Advanced P-3(BOMC-2)

[sourav das]

Most positive integers can be expressed as a sum of two or more consecutive
positive integers. For example, $24 = 7 + 8 + 9$ and $51 = 25 + 26$. A
positive integer that cannot be expressed as a sum of two or more consecutive
positive integers is therefore interesting. What are all the interesting
integers?

[*Mahi*]

Sharing Hints:
First and most basic idea:

$2n+1=n+(n+1)$

Then:

$\text{odd+even=odd}$

Then:

$(n-k)+(n-k+1)+ \cdots + n+ \cdots +(n+k) =(2k+1)n$

And then... problem solved.
(Open the hints in three stages, not all in once. This is because it is easier to get the later hints when you know the earlier ones.)

[Niloy Da Fermat]

same as mahi.the idea crossed my mind through

experimenting with first few numbers

[nafistiham]

I think the numbers are
$2^k$ where $k \in \mathbb {N}$ or $k=0$

[sowmitra]

I agree with Tiham. The solutions are indeed of the form $2^k$, where, $k\in\mathbb{N^0}
$.
I am actually familiar with this problem and came across it a few days ago. The solution said that all $n$ except which are powers of $2$ can be expressed as a sum of at least $2$ consecutive natural numbers.

[nafistiham]

I did not have the proof yesterday.today, I think I do.

the numbers which are not power of $2$ can be divided again and again by $2$. Thus, at last we will surely, get an odd number.
suppose, $n=p \cdot 2^k$
where $p$ is an odd greter than $1$.
\[p=m+(m+1)=(m-1)+(m+2)=(m-2)+(m+3)=(m-3)+(m+4)=\cdot \cdot \cdot\]
adding all them we can get $n$

[sourav das]

I did not have the proof yesterday.today, I think I do.

the numbers which are not power of $2$ can be divided again and again by $2$. Thus, at last we will surely, get an odd number.
suppose, $n=p \cdot 2^k$
where $p$ is an odd greter than $1$.
\[p=m+(m+1)=(m-1)+(m+2)=(m-2)+(m+3)=(m-3)+(m+4)=\cdot \cdot \cdot\]
adding all them we can get $n$

Really nice trick . But i think there is a bug. Try to find it out.

[zadid xcalibured]

nice thinking tiham.but it holds for very large numbers only.

[nafistiham]

thanks to zadid, found out that the numbers which has a large divisor in the form $2^k$ will have problem.because in those cases.$(m-s)$ might be negative.

[*Mahi*]

@Tiham:
Add these

$(n-k)+(n-k+1)+ \cdots + n+ \cdots +(n+k) =(2k+1)n$

To you solution , and see what you get. It improves the precision.