2006 AMC 12B Problems/Problem 25

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. (Also, $a_{2006}$ cannot be 1 in this case since $d$ does not divide $1$, so we do not care about these $a_2$ at all.)

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 $i \leq 2000$.

Proof: Suppose $(a_n)$ completed at some $i > 2000$ or not at all. Then by the second lemma and the fact that neither $999$ nor $a_2$ are $0$, 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 \[\max(a_1, a_2) > \max(a_3, a_4) > \cdots > \max(a_{1999}, a_{2000})  > 0,\] 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 $1$ (since if $(a_n)$ completes at $i$, then $a_k$ must be $0$ or $1$ for all $k \geq i$).


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 $\boxed{\text{B}}$.

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Question
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS