Help in solving a problem posed in International Mathematical Olympiad 1983
$begingroup$
There was this question posed in IMO 1983:
"Can you choose 1983 pairwise distinct positive integers less than $10^5$, such that no three are in Arithmetic Progression?"
A close look at the sequence reveals the following things as mentioned in Problem Solving Strategies by Arthur Engel.
When the numbers which belong to the sequence are written in the ternary system, all the digits in any of the number is either 0 or 1
A further look suggests that the $nth$ term of the sequence has the same representation in the ternary system as $n$ has in the binary system
These help me to solve the original problem However, I am unsure as to how to prove the above two points
Any help please!!
sequences-and-series elementary-number-theory
asked Dec 13 '18 at 10:04
saisanjeevsaisanjeev
1,073312
$endgroup$
add a comment |
$begingroup$
There was this question posed in IMO 1983:
"Can you choose 1983 pairwise distinct positive integers less than $10^5$, such that no three are in Arithmetic Progression?"
A close look at the sequence reveals the following things as mentioned in Problem Solving Strategies by Arthur Engel.
When the numbers which belong to the sequence are written in the ternary system, all the digits in any of the number is either 0 or 1
A further look suggests that the $nth$ term of the sequence has the same representation in the ternary system as $n$ has in the binary system
These help me to solve the original problem However, I am unsure as to how to prove the above two points
Any help please!!
sequences-and-series elementary-number-theory
asked Dec 13 '18 at 10:04
saisanjeevsaisanjeev
1,073312
$endgroup$
$begingroup$
You don't have to prove they must be the numbers. You just have to prove those numbers work - that no three are in arithmetic progression.
$endgroup$
– Empy2
Dec 13 '18 at 10:08
$begingroup$
Can you help me to prove that please
$endgroup$
– saisanjeev
Dec 13 '18 at 10:34
$begingroup$
Given a,b in the sequence, prove the third term in arithmetic sequence must contain a 2 in ternary.
$endgroup$
– Empy2
Dec 13 '18 at 10:40
$begingroup$
ok. I tried it like this. Take a string of 1's and 0's in base 3, I multiply it by 2 to give a string of 2's and 0's with the 1's replaced by 2's. Now how do I prove that no matter what I subtract from the new number, I will always end up with a two in my answer?
$endgroup$
– saisanjeev
Dec 13 '18 at 10:54
add a comment |
$begingroup$
There was this question posed in IMO 1983:
"Can you choose 1983 pairwise distinct positive integers less than $10^5$, such that no three are in Arithmetic Progression?"
A close look at the sequence reveals the following things as mentioned in Problem Solving Strategies by Arthur Engel.
When the numbers which belong to the sequence are written in the ternary system, all the digits in any of the number is either 0 or 1
A further look suggests that the $nth$ term of the sequence has the same representation in the ternary system as $n$ has in the binary system
These help me to solve the original problem However, I am unsure as to how to prove the above two points
Any help please!!
sequences-and-series elementary-number-theory
asked Dec 13 '18 at 10:04
saisanjeevsaisanjeev
1,073312
$endgroup$
There was this question posed in IMO 1983:
"Can you choose 1983 pairwise distinct positive integers less than $10^5$, such that no three are in Arithmetic Progression?"
A close look at the sequence reveals the following things as mentioned in Problem Solving Strategies by Arthur Engel.
When the numbers which belong to the sequence are written in the ternary system, all the digits in any of the number is either 0 or 1
A further look suggests that the $nth$ term of the sequence has the same representation in the ternary system as $n$ has in the binary system
These help me to solve the original problem However, I am unsure as to how to prove the above two points
Any help please!!
sequences-and-series elementary-number-theory
sequences-and-series elementary-number-theory
asked Dec 13 '18 at 10:04
saisanjeevsaisanjeev
1,073312
asked Dec 13 '18 at 10:04
saisanjeevsaisanjeev
1,073312
asked Dec 13 '18 at 10:04
saisanjeevsaisanjeev
1,073312
asked Dec 13 '18 at 10:04
saisanjeevsaisanjeev
1,073312
asked Dec 13 '18 at 10:04
saisanjeevsaisanjeev
1,073312
1,073312
$begingroup$
You don't have to prove they must be the numbers. You just have to prove those numbers work - that no three are in arithmetic progression.
$endgroup$
– Empy2
Dec 13 '18 at 10:08
$begingroup$
Can you help me to prove that please
$endgroup$
– saisanjeev
Dec 13 '18 at 10:34
$begingroup$
Given a,b in the sequence, prove the third term in arithmetic sequence must contain a 2 in ternary.
$endgroup$
– Empy2
Dec 13 '18 at 10:40
$begingroup$
ok. I tried it like this. Take a string of 1's and 0's in base 3, I multiply it by 2 to give a string of 2's and 0's with the 1's replaced by 2's. Now how do I prove that no matter what I subtract from the new number, I will always end up with a two in my answer?
$endgroup$
– saisanjeev
Dec 13 '18 at 10:54
add a comment |
$begingroup$
You don't have to prove they must be the numbers. You just have to prove those numbers work - that no three are in arithmetic progression.
$endgroup$
– Empy2
Dec 13 '18 at 10:08
$begingroup$
Can you help me to prove that please
$endgroup$
– saisanjeev
Dec 13 '18 at 10:34
$begingroup$
Given a,b in the sequence, prove the third term in arithmetic sequence must contain a 2 in ternary.
$endgroup$
– Empy2
Dec 13 '18 at 10:40
$begingroup$
ok. I tried it like this. Take a string of 1's and 0's in base 3, I multiply it by 2 to give a string of 2's and 0's with the 1's replaced by 2's. Now how do I prove that no matter what I subtract from the new number, I will always end up with a two in my answer?
$endgroup$
– saisanjeev
Dec 13 '18 at 10:54
$begingroup$
You don't have to prove they must be the numbers. You just have to prove those numbers work - that no three are in arithmetic progression.
$endgroup$
– Empy2
Dec 13 '18 at 10:08
$begingroup$
You don't have to prove they must be the numbers. You just have to prove those numbers work - that no three are in arithmetic progression.
$endgroup$
– Empy2
Dec 13 '18 at 10:08
$begingroup$
Can you help me to prove that please
$endgroup$
– saisanjeev
Dec 13 '18 at 10:34
$begingroup$
Can you help me to prove that please
$endgroup$
– saisanjeev
Dec 13 '18 at 10:34
$begingroup$
Given a,b in the sequence, prove the third term in arithmetic sequence must contain a 2 in ternary.
$endgroup$
– Empy2
Dec 13 '18 at 10:40
$begingroup$
Given a,b in the sequence, prove the third term in arithmetic sequence must contain a 2 in ternary.
$endgroup$
– Empy2
Dec 13 '18 at 10:40
$begingroup$
ok. I tried it like this. Take a string of 1's and 0's in base 3, I multiply it by 2 to give a string of 2's and 0's with the 1's replaced by 2's. Now how do I prove that no matter what I subtract from the new number, I will always end up with a two in my answer?
$endgroup$
– saisanjeev
Dec 13 '18 at 10:54
$begingroup$
ok. I tried it like this. Take a string of 1's and 0's in base 3, I multiply it by 2 to give a string of 2's and 0's with the 1's replaced by 2's. Now how do I prove that no matter what I subtract from the new number, I will always end up with a two in my answer?
$endgroup$
– saisanjeev
Dec 13 '18 at 10:54
add a comment |
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$begingroup$
You don't have to prove they must be the numbers. You just have to prove those numbers work - that no three are in arithmetic progression.
$endgroup$
– Empy2
Dec 13 '18 at 10:08
$begingroup$
Can you help me to prove that please
$endgroup$
– saisanjeev
Dec 13 '18 at 10:34
$begingroup$
Given a,b in the sequence, prove the third term in arithmetic sequence must contain a 2 in ternary.
$endgroup$
– Empy2
Dec 13 '18 at 10:40
$begingroup$
ok. I tried it like this. Take a string of 1's and 0's in base 3, I multiply it by 2 to give a string of 2's and 0's with the 1's replaced by 2's. Now how do I prove that no matter what I subtract from the new number, I will always end up with a two in my answer?
$endgroup$
– saisanjeev
Dec 13 '18 at 10:54