Difference between revisions of "2005 AIME I Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | The equation <math> 2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1 </math> has three real | + | The [[equation]] <math> 2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1 </math> has three [[real]] [[root]]s. Given that their sum is <math> \frac mn </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math> |
== Solution == | == Solution == | ||
| + | Let <math>y = 2^{111x}</math>. Then our equation reads <math>\frac{1}{4}y^3 + 4y = 2y^2 + 1</math> or <math>y^3 - 8y^2 + 16y - 4 = 0</math>. Thus, if this equation has roots <math>r_1, r_2</math> and <math>r_3</math>, we have <math>r_1\cdot r_2\cdot r_3 = 4</math>. Let the corresponding values of <math>x</math> be <math>x_1, x_2</math> and <math>x_3</math>. Then the previous statement says that <math>2^{111\cdot(x_1 + x_2 + x_3)} = 4</math> so that taking a [[logarithm]] gives <math>111(x_1 + x_2 + x_3) = 2</math> and <math>x_1 + x_2 + x_3 = \frac{2}{111}</math>. Thus the answer is <math>111 + 2 = 113</math>. | ||
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== See also == | == See also == | ||
| + | * [[2005 AIME I Problems/Problem 7 | Previous problem]] | ||
| + | * [[2005 AIME I Problems/Problem 9 | Next problem]] | ||
* [[2005 AIME I Problems]] | * [[2005 AIME I Problems]] | ||
| + | * [[Exponent]] | ||
Revision as of 13:06, 17 January 2007
Problem
The equation 
has three real roots. Given that their sum is 
where 
and 
are relatively prime positive integers, find 
Solution
Let 
. Then our equation reads 
or 
. Thus, if this equation has roots 
and 
, we have 
. Let the corresponding values of 
be 
and 
. Then the previous statement says that 
so that taking a logarithm gives 
and 
. Thus the answer is 
.
