### In how many ways can you represent a number as the sum of consecutive numbers.

To solve this, we can use the concept of A.P. We know that sum of A.P. is `n = (m/2)*(2*a + (m - 1)*d)`

if n is the sum and m is the number of integers with common difference d in the A.P. In our problem, n is the given number and m is the number of consecutive integers, obviously d is 1. Now we can derive two conclusions from above formula:

Manipulating the above formula as

`n/m = m/2 + a - 1/2`

we can see that n/m is greater than m/2 because`a - 1/2`

is always positive as 'a' belongs to the range [1, INF). Therefore, the conclusion is`n/m > m/2`

=>`m < sqrt(2n)`

.Above formula can also be written as

`a = n/m - m/2 + 1/2`

.From here we can conclude that`n/m - m/2 + 1/2`

is an integer as a is integer.

So if we iterate over m from 2 to sqrt(2n) and check for every such m that `n/m - m/2 + 1/2`

is integer or not. If we count the number of m's for which `n/m - m/2 + 1/2`

is integer then that count will be the number of ways in which we can represent n as sum of consecutive numbers.

```
int count = 0;
for(int i = 2;i < sqrt(2n);i++)
{
//If n/m - m/2 + 1/2 is integer: count++
}
count is the no. of ways.
```