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Check if a large number is divisible by 6 or not

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  • Difficulty Level : Basic
  • Last Updated : 04 Jul, 2022

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

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")

Javascript




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

Output

No

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
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 ")
     
# This article is contributed by Nikita Tiwari.

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 . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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