BdMO 2014: Secondary - Q5
For a positive integer \(n\), find the greatest \(p\) such that \(n\) can be rewritten as sum of \(p\) consecutive positive integers.
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
-
sadman sakib
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Re: BdMO 2014: Secondary - Q5
Let $a$ be the first term of the sequence . Then ,
$a + (a + 1) +.........+ (a + p - 1) = n$
$ \Rightarrow \frac p2 (a + a + p -1) = n$
$ \Rightarrow p(p + 2a -1) = 2n$
Here both $p$ and $p + 2a -1$ can't be odd or even , one of them have to be odd and another even . We want to minimize $a$ to get the greatest value of $p$ . If $ p(p-1) = 2n$ has an integer solution ,then we are done right here .
If not , then $ p < (p + 2a - 1)$ . Now call a odd factor of $n$ , $m$ if $ |m - \sqrt n|$ is the smallest possible . Thus $ p = m $ , if $ m < \sqrt n$ or $ p = \frac nm$ if $ m > \sqrt n$ .
What happens if there are two choices for $m$ such that $ |m - \sqrt n|$ is the smallest possible ? Name them $m_1$ and $m_2$ where $ m_1 < m_2$ . In this case $ m_1 < \frac{n}{m_2} < \sqrt n$ (can you prove why). So , our required $ p = \frac{n}{m_2}$ .
$a + (a + 1) +.........+ (a + p - 1) = n$
$ \Rightarrow \frac p2 (a + a + p -1) = n$
$ \Rightarrow p(p + 2a -1) = 2n$
Here both $p$ and $p + 2a -1$ can't be odd or even , one of them have to be odd and another even . We want to minimize $a$ to get the greatest value of $p$ . If $ p(p-1) = 2n$ has an integer solution ,then we are done right here .
If not , then $ p < (p + 2a - 1)$ . Now call a odd factor of $n$ , $m$ if $ |m - \sqrt n|$ is the smallest possible . Thus $ p = m $ , if $ m < \sqrt n$ or $ p = \frac nm$ if $ m > \sqrt n$ .
What happens if there are two choices for $m$ such that $ |m - \sqrt n|$ is the smallest possible ? Name them $m_1$ and $m_2$ where $ m_1 < m_2$ . In this case $ m_1 < \frac{n}{m_2} < \sqrt n$ (can you prove why). So , our required $ p = \frac{n}{m_2}$ .