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

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

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// 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;
}

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Java














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// 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. 



















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Python3














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# 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. 



















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C#














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// 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. 



















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PHP














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<?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. 


?> 



















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Output:


Yes




          
          
          
          

            

    

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Improved By :  nitin mittal, jit_t