Check if a large number is divisible by 3 or not

Given a number, the task is that we divide number by 3. The input number may be large and it may not be possible to store even if we use long long int.
Examples: 

Input  : n = 769452
Output : Yes

Input  : n = 123456758933312
Output : No

Input  : n = 3635883959606670431112222
Output : Yes

Since input number may be very large, we cannot use n % 3 to check if a number is divisible by 3 or not, especially in languages like C/C++. The idea is based on following fact.

A number is divisible by 3 if sum of its digits is divisible by 3.

Illustration:

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

How does this work?

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Below is the implementation of the above fact :

C++

// C++ program to find if a number is divisible by
// 3 or not
#include<bits/stdc++.h>
using namespace std;

// Function to find that number divisible by 3 or not
int check(string str)
{
    // Compute sum of digits
    int n = str.length();
    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 3 or not
import java.io.*;
class IsDivisible
{
    // Function to find that number
    // divisible by 3 or not
    static boolean check(String str)
    {
        // Compute sum of digits
        int n = str.length();
        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 program to find if a number is
# divisible by 3 or not

# Function to find that number
# divisible by 3 or not
def check(num) :

    # Compute sum of digits
    digitSum = 0
    while num > 0 :
        rem = num % 10
        digitSum = digitSum + rem
        num = num // 10

    # Check if sum of digits is
    # divisible by 3.
    return (digitSum % 3 == 0)

# main function
num = 1332
if(check(num)) :
    print ("Yes")
else :
    print ("No")

# This code is contributed by Nikita Tiwari.

C#

// C# program to find if a number is
// divisible by 3 or not
using System;

class GFG
{
    // Function to find that number
    // divisible by 3 or not
    static bool check(string str)
    {
        // Compute sum of digits
        int n = str.Length;
        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.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find if a
// number is divisible by
// 3 or not

// Function to find that
// number divisible by 3 or not
function check($str)
{

    // Compute sum of digits
    $n = strlen($str);
    $digitSum = 0;
    for ($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";
$x = check($str) ? "Yes" : "No ";
echo($x);

// This code is contributed by Ajit.
?>

Javascript

<script>
// Javascript program to find if a
// number is divisible by
// 3 or not

// Function to find that
// number divisible by 3 or not
function check(str)
{

    // Compute sum of digits
    let n = str.length;
    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";
let x = check(str) ? "Yes" : "No ";
document.write(x);

// This code is contributed by _saurabh_jaiswal.
</script>

Output

Yes

Time Complexity: O(n), where n is the number of digits in the input string. This is because the for loop is used to sum up all the digits in the string, and the loop runs for n iterations.
Auxiliary Space: O(1), as we are not using any extra space. 

Method 2: Checking given number is divisible by 3 or not by using the modulo division operator “%”. 

C++

#include <iostream>
using namespace std;
int main()
{
    //input
    long long int n=769452;

    // finding given number is divisible by 3 or not
    if (n % 3 ==0)
    {
        cout << "Yes";
    }
    else
    {
        cout << "No";
    }

    return 0;
}

// This code is contributed by satwik4409.

Java

/*package whatever //do not write package name here */
import java.io.*;

class GFG {
  public static void main(String[] args)
  {

    // input
    long n = 769452;


    // finding given number is
    // divisible by 3 or not
    if (n % 3 == 0) {
      System.out.println("Yes");
    }
    else {
      System.out.println("No");
    }
  }
}

// This code is contributed by laxmigangarajula03

Python3

# Python code
# To check whether the given number is divisible by 3 or not

#input
n=769452
# the above input can also be given as n=input() -> taking input from user
# finding given number is divisible by 3 or not
if int(n)%3==0:
  print("Yes")
else:
  print("No")

  # this code is contributed by gangarajula laxmi

C#

using System;

public class GFG{

  static public void Main (){

    //input
    long n = 769452;


    // finding given number is divisible by 3 or not
    if (n % 3 == 0)
    {
      Console.Write("Yes");
    }
    else
    {
      Console.Write("No");
    }

  }
}

// This code is contributed by laxmigangarajula03

Javascript

<script>
        // JavaScript code for the above approach
        // To check whether the given number is divisible by 3 or not

        //input
        var n = 769452

        // finding given number is divisible by 3 or not
        if (n % 3 == 0)
            document.write("Yes")
        else
            document.write("No")

    // This code is contributed by Potta Lokesh
    </script>

PHP

<?php
    $num = 769452;
    // checking if the given number is divisible by 3 or
    // not using modulo division operator if the output of
    // num%3 is equal to 0 then given number is divisible
    // by 3 otherwise not divisible by 3
    if ($num % 3 == 0) {
        echo "true";
    }
    else {
        echo "false";
    }

?>

Output

Yes

Time Complexity: O(1) as it is doing constant operations
Auxiliary Space: O(1) 

Method 3: Using regular expressions

The given code checks if a number is divisible by 3 by using a regular expression pattern. The pattern matches strings that consist of an optional plus or minus sign, followed by zero or more digits, and zero or more even digits (0, 2, 4, 6, 8). If the input number matches this pattern, it is divisible by 3.

Python3

import re

def check_divisible_by_3(num):
    pattern = r'^[+-]?[0-9]*[02468]*$'
    return re.match(pattern, str(num)) is not None

# Example usage
num = 1332
if check_divisible_by_3(num):
    print("Yes")
else:
    print("No")

Output

Yes

Time complexity:
The time complexity of this code is O(n), where n is the number of digits in the input number. This is because the regular expression pattern needs to be matched against the entire string representation of the input number.

Auxiliary space:
The auxiliary space used by this code is O(1), as only a single regular expression pattern is used, and the matching is done in place without creating any additional data structures.

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