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

(→‎Solution 3: Remove. 1, 2, 5, 21, ... is not Catalan, and the rest of the argument is wrong as well. Two wrongs make a right, I guess)
(→‎Solution 2: revise and add new solution)
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==Solution 2==
 
==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>:
+
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>.
−
<cmath>a_0=0,a_1=1\dots\implies 0,1,1,3,5,11,2,5,9,0,18,18,16,14,8,17,14,10,0\dots</cmath>
 
−
Listing the numbers out is expedited if you notice <math>a_{n+1}=2a_n+(-1)^n\pmod {19}</math>. 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>. (NOTE: This solution proves 17 is the upper bound, but since 17  is the lowest answer choice, it is correct. To rigorously prove it, you will have to add up the mods listed until you get <math>0\pmod{19}</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>
+
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,-2,-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:</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.
 +
 +
==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]]
  
 
==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 22:54, 6 November 2021

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, 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 \[0,1,1,3,5,11,2,5,9,0,-1,-1,-2,-5,-11,-2,-5,-9,0,\dots\] 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: 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.

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

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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