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Difference between revisions of "1985 AHSME Problems/Problem 27"
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==Problem==
 
==Problem==
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Consider a sequence <math> x_1, x_2, x_3, \cdots </math> defined by
+
Consider a sequence <math>x_1,x_2,x_3,\dotsc</math> defined by:
−
 
+
<cmath>\begin{align*}&x_1 = \sqrt[3]{3}, \\ &x_2 = \left(\sqrt[3]{3}\right)^{\sqrt[3]{3}},\end{align*}</cmath>
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<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>.
+
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>
+
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.
 
−
 
 
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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.
 
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−
We now get rid of the cubed roots by introducing fractions into the exponents.
 
−
 
 
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<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 integeral 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>.
+
We now rewrite our formula for <math>x_n</math> as follows:
 +
<cmath>\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*}</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 (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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