Check if a large number is divisible by 6 or not

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

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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).