Check if a large number is divisible by 11 or not

Given a number, the task is to check if the number is divisible by 11 or not. The input number may be large and it may not be possible to store it even if we use long long int.

Examples:

Input : n = 76945
Output : Yes

Input  : n = 1234567589333892
Output : Yes

Input  : n = 363588395960667043875487
Output : No



Since input number may be very large, we cannot use n % 11 to check if a number is divisible by 11 or not, especially in languages like C/C++. The idea is based on following fact.

A number is divisible by 11 if difference of following two is divisible by 11.

  1. Sum of digits at odd places.
  2. Sum of digits at even places.

Illustration:

For example, let us consider 76945 
Sum of digits at odd places  : 7 + 9 + 5
Sum of digits at even places : 6 + 4 
Difference of two sums = 21 - 10 = 11
Since difference is divisible by 11, the
number 7945 is divisible by 11.

How does this work?

Let us consider 7694, we can write it as
7694 = 7*1000 + 6*100 + 9*10 + 4

The proof is based on below observation:
Remainder of 10i divided by 11 is 1 if i is even
Remainder of 10i divided by 11 is -1 if i is odd

So the powers of 10 only result in values either 1 
or -1. 

Remainder of "7*1000 + 6*100 + 9*10 + 4"
divided by 11 can be written as : 
7*(-1) + 6*1 + 9*(-1) + 4*1

The above expression is basically difference 
between sum of even digits and odd digits.

Below is implementation of above fact :

C++

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// C++ program to find if a number is divisible by
// 11 or not
#include<bits/stdc++.h>
using namespace std;
  
// Function to find that number divisible by 11 or not
int check(string str)
{
    int n = str.length();
  
    // Compute sum of even and odd digit
    // sums
    int oddDigSum = 0, evenDigSum = 0;
    for (int i=0; i<n; i++)
    {
        // When i is even, position of digit is odd
        if (i%2 == 0)
            oddDigSum += (str[i]-'0');
        else
            evenDigSum += (str[i]-'0');
    }
  
    // Check its difference is divisible by 11 or not
    return ((oddDigSum - evenDigSum) % 11 == 0);
}
  
// Driver code
int main()
{
    string str = "76945";
    check(str)?  cout << "Yes" : cout << "No ";
    return 0;
}

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Java

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// Java program to find if a number is
// divisible by 11 or not
class IsDivisible
{
    // Function to find that number divisible by 11 or not
    static boolean check(String str)
    {
        int n = str.length();
       
        // Compute sum of even and odd digit
        // sums
        int oddDigSum = 0, evenDigSum = 0;
        for (int i=0; i<n; i++)
        {
            // When i is even, position of digit is odd
            if (i%2 == 0)
                oddDigSum += (str.charAt(i)-'0');
            else
                evenDigSum += (str.charAt(i)-'0');
        }
       
        // Check its difference is divisible by 11 or not
        return ((oddDigSum - evenDigSum) % 11 == 0);
    }
      
    // main function
    public static void main (String[] args) 
    {
        String str = "76945";
        if(check(str))
            System.out.println("Yes");
        else
            System.out.println("No");
    }

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Python3

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# Python 3 code program to find if a number 
# is divisible by 11 or not
  
  
# Function to find that number divisible by
#  11 or not
def check(st) :
    n = len(st) 
  
    # Compute sum of even and odd digit
    # sums
    oddDigSum = 0
    evenDigSum = 0
    for i in range(0,n) :
        # When i is even, position of digit is odd
        if (i % 2 == 0) :
            oddDigSum = oddDigSum + ((int)(st[i]))
        else:
            evenDigSum = evenDigSum + ((int)(st[i]))
      
      
    # Check its difference is divisible by 11 or not
    return ((oddDigSum - evenDigSum) % 11 == 0)
  
# Driver code
st = "76945"
if(check(st)) :
    print( "Yes")
else
    print("No ")
      
# This code is contributed by Nikita tiwari.

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C#

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// C# program to find if a number is
// divisible by 11 or not
using System;
  
class GFG
{
    // Function to find that number 
    // divisible by 11 or not
    static bool check(string str)
    {
        int n = str.Length;
      
        // Compute sum of even and odd digit
        // sums
        int oddDigSum = 0, evenDigSum = 0;
          
        for (int i = 0; i < n; i++)
        {
            // When i is even, position of
            // digit is odd
            if (i % 2 == 0)
                oddDigSum += (str[i] - '0');
            else
                evenDigSum += (str[i] - '0');
        }
      
        // Check its difference is
        // divisible by 11 or not
        return ((oddDigSum - evenDigSum) 
                                % 11 == 0);
    }
      
    // main function
    public static void Main () 
    {
        String str = "76945";
          
        if(check(str))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find if a 
// number is divisible by
// 11 or not
  
// Function to find that number
// divisible by 11 or not
function check($str)
{
    $n = strlen($str);
  
    // Compute sum of even 
    // and odd digit sums
    $oddDigSum = 0; $evenDigSum = 0;
    for ($i = 0; $i < $n; $i++)
    {
          
        // When i is even, position
        // of digit is odd
        if ($i % 2 == 0)
            $oddDigSum += ($str[$i] - '0');
        else
            $evenDigSum += ($str[$i] - '0');
    }
  
    // Check its difference 
    // is divisible by 11 or not
    return (($oddDigSum - $evenDigSum
                            % 11 == 0);
}
  
// Driver code
$str = "76945";
$x = check($str)? "Yes" : "No ";
echo($x);
  
// This code is contributed by Ajit.
?>

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Output:

Yes

This article is contributed by DANISH_RAZA . 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.

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Improved By : jit_t, ManasChhabra2