Check if a large number is divisible by 11 or not (original) (raw)

Last Updated : 19 Jun, 2025

Given a number in the form of string **s, 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: s = "76945"
**Output: true
**Explanation: s when divided by 11 gives 0 as remainder.

I**nput: s = "7695"
**Output: false
**Explanation: s does not give 0 as remainder when divided by 11.

**Input: s = "1234567589333892"
**Output: true
**Explanation: s when divided by 11 gives 0 as remainder.

Try It Yourself

Table of Content

[Naive Approach]: Modulo Division Method
[Expected Approach] - Even-Odd Digit Sum for Large String Input

[Naive Approach]: Modulo Division Method

Checking given number is divisible by 11 or not by using the modulo **division operator "%".
**Note: This approach may not work in some programming languages for very large input strings, due to limitations such as integer overflow or memory constraints..

C++ `

#include #include using namespace std;

bool divBy11(string &s) {

// Convert string to int
int n = stoi(s); 
return n % 11 == 0;

}

int main() { string s = "76945";

if (divBy11(s))
    cout << "true" << endl;
else
    cout << "false" << endl;

return 0;

}

Java

class GfG {

public static boolean divBy11(String s) {
    
    // Convert string to int
    int n = Integer.parseInt(s);
    return n % 11 == 0;
}

public static void main(String[] args) {
    String s = "76945";

    if (divBy11(s))
        System.out.println("true");
    else
        System.out.println("false");
}

}

Python

def divBy11(s):

# Convert string to integer
n = int(s)
return n % 11 == 0

if name == "main": s = "76945"

if divBy11(s):
    print("true")
else:
    print("false")

C#

using System;

class GfG { static bool divBy11(string s) { // Convert string to int int n = int.Parse(s); return n % 11 == 0; }

static void Main()
{
    string s = "76945";

    if (divBy11(s))
        Console.WriteLine("true");
    else
        Console.WriteLine("false");
}

}

JavaScript

function divBy11(s) {

// Convert string to int
let n = parseInt(s);
return n % 11 === 0;

}

// Driver Code const s = "76945";

if (divBy11(s)) { console.log("true"); } else { console.log("false"); }

**Time Complexity: O(n), where n is length of s.
**Auxiliary Space: O(1)

[Expected Approach] - Even-Odd Digit Sum for Large String Input

Since input number may be very large, we cannot use n % 11 to check if a number is divisible by 11 or not. The idea is based on following fact.
A number is divisible by 11 if difference of following two is divisible by 11.

Sum of digits at odd places.
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 76945 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 = 1 if i is **even
Remainder of 10i divided by 11 = 10 ≡ -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.

C++ `

#include #include using namespace std;

int divBy11(string &s){ int n = s.length(); int oddDigSum = 0, evenDigSum = 0;

for (int i = 0; i < n; i++){ 
    
    // check for even index
    if (i % 2 == 0)
        oddDigSum += (s[i] - '0');
    else
        evenDigSum += (s[i] - '0');
}

// Check divisibility by 11
// using digit sum difference
return ((oddDigSum - evenDigSum) % 11 == 0);

}

int main(){ string s = "76945"; cout << (divBy11(s) ? "true" : "false"); return 0; }

Java

class GfG {

public static boolean divBy11(String s) {
    int n = s.length();
    int oddDigSum = 0, evenDigSum = 0;

    for (int i = 0; i < n; i++) {
        int digit = s.charAt(i) - '0';
        
        // check for even index
        if (i % 2 == 0)
            oddDigSum += digit;
        else
            evenDigSum += digit;
    }
    // Check divisibility by 11 
    // using digit sum difference
    return (oddDigSum - evenDigSum) % 11 == 0;
}

public static void main(String[] args) {
    String s = "76945";
    System.out.println(divBy11(s) ? "true" : "false");
}

}

Python

def divBy11(s): odd_dig_sum = 0 even_dig_sum = 0

for i in range(len(s)):
    digit = int(s[i])
    
    # check for even index
    if i % 2 == 0:
        odd_dig_sum += digit
    else:
        even_dig_sum += digit

# Check divisibility by 11 
# using digit sum difference
return (odd_dig_sum - even_dig_sum) % 11 == 0

if name == "main": s = "76945" print("true" if divBy11(s) else "false")

C#

using System;

class GfG{ static bool divBy11(string s){ int oddDigSum = 0, evenDigSum = 0;

    for (int i = 0; i < s.Length; i++){
        int digit = s[i] - '0';
        
        // check for even index
        if (i % 2 == 0)
            oddDigSum += digit;
        else
            evenDigSum += digit;
    }
    
    // Check divisibility by 11 
    // using digit sum difference
    return (oddDigSum - evenDigSum) % 11 == 0;
}

static void Main(string[] args){
    string s = "76945";
    Console.WriteLine(divBy11(s) ? "true" : "false");
}

}

JavaScript

function divBy11(s) { let oddDigSum = 0, evenDigSum = 0;

for (let i = 0; i < s.length; i++) {
    let digit = parseInt(s[i]);
    
    // check for even index
    if (i % 2 === 0)
        oddDigSum += digit;
    else
        evenDigSum += digit;
}

// Check divisibility by 11 
// using digit sum difference
return (oddDigSum - evenDigSum) % 11 === 0;

}

// Driver Code const s = "76945"; console.log(divBy11(s) ? "true" : "false");

**Time Complexity: O(n), where n is length of s.
**Auxiliary Space: O(1)