2006 AMC 12B Problems/Problem 25

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Problem

A sequence $a_1,a_2,\dots$ of non-negative integers is defined by the rule $a_{n+2}=|a_{n+1}-a_n|$ for $n\geq 1$ . If $a_1=999$ , $a_2<999$ and $a_{2006}=1$ , how many different values of $a_2$ are possible?

$\mathrm{(A)}\ 165 \qquad \mathrm{(B)}\ 324 \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 499 \qquad \mathrm{(E)}\ 660$

Solution

We say the sequence $(a_n)$ completes at $i$ if $i$ is the minimal positive integer such that $a_i = a_{i + 1} = 1$ . Otherwise, we say $(a_n)$ does not complete.

Note that if $d = \gcd(999, a_2) \neq 1$ , then $d|a_n$ for all $n \geq 1$ , and $d$ does not divide $1$ , so if $\gcd(999, a_2) \neq 1$ , then $(a_n)$ does not complete.

From now on, suppose $\gcd(999, a_2) = 1$ .

We will now show that $(a_n)$ completes at $i$ for some $i \leq 2006$ . We will do this with 3 lemmas.

Lemma: If $a_j \neq a_{j + 1}$ , and neither value is $0$ , then $\max(a_j, a_{j + 1}) > \max(a_{j + 2}, a_{j + 3})$ .

Proof: There are 2 cases to consider.

If $a_j > a_{j + 1}$ , then $a_{j + 2} = a_j - a_{j + 1}$ , and $a_{j + 3} = |a_j - 2a_{j + 1}|$ . So $a_j > a_{j + 2}$ and $a_j > a_{j + 3}$ .

If $a_j < a_{j + 1}$ , then $a_{j + 2} = a_{j + 1} - a_j$ , and $a_{j + 3} = a_j$ . So $a_{j + 1} > a_{j + 2}$ and $a_{j + 1} > a_{j + 3}$ .

In both cases, $\max(a_j, a_{j + 1}) = \max(a_{j + 2}, a_{j + 3})$ , as desired.

Lemma: If $a_i = a_{i + 1}$ , then $a_i = 1$ . Moreover, if instead we have $a_i = 0$ for some $i > 2$ , then $a_{i - 1} = a_{i - 2} = 1$ .

Proof: By the way $(a_n)$ is constructed in the problem statement, having two equal consecutive terms $a_i = a_{i + 1}$ implies that $a_i$ divides every term in the sequence. So $a_i | 999$ and $a_i | a_2$ , so $a_i | \gcd(999, a_2) = 1$ , so $a_i = 1$ . For the proof of the second result, note that if $a_i = 0$ , then $a_{i - 1} = a_{i - 2}$ , so by the first result we just proved, $a_{i - 2} = a_{i - 1} = 1$ .

Lemma: $(a_n)$ completes at $i$ for some $i \leq 2006$ .

Proof: Suppose $(a_n)$ completed at some $i > 2000$ or not at all. Then by the second lemma, none of the pairs $(a_1, a_2), ..., (a_{1999}, a_{2000})$ can have a $0$ or be equal to $(1, 1)$ . So the first lemma implies $\max(a_1, a_2) > \max(a_3, a_4) > \cdots > \max(a_{1999}, a_{2000}),$ so $999 = \max(a_1, a_2) \geq 1000$ , a contradiction. Hence $(a_n)$ completes at $i$ for some $i \leq 2000$ .

Now we're ready to find exactly which values of $a_2$ we want to count.

Let's keep in mind that $2006 \equiv 2 \pmod 3$ and that $a_1 = 999$ is odd. We have two cases to consider.

Case 1: If $a_2$ is odd, then $a_3$ is even, so $a_4$ is odd, so $a_5$ is odd, so $a_6$ is even, and this pattern must repeat every three terms because of the recursive definition of $(a_n)$ , so the terms of $(a_n)$ reduced modulo 2 are $1, 1, 0, 1, 1, 0, ...,$ so $a_{2006}$ is odd and hence $1$ .

Case 2: If $a_2$ is even, then $a_3$ is odd, so $a_4$ is odd, so $a_5$ is even, so $a_6$ is odd, and this pattern must repeat every three terms, so the terms of $(a_n)$ reduced modulo 2 are $1, 0, 1, 1, 0, 1, ...,$ so $a_{2006}$ is even, and hence $0$ .

We have found that $a_{2006} = 1$ is true precisely when $\gcd(999, a_2) = 1$ and $a_2$ is odd. This tells us what we need to count.

There are $\phi(999) = 648$ numbers less than $999$ and relatively prime to it ( $\phi$ is the Euler totient function). We want to count how many of these are even. Note that $t \mapsto 999 - t$ is a 1-1 correspondence between the odd and even numbers less than and relatively prime to $999$ . So our final answer is $648/2 = 324$ , or $\boxed{B}$ .

2006 AMC 12B (Problems • Answer Key • Resources)
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2006 AMC 12B Problems/Problem 25

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