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Express a number as sum of consecutive numbers
  • Difficulty Level : Easy
  • Last Updated : 27 Apr, 2021

Given a number N, write a function to express N as sum of two or more consecutive positive numbers. If there is no solution, output -1. If there are multiple solution, then print one of them.
Examples: 
 

Input : N = 10
Output : 4 + 3 + 2 + 1

Input  : N = 8
Output : -1

Input  : N = 24
Output : 9 + 8 + 7

 

 

Sum of first n natural numbers = n * (n + 1)/2

Sum of first (n + k) numbers = (n + k) * (n + k + 1)/2

If N is sum of k consecutive numbers, then
following must be true.

N = [(n+k)(n+k+1) - n(n+1)] / 2

OR 

2 * N = [(n+k)(n+k+1) - n(n+1)]

Below is the implementation based on above idea.
 

C++




// C++ program to print a consecutive sequence
// to express N if possible.
#include <bits/stdc++.h>
using namespace std;
 
// Print consecutive numbers from
// last to first
void printConsecutive(int last, int first)
{
    cout << first++;
    for (int x = first; x<= last; x++)
        cout << " + " << x;
}
 
void findConsecutive(int N)
{
    for (int last=1; last<N; last++)
    {
        for (int first=0; first<last; first++)
        {
            if (2*N == (last-first)*(last+first+1))
            {
                cout << N << " = ";
                printConsecutive(last, first+1);
                return;
            }
        }
    }
    cout << "-1";
}
 
// Driver code
int main()
{
    int n = 12;
    findConsecutive(n);
    return 0;
}

Java 




// Java program to print a consecutive sequence
// to express N if possible.
class GfG
{
    // Print consecutive numbers from
    // last to first
    static void printConsecutive(int last, int first)
    {
        System.out.print(first++);
        for (int x = first; x<= last; x++)
            System.out.print(" + "+x);
    }  
    
    static void findConsecutive(int N)
    {
        for (int last=1; last<N; last++)
        {
            for (int first=0; first<last; first++)
            {
                if (2*N == (last-first)*(last+first+1))
                {
                    System.out.printf(N + " = ");
                    printConsecutive(last, first+1);
                    return;
                }
            }
        }
        System.out.print("-1");
    }
    
    // main function
    public static void main (String[] args) 
    {
        int n = 12;
        findConsecutive(n);
    }
}

Python3




# Python3 program to print a consecutive
# sequence to express N if possible.
 
# Print consecutive numbers
# from last to first
def printConsecutive(last, first):
    print (first, end = "")
    first += 1
    for x in range(first, last + 1):
        print (" +", x, end = "")
 
def findConsecutive(N):
    for last in range(1, N):
         
        for first in range(0, last):
             
            if 2 * N == (last - first) * (last + first + 1):
                print (N, "= ", end = "")
                printConsecutive(last, first + 1)
                return
    print ("-1")
 
# Driver code
n = 12
findConsecutive(n)
 
# This code is contributed by Shreyanshi Arun.

C#




// C# program to print a consecutive sequence
// to express N if possible.
using System;
 
class GfG
{
    // Print consecutive numbers from
    // last to first
    static void printConsecutive(int last, int first)
    {
        Console.Write(first++);
        for (int x = first; x <= last; x++)
            Console.Write(" + "+x);
    }
     
    static void findConsecutive(int N)
    {
        for (int last = 1; last < N; last++)
        {
            for (int first = 0; first < last; first++)
            {
                if (2 * N == (last - first)
                    * (last + first + 1))
                {
                    Console.Write(N + " = ");
                    printConsecutive(last, first + 1);
                    return;
                }
            }
        }
        Console.Write("-1");
    }
     
    // Driver code
    public static void Main ()
    {
        int n = 12;
        findConsecutive(n);
    }
}
 
// This code is contributed by vt_m

PHP




<?php
// PHP program to print a consecutive
// sequence to express N if possible.
 
// Print consecutive numbers from
// last to first
function printConsecutive($last, $first)
{
    echo $first++;
    for ($x = $first; $x<= $last; $x++)
        echo " + " , $x;
}
 
function findConsecutive($N)
{
    for ($last = 1; $last < $N; $last++)
    {
        for ($first = 0; $first < $last; $first++)
        {
            if (2 * $N == ($last - $first) *
                      ($last + $first + 1))
            {
                 echo $N , " = ";
                printConsecutive($last, $first + 1);
                return;
            }
        }
    }
    echo "-1";
}
 
    // Driver Code
    $n = 12;
    findConsecutive($n);
     
// This code is contributed by nitin mittal
?>

Javascript




<script>
    // Javascript program to print a consecutive
// sequence to express N if possible.
   
// Print consecutive numbers from
// last to first
function printConsecutive(last, first)
{
    document.write(first++);
    for (let x = first; x<= last; x++)
        document.write( " + " + x);
}
   
function findConsecutive(N)
{
    for (let last = 1; last < N; last++)
    {
        for (let first = 0; first < last; first++)
        {
            if (2 * N == (last - first) *
                      (last + first + 1))
            {
                 document.write(N + " = ");
                printConsecutive(last, first + 1);
                return;
            }
        }
    }
    document.write("-1");
}
   
    // Driver Code
    let n = 12;
    findConsecutive(n);
       
// This code is contributed by _saurabh_jaiswal
</script>

Output: 
 

12 = 3 + 4 + 5

Reference : 
https://math.stackexchange.com/questions/139842/in-how-many-ways-can-a-number-be-expressed-as-a-sum-of-consecutive-numbers
This article is contributed by Roshni Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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