Difference between revisions of "2006 AMC 12B Problems/Problem 25"
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== Problem == | == Problem == | ||
− | {{ | + | A sequence <math>a_1,a_2,\dots</math> of non-negative integers is defined by the rule <math>a_{n+2}=|a_{n+1}-a_n|</math> for <math>n\geq 1</math>. If <math>a_1=999</math>, <math>a_2<999</math> and <math>a_{2006}=1</math>, how many different values of <math>a_2</math> are possible? |
− | == Solution == | + | <math> |
− | {{ | + | \mathrm{(A)}\ 165 |
+ | \qquad | ||
+ | \mathrm{(B)}\ 324 | ||
+ | \qquad | ||
+ | \mathrm{(C)}\ 495 | ||
+ | \qquad | ||
+ | \mathrm{(D)}\ 499 | ||
+ | \qquad | ||
+ | \mathrm{(E)}\ 660 | ||
+ | </math> | ||
+ | |||
+ | == Solution 1== | ||
+ | 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. | ||
+ | |||
+ | |||
+ | Note that if <math>d = \gcd(999, a_2) \neq 1</math>, then <math>d|a_n</math> for all <math>n \geq 1</math>, and <math>d</math> does not divide <math>1</math>, so if <math>\gcd(999, a_2) \neq 1</math>, then <math>(a_n)</math> does not complete. (Also, <math>a_{2006}</math> cannot be 1 in this case since <math>d</math> does not divide <math>1</math>, so we do not care about these <math>a_2</math> at all.) | ||
+ | |||
+ | From now on, suppose <math>\gcd(999, a_2) = 1</math>. | ||
+ | |||
+ | |||
+ | We will now show that <math>(a_n)</math> completes at <math>i</math> for some <math>i \leq 2006</math>. We will do this with 3 lemmas. | ||
+ | |||
+ | |||
+ | '''Lemma:''' If <math>a_j \neq a_{j + 1}</math>, and neither value is <math>0</math>, then <math>\max(a_j, a_{j + 1}) > \max(a_{j + 2}, a_{j + 3})</math>. | ||
+ | |||
+ | '''Proof:''' There are 2 cases to consider. | ||
+ | |||
+ | If <math>a_j > a_{j + 1}</math>, then <math>a_{j + 2} = a_j - a_{j + 1}</math>, and <math>a_{j + 3} = |a_j - 2a_{j + 1}|</math>. So <math>a_j > a_{j + 2}</math> and <math>a_j > a_{j + 3}</math>. | ||
+ | |||
+ | If <math>a_j < a_{j + 1}</math>, then <math>a_{j + 2} = a_{j + 1} - a_j</math>, and <math>a_{j + 3} = a_j</math>. So <math>a_{j + 1} > a_{j + 2}</math> and <math>a_{j + 1} > a_{j + 3}</math>. | ||
+ | |||
+ | In both cases, <math>\max(a_j, a_{j + 1}) > \max(a_{j + 2}, a_{j + 3})</math>, as desired. | ||
+ | ---------------------- | ||
+ | |||
+ | '''Lemma:''' If <math>a_i = a_{i + 1}</math>, then <math>a_i = 1</math>. Moreover, if instead we have <math>a_i = 0</math> for some <math>i > 2</math>, then <math>a_{i - 1} = a_{i - 2} = 1</math>. | ||
+ | |||
+ | '''Proof:''' By the way <math>(a_n)</math> is constructed in the problem statement, having two equal consecutive terms <math>a_i = a_{i + 1}</math> implies that <math>a_i</math> divides every term in the sequence. So <math>a_i | 999</math> and <math>a_i | a_2</math>, so <math>a_i | \gcd(999, a_2) = 1</math>, so <math>a_i = 1</math>. For the proof of the second result, note that if <math>a_i = 0</math>, then <math>a_{i - 1} = a_{i - 2}</math>, so by the first result we just proved, <math>a_{i - 2} = a_{i - 1} = 1</math>. | ||
+ | ---------------------- | ||
+ | |||
+ | '''Lemma:''' <math>(a_n)</math> completes at <math>i</math> for some <math>i \leq 2000</math>. | ||
+ | |||
+ | '''Proof:''' Suppose <math>(a_n)</math> completed at some <math>i > 2000</math> or not at all. Then by the second lemma and the fact that neither <math>999</math> nor <math>a_2</math> are <math>0</math>, none of the pairs <math>(a_1, a_2), ..., (a_{1999}, a_{2000})</math> can have a <math>0</math> or be equal to <math>(1, 1)</math>. So the first lemma implies | ||
+ | <cmath>\max(a_1, a_2) > \max(a_3, a_4) > \cdots > \max(a_{1999}, a_{2000}) > 0,</cmath> | ||
+ | so <math>999 = \max(a_1, a_2) \geq 1000</math>, a contradiction. Hence <math>(a_n)</math> completes at <math>i</math> for some <math>i \leq 2000</math>. | ||
+ | ---------------------- | ||
+ | |||
+ | |||
+ | Now we're ready to find exactly which values of <math>a_2</math> we want to count. | ||
+ | |||
+ | Let's keep in mind that <math>2006 \equiv 2 \pmod 3</math> and that <math>a_1 = 999</math> is odd. We have two cases to consider. | ||
+ | |||
+ | |||
+ | '''Case 1:''' If <math>a_2</math> is odd, then <math>a_3</math> is even, so <math>a_4</math> is odd, so <math>a_5</math> is odd, so <math>a_6</math> is even, and this pattern must repeat every three terms because of the recursive definition of <math>(a_n)</math>, so the terms of <math>(a_n)</math> reduced modulo 2 are | ||
+ | <cmath>1, 1, 0, 1, 1, 0, ...,</cmath> | ||
+ | so <math>a_{2006}</math> is odd and hence <math>1</math> (since if <math>(a_n)</math> completes at <math>i</math>, then <math>a_k</math> must be <math>0</math> or <math>1</math> for all <math>k \geq i</math>). | ||
+ | --------------------- | ||
+ | |||
+ | '''Case 2:''' If <math>a_2</math> is even, then <math>a_3</math> is odd, so <math>a_4</math> is odd, so <math>a_5</math> is even, so <math>a_6</math> is odd, and this pattern must repeat every three terms, so the terms of <math>(a_n)</math> reduced modulo 2 are | ||
+ | <cmath>1, 0, 1, 1, 0, 1, ...,</cmath> | ||
+ | so <math>a_{2006}</math> is even, and hence <math>0</math>. | ||
+ | --------------------- | ||
+ | |||
+ | |||
+ | We have found that <math>a_{2006} = 1</math> is true precisely when <math>\gcd(999, a_2) = 1</math> and <math>a_2</math> is odd. This tells us what we need to count. | ||
+ | |||
+ | |||
+ | There are <math>\phi(999) = 648</math> numbers less than <math>999</math> and relatively prime to it (<math>\phi</math> is the Euler totient function). We want to count how many of these are odd. Note that | ||
+ | <cmath>t \mapsto 999 - t</cmath> | ||
+ | is a 1-1 correspondence between the odd and even numbers less than and relatively prime to <math>999</math>. So our final answer is <math>648/2 = 324</math>, or <math>\boxed{\text{B}}</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=24|after=Last Question}} | {{AMC12 box|year=2006|ab=B|num-b=24|after=Last Question}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:02, 15 June 2022
Problem
A sequence of non-negative integers is defined by the rule
for
. If
,
and
, how many different values of
are possible?
Solution 1
We say the sequence completes at
if
is the minimal positive integer such that
. Otherwise, we say
does not complete.
Note that if , then
for all
, and
does not divide
, so if
, then
does not complete. (Also,
cannot be 1 in this case since
does not divide
, so we do not care about these
at all.)
From now on, suppose .
We will now show that completes at
for some
. We will do this with 3 lemmas.
Lemma: If , and neither value is
, then
.
Proof: There are 2 cases to consider.
If , then
, and
. So
and
.
If , then
, and
. So
and
.
In both cases, , as desired.
Lemma: If , then
. Moreover, if instead we have
for some
, then
.
Proof: By the way is constructed in the problem statement, having two equal consecutive terms
implies that
divides every term in the sequence. So
and
, so
, so
. For the proof of the second result, note that if
, then
, so by the first result we just proved,
.
Lemma: completes at
for some
.
Proof: Suppose completed at some
or not at all. Then by the second lemma and the fact that neither
nor
are
, none of the pairs
can have a
or be equal to
. So the first lemma implies
so
, a contradiction. Hence
completes at
for some
.
Now we're ready to find exactly which values of we want to count.
Let's keep in mind that and that
is odd. We have two cases to consider.
Case 1: If is odd, then
is even, so
is odd, so
is odd, so
is even, and this pattern must repeat every three terms because of the recursive definition of
, so the terms of
reduced modulo 2 are
so
is odd and hence
(since if
completes at
, then
must be
or
for all
).
Case 2: If is even, then
is odd, so
is odd, so
is even, so
is odd, and this pattern must repeat every three terms, so the terms of
reduced modulo 2 are
so
is even, and hence
.
We have found that is true precisely when
and
is odd. This tells us what we need to count.
There are numbers less than
and relatively prime to it (
is the Euler totient function). We want to count how many of these are odd. Note that
is a 1-1 correspondence between the odd and even numbers less than and relatively prime to
. So our final answer is
, or
.
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
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