1985 AHSME Problems/Problem 27
Problem
Consider a sequence 
defined by:
![\begin{align*}&x_1 = \sqrt[3]{3}, \\ &x_2 = \left(\sqrt[3]{3}\right)^{\sqrt[3]{3}},\end{align*}](http://latex.artofproblemsolving.com/e/d/5/ed51c2019aa1b7eeb8ca1aebea158c707113bf1d.png)
and in general
![\[x_n = \left(x_{n-1}\right)^{\sqrt[3]{3}} \text{ for } n > 1.\]](http://latex.artofproblemsolving.com/5/9/a/59a772cd4c0401874f7fa0d0ccabae36e2d68b78.png)
What is the smallest value of 
for which 
is an integer?
Solution
Firstly, we will show by induction that ![\[x_n = \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}.\]](http://latex.artofproblemsolving.com/f/e/0/fe006ce0e854fad53a9fb5a8b63f9dc132c01091.png)
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*}](http://latex.artofproblemsolving.com/a/d/1/ad14827c6054c70a499b6fc9ee0f3e44dea47af1.png)
and for the inductive step, if our claim is true for , 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*}](http://latex.artofproblemsolving.com/f/0/9/f096001a8b35111135baad7d384593b50eb4711c.png)
which completes the proof.
We now rewrite our formula for 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*}](http://latex.artofproblemsolving.com/4/a/8/4a8e04d8f3e268a055bc66753110d434e394e1d9.png)
and as 
is not a perfect power, we deduce that is an integer if and only if the exponent, 
, is itself an integer. By precisely the same argument, this reduces to 
being an integer, so the smallest possible (positive) value of is 
.
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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