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Difference between revisions of "2017 IMO Problems/Problem 1"

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When we start with <math>a_0=12</math>, we get <math>a_1=15</math>, <math>a_2=15</math>,..., <math>a_8=36</math>, <math>a_9=6</math>, <math>a_{10}=9</math>, <math>a_{11}=3</math> and the pattern repeats.
 
When we start with <math>a_0=12</math>, we get <math>a_1=15</math>, <math>a_2=15</math>,..., <math>a_8=36</math>, <math>a_9=6</math>, <math>a_{10}=9</math>, <math>a_{11}=3</math> and the pattern repeats.
  
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When this pattern <math>3,6,9</math> repeats, this means that  
+
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>.
  
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So,  
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When we start with any number <math>a_0\not\equiv 0 (mod 3), we don't see a repeating pattern.
  
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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.
  
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When we start with </math>a_0=3k<math>, the next term if it is not a square is </math>3k+3<math>, then </math>3k+6<math> and so on until we get </math>3k+3p<math> where </math>p<math> is an integer and </math>(k+p)=3q^2<math> where </math>q<math> is an integer.  Then the next term will be </math>\sqrt(9q^2)=3q<math> and the pattern repeats again when </math>q=k<math> or when </math>q=3<math> or </math>6<math>.  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.
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So, the answer to this problem is </math>a_0=3k\;\forall k \in \mathbb{Z}^{+}$
  
 
~Tomas Diaz. orders@tomasdiaz.com
 
~Tomas Diaz. orders@tomasdiaz.com

Revision as of 02:42, 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$ (Error compiling LaTeX. Unknown error_msg)a_0=3k$where$k$is a positive integer and we need to prove this claim.

When we start with$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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
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