Difference between revisions of "2010 USAJMO Problems/Problem 2"
Sixoneeight (talk | contribs) (Added a third solution that I think is nice.) |
|||
(11 intermediate revisions by 4 users not shown) | |||
Line 14: | Line 14: | ||
The sequence is <math>2, 4, 6, \ldots, 2n-2</math>. | The sequence is <math>2, 4, 6, \ldots, 2n-2</math>. | ||
− | ===Proof=== | + | ===Proof 1=== |
We will prove that any sequence <math>x_1, \ldots, x_{n-1}</math>, that satisfies | We will prove that any sequence <math>x_1, \ldots, x_{n-1}</math>, that satisfies | ||
the given conditions, is an | the given conditions, is an | ||
Line 42: | Line 42: | ||
x_{n-2} = x_{n-1}</math>. | x_{n-2} = x_{n-1}</math>. | ||
+ | Now for the induction step on all values of <math>m</math>. | ||
Suppose we have shown that for all <math>i \le m</math>, <math>x_1 + x_{n-1-i} = | Suppose we have shown that for all <math>i \le m</math>, <math>x_1 + x_{n-1-i} = | ||
x_{n-i}</math>. If <math>m = n-2</math> we are done, otherwise <math>m < n-2</math>, and by | x_{n-i}</math>. If <math>m = n-2</math> we are done, otherwise <math>m < n-2</math>, and by | ||
Line 53: | Line 54: | ||
which completes the proof. | which completes the proof. | ||
+ | ===Proof 2=== | ||
+ | Let <math>S=\{x_1,x_2,...,x_{n-1}\}</math>. | ||
+ | Notice that <cmath>x_1<x_1+x_1<x_1+x_2<\dots <x_1+x_{n-2}<2n.</cmath> Then by condition (c), we must have <math>x_1,x_1+x_1,...,x_1+x_{n-2}\in S</math>. This implies that <math>x_1=x_1,x_1+x_1=x_2,...,x_1+x_{n-2}=x_{n-1}</math>, or that <math>x_k=kx_1</math>. Then we have <math>x_1+x_{n-1}=n(x_1)=2n\rightarrow x_1=2</math>, and the rest is trivial. | ||
+ | |||
+ | ==Solution 2== | ||
+ | The claim is that in this sequence, if there are <math>2</math> elements <math>a,b</math> where <math>a,b<n</math>, such that <math>\gcd(a,b)=1</math>, then the sequence contains every number less than <math>2n</math>. | ||
+ | |||
+ | Proof: Let <math>a</math> and <math>b</math> be the numbers less than <math>n</math> such that <math>\gcd(a,b)=1</math>. | ||
+ | We take this sequence modulo <math>n</math>. This means that if <math>x_i</math> is an element in this sequence then <math>-x_i</math> is as well. | ||
+ | <math>a,b,2n-a,2n-b</math> are all elements in the sequence. Clearly, one of <math>2n-a+b</math> and <math>2n-b+a</math> is less than <math>2n</math>, which means that <math>\pm (a-b)</math> are in this sequence modulo <math>n</math>. | ||
+ | Now we want to show every number is achievable. We have already established that <math>a</math> and <math>b</math> are relatively prime, so by euclidean algorithm, if we take the positive difference of <math>a'</math> and <math>b'</math> every time, we will get that <math>1</math> is in our sequence. Then, we can simply add or subtract <math>1</math> as many times from <math>a</math> as desired to get every single number. | ||
+ | |||
+ | We have proved that there are no two numbers that can be relatively prime in our sequence, implying that no two consecutive numbers can be in this sequence. Because our sequence has <math>n-1</math> terms, our sequence must be one of <math>2,4,6,..,2(n-1)</math> or <math>1,3,5,...</math>, the latter obviously fails, so <math>2,4,6...2(n-1)</math> is our only possible sequence. | ||
+ | |||
+ | ==Solution 3== | ||
+ | We can add <math>x_1</math> to every expression in property (a) to get <cmath>2x_1<x_1+x_2<\dots<x_1+x_{n-1}=2n.</cmath> Therefore, we have <math>n-2</math> distinct (because all the <math>x_i</math> are distinct) expressions of the form <math>x_1+x_i</math> for <math>i = 1, 2, \dots n-1</math> that are all less than <math>2n</math>, which means these <math>n-2</math> expressions are also equal to some <math>x_k</math>. | ||
+ | |||
+ | Now, we have that the <math>x_i</math> are all positive, so <math>x_1>0</math>. Adding <math>x_1</math> to both sides, <math>2x_1>x_1</math>. Therefore, since the <math>n-2</math> expressions are all at least <math>2x_1</math>, they have to be equal to <math>x_2, x_3, \dots, x_{n-1}</math>. | ||
+ | |||
+ | Since there are <math>n-2</math> distinct expressions <math>x_1+x_i</math> equal to <math>n-2</math> distinct expressions <math>x_j</math>, we have that each <math>x_1+x_i</math> is equal to one <math>x_j</math>. Using the orders of the <math>x_1+x_i</math> and the <math>x_j</math>, we find that | ||
+ | <cmath>\begin{align*} | ||
+ | x_1+x_1 &= x_2 \\ | ||
+ | x_1+x_2 &= x_3 \\ | ||
+ | ...& \\ | ||
+ | x_1+x_{n-2} &= x_{n-1}. | ||
+ | \end{align*}</cmath> | ||
+ | This gives us | ||
+ | <cmath>\begin{align*} | ||
+ | x_1 &= x_1 \\ | ||
+ | x_2 &= 2x_1 \\ | ||
+ | x_3 &= 3x_1 \\ | ||
+ | &\text{ }...\\ | ||
+ | x_{n-1} &= (n-1)x_1\\ | ||
+ | \end{align*}</cmath> | ||
+ | Now, we can use property (b) with <math>i=1</math>, so <math>x_1 + x_{n-1} = 2n</math>, which means <math>x_1 = 2</math>. This gives the sequence <cmath>\boxed{2, 4, 6, \dots, 2(n-1)}</cmath> | ||
+ | -sixeoneeight | ||
== See Also == | == See Also == | ||
{{USAJMO newbox|year=2010|num-b=1|num-a=3}} | {{USAJMO newbox|year=2010|num-b=1|num-a=3}} |
Latest revision as of 23:09, 20 November 2022
Problem
Let be an integer. Find, with proof, all sequences
of positive integers with the following
three properties:
- (a).
;
- (b).
for all
;
- (c). given any two indices
and
(not necessarily distinct) for which
, there is an index
such that
.
Solution
The sequence is .
Proof 1
We will prove that any sequence , that satisfies
the given conditions, is an
arithmetic progression with
as both the first term and the
increment. Once this is proved, condition (b) implies that
. Therefore
,
and the sequence is just the even numbers from
to
. The
sequence of successive even numbers clearly satisfies all three conditions,
and we are done.
First a degenerate case.
If , there is only one element
, and condition (b) gives
or
. Conditions (a) and (c) are vacuously
true.
Otherwise, for , we will prove by induction on
that the
difference
for all
,
which makes all the differences
, i.e. the sequence is an arithmetic progression with
as the first term and increment as promised.
So first the case. With
,
exists and is less
than
by condition (a). Now since by condition (b)
, we conclude that
, and therefore
by condition (c)
for some
. Now, since
,
and can only be
. So
.
Now for the induction step on all values of .
Suppose we have shown that for all
,
. If
we are done, otherwise
, and by
condition (c)
for some
. This
is
larger than
, but smaller than
by the inductive hypothesis. It then follows that
, the only element of the sequence between
and
. This establishes the result for
.
So, by induction for all
,
which completes the proof.
Proof 2
Let .
Notice that
Then by condition (c), we must have
. This implies that
, or that
. Then we have
, and the rest is trivial.
Solution 2
The claim is that in this sequence, if there are elements
where
, such that
, then the sequence contains every number less than
.
Proof: Let and
be the numbers less than
such that
.
We take this sequence modulo
. This means that if
is an element in this sequence then
is as well.
are all elements in the sequence. Clearly, one of
and
is less than
, which means that
are in this sequence modulo
.
Now we want to show every number is achievable. We have already established that
and
are relatively prime, so by euclidean algorithm, if we take the positive difference of
and
every time, we will get that
is in our sequence. Then, we can simply add or subtract
as many times from
as desired to get every single number.
We have proved that there are no two numbers that can be relatively prime in our sequence, implying that no two consecutive numbers can be in this sequence. Because our sequence has terms, our sequence must be one of
or
, the latter obviously fails, so
is our only possible sequence.
Solution 3
We can add to every expression in property (a) to get
Therefore, we have
distinct (because all the
are distinct) expressions of the form
for
that are all less than
, which means these
expressions are also equal to some
.
Now, we have that the are all positive, so
. Adding
to both sides,
. Therefore, since the
expressions are all at least
, they have to be equal to
.
Since there are distinct expressions
equal to
distinct expressions
, we have that each
is equal to one
. Using the orders of the
and the
, we find that
This gives us
Now, we can use property (b) with
, so
, which means
. This gives the sequence
-sixeoneeight
See Also
2010 USAJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.