IMO 2005-4

Discussion on International Mathematical Olympiad (IMO)
User avatar
Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
Location: Dhaka,Bangladesh

IMO 2005-4

Unread post by Masum » Thu Dec 09, 2010 4:43 am

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 $0$.

User avatar
Zzzz
Posts: 172
Joined: Tue Dec 07, 2010 6:28 am
Location: 22° 48' 0" N / 89° 33' 0" E

Re: Imo 2005-4

Unread post by Zzzz » Thu Dec 09, 2010 1:19 pm

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 :P 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.
(Important: Please make sure that you have read about the Rules, Posting Permissions and Forum Language)

User avatar
Moon
Site Admin
Posts: 751
Joined: Tue Nov 02, 2010 7:52 pm
Location: Dhaka, Bangladesh
Contact:

Re: Imo 2005-4

Unread post by Moon » Thu Dec 09, 2010 1:27 pm

আসলেই সেইরকম সমস্যা। one liner with unique crux move. :)
তবে আইডিয়াটা পাওয়া কঠিন ব্যাপার। :)
"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.

User avatar
Moon
Site Admin
Posts: 751
Joined: Tue Nov 02, 2010 7:52 pm
Location: Dhaka, Bangladesh
Contact:

Re: Imo 2005-4

Unread post by Moon » Thu Dec 09, 2010 1:56 pm

ওহ ভুলেই গেছিলাম...

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.

User avatar
Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
Location: Dhaka,Bangladesh

Re: Imo 2005-4

Unread post by Masum » Thu Dec 09, 2010 3:21 pm

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 $0$.

User avatar
TIUrmi
Posts: 61
Joined: Tue Dec 07, 2010 12:13 am
Location: Dinajpur, Bangladesh
Contact:

Re: Imo 2005-4

Unread post by TIUrmi » Thu Dec 09, 2010 9:05 pm

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

User avatar
Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
Location: Dhaka,Bangladesh

Re: Imo 2005-4

Unread post by Masum » Thu Dec 09, 2010 11:57 pm

But Zzzz,what do you mean by 'determine all positive integers'-$n$ or $a_n$
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 :P আমি খালি সমাধান মনে করে লিখতেছি না, কেউ যদি এখন সমাধান করে তাহলে সে সমাধান দিক :)
TIURMI,I didn't undestand your comment too.
One one thing is neutral in the universe, that is $0$.

User avatar
Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
Location: Dhaka,Bangladesh

Re: Imo 2005-4

Unread post by Masum » Fri Dec 10, 2010 12:01 am

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 $0$.

User avatar
Zzzz
Posts: 172
Joined: Tue Dec 07, 2010 6:28 am
Location: 22° 48' 0" N / 89° 33' 0" E

Re: Imo 2005-4

Unread post by Zzzz » Fri Dec 10, 2010 5:15 am

\[a_1=6^1+2^1+3^1-1=10\]
\[a_2=6^2+2^2+3^2-1=48\]
\[a_3=6^3+2^3+3^3-1=250\]
\[.\]
\[.\]
\[.\]

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 !
Every logical solution to a problem has its own beauty.
(Important: Please make sure that you have read about the Rules, Posting Permissions and Forum Language)

User avatar
TIUrmi
Posts: 61
Joined: Tue Dec 07, 2010 12:13 am
Location: Dinajpur, Bangladesh
Contact:

Re: Imo 2005-4

Unread post by TIUrmi » Fri Dec 10, 2010 10:57 am

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

Post Reply