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 : n = 15 Output : 3 15 can be represented as: 1 + 2 + 3 + 4 + 5 4 + 5 + 6 7 + 8 Input :10 Output :2 10 can only be represented as: 1 + 2 + 3 + 4

We have already discussed one approach in below post.

Count ways to express a number as sum of consecutive numbers

Here a new approach is discussed. Suppose that we are talking about the sum of numbers from X to Y ie [X, X+1, …, Y-1, Y]

Then the arithmetic sum is

(Y+X)(Y-X+1)/2

If this should be N, then

2N = (Y+X)(Y-X+1)

Note that one of the factors should be even and the other should be odd because Y-X+1 and Y+X should have opposite parity because Y-X and Y+X have the same parity. Since 2N is anyways even, we find the number of odd factors of N.

For example, n = 15 all odd factors of 15 are 1 3 and 5 so the answer is 3.

## C++

`// C++ program to count number of ways to express` `// N as sum of consecutive numbers.` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// returns the number of odd factors` `int` `countOddFactors(` `long` `long` `n)` `{` ` ` `int` `odd_factors = 0;` ` ` ` ` `for` `(` `int` `i = 1; 1ll * i * i <= n; i++) {` ` ` `if` `(n % i == 0) {` ` ` ` ` `// If i is an odd factor and` ` ` `// n is a perfect square` ` ` `if` `(1ll * i * i == n) {` ` ` `if` `(i & 1)` ` ` `odd_factors++;` ` ` `}` ` ` ` ` `// If n is not perfect square` ` ` `else` `{` ` ` `if` `(i & 1)` ` ` `odd_factors++;` ` ` ` ` `int` `factor = n / i;` ` ` `if` `(factor & 1)` ` ` `odd_factors++;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `odd_factors - 1;` `}` ` ` `// Driver Code` `int` `main()` `{` ` ` `// N as sum of consecutive numbers` ` ` `long` `long` `int` `N = 15;` ` ` `cout << countOddFactors(N) << endl;` ` ` ` ` `N = 10;` ` ` `cout << countOddFactors(N) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java program to count number ` `// of ways to express N as sum ` `// of consecutive numbers.` `import` `java.io.*;` ` ` `class` `GFG` `{` `// returns the number` `// of odd factors` `static` `int` `countOddFactors(` `long` `n)` `{` ` ` `int` `odd_factors = ` `0` `;` ` ` ` ` `for` `(` `int` `i = ` `1` `; ` `1` `* i * i <= n; i++) ` ` ` `{` ` ` `if` `(n % i == ` `0` `) ` ` ` `{` ` ` ` ` `// If i is an odd factor and` ` ` `// n is a perfect square` ` ` `if` `(` `1` `* i * i == n) ` ` ` `{` ` ` `if` `((i & ` `1` `) == ` `1` `)` ` ` `odd_factors++;` ` ` `}` ` ` ` ` `// If n is not ` ` ` `// perfect square` ` ` `else` `{` ` ` `if` `((i & ` `1` `) == ` `1` `)` ` ` `odd_factors++;` ` ` ` ` `int` `factor = (` `int` `)n / i;` ` ` `if` `((factor & ` `1` `) == ` `1` `)` ` ` `odd_factors++;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `odd_factors - ` `1` `;` `}` ` ` `// Driver Code` `public` `static` `void` `main(String args[])` `{` ` ` `// N as sum of consecutive numbers` ` ` `long` `N = ` `15` `;` ` ` `System.out.println(countOddFactors(N));` ` ` ` ` `N = ` `10` `;` ` ` `System.out.println(countOddFactors(N));` `}` `}` ` ` `// This code is contributed by ` `// Manish Shaw(manishshaw1)` |

## Python3

`# Python3 program to count number ` `# of ways to express N as sum` `# of consecutive numbers.` ` ` `# returns the number` `# of odd factors` `def` `countOddFactors(n) :` ` ` `odd_factors ` `=` `0` ` ` `i ` `=` `1` ` ` `while` `((` `1` `*` `i ` `*` `i) <` `=` `n) :` ` ` `if` `(n ` `%` `i ` `=` `=` `0` `) :` ` ` ` ` `# If i is an odd factor and` ` ` `# n is a perfect square` ` ` `if` `(` `1` `*` `i ` `*` `i ` `=` `=` `n) :` ` ` `if` `(i & ` `1` `) :` ` ` `odd_factors ` `=` `odd_factors ` `+` `1` ` ` ` ` `# If n is not perfect square` ` ` `else` `:` ` ` `if` `((i & ` `1` `)) :` ` ` `odd_factors ` `=` `odd_factors ` `+` `1` ` ` ` ` `factor ` `=` `int` `(n ` `/` `i)` ` ` `if` `(factor & ` `1` `) :` ` ` `odd_factors ` `=` `odd_factors ` `+` `1` ` ` `i ` `=` `i ` `+` `1` ` ` `return` `odd_factors ` `-` `1` ` ` `# Driver Code` ` ` `# N as sum of consecutive numbers` `N ` `=` `15` `print` `(countOddFactors(N))` ` ` `N ` `=` `10` `print` `(countOddFactors(N))` ` ` `# This code is contributed by ` `# Manish Shaw(manishshaw1)` |

## C#

`// C# program to count number of ` `// ways to express N as sum of ` `// consecutive numbers.` `using` `System;` ` ` `class` `GFG` `{` ` ` `// returns the number` ` ` `// of odd factors` ` ` `static` `int` `countOddFactors(` `long` `n)` ` ` `{` ` ` `int` `odd_factors = 0;` ` ` ` ` `for` `(` `int` `i = 1; 1 * i * i <= n; i++) ` ` ` `{` ` ` `if` `(n % i == 0) ` ` ` `{` ` ` ` ` `// If i is an odd factor and` ` ` `// n is a perfect square` ` ` `if` `(1 * i * i == n) ` ` ` `{` ` ` `if` `((i & 1) == 1)` ` ` `odd_factors++;` ` ` `}` ` ` ` ` `// If n is not ` ` ` `// perfect square` ` ` `else` `{` ` ` `if` `((i & 1) == 1)` ` ` `odd_factors++;` ` ` ` ` `int` `factor = (` `int` `)n / i;` ` ` `if` `((factor & 1) == 1)` ` ` `odd_factors++;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `odd_factors - 1;` ` ` `}` ` ` ` ` `// Driver Code` ` ` `static` `void` `Main()` ` ` `{` ` ` `// N as sum of consecutive numbers` ` ` `long` `N = 15;` ` ` `Console.WriteLine(countOddFactors(N));` ` ` ` ` `N = 10;` ` ` `Console.WriteLine(countOddFactors(N));` ` ` `}` `}` |

## PHP

`<?php` `// PHP program to count number ` `// of ways to express N as sum` `// of consecutive numbers.` ` ` `// returns the number` `// of odd factors` `function` `countOddFactors(` `$n` `)` `{` ` ` `$odd_factors` `= 0;` ` ` ` ` `for` `(` `$i` `= 1; 1 * ` `$i` `* ` `$i` `<= ` `$n` `; ` `$i` `++) ` ` ` `{` ` ` `if` `(` `$n` `% ` `$i` `== 0) ` ` ` `{` ` ` ` ` `// If i is an odd factor and` ` ` `// n is a perfect square` ` ` `if` `(1 * ` `$i` `* ` `$i` `== ` `$n` `) ` ` ` `{` ` ` `if` `(` `$i` `& 1)` ` ` `$odd_factors` `++;` ` ` `}` ` ` ` ` `// If n is not perfect square` ` ` `else` `{` ` ` `if` `(` `$i` `& 1)` ` ` `$odd_factors` `++;` ` ` ` ` `$factor` `= ` `$n` `/ ` `$i` `;` ` ` `if` `(` `$factor` `& 1)` ` ` `$odd_factors` `++;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `$odd_factors` `- 1;` `}` ` ` `// Driver Code` ` ` `// N as sum of consecutive numbers` `$N` `= 15;` `echo` `(countOddFactors(` `$N` `) . (` `"\n"` `));` ` ` `$N` `= 10;` `echo` `(countOddFactors(` `$N` `));` ` ` `// This code is contributed by ` `// Manish Shaw(manishshaw1)` `?>` |

**Output :**

3 1

The Time complexity for this program is O(N^0.5).

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