AIME 2005 I · 第 8 题

Solution

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 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 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 $111 + 2 = \boxed{113}$.