Difference between revisions of "1995 AIME Problems/Problem 2"
m (→Solution) |
|||
Line 3: | Line 3: | ||
<math>\sqrt{1995}x^{\log_{1995}x}=x^2</math>. | <math>\sqrt{1995}x^{\log_{1995}x}=x^2</math>. | ||
− | |||
===Solution 1=== | ===Solution 1=== | ||
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>. | 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>. |
Revision as of 20:38, 22 April 2016
Contents
[hide]Problem
Find the last three digits of the product of the positive roots of
.
Solution 1
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
.
Solution 2
Instead of taking , we take
of both sides and simplify:
Hrm... we know that and
are reciprocals, so let
. Then we have
. Multiplying by
and simplifying gives us
, as shown above.
By Vieta's formulas, the sum of the possible values of is
. This means that the roots
and
that satisfy the original equation also satisfy
We can combine these logs to get
, or
. Finally, we find this value mod
, which is easy.
, so our answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.