Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 4: Line 4:
 
<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>
  
−
==Solution==
+
==Solution 1==
  
 
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>.  
 
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>.  
Line 24: Line 24:
  
 
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>.
 +
 +
==Solution 2==
 +
Since the product <math>a_1a_2\cdots a_k</math> is an integer, the sum of the logarithms <math>\log _2 a_k</math> must be an integer. Multiply all of these logarithms by <math>19</math>, so that the sum must be a multiple of <math>19</math>. We take these vales modulo <math>19</math> to save calculation time. Using the recursion <math>a_n=a_{n-1}a_{n-2}^2</math>:
 +
<cmath>a_0=0,a_1=1\dots\implies 0,1,1,3,5,11,2,5,9,0,18,18,16,14,10,17,14,10,0\dots</cmath>
 +
Notice that <math>a_k+a_{k+9}\equiv 0\text{ mod }19</math>. The cycle repeats every <math>9+9=18</math> terms. Since <math>a_0=0</math> and <math>a_{18}=0</math>, we only need the first <math>17</math> terms to sum up to a multiple of <math>19</math>: <math>\boxed{\textbf{(A) }17}</math>.
 +
 +
 +
<math>a_1a_2a_3a_4a_5a_6a_7a_8a_9a_{10}a_{11}a_{12}a_{13}a_{14}a_{15}a_{16}a_{17}=2^{87381/19}=2^{4599}\approx 2.735\cdot 10{1384}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|after=Last Problem|num-b=24}}
 
{{AMC12 box|year=2016|ab=B|after=Last Problem|num-b=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:44, 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 1

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 $\textbf{(A) } 17$.

Solution 2

Since the product $a_1a_2\cdots a_k$ is an integer, the sum of the logarithms $\log _2 a_k$ must be an integer. Multiply all of these logarithms by $19$, so that the sum must be a multiple of $19$. We take these vales modulo $19$ to save calculation time. Using the recursion $a_n=a_{n-1}a_{n-2}^2$: \[a_0=0,a_1=1\dots\implies 0,1,1,3,5,11,2,5,9,0,18,18,16,14,10,17,14,10,0\dots\] Notice that $a_k+a_{k+9}\equiv 0\text{ mod }19$. The cycle repeats every $9+9=18$ terms. Since $a_0=0$ and $a_{18}=0$, we only need the first $17$ terms to sum up to a multiple of $19$: $\boxed{\textbf{(A) }17}$.


$a_1a_2a_3a_4a_5a_6a_7a_8a_9a_{10}a_{11}a_{12}a_{13}a_{14}a_{15}a_{16}a_{17}=2^{87381/19}=2^{4599}\approx 2.735\cdot 10{1384}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_25&oldid=76893"