1995 AIME Problems/Problem 2

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:

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

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

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

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 $\boxed{025}$ .

1995 AIME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1995 AIME Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

See also