Difference between revisions of "1995 AIME Problems/Problem 2"
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== Problem == | == Problem == | ||
− | Find the last three digits of the product of the positive | + | Find the last three digits of the product of the [[positive root]]s of |
− | <math>\sqrt{1995}x^{\log_{1995}x}=x^2 | + | <math>\sqrt{1995}x^{\log_{1995}x}=x^2</math>. |
== Solution == | == Solution == | ||
+ | Taking the <math>\log_{1995}</math> ([[logarithm]]) of both sides and then moving to one side yields a [[quadratic equation]]: <math>2(\log_{1995}x)^2 - 4(\log_{1995}x) + 1 = 0</math>. Applying the [[quadratic formula]] yields that <math>\log_{1995}x = \frac{2 \pm \sqrt{2}}{2}</math>. Thus, the product of the two roots is <math>= 1995^2</math>, making the solution <math>025</math>. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1995|num-b=1|num-a=3}} | |
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Revision as of 21:29, 13 February 2007
Problem
Find the last three digits of the product of the positive roots of .
Solution
Taking the (logarithm) of both sides and then moving to one side yields a quadratic equation: . Applying the quadratic formula yields that . Thus, the product of the two roots is , making the solution .
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |