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

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 $(2000-5)^2 \equiv \boxed{025} \pmod{1000}$.

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