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Difference between revisions of "2016 AMC 12B Problems/Problem 25"

(Created page with "==Problem== The sequence <math>(a_n)</math> is defined recursively by <math>a_0=1</math>, <math>a_1=\sqrt[19]{2}</math>, and <math>a_n=a_{n-1}a_{n-2}^2</math> for <math>n\geq ...")
 
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<math>\textbf{(A)}\ 17\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 21</math>
 
<math>\textbf{(A)}\ 17\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 21</math>
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==Solution==
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Let <math>b_i=19\text{log}_2a_i</math>. Then <math>b_0=0, b_1=1,</math> and <math>b_n=b_{n-1}+2b_{n-2}</math> for all <math>n\geq 2</math>. The characteristic polynomial of this linear recurrence is <math>x^2-x-2=0</math>, which has roots <math>2</math> and <math>-1</math>. Therefore, <math>b_n=k_12^{n}+k_2(-1)^n</math> for constants to be determined <math>k_1, k_2</math>. Using the fact that <math>b_0=0, b_1=1,</math> we can solve a pair of linear equations for <math>k_1, k_2</math>:
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<math>k_1+k_2=0</math>
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<math>2k_1-k_2=1</math>.
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Thus <math>k_1=\frac{1}{3}</math>, <math>k_2=-\frac{1}{3}</math>, and <math>b_n=\frac{2^n-(-1)^n}{3}</math>.
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Now, <math>a_1a_2\cdots a_k=2^{\frac{(b_1+b_2+\cdots+b_k)}{19}}</math>, so we are looking for the least value of <math>k</math> so that

Revision as of 11:12, 21 February 2016

Problem

The sequence $(a_n)$ is defined recursively by $a_0=1$, $a_1=\sqrt[19]{2}$, and $a_n=a_{n-1}a_{n-2}^2$ for $n\geq 2$. What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?

$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 21$

Solution

Let $b_i=19\text{log}_2a_i$. Then $b_0=0, b_1=1,$ and $b_n=b_{n-1}+2b_{n-2}$ for all $n\geq 2$. The characteristic polynomial of this linear recurrence is $x^2-x-2=0$, which has roots $2$ and $-1$. Therefore, $b_n=k_12^{n}+k_2(-1)^n$ for constants to be determined $k_1, k_2$. Using the fact that $b_0=0, b_1=1,$ we can solve a pair of linear equations for $k_1, k_2$:

$k_1+k_2=0$ $2k_1-k_2=1$.

Thus $k_1=\frac{1}{3}$, $k_2=-\frac{1}{3}$, and $b_n=\frac{2^n-(-1)^n}{3}$.

Now, $a_1a_2\cdots a_k=2^{\frac{(b_1+b_2+\cdots+b_k)}{19}}$, so we are looking for the least value of $k$ so that

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