# 1995 AIME Problems/Problem 2

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## 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 $\boxed{025}$.

## Solution 3

Let $y=\log_{1995}x$. Rewriting the equation in terms of $y$, we have $$\sqrt{1995}\left(1995^y\right)^y=1995^{2y}$$ $$1995^{y^2+\frac{1}{2}}=1995^{2y}$$ $$y^2+\frac{1}{2}=2y$$ $$2y^2-4y+1=0$$ $$y=\frac{4\pm\sqrt{16-\left(4\right)\left(2\right)\left(1\right)}}{4}=\frac{4\pm\sqrt{8}}{4}=\frac{2\pm\sqrt{8}}{2}={1\pm\sqrt{2}}$$ Thus, the product of the positive roots is $\left(1995^{\frac{2+\sqrt{8}}{2}}\right)\left(1995^{\frac{2-\sqrt{8}}{2}}\right)=1995^2=\left(2000-5\right)^2$, so the last three digits are $\boxed{025}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 