Divisibility by 12 for a large number

Given a large Number the task is to check the number is divisible by 12 or not .

Examples :

Input : 12244824607284961224
Output : Yes

Input : 92387493287593874594898678979792
Output : No

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

This is very simple approach. if a number is divisible by 4 and 3 then number is divisible by 12.
Point 1. If the last two digit of the number is divisible by 4 then number is divisible by 4. Please see divisibility by 4 for large numbers for details.
Point 2. if sum of all digit of a number divide by 3 then number is divisible by 3. Please see divisibility by 3 for large numbers for details.

CPP

// C++ Program to check if 
// number is divisible by 12 
#include <iostream> 
using namespace std; 
  
bool isDvisibleBy12(string num) 
{ 
    // if number greater then 3 
    if (num.length() >= 3) {  
  
        // fiind last digit 
        int d1 = (int)num[num.length() - 1]; 
  
        // no is odd 
        if (d1 % 2 != 0) 
            return (0); 
  
        // find second last digit 
        int d2 = (int)num[num.length() - 2]; 
  
        // find sum of all digits 
        int sum = 0; 
        for (int i = 0; i < num.length(); i++) 
            sum += num[i];             
          
        return (sum % 3 == 0 && (d2 * 10 + d1) % 4 == 0);             
    } 
      
    else { 
          
        // if number is less then 
        // or equal to 100 
        int number = stoi(num); 
        return (number % 12 == 0); 
    } 
} 
  
// Driver function 
int main() 
{ 
    string num = "12244824607284961224";   
    if (isDvisibleBy12(num)) 
        cout << "Yes" << endl; 
    else
        cout << "No" << endl; 
    return 0; 
} 

Java

// Java Program to check if 
// number is divisible by 12 
  
import java.io.*; 
  
class GFG { 
static boolean isDvisibleBy12(String num) 
{ 
    // if number greater then 3 
    if (num.length() >= 3) {  
   
        // fiind last digit 
        int d1 = (int)num.charAt(num.length() - 1); 
   
        // no is odd 
        if (d1 % 2 != 0) 
            return false; 
   
        // find second last digit 
        int d2 = (int)num.charAt(num.length() - 2); 
   
        // find sum of all digits 
        int sum = 0; 
        for (int i = 0; i < num.length(); i++) 
            sum += num.charAt(i);             
           
        return (sum % 3 == 0 && 
               (d2 * 10 + d1) % 4 == 0);             
    } 
       
    else { 
           
        // if number is less then 
        // or equal to 100 
        int number = Integer.parseInt(num); 
        return (number % 12 == 0); 
    } 
  
// driver function 
} 
    public static void main (String[] args) { 
  
    String num = "12244824607284961224";   
    if (isDvisibleBy12(num)) 
        System.out.print("Yes"); 
    else
        System.out.print("No"); 
          
    } 
} 
  
// This code is contributed by Gitanjali. 

Python3

# Python Program to check if 
# number is divisible by 12  
  
import math  
  
def isDvisibleBy12( num): 
  
    # if number greater then 3 
    if (len(num) >= 3): 
   
        # fiind last digit 
        d1 = int(num[len(num) - 1]); 
   
        # no is odd 
        if (d1 % 2 != 0): 
            return False
   
        # find second last digit 
        d2 = int(num[len(num) - 2]) 
   
        # find sum of all digits 
        sum = 0
        for  i in range(0, len(num) ): 
            sum += int(num[i])            
           
        return (sum % 3 == 0 and
               (d2 * 10 + d1) % 4 == 0)             
  
       
    else : 
           
        # f number is less then 
        # r equal to 100 
        number = int(num) 
        return (number % 12 == 0) 
      
  
num = "12244824607284961224"  
if(isDvisibleBy12(num)): 
       print("Yes") 
else: 
       print("No") 
  
# This code is contributed by Gitanjali. 

C#

// C# Program to check if 
// number is divisible by 12 
using System; 
  
class GFG  
{ 
static bool isDvisibleBy12(string num) 
{ 
    // if number greater then 3 
    if (num.Length >= 3) {  
  
        // find last digit 
        int d1 = (int)num[num.Length - 1]; 
  
        // no is odd 
        if (d1 % 2 != 0) 
            return false; 
  
        // find second last digit 
        int d2 = (int)num[num.Length - 2]; 
  
        // find sum of all digits 
        int sum = 0; 
        for (int i = 0; i < num.Length; i++) 
            sum += num[i];          
          
        return (sum % 3 == 0 && 
            (d2 * 10 + d1) % 4 == 0);          
    } 
      
    else { 
          
        // if number is less then 
        // or equal to 100 
        int number = int.Parse(num); 
        return (number % 12 == 0); 
    } 
} 
  
    // Driver function 
    public static void Main ()  
    { 
       String num = "12244824607284961224";  
       if (isDvisibleBy12(num)) 
          Console.Write("Yes"); 
       else
          Console.Write("No"); 
    } 
} 
  
// This code is contributed by nitin mittal. 

PHP

<?php 
// PHP Program to check if 
// number is divisible by 12 
  
function isDvisibleBy12($num) 
{ 
      
    // if number greater then 3 
    if (strlen($num) >= 3)  
    {  
  
        // find last digit 
        $d1 = (int)$num[strlen($num) - 1]; 
  
        // no is odd 
        if ($d1 % 2 != 0) 
            return (0); 
  
        // find second last digit 
        $d2 = (int)$num[strlen($num) - 2]; 
  
        // find sum of all digits 
        $sum = 0; 
        for ($i = 0; $i < strlen($num); $i++) 
            $sum += $num[$i];      
          
        return ($sum % 3 == 0 &&  
               ($d2 * 10 + $d1) % 4 == 0);          
    } 
      
    else { 
          
        // if number is less then 
        // or equal to 100 
        $number = stoi($num); 
        return ($number % 12 == 0); 
    } 
} 
  
// Driver Code 
$num = "12244824607284961224";  
if (isDvisibleBy12($num)) 
    echo("Yes"); 
else
    echo("No"); 
  
// This code is contributed by Ajit. 
?> 

Output:

Yes

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

CPP

Java

Python3

C#

PHP

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