Difference between revisions of "2017 IMO Problems/Problem 1"
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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 | + | 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. | ||
− | 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 | + | 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. |
So, the answer to this problem is <math>a_0=3k\;\forall k \in \mathbb{Z}^{+}</math> | So, the answer to this problem is <math>a_0=3k\;\forall k \in \mathbb{Z}^{+}</math> |
Revision as of 01:43, 19 November 2023
Problem
For each integer , define the sequence for as Determine all values of such that there exists a number such that for infinitely many values of .
Solution
First we notice the following:
When we start with , we get , , and the pattern repeats.
When we start with , we get , , and the pattern repeats.
When we start with , we get , , and the pattern repeats.
When we start with , we get , ,..., , , , and the pattern repeats.
When this pattern repeats, this means that there exists a number such that for infinitely many values of and that number is either or .
When we start with any number , we don't see a repeating pattern.
Therefore the claim is that where is a positive integer and we need to prove this claim.
When we start with , the next term if it is not a square is , then and so on until we get where is an integer and where is an integer. Then the next term will be and the pattern repeats again when or when or . 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
~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 |