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

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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>
  
−
==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>. 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>
Line 15: Line 17:
 
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>\boxed{\textbf{(A) } 17}</math>.
 +
 
 +
==Solution 2==
 +
Since the product <math>a_1a_2\cdots a_k</math> is an integer, it must be a power of <math>2</math>, so the sum of the base-<math>2</math> logarithms must be an integer. Multiply all of these logarithms by <math>19</math> (to make them integers), so the sum must be a multiple of <math>19</math>.
 +
 
 +
The logarithms are <math>b_n = 19\log_2 a_n</math>. Using the recursion <math>b_0 = 0, b_1 = 1, b_n = b_{n-1}+2b_{n-2}</math> (modulo <math>19</math> to save calculation time), we get the sequence
 +
<cmath>0,1,1,3,5,11,2,5,9,0,-1,-1,-3,-5,-11,-2,-5,-9,0,\dots</cmath>
 +
Listing the numbers out is expedited if you notice <math>b_{n+1}=2b_n+(-1)^n</math>.
 +
 
 +
The cycle repeats every <math>9+9=18</math> terms. Notice that since <math>b_n+b_{n+9} \equiv 0 \pmod{19}</math>, the first <math>18</math> terms sum up to a multiple of <math>19</math>. Since <math>b_{18}=0</math>, we only need at most the first <math>\boxed{\textbf{(A)}\ 17}</math> terms to sum up to a multiple of <math>19</math>, and this is the lowest answer choice.
 +
 
 +
<b>Note 1:</b> To rigorously prove this is the smallest value, you will have to keep a running sum of the terms and check that it is never a multiple of <math>19</math> before the <math>17</math>th term.
 +
 
 +
<b>Note 2:</b> In response to note 1, it can be proven that <math>b_{n+2} = 2S + 1</math>, where <math>S = \sum^{n}_{i=1} b_i</math>. Since <math>S</math> is a multiple of <math>19</math>, it suffices to find the minimal <math>n \geq 1</math> such that <math>b_{n+2} \equiv 1 \pmod {19} </math>. In this case, <math>n = 17</math> happens to be minimal such <math>n</math>, so the answer would be <math>17</math>.
 +
 
 +
The relation <math>b_{n+2} = 2S + 1</math> can be proven by rearranging the relation <math>b_{i+2} = b_{i+1} + 2b_i</math> to <math>b_{i+2} - b_{i+1} = 2b_i</math> for all integers <math>0 \leq i \leq n</math>, then adding those <math>n+1</math> equations together. The LHS telescopes into <math>b_{n+2} - 1</math>, and the RHS becomes <math>2S</math>. Therefore, if you don't find a cleaner solution involving the relation <math>b_n+b_{n+9} \equiv 0 \pmod{19}</math>, you can always solve the problem just by considering the value of <math>b_{n+2}</math> rather than keeping a running sum.
 +
 
 +
==Solution 3==
 +
Like in [[#Solution 2|Solution 2]], calculate the first few terms of the sequence, but also keep a running sum <math>c_n</math> of the logarithms (not modulo <math>19</math> here):
 +
<cmath>0,1,2,5,10,21,42,\dots</cmath>
 +
Notice that <math>c_n=2c_{n-1}+1</math> for odd <math>n</math> and <math>c_n=2c_{n-1}</math> for even <math>n</math>. Since <math>2</math> is relatively prime to <math>19</math>, we can ignore even <math>n</math> and calculate odd <math>n</math> using <math>c_1 = 1, c_{n} = 4c_{n-2}+1</math> (modulo <math>19</math>):
 +
<cmath>,1,,5,,2,,9,,-1,,-3,,8,,-5,,0,\dots</cmath>
 +
<math>c_n</math> is first a multiple of <math>19</math> at <math>n = \boxed{\textbf{(A)}\ 17}</math>. ~[[User:emerald_block|emerald_block]]
 +
 
 +
==Solution 4 (Using a formula)==
 +
 
 +
Consider the product <math>a_1a_2\cdots a_k</math> (will finish tommorow)
 +
 
 +
==Video Solution by CanadaMath (Problem 21-25)==
 +
Fast Forward to 26:01 for problem 25
 +
https://www.youtube.com/watch?v=P3jJDLGyF2w&t=1546s
 +
 
 +
~THEMATHCANADIAN
 +
 
 +
==See Also==
 +
{{AMC12 box|year=2016|ab=B|after=Last Problem|num-b=24}}
 +
{{MAA Notice}}

Latest revision as of 23:43, 9 November 2024

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 $\boxed{\textbf{(A) } 17}$.

Solution 2

Since the product $a_1a_2\cdots a_k$ is an integer, it must be a power of $2$, so the sum of the base-$2$ logarithms must be an integer. Multiply all of these logarithms by $19$ (to make them integers), so the sum must be a multiple of $19$.

The logarithms are $b_n = 19\log_2 a_n$. Using the recursion $b_0 = 0, b_1 = 1, b_n = b_{n-1}+2b_{n-2}$ (modulo $19$ to save calculation time), we get the sequence \[0,1,1,3,5,11,2,5,9,0,-1,-1,-3,-5,-11,-2,-5,-9,0,\dots\] Listing the numbers out is expedited if you notice $b_{n+1}=2b_n+(-1)^n$.

The cycle repeats every $9+9=18$ terms. Notice that since $b_n+b_{n+9} \equiv 0 \pmod{19}$, the first $18$ terms sum up to a multiple of $19$. Since $b_{18}=0$, we only need at most the first $\boxed{\textbf{(A)}\ 17}$ terms to sum up to a multiple of $19$, and this is the lowest answer choice.

Note 1: To rigorously prove this is the smallest value, you will have to keep a running sum of the terms and check that it is never a multiple of $19$ before the $17$th term.

Note 2: In response to note 1, it can be proven that $b_{n+2} = 2S + 1$, where $S = \sum^{n}_{i=1} b_i$. Since $S$ is a multiple of $19$, it suffices to find the minimal $n \geq 1$ such that $b_{n+2} \equiv 1 \pmod {19}$. In this case, $n = 17$ happens to be minimal such $n$, so the answer would be $17$.

The relation $b_{n+2} = 2S + 1$ can be proven by rearranging the relation $b_{i+2} = b_{i+1} + 2b_i$ to $b_{i+2} - b_{i+1} = 2b_i$ for all integers $0 \leq i \leq n$, then adding those $n+1$ equations together. The LHS telescopes into $b_{n+2} - 1$, and the RHS becomes $2S$. Therefore, if you don't find a cleaner solution involving the relation $b_n+b_{n+9} \equiv 0 \pmod{19}$, you can always solve the problem just by considering the value of $b_{n+2}$ rather than keeping a running sum.

Solution 3

Like in Solution 2, calculate the first few terms of the sequence, but also keep a running sum $c_n$ of the logarithms (not modulo $19$ here): \[0,1,2,5,10,21,42,\dots\] Notice that $c_n=2c_{n-1}+1$ for odd $n$ and $c_n=2c_{n-1}$ for even $n$. Since $2$ is relatively prime to $19$, we can ignore even $n$ and calculate odd $n$ using $c_1 = 1, c_{n} = 4c_{n-2}+1$ (modulo $19$): \[,1,,5,,2,,9,,-1,,-3,,8,,-5,,0,\dots\] $c_n$ is first a multiple of $19$ at $n = \boxed{\textbf{(A)}\ 17}$. ~emerald_block

Solution 4 (Using a formula)

Consider the product $a_1a_2\cdots a_k$ (will finish tommorow)

Video Solution by CanadaMath (Problem 21-25)

Fast Forward to 26:01 for problem 25 https://www.youtube.com/watch?v=P3jJDLGyF2w&t=1546s

~THEMATHCANADIAN

See Also

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

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