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:

1. Manipulating the above formula as n/m = m/2 + (a - 1/2) we can see that n/m is definitely 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).

2. 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.