Difference between revisions of "1985 AHSME Problems/Problem 27"

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==Problem==
 
==Problem==
Consider a sequence <math> x_1, x_2, x_3, \cdots </math> defined by
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Consider a sequence <math>x_1,x_2,x_3,\dotsc</math> defined by:
 
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<cmath>\begin{align*}&x_1 = \sqrt[3]{3}, \ &x_2 = \left(\sqrt[3]{3}\right)^{\sqrt[3]{3}},\end{align*}</cmath>
<math> x_1=\sqrt[3]{3} </math>
 
 
 
<math> x_2=(\sqrt[3]{3})^{\sqrt[3]{3}} </math>
 
 
 
 
and in general
 
and in general
 +
<cmath>x_n = \left(x_{n-1}\right)^{\sqrt[3]{3}} \text{ for } n > 1.</cmath>
  
<math> x_n=(x_{n-1})^{\sqrt[3]{3}} </math> for <math> n>1 </math>.
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What is the smallest value of <math>n</math> for which <math>x_n</math> is an integer?
 
 
What is the smallest value of <math> n </math> for which <math> x_n </math> is an [[integer]]?
 
  
 
<math> \mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \  } 4 \qquad \mathrm{(D) \  } 9 \qquad \mathrm{(E) \  }27 </math>
 
<math> \mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \  } 4 \qquad \mathrm{(D) \  } 9 \qquad \mathrm{(E) \  }27 </math>
  
 
==Solution==
 
==Solution==
First, we will use induction to prove that <math> x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)} </math>
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Firstly, we will show by induction that <cmath>x_n = \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}.</cmath> For the base case, we indeed have <cmath>\begin{align*}x_1 &= \sqrt[3]{3} \ &= \left(\sqrt[3]{3}\right)^1 \ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^0\right)},\end{align*}</cmath> and for the inductive step, if our claim is true for <math>x_n</math>, then <cmath>\begin{align*}x_{n+1} &= \left(x_n\right)^{\sqrt[3]{3}} \ &= \left(\left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}\right)^{\sqrt[3]{3}} \ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\cdot\sqrt[3]{3}\right)} \ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^n\right)},\end{align*}</cmath> which completes the proof.
 
 
We see that <math> x_1=\sqrt[3]{3}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^0\right)} </math>. This is our base case.
 
 
 
Now, we have <math> x_n=(x_{n-1})^{\sqrt[3]{3}}=\left(\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-2}\right)}\right)^{\sqrt[3]{3}}=\sqrt[3]{3}^{\left(\sqrt[3]{3}\cdot\sqrt[3]{3}^{n-2}\right)}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)} </math>. Thus the induction is complete.
 
 
 
We now get rid of the cube roots by introducing fractions into the exponents.
 
 
 
<math> x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}=\sqrt[3]{3}^{\left(3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(\frac{1}{3}\cdot3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)} </math>.
 
  
Notice that since <math> 3 </math> isn't a perfect power, <math> x_n </math> is integral if and only if the exponent, <math> 3^{\left(\frac{n-4}{3}\right)} </math>, is integral. By the same logic, this is integral if and only if <math> \frac{n-4}{3} </math> is integral. We can now clearly see that the smallest positive value of <math> n </math> for which this is integral is <math> 4, \boxed{\text{C}} </math>.
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We now rewrite our formula for <math>x_n</math> as follows:
 +
<cmath>xn=(33)((33)n1)=(33)(3n13)=3(133n13)=3(3(n131))=3(3(n43)),</cmath>
 +
and as <math>3</math> is not a perfect power, we deduce that <math>x_n</math> is an integer if and only if the exponent, <math>3^{\left(\frac{n-4}{3}\right)}</math>, is itself an integer. By precisely the same argument, this reduces to <math>\frac{n-4}{3}</math> being an integer, so the smallest possible (positive) value of <math>n</math> is <math>\boxed{\text{(C)} \ 4}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=26|num-a=28}}
 
{{AHSME box|year=1985|num-b=26|num-a=28}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:06, 20 March 2024

Problem

Consider a sequence $x_1,x_2,x_3,\dotsc$ defined by: \begin{align*}&x_1 = \sqrt[3]{3}, \\ &x_2 = \left(\sqrt[3]{3}\right)^{\sqrt[3]{3}},\end{align*} and in general \[x_n = \left(x_{n-1}\right)^{\sqrt[3]{3}} \text{ for } n > 1.\]

What is the smallest value of $n$ for which $x_n$ is an integer?

$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \  } 4 \qquad \mathrm{(D) \  } 9 \qquad \mathrm{(E) \  }27$

Solution

Firstly, we will show by induction that \[x_n = \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}.\] For the base case, we indeed have \begin{align*}x_1 &= \sqrt[3]{3} \\ &= \left(\sqrt[3]{3}\right)^1 \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^0\right)},\end{align*} and for the inductive step, if our claim is true for $x_n$, then \begin{align*}x_{n+1} &= \left(x_n\right)^{\sqrt[3]{3}} \\ &= \left(\left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}\right)^{\sqrt[3]{3}} \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\cdot\sqrt[3]{3}\right)} \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^n\right)},\end{align*} which completes the proof.

We now rewrite our formula for $x_n$ as follows: \begin{align*}x_n &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)} \\ &= \left(\sqrt[3]{3}\right)^{\left(3^{\frac{n-1}{3}}\right)} \\ &=3^{\left(\frac{1}{3} \cdot 3^{\frac{n-1}{3}}\right)} \\ &= 3^{\left(3^{\left(\frac{n-1}{3}-1\right)}\right)} \\ &= 3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)},\end{align*} and as $3$ is not a perfect power, we deduce that $x_n$ is an integer if and only if the exponent, $3^{\left(\frac{n-4}{3}\right)}$, is itself an integer. By precisely the same argument, this reduces to $\frac{n-4}{3}$ being an integer, so the smallest possible (positive) value of $n$ is $\boxed{\text{(C)} \ 4}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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