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Difference between revisions of "1995 AIME Problems/Problem 2"
 
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== Problem ==
 
== Problem ==
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Find the last three digits of the product of the positive roots of
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Find the last three digits of the product of the [[positive root]]s of
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<math>\sqrt{1995}x^{\log_{1995}x}=x^2.</math>
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<math>\sqrt{1995}x^{\log_{1995}x}=x^2</math>.
  
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== Solution ==
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==Solution 1==
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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>.
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==Solution 2==
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Instead of taking <math>\log_{1995}</math>, we take <math>\log_x</math> of both sides and simplify:
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 +
<math> \log_x(\sqrt{1995}x^{\log_{1995}x})=\log_x(x^{2}) </math>
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<math> \log_x\sqrt{1995}+\log_x x^{\log_{1995}x}=2 </math>
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<math>\dfrac{1}{2} \log_x 1995 + \log_{1995} x = 2</math>
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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.
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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>
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<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>.
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==Solution 3==
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Let <math>y=\log_{1995}x</math>. Rewriting the equation in terms of <math>y</math>, we have
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<cmath> \sqrt{1995}\left(1995^y\right)^y=1995^{2y}</cmath>
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<cmath> 1995^{y^2+\frac{1}{2}}=1995^{2y}</cmath>
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<cmath> y^2+\frac{1}{2}=2y</cmath>
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<cmath> 2y^2-4y+1=0</cmath>
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<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>
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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 ==
 
== See also ==
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* [[1995_AIME_Problems/Problem_1|Previous Problem]]
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{{AIME box|year=1995|num-b=1|num-a=3}}
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* [[1995_AIME_Problems/Problem_3|Next Problem]]
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* [[1995 AIME Problems]]
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[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

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 $\boxed{025}$.

See also

1995 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions

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