Open In App
Related Articles

Check if a large number is divisible by 6 or not

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Report issue
Report

Given a number, the task is to check if a number is divisible by 6 or not. The input number may be large and it may not be possible to store even if we use long long int.

Examples: 

Input  : n = 2112
Output: Yes

Input : n = 1124
Output : No

Input  : n = 363588395960667043875487
Output : No

C++

#include <iostream>
using namespace std;
  
int main() {
  
    long n = 36358839596;
    
      // finding given number is divisible by 6 or not
    if (n % 6 == 0)
        cout << "Yes";
    else
        cout << "No";
    return 0;
}
  
// This code is contributed by lokesh

                    

Java

// Java code for the above approach
// To check whether the given number is divisible by 6 or not
import java.io.*;
  
class GFG {
    public static void main (String[] args) {
        long n = 36358839596L;
   
        // finding given number is divisible by 6 or not
        if (n % 6 == 0)
              System.out.print("Yes");
        else
              System.out.print("No");
    }
}
  
// This code is contributed by lokesh

                    

Python3

# Python code 
# To check whether the given number is divisible by 6 or not
  
#input 
n=363588395960667043875487 
# the above input can also be given as n=input() -> taking input from user
# finding given number is divisible by 6 or not
if int(n)%6==0:
  print("Yes"
else
  print("No")

                    

C#

using System;
public class GFG {
  
  static public void Main()
  {
  
    // C# code for the above approach
    // To check whether the given number is divisible by 6 or not
  
    //input
    long n = 36358839596;
  
    // finding given number is divisible by 6 or not
    if (n % 6 == 0)
      Console.Write("Yes");
    else
      Console.Write("No");
  
  
}
  
// This code is contributed by ksrikanth0498

                    

Javascript

<script>
        // JavaScript code for the above approach
        // To check whether the given number is divisible by 6 or not
  
        //input 
        var n = 363588395960667043875487
          
        // finding given number is divisible by 6 or not
        if (n % 6 == 0)
            document.write("Yes")
        else
            document.write("No")
    // This code is contributed by Potta Lokesh
    </script>

                    

PHP

<?php
// PHP program to check 
// if a large number is 
// divisible by 6.
  
  // Driver Code
  // input number
$num = 363588395960667043875487;
  
// finding given number is divisible by 6 or not
if ( $num %6 == 0)
    echo "Yes";
else
    echo "No";
  
// This code is contributed by satwik4409.
?>

                    

Output
No

Time complexity: O(1) as it is performing constant operations
Auxiliary Space: O(1) as it is using constant space for variables

Since input number may be very large, we cannot use n % 6 to check if a number is divisible by 6 or not, especially in languages like C/C++. The idea is based on following fact. 

A number is divisible by 6 it's divisible by 2 and 3. 
a)  A number is divisible by 2 if its last digit is divisible by 2.
b)  A number is divisible by 3 if sum of digits is divisible by 3.

Below is the implementation based on above steps. 

C++

// C++ program to find if a number is divisible by
// 6 or not
#include<bits/stdc++.h>
using namespace std;
  
// Function to find that number divisible by 6 or not
bool check(string str)
{
    int n = str.length();
  
    // Return false if number is not divisible by 2.
    if ((str[n-1]-'0')%2 != 0)
       return false;
  
    // If we reach here, number is divisible by 2.
    // Now check for 3.
  
    // Compute sum of digits
    int digitSum = 0;
    for (int i=0; i<n; i++)
        digitSum += (str[i]-'0');
  
    // Check if sum of digits is divisible by 3
    return (digitSum % 3 == 0);
}
  
// Driver code
int main()
{
    string str = "1332";
    check(str)?  cout << "Yes" : cout << "No ";
    return 0;
}

                    

Java

// Java program to find if a number is
// divisible by 6 or not
import java.io.*;
class IsDivisible
{
    // Function to find that number divisible by 6 or not
    static boolean check(String str)
    {
        int n = str.length();
       
        // Return false if number is not divisible by 2.
        if ((str.charAt(n-1) -'0')%2 != 0)
           return false;
       
        // If we reach here, number is divisible by 2.
        // Now check for 3.
       
        // Compute sum of digits
        int digitSum = 0;
        for (int i=0; i<n; i++)
            digitSum += (str.charAt(i)-'0');
       
        // Check if sum of digits is divisible by 3
        return (digitSum % 3 == 0);
    }
      
    // main function
    public static void main (String[] args) 
    {
        String str = "1332";
        if(check(str))
            System.out.println("Yes");
        else
            System.out.println("No");
    }

                    

Python3

# Python 3 program to find 
# if a number is divisible
# by 6 or not
  
# Function to find that number 
# is divisible by 6 or not
def check(st) :
    n = len(st)
      
      
    # Return false if number
    # is not divisible by 2.
    if (((int)(st[n-1])%2) != 0) :
        return False
   
    # If we reach here, number 
    # is divisible by 2. Now
    # check for 3.
   
    # Compute sum of digits
    digitSum = 0
    for i in range(0, n) :
        digitSum = digitSum + (int)(st[i])
   
    # Check if sum of digits
    # is divisible by 3
    return (digitSum % 3 == 0)
  
  
# Driver code
st = "1332"
if(check(st)) :
    print("Yes")
else :
    print("No ")
      

                    

C#

// C# program to find if a number is
// divisible by 6 or not
using System;
  
class GFG {
      
    // Function to find that number
    // divisible by 6 or not
    static bool check(String str)
    {
        int n = str.Length;
      
        // Return false if number is
        // not divisible by 2.
        if ((str[n-1] -'0') % 2 != 0)
            return false;
      
        // If we reach here, number is
        // divisible by 2.
        // Now check for 3.
      
        // Compute sum of digits
        int digitSum = 0;
        for (int i = 0; i < n; i++)
            digitSum += (str[i] - '0');
      
        // Check if sum of digits is
        // divisible by 3
        return (digitSum % 3 == 0);
    }
      
    // main function
    public static void Main () 
    {
        String str = "1332";
          
        if(check(str))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
  
// This code is contributed by parashar.

                    

PHP

<?php
// PHP program to find if a 
// number is divisible by
// 6 or not
  
// Function to find that number 
// divisible by 6 or not
function check($str)
{
    $n = strlen($str);
  
    // Return false if number is
    // not divisible by 2.
    if (($str[$n - 1] - '0') % 2 != 0)
    return false;
  
    // If we reach here, number 
    // is divisible by 2.
    // Now check for 3.
    // Compute sum of digits
    $digitSum = 0;
    for ($i = 0; $i < $n; $i++)
        $digitSum += ($str[$i] - '0');
  
    // Check if sum of digits
    // is divisible by 3
    return ($digitSum % 3 == 0);
}
  
    // Driver code
    $str = "1332";
    if(check($str))
        echo "Yes" ;
    else
        echo " No ";
    return 0;
  
// This code is contributed by nitin mittal.
?>

                    

Javascript

<script>
  
// JavaScript program for the above approach
  
    // Function to find that number 
    // divisible by 6 or not 
    function check(str) 
    
        let n = str.length; 
        
        // Return false if number is 
        // not divisible by 2. 
        if ((str[n-1] -'0') % 2 != 0) 
            return false
        
        // If we reach here, number is 
        // divisible by 2. 
        // Now check for 3. 
        
        // Compute sum of digits 
        let digitSum = 0; 
        for (let i = 0; i < n; i++) 
            digitSum += (str[i] - '0'); 
        
        // Check if sum of digits is 
        // divisible by 3 
        return (digitSum % 3 == 0); 
    
  
// Driver Code
     let str = "1332"
     if(check(str)) 
        document.write("Yes"); 
     else
        document.write("No");
  
// This code is contributed by splevel62.
</script>

                    

Output
Yes


Time Complexity: O(logN) where N is the given number
Auxiliary Space: O(1) since no extra space is being used

This article is contributed by DANISH_RAZA .

 

Approach#3: using the fact that it is divisible by 3 and 2 consecutive even numbers

We can check if a number is divisible by 6 by checking if it is divisible by 3 and any 2 consecutive even numbers (i.e., 2 and 4 or 4 and 6, etc.). We can use the modulo operation to find the remainder of the number when divided by 3 and check if it is divisible by any 2 consecutive even numbers.

Algorithm

1. Check if the remainder of the number when divided by 3 is 0.
2. Check if the last digit of the number is even.
3. If the last digit of the number is even, check if the second-last digit of the number is odd.
4. If the second-last digit of the number is odd, print “Yes”, otherwise print “No”.

C++

// C++ program for the above approach
#include <iostream>
using namespace std;
  
// Function to check divisibility by 6
void check_divisibility_by_6(int n)
{
    if (n % 3 == 0 && n % 10 % 2 == 0
        && (n / 10) % 10 % 2 != 0) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
}
  
// Driver Code
int main()
{
    int n1 = 2112;
    check_divisibility_by_6(n1);
  
    int n2 = 1124;
    check_divisibility_by_6(n2);
  
    return 0;
}

                    

Java

public class Main {
  
    // Function to check divisibility by 6
    public static void check_divisibility_by_6(int n) {
        if (n % 3 == 0 && n % 10 % 2 == 0 && (n / 10) % 10 % 2 != 0) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
  
    // Driver Code
    public static void main(String[] args) {
        int n1 = 2112;
        check_divisibility_by_6(n1);
  
        int n2 = 1124;
        check_divisibility_by_6(n2);
    }
}

                    

Python3

def check_divisibility_by_6(n):
    if n % 3 == 0 and n % 10 % 2 == 0 and (n // 10) % 10 % 2 != 0:
        print("Yes")
    else:
        print("No")
  
  
n = 2112
check_divisibility_by_6(n)
  
n = 1124
check_divisibility_by_6(n)

                    

Javascript

function check_divisibility_by_6(n) {
  if (n % 3 === 0 && n % 10 % 2 === 0 && Math.floor(n / 10) % 10 % 2 !== 0) {
    console.log("Yes");
  } else {
    console.log("No");
  }
}
  
let n = 2112;
check_divisibility_by_6(n);
  
n = 1124;
check_divisibility_by_6(n);

                    

C#

using System;
  
class Program
{
    // Function to check divisibility by 6
    static void CheckDivisibilityBy6(int n)
    {
        if (n % 3 == 0 && n % 10 % 2 == 0
            && (n / 10) % 10 % 2 != 0)
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
  
    // Driver Code
    static void Main(string[] args)
    {
        int n1 = 2112;
        CheckDivisibilityBy6(n1);
  
        int n2 = 1124;
        CheckDivisibilityBy6(n2);
  
        Console.ReadKey();
    }
}

                    

Output
Yes
No

Time Complexity: O(log n) where n is the value of the number.
Auxiliary Space: O(1).



Last Updated : 13 Sep, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads