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 "2017 IMO Problems/Problem 1"

(→‎Solution)
(→‎Solution)
Line 23: Line 23:
 
When this pattern <math>3,6,9</math> repeats, this means that there exists a number <math>A</math> such that <math>a_n = A</math> for infinitely many values of <math>n</math> and that number <math>A</math> is either <math>3,6,</math> or <math>9</math>.
 
When this pattern <math>3,6,9</math> repeats, this means that there exists a number <math>A</math> such that <math>a_n = A</math> for infinitely many values of <math>n</math> and that number <math>A</math> is either <math>3,6,</math> or <math>9</math>.
  
−
When we start with any number <math>a_0\not\equiv 0 mod 3</math>, we don't see a repeating pattern.
+
When we start with any number <math>a_0\not\equiv 0\; mod\; 3</math>, we don't see a repeating pattern.
  
 
Therefore the claim is that <math>a_0=3k</math> where <math>k</math> is a positive integer and we need to prove this claim.
 
Therefore the claim is that <math>a_0=3k</math> where <math>k</math> is a positive integer and we need to prove this claim.

Revision as of 02:44, 19 November 2023

Problem

For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, \ldots$ for $n \geq 0$ as \[a_{n+1} = \begin{cases} \sqrt{a_n} & \text{if } \sqrt{a_n} \text{ is an integer,} \\ a_n + 3 & \text{otherwise.} \end{cases}\]Determine all values of $a_0$ such that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$.

Solution

First we notice the following:

When we start with $a_0=3$, we get $a_1=6$, $a_2=9$, $a_3=3$ and the pattern repeats.

When we start with $a_0=6$, we get $a_1=9$, $a_2=3$, $a_3=6$ and the pattern repeats.

When we start with $a_0=9$, we get $a_1=3$, $a_2=6$, $a_3=9$ and the pattern repeats.

When we start with $a_0=12$, we get $a_1=15$, $a_2=15$,..., $a_8=36$, $a_9=6$, $a_{10}=9$, $a_{11}=3$ and the pattern repeats.

When this pattern $3,6,9$ repeats, this means that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$ and that number $A$ is either $3,6,$ or $9$.

When we start with any number $a_0\not\equiv 0\; mod\; 3$, we don't see a repeating pattern.

Therefore the claim is that $a_0=3k$ where $k$ is a positive integer and we need to prove this claim.

When we start with $a_0=3k$, the next term if it is not a square is $3k+3$, then $3k+6$ and so on until we get $3k+3p$ where $p$ is an integer and $(k+p)=3q^2$ where $q$ is an integer. Then the next term will be $\sqrt{9q^2}=3q$ and the pattern repeats again when $q=k$ or when $q=3$ or $6$. In order for these patterns to repeat, any square in the sequence need to be a multiple of 3. This will not work with any number or square that is not a multiple of 3.

So, the answer to this problem is $a_0=3k\;\forall k \in \mathbb{Z}^{+}$

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

2017 IMO (Problems) • Resources
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_IMO_Problems/Problem_1&oldid=204616"