5 + 8 x 4 => 37?

35 seems more appropriate with pattern a + a x b ]]>

reasoning: the answer sought is not for the next in the (finite) series of equations, but the one after that. the next equation would have been

4+7= 102 (eleven in base 3), and would be followed by

5+8=1101 (thirteen in base 2). the series would then end with

6+9=111111111111111 .

37 is the answer. 1+4(x1)

2+5(x2)

3+6(x2)

Etc.

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