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2016 AMC 12B Problems/Problem 25: Difference between revisions

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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
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


<math>b_1+b_2+\cdots+b_k \equiv 0 \pmod{19}</math>
<math>b_1+b_2+\cdots+b_k \equiv 0 \pmod{19}</math>. Note that we can multiply all <math>b_i</math> by three for convenience, as the <math>b_i</math> are always integers, and it does not affect divisibility by <math>19</math>.
.
Now, for all even <math>k</math> the sum (adjusted by a factor of three) is <math>2^1+2^2+\cdots+2^k=2^{k+1}-2</math>. The smallest <math>k</math> for which this is a multiple of <math>19</math> is <math>k=18</math> by Fermat's Little Theorem, as it is seen with further testing that <math>2</math> is a primitive root <math>\pmod{19}</math>. 
 
Now, assume <math>k</math> is odd. Then the sum (again adjusted by a factor of three) is <math>2^1+2^2+\cdots+2^k+1=2^{k+1}-1</math>. The smallest <math>k</math> for which this is a multiple of <math>19</math> is <math>k=17</math>, by the same reasons. Thus, the minimal value of <math>k</math> is <math>\textbf{(A)} 19</math>.

Revision as of 11:21, 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

$b_1+b_2+\cdots+b_k \equiv 0 \pmod{19}$. Note that we can multiply all $b_i$ by three for convenience, as the $b_i$ are always integers, and it does not affect divisibility by $19$. . Now, for all even $k$ the sum (adjusted by a factor of three) is $2^1+2^2+\cdots+2^k=2^{k+1}-2$. The smallest $k$ for which this is a multiple of $19$ is $k=18$ by Fermat's Little Theorem, as it is seen with further testing that $2$ is a primitive root $\pmod{19}$.

Now, assume $k$ is odd. Then the sum (again adjusted by a factor of three) is $2^1+2^2+\cdots+2^k+1=2^{k+1}-1$. The smallest $k$ for which this is a multiple of $19$ is $k=17$, by the same reasons. Thus, the minimal value of $k$ is $\textbf{(A)} 19$.

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