Bangladesh TST(Team selection test) 2011 Exam 4 Problem 1

Discussion on Bangladesh National Math Camp
sourav das
Posts:461
Joined:Wed Dec 15, 2010 10:05 am
Location:Dhaka
Contact:
Bangladesh TST(Team selection test) 2011 Exam 4 Problem 1

Unread post by sourav das » Thu Dec 15, 2011 7:34 pm

Find for which positive integers $n$ there exists real number $x$ such that:
\[\sum_{i=1}^{n}\left \lfloor ix \right \rfloor= n\]
Where $\left \lfloor x \right \rfloor=$greatest integer which is less or equal to $x$
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

mahathir
Posts:24
Joined:Tue Feb 15, 2011 11:01 pm

Re: Bangladesh TST(Team selection test) 2011 Exam 4 Problem

Unread post by mahathir » Tue Nov 06, 2012 1:03 am

Is it all $n$ which are divisible by $3$ ?

User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:

Re: Bangladesh TST(Team selection test) 2011 Exam 4 Problem

Unread post by Phlembac Adib Hasan » Tue Nov 06, 2012 10:41 am

mahathir wrote:Is it all $n$ which are divisible by $3$ ?
$1$-ও তো হয়। $x=1.5$ নেন।
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

Re: Bangladesh TST(Team selection test) 2011 Exam 4 Problem

Unread post by *Mahi* » Tue Nov 06, 2012 12:20 pm

Hint:
Try bounding :)
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:

Re: Bangladesh TST(Team selection test) 2011 Exam 4 Problem

Unread post by Phlembac Adib Hasan » Tue Apr 09, 2013 10:26 am

Any $x$ in the interval $\dfrac 1 k>x\ge \dfrac {2}{2k+1}$ satisfies the equation if $n=3k$. If $n=3k+1$, take any $x$ from the interval $\dfrac 2 {2k+1}>x\ge \dfrac 1 {k+1}$. And if $3\mid n-2$, it is not that hard to prove there is no such $x$. Just keep in mind $2a\ge 2\Longrightarrow a\ge 1$.

Post Reply