2016 AMC 12B Problem 25

Below is the professionally curated solution for Problem 25 of the 2016 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 12B solutions, or check the answer key.

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Concepts:recursion modular arithmetic multiplicative order

Difficulty rating: 2650

25.

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 $n\ge2.$ What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?

$17$

$18$

$19$

$20$

$21$

Solution:

Write $a_n=2^{b_n/19}.$ The recursion becomes $b_0=0,$ $b_1=1,$ $b_n=b_{n-1}+2b_{n-2},$ solved by $b_n=\tfrac13\bigl(2^n-(-1)^n\bigr).$ The product $a_1\cdots a_k$ is an integer exactly when $19\mid b_1+\cdots+b_k.$ For odd $k,$ this sum equals $\tfrac13(2^{k+1}-1),$ which is a multiple of $19$ exactly when $19\mid2^{k+1}-1,$ i.e. when the order of $2$ modulo $19,$ namely $18,$ divides $k+1.$ The smallest such odd $k$ is $k+1=18,$ so $k=17.$

Thus, the correct answer is A.

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