Check if a large number is divisible by 2, 3 and 5 or not

Given a number, the task is to check if a number is divisible by 2, 3, and 5 or not. The input number may be large and it may not be possible to store even if we use long long int, so the number is taken as a string.
Examples:

Input : str = "725" 
Output : NO

Input : str = "263730746028908374890"
Output : YES

A number is divisible by 2 if it’s right most digit is even and also a number is divisible by 5 if it’s right most digit is zero or five.
So, from above two observations, one can conclude that for the number to be divisible by both 2 and 5 the rightmost digit of the number must be zero.
Now, a number is divisible by 3 if the sum of its digits is divisible by three.
Therefore, a number will be divisible by all of 2, 3, and 5 if: 

• Its rightmost digit is zero.
• Sum of all of its digits is divisible by 3.

Below is the implementation of the above approach: 

C++

// CPP program to Check if a large number is
// divisible by 2, 3 and 5 or not.
#include <bits/stdc++.h>
using namespace std;

// function to return sum of digits of
// a number
int SumOfDigits(string str, int n)
{
    int sum = 0;

    for (int i = 0; i < n; i++)
        sum += (int)(str[i] - '0');

    return sum;
}

// function to Check if a large number is
// divisible by 2, 3 and 5 or not
bool Divisible(string str, int n)
{
    if (SumOfDigits(str, n) % 3 == 0 and str[n - 1] == '0')
        return true;

    return false;
}

// Driver code
int main()
{
    string str = "263730746028908374890";

    int n = str.size();

    if (Divisible(str, n))
        cout << "YES";
    else
        cout << "NO";

    return 0;
}

C

// C program to Check if a large number is
// divisible by 2, 3 and 5 or not.
#include <stdio.h>
#include <stdbool.h>
#include <string.h>

// function to return sum of digits of
// a number
int SumOfDigits(char str[], int n)
{
    int sum = 0;

    for (int i = 0; i < n; i++)
        sum += (int)(str[i] - '0');

    return sum;
}

// function to Check if a large number is
// divisible by 2, 3 and 5 or not
bool Divisible(char str[], int n)
{
    if (SumOfDigits(str, n) % 3 == 0 && str[n - 1] == '0')
        return true;

    return false;
}

// Driver code
int main()
{
    char str[] = "263730746028908374890";

    int n = strlen(str);

    if (Divisible(str, n))
        printf("YES");
    else
        printf("NO");

    return 0;
}

// This code is contributed by kothavvsaakash.

Java

// Java program to Check if a large
// number is divisible by 2, 3 and
// 5 or not.
import java.io.*;
class GFG
{
// function to return sum of
// digits of a number
static int SumOfDigits(String str,
                       int n)
{
    int sum = 0;

    for (int i = 0; i < n; i++)
        sum += (int)(str.charAt(i) - '0');

    return sum;
}

// function to Check if a large number
// is divisible by 2, 3 and 5 or not
static boolean Divisible(String str,
                         int n)
{
    if (SumOfDigits(str, n) % 3 == 0 &&
        str.charAt(n - 1) == '0')
        return true;

    return false;
}

// Driver code
public static void main(String []args)
{
    String str = "263730746028908374890";

    int n = str.length();

    if (Divisible(str, n))
        System.out.println("YES");
    else
        System.out.println("NO");
}
}

// This code is contributed by ihritik

Python 3

# Python 3 program to Check if
# a large number is
# divisible by 2, 3 and 5 or not.

# function to return sum of digits of
# a number
def SumOfDigits(str, n):

    sum = 0
    for i in range(0,n):
        sum += int(ord(str[i] )- ord('0'))

    return sum

# function to Check if a large number is
# divisible by 2, 3 and 5 or not
def Divisible(str, n):
    if ((SumOfDigits(str, n) % 3 == 0 and
        str[n - 1] == '0')):
        return True

    return False

# Driver code
if __name__ == "__main__":
    str = "263730746028908374890"

    n = len(str)

    if (Divisible(str, n)):
        print("YES")
    else:
        print("NO")

# this code is contributed by
# ChitraNayal

C#

// C# program to Check if a large number
// is divisible by 2, 3 and 5 or not.
using System;

class GFG
{
// function to return sum of digits
// of a number
static int SumOfDigits(String str,
                       int n)
{
    int sum = 0;

    for (int i = 0; i < n; i++)
        sum += (int)(str[i] - '0');

    return sum;
}

// function to Check if a large number
// is divisible by 2, 3 and 5 or not
static bool Divisible(String str, int n)
{
    if (SumOfDigits(str, n) % 3 == 0 &&
                    str[n - 1] == '0')
        return true;

    return false;
}

// Driver code
public static void Main()
{
    String str = "263730746028908374890";

    int n = str.Length;

    if (Divisible(str, n))
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}

// This code is contributed by ihritik

PHP

<?php
// PHP program to Check if a large number
// is divisible by 2, 3 and 5 or not.

// function to return sum of digits
// of a number
function SumOfDigits($str, $n)
{
    $sum = 0;

    for ($i = 0; $i < $n; $i++)
        $sum += (int)($str[$i] - '0');

    return $sum;
}

// function to Check if a large number
// is divisible by 2, 3 and 5 or not
function Divisible($str, $n)
{
    if (SumOfDigits($str, $n) % 3 == 0 and
                    $str[$n - 1] == '0')
        return true;

    return false;
}

// Driver code
$str = "263730746028908374890";

$n = strlen($str);

if (Divisible($str, $n))
    echo "YES";
else
    echo "NO";

// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

<script>

// JavaScript program to Check if a large
// number is divisible by 2, 3 and
// 5 or not.

// Function to return sum of
// digits of a number
function SumOfDigits(str, n)
{
    var sum = 0;

    for(var i = 0; i < n; i++)
        sum += (str.charAt(i) - '0');

    return sum;
}

// Function to Check if a large number
// is divisible by 2, 3 and 5 or not
function Divisible(str, n)
{
    if (SumOfDigits(str, n) % 3 == 0 &&
        str.charAt(n - 1) == '0')
        return true;

    return false;
}

// Driver code
var str = "263730746028908374890";
var n = str.length;

if (Divisible(str, n))
    document.write("YES");
else
    document.write("NO");

// This code is contributed by Ankita saini

</script>

Output:

YES

Time Complexity: O(n), where n is the size of the given string str
Auxiliary Space: O(1), as no extra space is required

Another approach:

Approach:

1. Check if the last digit of the number is even. If it is, then the number is divisible by 2.
2. Check if the sum of the digits of the number is divisible by 3. If it is, then the number is divisible by 3.
3. Check if the last digit of the number is 0 or 5. If it is, then the number is divisible by 5.

C++

#include <iostream>

using namespace std;

int main()
{
    long long int n = 1234567890123456789; // example number
    int sum = 0;
    int last_digit = n % 10;
    // Check for divisibility by 2
    if (last_digit % 2 == 0) {
        cout << n << " is divisible by 2\n";
    }
    else {
        cout << n << " is not divisible by 2\n";
    }

    // Check for divisibility by 3
    long long int temp_n = n;
    while (temp_n > 0) {
        sum += temp_n % 10;
        temp_n /= 10;
    }
    if (sum % 3 == 0) {
        cout << n << " is divisible by 3\n";
    }
    else {
        cout << n << " is not divisible by 3\n";
    }

    // Check for divisibility by 5
    if (last_digit == 0 || last_digit == 5) {
        cout << n << " is divisible by 5\n";
    }
    else {
        cout << n << " is not divisible by 5\n";
    }

    return 0;
}

C

#include <stdio.h>

int main() {
    long long int n = 1234567890123456789; // example number
    int sum = 0;
    int last_digit = n % 10;

    // Check for divisibility by 2
    if (last_digit % 2 == 0) {
        printf("%lld is divisible by 2\n", n);
    } else {
        printf("%lld is not divisible by 2\n", n);
    }

    // Check for divisibility by 3
    while (n > 0) {
        sum += n % 10;
        n /= 10;
    }
    if (sum % 3 == 0) {
        printf("%lld is divisible by 3\n", n);
    } else {
        printf("%lld is not divisible by 3\n", n);
    }

    // Check for divisibility by 5
    if (last_digit == 0 || last_digit == 5) {
        printf("%lld is divisible by 5\n", n);
    } else {
        printf("%lld is not divisible by 5\n", n);
    }

    return 0;
}

Java

// Java Equivalent
public class Main {
    public static void main(String[] args) {
        long n = 1234567890123456789L; // example number
        int sum = 0;
        int last_digit = (int) n % 10;

        // Check for divisibility by 2
        if (last_digit % 2 == 0) {
            System.out.println(n + " is divisible by 2");
        } else {
            System.out.println(n + " is not divisible by 2");
        }

        // Check for divisibility by 3
        long temp_n = n;
        while (temp_n > 0) {
            sum += temp_n % 10;
            temp_n /= 10;
        }
        if (sum % 3 == 0) {
            System.out.println(n + " is divisible by 3");
        } else {
            System.out.println(n + " is not divisible by 3");
        }

        // Check for divisibility by 5
        if (last_digit == 0 || last_digit == 5) {
            System.out.println(n + " is divisible by 5");
        } else {
            System.out.println(n + " is not divisible by 5");
        }
    }
}

Python3

n = 1234567890123456789  # example number
sum = 0
last_digit = n % 10

# Check for divisibility by 2
if last_digit % 2 == 0:
    print(n, "is divisible by 2")
else:
    print(n, "is not divisible by 2")

# Check for divisibility by 3
temp_n = n
while temp_n > 0:
    sum += temp_n % 10
    temp_n //= 10
if sum % 3 == 0:
    print(n, "is divisible by 3")
else:
    print(n, "is not divisible by 3")

# Check for divisibility by 5
if last_digit == 0 or last_digit == 5:
    print(n, "is divisible by 5")
else:
    print(n, "is not divisible by 5")

C#

using System;

class Gfg {
  public static void Main() {
    long n = 1234567890123456789; // example number
    int sum = 0;
    int last_digit = (int)(n % 10);

    // Check for divisibility by 2
    if (last_digit % 2 == 0) {
      Console.WriteLine($"{n} is divisible by 2");
    } else {
      Console.WriteLine($"{n} is not divisible by 2");
    }

    // Check for divisibility by 3
    long temp_n = n;
    while (temp_n > 0) {
      sum += (int)(temp_n % 10);
      temp_n /= 10;
    }
    if (sum % 3 == 0) {
      Console.WriteLine($"{n} is divisible by 3");
    } else {
      Console.WriteLine($"{n} is not divisible by 3");
    }

    // Check for divisibility by 5
    if (last_digit == 0 || last_digit == 5) {
      Console.WriteLine($"{n} is divisible by 5");
    } else {
      Console.WriteLine($"{n} is not divisible by 5");
    }
  }
}

Javascript

let n = 1234567890123456789; // example number

// Check for divisibility by 2
if (n % 2 === 0) {
  console.log(`${n} is divisible by 2`);
} else {
  console.log(`${n} is not divisible by 2`);
}

// Check for divisibility by 3
let sum = 0;
let temp_n = n;
while (temp_n > 0) {
  sum += temp_n % 10;
  temp_n = Math.floor(temp_n / 10);
}
if (sum % 3 === 0) {
  console.log(`${n} is divisible by 3`);
} else {
  console.log(`${n} is not divisible by 3`);
}

// Check for divisibility by 5
let last_digit = n % 10;
if (last_digit === 0 || last_digit === 5) {
  console.log(`${n} is divisible by 5`);
} else {
  console.log(`${n} is not divisible by 5`);
}

Output

1234567890123456789 is not divisible by 2
0 is divisible by 3
0 is not divisible by 5

Time Complexity: O(log10(n)), where n is the number being checked, because we iterate over each digit of the number once.

Space Complexity: O(1), because we only need a few variables to store the state of the algorithm