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

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==Solution==
 
==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>:
+
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>:
  
 
<math>k_1+k_2=0</math>
 
<math>k_1+k_2=0</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
 
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>. 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>.
+
<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, 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) } 17</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) } 17</math>.
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 +
==See Also==
 +
{{AMC12 box|year=2016|ab=B|after=Last Problem|num-b=24}}
 +
{{MAA Notice}}

Revision as of 11:25, 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 $\textbf{(A) } 17$.

See Also

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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
All AMC 12 Problems and Solutions

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