Check if a large number is divisible by 6 or not

Last Updated : 08 Jul, 2024

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

Python

# 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")

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# 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.

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>

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.
?>

Output

Yes

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

Here is a bonus fact for the readers. The product of 3 consecutive numbers is always a multiple of 6. The reason is simple,

1) Among three consecutive numbers, one of the numbers must be a multiple of 3 (Note every third number is a multiple of 3 starting from
2) Due to the same reason as above, at least one of the three numbers must be a multiple of 2.

Check divisibility by 7

Nikita Tiwari

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Check if a large number is divisible by 6 or not

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