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

Latest revision as of 23:40, 28 January 2021

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 $\boxed{025}$ .

@@ Line 1: / Line 1: @@
 == Problem ==
+Find the last three digits of the product of the [[positive root]]s of
+<math>\sqrt{1995}x^{\log_{1995}x}=x^2</math>.
-== Solution ==
+==Solution 1==
+Taking the <math>\log_{1995}</math> ([[logarithm]]) of both sides and then moving to one side yields the [[quadratic equation]] <math>2(\log_{1995}x)^2 - 4(\log_{1995}x)  + 1 = 0</math>. Applying the [[quadratic formula]] yields that <math>\log_{1995}x = 1 \pm \frac{\sqrt{2}}{2}</math>. Thus, the product of the two roots (both of which are positive) is <math>1995^{1+\sqrt{2}/2} \cdot 1995^{1 - \sqrt{2}/2} = 1995^2</math>, making the solution <math>(2000-5)^2 \equiv \boxed{025} \pmod{1000}</math>.
+==Solution 2==
+Instead of taking <math>\log_{1995}</math>, we take <math>\log_x</math> of both sides and simplify:
+<math> \log_x(\sqrt{1995}x^{\log_{1995}x})=\log_x(x^{2}) </math>
+<math> \log_x\sqrt{1995}+\log_x x^{\log_{1995}x}=2 </math>
+<math>\dfrac{1}{2} \log_x 1995 + \log_{1995} x = 2</math>
+We know that <math>\log_x 1995</math> and <math>\log_{1995} x</math> are reciprocals, so let <math>a=\log_{1995} x</math>.  Then we have <math>\dfrac{1}{2}\left(\dfrac{1}{a}\right) + a = 2</math>.  Multiplying by <math>2a</math> and simplifying gives us <math>2a^2-4a+1=0</math>, as shown above.
+Because <math>a=\log_{1995} x</math>, <math>x=1995^a</math>. By the quadratic formula, the two roots of our equation are <math>a=\frac{2\pm\sqrt2}{2}</math>. This means our two roots in terms of <math>x</math> are <math>1995^\frac{2+\sqrt2}{2}</math> and <math>1995^\frac{2-\sqrt2}{2}.</math> Multiplying these gives <math>1995^2</math>
+<math>1995^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}</math>, so our answer is <math>\boxed{025}</math>.
+==Solution 3==
+Let <math>y=\log_{1995}x</math>. Rewriting the equation in terms of <math>y</math>, we have
+<cmath> \sqrt{1995}\left(1995^y\right)^y=1995^{2y}</cmath>
+<cmath> 1995^{y^2+\frac{1}{2}}=1995^{2y}</cmath>
+<cmath> y^2+\frac{1}{2}=2y</cmath>
+<cmath> 2y^2-4y+1=0</cmath>
+<cmath> 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}}</cmath>
+Thus, the product of the positive roots is <math>\left(1995^{\frac{2+\sqrt{8}}{2}}\right)\left(1995^{\frac{2-\sqrt{8}}{2}}\right)=1995^2=\left(2000-5\right)^2</math>, so the last three digits are <math>\boxed{025}</math>.
 == See also ==
-* [[1995 AIME Problems]]
+{{AIME box|year=1995|num-b=1|num-a=3}}
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

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