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

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Revision as of 18:29, 4 July 2013

Problem

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

Solution

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:

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

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

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

Hrm... 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.

By Vieta's formulas, the sum of the possible values of $a$ is $2$. This means that the roots $x_1$ and $x_2$ that satisfy the original equation also satisfy $\log_{1995} x_1 + \log_{1995} x_2 = 2.$ We can combine these logs to get $\log_{1995}x_1x_2=2$, or $x_1x_2=1995^2$. Finally, we find this value mod $1000$, which is easy.

$1995^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}$, 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

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