Difference between revisions of "1995 AIME Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | Taking the <math>\log_{1995}</math> ([[logarithm]]) of both sides and then moving to one side yields | + | Taking the <math>\log_{1995}</math> ([[logarithm]]) of both sides and then moving to one side yields the [[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 = 1 \pm \frac{\sqrt{2}}{2}</math>. Thus, the product of the two roots (both of which are positive) is <math>1995^{1+\sqrt{2}/2} \cdot 1995^{1 - \sqrt{2}/2} = 1995^2</math>, making the solution <math>(2000-5)^2 \equiv \boxed{025} \pmod{1000}</math>. |
== See also == | == See also == | ||
{{AIME box|year=1995|num-b=1|num-a=3}} | {{AIME box|year=1995|num-b=1|num-a=3}} | ||
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| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 17:56, 29 July 2008
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 the quadratic equation 
. Applying the quadratic formula yields that 
. Thus, the product of the two roots (both of which are positive) 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 | ||
