IMO 2005-4
Let $a_n=2^n+6^n+3^n-1^n$,find all $n$ such that $a_n$ is coprime to all other elements of this sequence.
One one thing is neutral in the universe, that is $0$.
Re: Imo 2005-4
Probably the problem statement should be:
Determine all positive integers such that they are coprime to all the terms of this sequence
\[a_n=2^n+6^n+3^n-1^n, \forall n \in N\]
আর যে এই সমস্যার সমাধান একবার দেখছে তার ভোলার কথা না আমি খালি সমাধান মনে করে লিখতেছি না, কেউ যদি এখন সমাধান করে তাহলে সে সমাধান দিক
Determine all positive integers such that they are coprime to all the terms of this sequence
\[a_n=2^n+6^n+3^n-1^n, \forall n \in N\]
আর যে এই সমস্যার সমাধান একবার দেখছে তার ভোলার কথা না আমি খালি সমাধান মনে করে লিখতেছি না, কেউ যদি এখন সমাধান করে তাহলে সে সমাধান দিক
Every logical solution to a problem has its own beauty.
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Re: Imo 2005-4
আসলেই সেইরকম সমস্যা। one liner with unique crux move.
তবে আইডিয়াটা পাওয়া কঠিন ব্যাপার।
তবে আইডিয়াটা পাওয়া কঠিন ব্যাপার।
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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Re: Imo 2005-4
ওহ ভুলেই গেছিলাম...
শুভ জন্মদিন, মাসুম
শুভ জন্মদিন, মাসুম
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
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Re: Imo 2005-4
Thanks.Now I am 17 .And this is the easiest problem of Imo(according to me).Just note that $a_n$ is even for all $n$.So $gcd(a_m,a_n)$ is at least $2$,therefore never coprime.
One one thing is neutral in the universe, that is $0$.
Re: Imo 2005-4
Masum's solution is correct if the problem is as stated by himself but if the problem is as stated by Zzzz (:P) then it's incorrect.
"Go down deep enough into anything and you will find mathematics." ~Dean Schlicter
Re: Imo 2005-4
But Zzzz,what do you mean by 'determine all positive integers'-$n$ or $a_n$
TIURMI,I didn't undestand your comment too.Zzzz wrote:Probably the problem statement should be:
Determine all positive integers such that they are coprime to all the terms of this sequence
\[a_n=2^n+6^n+3^n-1^n, \forall n \in N\]
আর যে এই সমস্যার সমাধান একবার দেখছে তার ভোলার কথা না আমি খালি সমাধান মনে করে লিখতেছি না, কেউ যদি এখন সমাধান করে তাহলে সে সমাধান দিক
One one thing is neutral in the universe, that is $0$.
Re: Imo 2005-4
Because never is $gcd(a_m,a_n)=1,$ I think there exists neither $n$ nor $a_n$ satisfying the condition.
One one thing is neutral in the universe, that is $0$.
Re: Imo 2005-4
\[a_1=6^1+2^1+3^1-1=10\]
\[a_2=6^2+2^2+3^2-1=48\]
\[a_3=6^3+2^3+3^3-1=250\]
\[.\]
\[.\]
\[.\]
So the sequence is $10,48, 250,...,\infty$
In the IMO question, they asked for all positive numbers which are coprime to all numbers of the sequence. For example, $1$ is such a number. $1$ is coprime to all the numbers of the sequence. Now look for others !
\[a_2=6^2+2^2+3^2-1=48\]
\[a_3=6^3+2^3+3^3-1=250\]
\[.\]
\[.\]
\[.\]
So the sequence is $10,48, 250,...,\infty$
In the IMO question, they asked for all positive numbers which are coprime to all numbers of the sequence. For example, $1$ is such a number. $1$ is coprime to all the numbers of the sequence. Now look for others !
Every logical solution to a problem has its own beauty.
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Re: Imo 2005-4
Well Masum your mistake is you thought that you are to find number "in the sequence" which are coprime to the other numbers of the sequence but the question states that you have to find "all" the positive integers which are "within" or "out of the sequence" that are coprime with any $a_n$ (i.e. all of the numbers in the sequence)
"Go down deep enough into anything and you will find mathematics." ~Dean Schlicter