1985 AHSME Problems/Problem 27
Problem
Consider a sequence 
defined by
$x_1=\sqrt{3]{3}$ (Error compiling LaTeX. Unknown error_msg)
$x_2=\sqrt[3]{3}^\sqrt[3]{3}$ (Error compiling LaTeX. Unknown error_msg)
and in general
$x_n=(x_{n-1})^\sqrt[3]{3}$ (Error compiling LaTeX. Unknown error_msg) for 
.
What is the smallest value of 
for which 
is an integer?
Solution
First, we will use induction to prove that ![$x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}$](http://latex.artofproblemsolving.com/f/c/8/fc852b0b4f87586b6ab4a45d191bda14d1b204f9.png)
We see that ![$x_1=\sqrt[3]{3}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^0\right)}$](http://latex.artofproblemsolving.com/7/d/2/7d28faccb93a237c5723f35d3f21ab491b8de825.png)
. This is our base case.
Now, we have ![$x_n=(x_{n-1})^{\sqrt[3]{3}}=\left(\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-2}\right)}\right)^{\sqrt[3]{3}}=\sqrt[3]{3}^{\left(\sqrt[3]{3}\cdot\sqrt[3]{3}^{n-2}\right)}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}$](http://latex.artofproblemsolving.com/3/1/6/316d1809e9ab0da2e5cdcb74370f9b0db4d4846f.png)
. Thus the induction is complete.
We now get rid of the cubed roots by introducing fractions into the exponents.
![$x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}=\sqrt[3]{3}^{\left(3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(\frac{1}{3}\cdot3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)}$](http://latex.artofproblemsolving.com/d/4/f/d4f0fd17945876cbc6caeea70816c5bc297616a5.png)
.
Notice that since 
isn't a perfect power, is integral if and only if the exponent, 
, is integral. By the same logic, this is integeral if and only if 
is integral. We can now clearly see that the smallest positive value of for which this is integral 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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