1995 AIME Problems/Problem 2

Revision as of 19:59, 5 November 2018 by Mp8148 (talk | contribs) (Solution 3)

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 $\boxed{025}$.

See also

1995 AIME (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS