Difference between revisions of "2005 AIME I Problems/Problem 8"

Revision as of 17:52, 4 March 2007

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 $r_3$ , 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 $111 + 2 = 113$ .

@@ Line 7: / Line 7: @@
 == See also ==
-* [[2005 AIME I Problems/Problem 7 | Previous problem]]
-* [[2005 AIME I Problems/Problem 9 | Next problem]]
-* [[2005 AIME I Problems]]
 * [[Exponent]]
+{{AIME box|year=2005|n=I|num-b=7|num-a=9}}

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Difference between revisions of "2005 AIME I Problems/Problem 8"

Revision as of 17:52, 4 March 2007

Problem

Solution

See also