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

Revision as of 11:15, 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

$b_1+b_2+\cdots+b_k \equiv 0 \pmod{19}$

Revision as of 11:12, 21 February 2016 view source Zhuangzhuang (talk \| contribs) editor 42 edits →Problem ← Older edit		Revision as of 11:15, 21 February 2016 view source Zhuangzhuang (talk \| contribs) editor 42 edits →Problem Newer edit →
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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>

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

Revision as of 11:15, 21 February 2016

Problem

Solution