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 "2012 USAMO Problems/Problem 3"
(→‎Solution that involves a non-elementary result)
(4 intermediate revisions by the same user not shown)
Line 9: Line 9:
  
 
==Solution that involves a non-elementary result==
 
==Solution that involves a non-elementary result==
 +
 +
(Since Bertrand's is well known and provable using elementary techniques, I see nothing wrong with this-tigershark22)
 +
 
For <math>n=2</math>, <math> |a_1| = 2 |a_2| = \cdots = 2^m |a_{2^m}| </math> implies that for any positive integer <math>m</math>, <math>|a_1| \ge 2^m</math>, which is impossible.
 
For <math>n=2</math>, <math> |a_1| = 2 |a_2| = \cdots = 2^m |a_{2^m}| </math> implies that for any positive integer <math>m</math>, <math>|a_1| \ge 2^m</math>, which is impossible.
  
Line 58: Line 61:
  
 
--[[User:Lightest|Lightest]] 21:24, 2 May 2012 (EDT)
 
--[[User:Lightest|Lightest]] 21:24, 2 May 2012 (EDT)
−
 
−
==Solution 2 (Bezout)==
 
−
 
−
I claim that when <math>n\geq 3</math> there exists an infinite sequence <math>a_i</math> satisfying the given condition.
 
−
 
−
n=2: Since <math>a_1=(-2)^k*a_k</math>, we have that <math>|a_1|</math> has no upper bound and thus no sequence exists.
 
−
 
−
n=4: Let <math>a_i=(-1)^{j+k}</math> where <math>j</math> is the number of factors of 2 in <math>i</math> and <math>k</math> is the number of factors of 3 in <math>i</math>
 
−
 
−
n=odd: Let <math>a_i=(\frac{-n+1}{2})^j</math> where <math>j</math> is the number of factors of <math>n</math> in <math>i</math>.
 
−
 
−
n=even<math>></math>4: Let <math>b</math> be either <math>n-3</math> or <math>n-1</math> such that <math>b</math> is not 0 mod 3. For <math>n=6</math> we have <math>b=5>n/2</math> and for even <math>n>6</math> we have <math>b\geq n-3>n/2</math>, so <math>b</math> has no multiples except for itself in the numbers 1 through n. Then we get <math>\gcd(b,3n/2)=1</math> and by Bezout we have <math>x*3n/2+y*b=1</math> for nonzero integer <math>x, y</math>. Then let <math>(3n/2+b-n(n+1)/2)x=x'</math> and similarly define <math>y'</math>. Now let <math>a_i=x'^j*y'^k</math> where <math>j</math> is the number of factors of <math>n/2</math> in <math>i</math> and <math>k</math> is the number of factors of <math>b</math> in <math>i</math>.
 
−
 
−
We can check that this indeed satisfies the problem conditions. For <math>n=4</math>, note that all the <math>(a_k, a_{2k}, a_{3k}, a_{4k})</math> are <math>(1,-1,-1,1)</math>, and <math>1-2-3+4=0</math>. Similarly, for odd <math>n</math>, we have <math>(a_k,a_{2k},...a_{nk})</math> is <math>(c, c, c,...,\frac{-n+1}{2}c)</math> which adds up to <math>cn(n-1)/2-cn(n-1)/2=0</math>. Note that <math>(a_k,a_{2k},...a_{nk})</math> is <math>(c, c, c,..., x'c, c, c,..., y', x'c)</math> or <math>(c, c, c,..., x'c, c, c,..., y', c, c, x'c)</math> depending on whether <math>b</math> is <math>n-3</math> or <math>n-1</math>, and where the first <math>x'c</math> is located at <math>a_{\frac{nk}{2}}</math>. We can check that this adds up to <math>c(n(n+1)/2-n/2-n-b+3nx'/2+by')=c(n(n+1)/2-3n/2-b+3n/2+b-n(n+1)/2)=0</math>.
 
−
 
−
-tigershark22
 
  
 
==See Also==
 
==See Also==

Revision as of 10:10, 11 July 2020

Problem

Determine which integers $n > 1$ have the property that there exists an infinite sequence $a_1$, $a_2$, $a_3$, $\dots$ of nonzero integers such that the equality \[a_k + 2a_{2k} + \dots + na_{nk} = 0\] holds for every positive integer $k$.

Partial Solution

For $n$ equal to any odd prime $p$, the sequence $\left\{a_i = \left(\frac{1-n}{2}\right)^{m_p\left(i\right)}\right\}$, where $p^{m_p\left(i\right)}$ is the greatest power of $p$ that divides $i$, gives a valid sequence. Therefore, the set of possible values for $n$ is at least the set of odd primes.


Solution that involves a non-elementary result

(Since Bertrand's is well known and provable using elementary techniques, I see nothing wrong with this-tigershark22)

For $n=2$, $|a_1| = 2 |a_2| = \cdots = 2^m |a_{2^m}|$ implies that for any positive integer $m$, $|a_1| \ge 2^m$, which is impossible.

We proceed to prove that the infinite sequence exists for all $n\ge 3$.

First, one notices that if we have $a_{xy} = a_x a_y$ for any integers $x$ and $y$, then it is suffice to define all $a_x$ for $x$ prime, and one only needs to verify the equation (*)

\[a_1+2a_2+\cdots+na_n=0\]

for the other equations will be automatically true.

To proceed with the construction, I need the following fact: for any positive integer $m>2$, there exists a prime $p$ such that $\frac{m}{2} <p \le m$.

To prove this, I am going to use Bertrand's Theorem ([1]) without proof. The Theorem states that, for any integer $n>1$, there exists a prime $p$ such that $n<p\le 2n-1$. In other words, for any positive integer $m>2$, if $m=2n$ with $n>1$, then there exists a prime $p$ such that $\frac{m}{2} < p < m$, and if $m=2n-1$ with $n>1$, then there exists a prime $p$ such that $\frac{m+1}{2} <p\le m$, both of which guarantees that for any integer $m>2$, there exists a prime $p$ such that $\frac{m}{2} <p \le m$.



Go back to the problem. Suppose $n\ge 3$. Let the largest two primes not larger than $n$ are $P$ and $Q$, and that $n\ge P > Q$. By the fact stated above, one can conclude that $2P > n$, and that $4Q = 2(2Q) \ge 2P > n$. Let's construct $a_n$:

Let $a_1=1$. There will be three cases: (i) $Q>\frac{n}{2}$, (ii) $\frac{n}{2} \ge Q > \frac{n}{3}$, and (iii) $\frac{n}{3} \ge Q > \frac{n}{4}$.

Case (i): $2Q>n$. Let $a_x = 1$ for all prime numbers $x<Q$, and $a_{xy}=a_xa_y$, then (*) becomes:

\[Pa_P + Qa_Q = C_1\]

Case (ii): $2Q\le n$ but $3Q > n$. In this case, let $a_2=-1$, and $a_x = 1$ for all prime numbers $2<x<Q$, and $a_{xy}=a_xa_y$, then (*) becomes:

\[Pa_P + Qa_Q - Qa_{2Q} = C_2\]

or

\[Pa_P - Qa_Q = C_2\]

Case (iii): $3Q\le n$. In this case, let $a_2=3$, $a_3=-2$, and $a_x = 1$ for all prime numbers $3<x<Q$, and $a_{xy}=a_xa_y$, then (*) becomes:

\[Pa_P + Qa_Q + 3Qa_{2Q} - 2Qa_{3Q} = C_3\]

or \[Pa_P + Qa_Q = C_3\]

In each case, by Bezout's Theorem, there exists non zero integers $a_P$ and $a_Q$ which satisfy the equation. For all other primes $p > P$, just let $a_p=1$ (or any other non-zero integer).

This construction is correct because, for any $k> 1$,

\[a_k + 2 a_{2k} + \cdots n a_{nk} = a_k (1 + 2 a_2 + \cdots n a_n ) = 0\]

Since Bertrand's Theorem is not elementary, we still need to wait for a better proof.

--Lightest 21:24, 2 May 2012 (EDT)

See Also

2012 USAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAMO 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=2012_USAMO_Problems/Problem_3&oldid=128051"