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

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When we start with <math>a_0=9</math>, we get <math>a_1=3</math>, <math>a_2=6</math>, <math>a_3=9</math> and the pattern repeats.
 
When we start with <math>a_0=9</math>, we get <math>a_1=3</math>, <math>a_2=6</math>, <math>a_3=9</math> and the pattern repeats.
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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.
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 +
When this pattern <math>3,6,9</math> repeats, this means that
  
 
So,  
 
So,  

Revision as of 01:30, 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

So,


~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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