Question

Consider a sequence $x_1,$ $x_2,$ $x_3,$ $\ldots,$ defined by $\begin{aligned} x_1&=\sqrt[3]{3},\\ x_2&=(\sqrt[3]{3})^{\sqrt[3]{3}}. \end{aligned}$ and in general $x_n=(x_{n-1})^{\sqrt[3]{3}}$ for $n\gt1.$ What is the smallest value of $n$ for which $x_n$ is an integer?

$2$

$3$

$4$

$9$

$27$

Answer: C

Concepts:exponent inequality radical recursion

Difficulty rating: 2380

Small Hint:

Let $c=\sqrt[3]{3}$ and track the exponent of $3$

Big Hint:

Show that $x_n=3^{\frac{c^{\,n-1}}{3}}$ and examine the first four terms

Solution:

Let $c=\sqrt[3]{3}.$ Repeated application of the recurrence gives $x_n=3^{\frac{c^{\,n-1}}{3}},$ so $x_4=3^{\frac{c^3}{3}}=3.$ It remains to rule out the first three terms. The sequence is strictly increasing. Also $c\lt\frac{3}{2},$ so $x_2=c^c \lt\left(\frac32\right)^{\frac{3}{2}} \lt2.$ Finally $x_3=3^{\frac{1}{c}}.$ Since $2^c\lt2^{\frac{3}{2}}\lt3,$ we have $x_3\gt2,$ while $\frac{1}{c}\lt1$ gives $x_3\lt3.$ Thus $x_1,x_2,x_3$ are not integers, and the first integral term is $x_4.$

Therefore the correct answer is C.

1985 AMC 12 Problem 27

Answer: C

Difficulty rating: 2380

Small Hint:

Big Hint:

Solution:

Problem 27 in Other Years