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

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(Solution)
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==Solution==
 
==Solution==
  
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First we notice the following:
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When we start with <math>a_0=3</math>, we get <math>a_1=6</math>, <math>a_2=9</math>, <math>a_3=3</math> and the pattern repeats.
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When we start with <math>a_0=6</math>, we get <math>a_1=9</math>, <math>a_2=3</math>, <math>a_3=6</math> and the pattern repeats.
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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.
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So,
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~Tomas Diaz. orders@tomasdiaz.com
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 02:23, 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.

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