2006 AMC 12B Problems/Problem 25
Contents
Problem
A sequence of non-negative integers is defined by the rule
for
. If
,
and
, how many different values of
are possible?
Solution 1
Define ; then
, and, by induction,
for all
. Since
, we have
, i.e.
is in the set,
, of positive integers less than
and relatively prime to
. We have
. Moreover,
must be odd, because if
is even, then every third term after
will be even; in particular,
, so
must be even, which is not the case. The subset of odd integers in
(denoted
) is in
--
correspondence with the subset of even integers in
via
; so
with
.
We will now show that whenever
. Suppose
is odd, and define
and
.
Since , we have
for all
. In fact, as long as
, we have
and it follows that
. Then
,
,
, and, in general,
. This implies that
for some
, i.e.
for some
. If
then
(say), and it follows that
, and, by induction,
for all
. Then
, so
. Therefore, the sub-sequence
will be
In particular,
will be either
or
; but since
is odd, every third term after
will be odd, so
must be odd, so it must equal
. We have shown that
implies
. So our final answer is
, or
.
Solution 2
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 even. 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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All AMC 12 Problems and Solutions |
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