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2005 AIME I Problems/Problem 8

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Problem

The equation 2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1 has three real roots. Given that their sum is \frac mn where m and n are relatively prime positive integers, find m+n.

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 that taking a logarithm 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 = 113.

See also

2005 AIME I (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
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