The [equation](https://artofproblemsolving.com/wiki/index.php/Equation) $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three [real](https://artofproblemsolving.com/wiki/index.php/Real) [roots](https://artofproblemsolving.com/wiki/index.php/Root). Given that their sum is $m/n$ where $m$ and $n$ are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime) [positive integers](https://artofproblemsolving.com/wiki/index.php/Positive_integer), find $m+n.$
Let $y = 2^{111x}$.  Then our equation reads $\frac{1}{4}y^3 + 4y = 2y^2 + 1$ or $y^3 - 8y^2 + 16y - 4 = 0$.  Thus, if this equation has roots $r_1, r_2$ and $r_3$, by [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas) we have $r_1\cdot r_2\cdot r_3 = 4$.  Let the corresponding values of $x$ be $x_1, x_2$ and $x_3$.  Then the previous statement says that $2^{111\cdot(x_1 + x_2 + x_3)} = 4$ so taking a [logarithm](https://artofproblemsolving.com/wiki/index.php/Logarithm) of that gives $111(x_1 + x_2 + x_3) = 2$ and $x_1 + x_2 + x_3 = \frac{2}{111}$.  Thus the answer is $111 + 2 = \boxed{113}$.