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

Revision as of 22:44, 31 August 2023

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 $0,1,1,3,5,11,2,5,9,0,-1,-1,-2,-5,-11,-3,-5,-9,0,\dots$ 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

@@ Line 29: / Line 29: @@
 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>
+<cmath>0,1,1,3,5,11,2,5,9,0,-1,-1,-2,-5,-11,-3,-5,-9,0,\dots</cmath>
 Listing the numbers out is expedited if you notice <math>b_{n+1}=2b_n+(-1)^n</math>.

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

Page

Toolbox

Search