Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 AMC 12B Problems/Problem 25"

m (→‎Solution)
(→‎Solution)
Line 14: Line 14:
 
</math>
 
</math>
  
−
== Solution ==
+
== Solution 1 ==
 +
Define <math>g:= \gcd(999,a_2)</math>; then <math>g\mid a_3</math>, and, by induction, <math>g\mid a_k</math> for all <math>k>0</math>. Since <math>a_{2006}=1</math>, we have <math>g=1</math>, i.e. <math>a_2</math> is in the set, <math>S</math>, of positive integers less than <math>999</math> and relatively prime to <math>999</math>. We have <math>|S|=\varphi(999)=648</math>. Moreover, <math>a_2</math> must be odd, because if <math>a_2</math> is even, then every third term after <math>a_2</math> will be even; in particular, <math>2006\equiv 2\pmod 3</math>, so <math>a_{2006}</math> must be even, which is not the case. The subset of odd integers in <math>S</math> (denoted <math>S_{\textrm{o}}</math>) is in <math>1</math>--<math>1</math> correspondence with the subset of even integers in <math>S</math> via <math>x \mapsto 999 - x</math>; so <math>a_2\in S_{\textrm{o}}</math> with <math>|S_{\textrm{o}}|=324</math>.
 +
 
 +
We will now show that <math>a_{2006}=1</math> whenever <math>a_2\in S_{\textrm{o}}</math>. Suppose <math>a_2\in S</math> is odd, and define <math>M_i=\max(a_{i}, a_{i+1})</math> and <math>m_i=\min(a_{i}, a_{i+1})</math>.
 +
 
 +
Since <math>a_{i+1}=M_{i-1}-m_{i-1} \le M_{i-1}</math>, we have <math>M_i \le M_{i-1}</math> for all <math>i>0</math>. In fact, as long as <math>m_{i-1}\neq 0</math>, we have <math>a_{i+1}< M_{i-1}</math> and it follows that <math>M_{i+1} < M_{i-1}</math>. Then <math>M_1=999</math>, <math>M_3\le 998</math>, <math>M_5\le 997</math>, and, in general, <math>M_{2k+1}\le 999-k</math>. This implies that <math>m_k=0</math> for some <math>k< 2000</math>, i.e. <math>a_k=0</math> for some <math>k\le 2000</math>. If <math>a_k=0</math> then <math>a_{k-2}=a_{k-1}=a</math> (say), and it follows that <math>a\mid a_{k-3}</math>, and, by induction, <math>a\mid a_j</math> for all <math>j=1,2,\ldots , k-1</math>. Then <math>a\mid \gcd(999,a_2)=1</math>, so <math>a=1</math>. Therefore, the sub-sequence <math>\{a_i\}_{i\ge k-2}</math> <math>(k\le 2000)</math> will be<cmath>1, 1, 0, 1, 1, 0, \ldots</cmath>In particular, <math>a_{2006}</math> will be either <math>0</math> or <math>1</math>; but since <math>a_2</math> is odd, every third term after <math>a_2</math> will be odd, so <math>a_{2006}</math> must be odd, so it must equal <math>1</math>. We have shown that <math>a_2\in S_{\textrm{o}}</math> implies <math>a_{2006}=1</math>. So our final answer is <math>|S_{\textrm{o}}|=324</math>, or <math>\boxed{\text{B}}</math>.
 +
 
 +
== Solution 2==
 
We say the sequence <math>(a_n)</math> completes at <math>i</math> if <math>i</math> is the minimal positive integer such that <math>a_i = a_{i + 1} = 1</math>. Otherwise, we say <math>(a_n)</math> does not complete.
 
We say the sequence <math>(a_n)</math> completes at <math>i</math> if <math>i</math> is the minimal positive integer such that <math>a_i = a_{i + 1} = 1</math>. Otherwise, we say <math>(a_n)</math> does not complete.
  

Revision as of 10:12, 14 June 2022

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 1

Define $g:= \gcd(999,a_2)$; then $g\mid a_3$, and, by induction, $g\mid a_k$ for all $k>0$. Since $a_{2006}=1$, we have $g=1$, i.e. $a_2$ is in the set, $S$, of positive integers less than $999$ and relatively prime to $999$. We have $|S|=\varphi(999)=648$. Moreover, $a_2$ must be odd, because if $a_2$ is even, then every third term after $a_2$ will be even; in particular, $2006\equiv 2\pmod 3$, so $a_{2006}$ must be even, which is not the case. The subset of odd integers in $S$ (denoted $S_{\textrm{o}}$) is in $1$--$1$ correspondence with the subset of even integers in $S$ via $x \mapsto 999 - x$; so $a_2\in S_{\textrm{o}}$ with $|S_{\textrm{o}}|=324$.

We will now show that $a_{2006}=1$ whenever $a_2\in S_{\textrm{o}}$. Suppose $a_2\in S$ is odd, and define $M_i=\max(a_{i}, a_{i+1})$ and $m_i=\min(a_{i}, a_{i+1})$.

Since $a_{i+1}=M_{i-1}-m_{i-1} \le M_{i-1}$, we have $M_i \le M_{i-1}$ for all $i>0$. In fact, as long as $m_{i-1}\neq 0$, we have $a_{i+1}< M_{i-1}$ and it follows that $M_{i+1} < M_{i-1}$. Then $M_1=999$, $M_3\le 998$, $M_5\le 997$, and, in general, $M_{2k+1}\le 999-k$. This implies that $m_k=0$ for some $k< 2000$, i.e. $a_k=0$ for some $k\le 2000$. If $a_k=0$ then $a_{k-2}=a_{k-1}=a$ (say), and it follows that $a\mid a_{k-3}$, and, by induction, $a\mid a_j$ for all $j=1,2,\ldots , k-1$. Then $a\mid \gcd(999,a_2)=1$, so $a=1$. Therefore, the sub-sequence $\{a_i\}_{i\ge k-2}$ $(k\le 2000)$ will be\[1, 1, 0, 1, 1, 0, \ldots\]In particular, $a_{2006}$ will be either $0$ or $1$; but since $a_2$ is odd, every third term after $a_2$ will be odd, so $a_{2006}$ must be odd, so it must equal $1$. We have shown that $a_2\in S_{\textrm{o}}$ implies $a_{2006}=1$. So our final answer is $|S_{\textrm{o}}|=324$, or $\boxed{\text{B}}$.

Solution 2

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 (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_25&oldid=174862"