Difference between revisions of "2006 AMC 12B Problems/Problem 25"
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− | == Solution == | + | == 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. | 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. | ||
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− | 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 | + | 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> | <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>. | 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>. |
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 |
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.