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 the implementation of above approach:

C++

// 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;
}

Java

// Java program to find if a number is
// divisible by 11 or not
import java.io.*;
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");
    }
}

Python3

# 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.

C#

// 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.

PHP

<?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.
?>

Javascript

<script>

// JavaScript program for the above approach

    // Function to find that number
    // divisible by 11 or not
    function check(str)
    {
        let n = str.length;

        // Compute sum of even and odd digit
        // sums
        let oddDigSum = 0, evenDigSum = 0;

        for (let 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
    let str = "76945";
    if(check(str))
        document.write("Yes");
    else
        document.write("No");

// This code is contributed by chinmoy1997pal.
</script>

Output

Yes

Time Complexity: O(n), where n is the given number.
Auxiliary Space: O(1), as we are not using any extra space.

Method: Checking given number is divisible by 11 or not by using the modulo division operator “%”.  

C++

// C++ code to check whether
// the given number is divisible by 11 or not
#include <bits/stdc++.h>
using namespace std;

int main()
{

    // input
    long long n = 1234567589333892;

    // the above input can also be given as n=input() ->
    // taking input from user finding given number is
    // divisible by 11 or not
    if (n % 11 == 0)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}

// This code is contributed by phasing17

Java

// Java code to check whether
// the given number is divisible by 11 or not

import java.io.*;

class GFG {
    public static void main(String[] args)
    {

        // input
        Long n = Long.parseUnsignedLong("1234567589333892");


        //finding given number is
        // divisible by 11 or not
        if (n % 11 == 0)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by phasing17

Python3

# Python code
# To check whether the given number is divisible by 11 or not

#input
n=1234567589333892
# the above input can also be given as n=input() -> taking input from user
# finding given number is divisible by 11 or not
if int(n)%11==0:
  print("Yes")
else:
  print("No")

  # this code is contributed by gangarajula laxmi

C#

// C# code to check whether
// the given number is divisible by 11 or not
using System;

class GFG {
    public static void Main(string[] args)
    {

        // input
        long n = 1234567589333892;

        // the above input can also be given as n=input() ->
        // taking input from user finding given number is
        // divisible by 11 or not
        if (n % 11 == 0)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by phasing17

Javascript

// JavaScript code to check whether
// the given number is divisible by 11 or not

// input
let n = 1234567589333892

// the above input can also be given as n=input() -> taking input from user
// finding given number is divisible by 11 or not
if (n % 11 == 0)
  console.log("Yes")
else
  console.log("No")

// this code is contributed by phasing17

PHP

<?php
    $num = 1234567589333892;
    // checking if the given number is divisible by 37 or
    // not using modulo division operator if the output of
    // num%37 is equal to 0 then given number is divisible
    // by 37 otherwise not divisible by 37
    if ($num % 11 == 0) {
        echo "Yes";
    }
    else {
        echo "No";
    }

?>

Output

Yes

Time Complexity: O(1) because it is performing constant operations
Auxiliary Space: O(1)

Method: Checking given number is divisible by 11 or not using modulo division.

C++

// C++ program to check if given number is divisible by 11
// or not using modulo division

#include <iostream>
using namespace std;

int main()
{

    // input number
    int num = 76945;
    // checking if the given number is divisible by 11 or
    // not using modulo division operator if the output of
    // num%11 is equal to 0 then given number is divisible
    // by 11 otherwise not divisible by 11
    if (num % 11 == 0) {
        cout << " divisible";
    }
    else {
        cout << " not divisible";
    }
    return 0;
}

// this code is contributed by gangarajula laxmi

Java

// java program to check if given number is divisible by 11
// or not using modulo division

import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        // input number
        int num = 76945;
        // checking if the given number is divisible by 11
        // or not
        // using modulo division operator if the output of
        // num%11 is equal to 0 then given number is
        // divisible by 11 otherwise not divisible by 11
        if (num % 11 == 0) {
            System.out.println(" divisible");
        }
        else {
            System.out.println(" not divisible");
        }
    }
}

// this code is contributed by gangarajula laxmi

Python3

# Python3 code for the above approach

# To check whether the given number is divisible by 11 or not

# input
n = 76945

# finding given number is divisible by 11 or not
if (n % 11 == 0):
    print("Yes")
else:
    print("No")

#  This code is contributed by phasing17

C#

using System;

public class GFG {

    static public void Main()
    {

        // input number
        double num = 76945;

        // checking if the given number is divisible by 11
        // or not using modulo division operator if the
        // output of num%11 is equal to 0 then given number
        // is divisible by 11 otherwise not divisible by 11
        if (num % 11 == 0) {
            Console.Write(" divisible");
        }
        else {
            Console.Write(" not divisible");
        }
    }

  // this code is contributed by gangarajula laxmi

Javascript

<script>
        // JavaScript code for the above approach

        // To check whether the given number is divisible by 11 or not

        // input
        var n = 76945

        // finding given number is divisible by 11 or not
        if (n % 11 == 0)
            document.write("Yes")
        else
            document.write("No")

// This code is contributed by laxmigangarajula03
    </script>

PHP

<?php
// PHP program to check
// if a large number is
// divisible by 11.

  // Driver Code
  // input number
$num = 76945;

// finding given number is divisible by 11 or not
if ( $num % 11 == 0)
    echo " divisible";
else
    echo "not divisible";

// This code is contributed by satwik4409.
?>

Output

 divisible

Time Complexity : O(1)

Space Complexity : O(1)

Method 4: 

1. Initialize two variables: alternating_sum to store the alternating sum of the digits and multiplier to keep track of whether to add or subtract each digit. Set alternating_sum to 0 and multiplier to 1.

2. Use a while loop to iterate over the digits of the number num. The loop will continue as long as num is greater than 0.

         a. Get the last digit of the number by using the modulo operator (%) with 10. Store this digit in a temporary variable.

         b. Update the alternating_sum by adding (or subtracting) the last digit, multiplied by multiplier.

         c. Change the value of multiplier to its opposite by multiplying it by -1.

         d. Remove the last digit of the number by using integer division (//) with 10.

3. After the loop, check if the alternating_sum is divisible by 11 by using the modulo operator (%) with 11. If it is, return True, otherwise return False.

C++

#include <iostream>

bool is_divisible_by_11(int num) {
    int alternating_sum = 0;
    int multiplier = 1;
    while (num > 0) {
        alternating_sum += multiplier * (num % 10);
        multiplier *= -1;
        num /= 10;
    }
    // checking if divisible by 11 or not
    return alternating_sum % 11 == 0;
}

int main() {
    std::cout << std::boolalpha; // enable printing of true/false instead of 1/0
    std::cout << is_divisible_by_11(11) << std::endl; // true
    std::cout << is_divisible_by_11(22) << std::endl; // true
    std::cout << is_divisible_by_11(121) << std::endl; // true
    std::cout << is_divisible_by_11(10) << std::endl; // false
    return 0;
}

Python3

def is_divisible_by_11(num):
    alternating_sum = 0
    multiplier = 1
    while num > 0:
        alternating_sum += multiplier * (num % 10)
        multiplier *= -1
        num //= 10
    #checking if divisible by 11 or not
    return alternating_sum % 11 == 0

print(is_divisible_by_11(11)) # True
print(is_divisible_by_11(22)) # True
print(is_divisible_by_11(121)) # True
print(is_divisible_by_11(10)) # False

Java

public class GFG {
    public static boolean isDivisibleBy11(int num)
    {
        int alternatingSum = 0;
        int multiplier = 1;
        while (num > 0) {
            alternatingSum += multiplier * (num % 10);
            multiplier *= -1;
            num /= 10;
        }
        // Checking if divisible by 11 or not
        return alternatingSum % 11 == 0;
    }

    public static void main(String[] args)
    {
        System.out.println(isDivisibleBy11(11)); // true
        System.out.println(isDivisibleBy11(22)); // true
        System.out.println(isDivisibleBy11(121)); // true
        System.out.println(isDivisibleBy11(10)); // false
    }
}

C#

using System;

class Program {
    static bool IsDivisibleBy11(int num)
    {
        int alternatingSum = 0;
        int multiplier = 1;
        while (num > 0) {
            alternatingSum += multiplier * (num % 10);
            multiplier *= -1;
            num /= 10;
        }
        // checking if divisible by 11 or not
        return alternatingSum % 11 == 0;
    }

    static void Main()
    {
        Console.WriteLine(IsDivisibleBy11(11)); // true
        Console.WriteLine(IsDivisibleBy11(22)); // true
        Console.WriteLine(IsDivisibleBy11(121)); // true
        Console.WriteLine(IsDivisibleBy11(10)); // false
    }
}

Javascript

function is_divisible_by_11(num)
{

    // Initialize an integer named "alternating_sum" to 0
    let alternating_sum = 0;

    // Initialize an integer named "multiplier" to 1
    let multiplier = 1;

    // Loop through the digits of the input integer from right to left
    while (num > 0)
    {

        // Add the product of the current digit and "multiplier" to "alternating_sum"
        alternating_sum += multiplier * (num % 10);

        // Toggle the value of "multiplier" between 1 and -1
        multiplier *= -1;

        // Remove the last digit from the input integer
        num = Math.floor(num / 10);
    }
    // Check if "alternating_sum" is divisible by 11
    return alternating_sum % 11 == 0;
}

// Enable printing of true/false instead of 1/0
console.log(is_divisible_by_11(11)); // true
console.log(is_divisible_by_11(22)); // true
console.log(is_divisible_by_11(121)); // true
console.log(is_divisible_by_11(10)); // false

Output

True
True
True
False

Time complexity: O(log(n))
Auxiliary space:  O(1)

Method: Using string manipulation 

• The input number is converted to a string, and then the digits are processed one by one in reverse order using a loop.
• The alternating sum is computed by adding each digit to the sum with the appropriate sign (+ or -) based on its position.
• Finally, the alternating sum is checked for divisibility by 11 using the modulo operator (%).

Python3

def is_divisible_by_11(number):
    # Compute the alternating sum of the digits from right to left
    alternating_sum = 0
    for i, digit in enumerate(reversed(str(number))):
        alternating_sum += (-1) ** i * int(digit)

    # If the alternating sum is divisible by 11, then the number is divisible by 11
    return alternating_sum % 11 == 0

st = "76945"
if(is_divisible_by_11(st)) :
    print( "Yes")
else :
    print("No ")

Output

Yes

Time complexity:

The time complexity of the function is O(n), where n is the number of digits in the input number.
This is because the function processes each digit of the input number once in the loop.
Auxiliary space:

The auxiliary space complexity of the function is O(1), because it uses a constant amount of extra memory to store the alternating sum and loop variables.
The amount of memory used does not depend on the size of the input number.

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