Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1995 AIME Problems/Problem 2"

m (→‎Solution)
(→‎Solution 2: Edited for clarity.)
Line 15: Line 15:
 
<math>\dfrac{1}{2} \log_x 1995 + \log_{1995} x = 2</math>
 
<math>\dfrac{1}{2} \log_x 1995 + \log_{1995} x = 2</math>
  
−
Hrm... we know that <math>\log_x 1995</math> and <math>\log_{1995} x</math> are reciprocals, so let <math>a=\log_{1995} x</math>.  Then we have <math>\dfrac{1}{2}\left(\dfrac{1}{a}\right) + a = 2</math>.  Multiplying by <math>2a</math> and simplifying gives us <math>2a^2-4a+1=0</math>, as shown above.
+
We know that <math>\log_x 1995</math> and <math>\log_{1995} x</math> are reciprocals, so let <math>a=\log_{1995} x</math>.  Then we have <math>\dfrac{1}{2}\left(\dfrac{1}{a}\right) + a = 2</math>.  Multiplying by <math>2a</math> and simplifying gives us <math>2a^2-4a+1=0</math>, as shown above.
  
−
By Vieta's formulas, the sum of the possible values of <math>a</math> is <math>2</math>. This means that the roots <math>x_1</math> and <math>x_2</math> that satisfy the original equation also satisfy <math>\log_{1995} x_1 + \log_{1995} x_2 = 2.</math> We can combine these logs to get <math>\log_{1995}x_1x_2=2</math>, or <math>x_1x_2=1995^2</math>.  Finally, we find this value mod <math>1000</math>, which is easy.
+
Because <math>a=\log_{1995} x</math>, <math>x=1995^a</math>. By the quadratic formula, the two roots of our equation are <math>a=\frac{2\pm\sqrt2}{2}</math>. This means our two roots in terms of <math>x</math> are <math>1995^\frac{2+\sqrt2}{2}</math> and <math>1995^\frac{2-\sqrt2}{2}.</math> Multiplying these gives <math>1995^2</math>
  
 
<math>1995^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}</math>, so our answer is <math>\boxed{025}</math>.
 
<math>1995^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}</math>, so our answer is <math>\boxed{025}</math>.

Revision as of 18:09, 6 October 2016

Problem

Find the last three digits of the product of the positive roots of $\sqrt{1995}x^{\log_{1995}x}=x^2$.

Solution 1

Taking the $\log_{1995}$ (logarithm) of both sides and then moving to one side yields the quadratic equation $2(\log_{1995}x)^2 - 4(\log_{1995}x) + 1 = 0$. Applying the quadratic formula yields that $\log_{1995}x = 1 \pm \frac{\sqrt{2}}{2}$. Thus, the product of the two roots (both of which are positive) is $1995^{1+\sqrt{2}/2} \cdot 1995^{1 - \sqrt{2}/2} = 1995^2$, making the solution $(2000-5)^2 \equiv \boxed{025} \pmod{1000}$.

Solution 2

Instead of taking $\log_{1995}$, we take $\log_x$ of both sides and simplify:

$\log_x(\sqrt{1995}x^{\log_{1995}x})=\log_x(x^{2})$

$\log_x\sqrt{1995}+\log_x x^{\log_{1995}x}=2$

$\dfrac{1}{2} \log_x 1995 + \log_{1995} x = 2$

We know that $\log_x 1995$ and $\log_{1995} x$ are reciprocals, so let $a=\log_{1995} x$. Then we have $\dfrac{1}{2}\left(\dfrac{1}{a}\right) + a = 2$. Multiplying by $2a$ and simplifying gives us $2a^2-4a+1=0$, as shown above.

Because $a=\log_{1995} x$, $x=1995^a$. By the quadratic formula, the two roots of our equation are $a=\frac{2\pm\sqrt2}{2}$. This means our two roots in terms of $x$ are $1995^\frac{2+\sqrt2}{2}$ and $1995^\frac{2-\sqrt2}{2}.$ Multiplying these gives $1995^2$

$1995^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}$, so our answer is $\boxed{025}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_2&oldid=80527"