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

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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.
 
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 cubed roots by introducing fractions into the exponents.  
+
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>.
 
<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>.
+
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>.
  
 
==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}}

Revision as of 00:14, 3 April 2018

Problem

Consider a sequence $x_1, x_2, x_3, \cdots$ defined by

$x_1=\sqrt[3]{3}$

$x_2=(\sqrt[3]{3})^{\sqrt[3]{3}}$

and in general

$x_n=(x_{n-1})^{\sqrt[3]{3}}$ 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

First, we will use induction to prove that $x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}$

We see that $x_1=\sqrt[3]{3}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^0\right)}$. This is our base case.

Now, we have $x_n=(x_{n-1})^{\sqrt[3]{3}}=\left(\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-2}\right)}\right)^{\sqrt[3]{3}}=\sqrt[3]{3}^{\left(\sqrt[3]{3}\cdot\sqrt[3]{3}^{n-2}\right)}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}$. Thus the induction is complete.

We now get rid of the cube roots by introducing fractions into the exponents.

$x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}=\sqrt[3]{3}^{\left(3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(\frac{1}{3}\cdot3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)}$.

Notice that since $3$ isn't a perfect power, $x_n$ is integral if and only if the exponent, $3^{\left(\frac{n-4}{3}\right)}$, is integral. By the same logic, this is integral if and only if $\frac{n-4}{3}$ is integral. We can now clearly see that the smallest positive value of $n$ for which this is integral is $4, \boxed{\text{C}}$.

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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