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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)
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<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>
 
<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>
  
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== Solution 3==
 
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We are trying to find the given product, i.e. the sum of the powers should be an integer. Another way of thinking is that if we take
 
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only the numerators and sum it should be divisible by 19. So if we generate the first few terms, we get the sequence <math>1,2,5,10,21...</math> which is the Catalan sequence given by <math>C_{n} = \frac{1}{n+1}*\binom{2n}{n}</math>. Now we expand out the factorial:
 
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<math>C_{n} = (2n)(2n-1)...(n+2)</math>
 
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So <math>n=17</math> or <math>\boxed{A}</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 12:06, 20 September 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, 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 $a_n=a_{n-1}a_{n-2}^2$: \[a_0=0,a_1=1\dots\implies 0,1,1,3,5,11,2,5,9,0,18,18,16,14,8,17,14,10,0\dots\] Listing the numbers out is expedited if you notice $a_{n+1}=2a_n+(-1)^n\pmod {19}$. 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 $19$: $\boxed{\textbf{(A) }17}$. (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 $0\pmod{19}$.

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


See Also

2016 AMC 12B (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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