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Count ways to express a number as sum of consecutive numbers
• Difficulty Level : Medium
• Last Updated : 28 Apr, 2021

Given a number N, find the number of ways to represent this number as a sum of 2 or more consecutive natural numbers.
Examples:

```Input :15
Output :3
15 can be represented as:
1+2+3+4+5
4+5+6
7+8

Input :10
Output :1
10 can only be represented as:
1+2+3+4```

The idea is to represent N as a sequence of length L+1 as:
N = a + (a+1) + (a+2) + .. + (a+L)
=> N = (L+1)*a + (L*(L+1))/2
=> a = (N- L*(L+1)/2)/(L+1)
We substitute the values of L starting from 1 till L*(L+1)/2 < N
If we get ‘a’ as a natural number then the solution should be counted.

## C++

 `// C++ program to count number of ways to express``// N as sum of consecutive numbers.``#include ``using` `namespace` `std;` `long` `int` `countConsecutive(``long` `int` `N)``{``    ``// constraint on values of L gives us the``    ``// time Complexity as O(N^0.5)``    ``long` `int` `count = 0;``    ``for` `(``long` `int` `L = 1; L * (L + 1) < 2 * N; L++)``    ``{``        ``float` `a = (1.0 * N-(L * (L + 1)) / 2) / (L + 1);``        ``if` `(a-(``int``)a == 0.0)``            ``count++;       ``    ``}``    ``return` `count;``}` `// Driver Code``int` `main()``{``    ``long` `int` `N = 15;``    ``cout << countConsecutive(N) << endl;``    ``N = 10;``    ``cout << countConsecutive(N) << endl;``    ``return` `0;``}`

## Java

 `// A Java program to count number of ways``// to express N as sum of consecutive numbers.``public` `class` `SumConsecutiveNumber``{   ``    ``// Utility method to compute number of ways``    ``// in which N can be represented as sum of``    ``// consecutive number``    ``static` `int` `countConsecutive(``int` `N)``    ``{``        ``// constraint on values of L gives us the``        ``// time Complexity as O(N^0.5)``        ``int` `count = ``0``;``        ``for` `(``int` `L = ``1``; L * (L + ``1``) < ``2` `* N; L++)``        ``{``            ``float` `a = (``float``) ((``1.0` `* N-(L * (L + ``1``)) / ``2``) / (L + ``1``));``            ``if` `(a-(``int``)a == ``0.0``)``                ``count++;       ``        ``}``        ``return` `count;``    ``}``    ` `    ``// Driver code to test above function``    ``public` `static` `void` `main(String[] args) {``        ``int` `N = ``15``;``        ``System.out.println(countConsecutive(N));``        ``N = ``10``;``        ``System.out.println(countConsecutive(N));``    ``}``}``// This code is contributed by Sumit Ghosh`

## Python

 `# Python program to count number of ways to``# express N as sum of consecutive numbers.` `def` `countConsecutive(N):``    ` `    ``# constraint on values of L gives us the``    ``# time Complexity as O(N^0.5)``    ``count ``=` `0``    ``L ``=` `1``    ``while``( L ``*` `(L ``+` `1``) < ``2` `*` `N):``        ``a ``=` `(``1.0` `*` `N ``-` `(L ``*` `(L ``+` `1``) ) ``/` `2``) ``/` `(L ``+` `1``)``        ``if` `(a ``-` `int``(a) ``=``=` `0.0``):``            ``count ``+``=` `1``        ``L ``+``=` `1``    ``return` `count` `# Driver code` `N ``=` `15``print` `countConsecutive(N)``N ``=` `10``print` `countConsecutive(N)` `# This code is contributed by Sachin Bisht`

## C#

 `// A C# program to count number of``// ways to express N as sum of``// consecutive numbers.``using` `System;` `public` `class` `GFG {``    ` `    ``// Utility method to compute``    ``// number of ways in which N``    ``// can be represented as sum``    ``// of consecutive number``    ``static` `int` `countConsecutive(``int` `N)``    ``{``        ` `        ``// constraint on values of L``        ``// gives us the time``        ``// Complexity as O(N^0.5)``        ``int` `count = 0;``        ``for` `(``int` `L = 1; L * (L + 1)``                         ``< 2 * N; L++)``        ``{``            ``float` `a = (``float``) ((1.0``                    ``* N-(L * (L + 1))``                     ``/ 2) / (L + 1));``                     ` `            ``if` `(a - (``int``)a == 0.0)``                ``count++;    ``        ``}``        ` `        ``return` `count;``    ``}``    ` `    ``// Driver code to test above``    ``// function``    ``public` `static` `void` `Main()``    ``{``        ``int` `N = 15;``        ``Console.WriteLine(``              ``countConsecutive(N));``        ` `        ``N = 10;``        ``Console.Write(``              ``countConsecutive(N));``    ``}``}` `// This code is contributed by``// nitin mittal.`

## PHP

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## Javascript

 ``

Output:

```3
1```

The Time complexity for this program is O(N^0.5), because of the condition in the for loop.
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