Check if a large number is divisible by 75 or not

Given a very large number in the form of a string, the task is to check if the number is divisible by 75 or not.

Examples:

Input: N = 175
Output: No 

Input: N = 100000000000000000754586672150
Output: Yes

Approach 1: A number is divisible by 75 only if it is divisible by 3(if the sum of the digit is divisible by 3) and divisible by 25 (if the last two digits is divisible by 25) both.

Below is the implementation to check whether the given number is divisible by 75 or not. 

C++

// C++ implementation to check the number
// is divisible by 75 or not
#include <bits/stdc++.h>
using namespace std;

// check divisibleBy3
bool divisibleBy3(string number)
{
    // to store sum of Digit
    int sumOfDigit = 0;

    // traversing through each digit
    for (int i = 0; i < number.length(); i++)
        // summing up Digit
        sumOfDigit += number[i] - '0';

    // check if sumOfDigit is divisibleBy3
    if (sumOfDigit % 3 == 0)
        return true;

    return false;
}

// check divisibleBy25
bool divisibleBy25(string number)
{
    // if a single digit number
    if (number.length() < 2)
        return false;

    // length of the number
    int length = number.length();

    // taking the last two digit
    int lastTwo = (number[length - 2] - '0') * 10
                  + (number[length - 1] - '0');

    // checking if the lastTwo digit is divisibleBy25
    if (lastTwo % 25 == 0)
        return true;

    return false;
}

// Function to check divisibleBy75 or not
bool divisibleBy75(string number)
{

    // check if divisibleBy3 and divisibleBy25
    if (divisibleBy3(number) && divisibleBy25(number))
        return true;

    return false;
}

// Drivers code
int main()
{
    string number = "754586672150";

    // divisible
    bool divisible = divisibleBy75(number);

    // if divisibleBy75
    if (divisible)
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

 // Java implementation to check the number
// is divisible by 75 or not

import java.io.*;

class GFG {


// check divisibleBy3
static boolean divisibleBy3(String number)
{
    // to store sum of Digit
    int sumOfDigit = 0;

    // traversing through each digit
    for (int i = 0; i < number.length(); i++)
        // summing up Digit
        sumOfDigit += number.charAt(i) - '0';

    // check if sumOfDigit is divisibleBy3
    if (sumOfDigit % 3 == 0)
        return true;

    return false;
}

// check divisibleBy25
static  boolean divisibleBy25(String number)
{
    // if a single digit number
    if (number.length() < 2)
        return false;

    // length of the number
    int length = number.length();

    // taking the last two digit
    int lastTwo = (number.charAt(length - 2) - '0') * 10
                + (number.charAt(length - 1) - '0');

    // checking if the lastTwo digit is divisibleBy25
    if (lastTwo % 25 == 0)
        return true;

    return false;
}

// Function to check divisibleBy75 or not
static boolean divisibleBy75(String number)
{

    // check if divisibleBy3 and divisibleBy25
    if (divisibleBy3(number) && divisibleBy25(number))
        return true;

    return false;
}

// Drivers code

    public static void main (String[] args) {
    String number = "754586672150";

    // divisible
    boolean divisible = divisibleBy75(number);

    // if divisibleBy75
    if (divisible)
        System.out.println( "Yes");
    else
        System.out.println( "No");
    }
}
// This code is contributed 
// by inder_verma..

Python3

# Python 3 implementation to check the 
# number is divisible by 75 or not

# check divisibleBy3
def divisibleBy3(number):
    
    # to store sum of Digit
    sumOfDigit = 0

    # traversing through each digit
    for i in range(0, len(number), 1):
        
        # summing up Digit
        sumOfDigit += ord(number[i]) - ord('0')

    # check if sumOfDigit is divisibleBy3
    if (sumOfDigit % 3 == 0):
        return True

    return False

# check divisibleBy25
def divisibleBy25(number):
    
    # if a single digit number
    if (len(number) < 2):
        return False

    # length of the number
    length = len(number)

    # taking the last two digit
    lastTwo = ((ord(number[length - 2]) - 
                ord('0')) * 10 + 
               (ord(number[length - 1]) - ord('0')))

    # checking if the lastTwo digit 
    # is divisibleBy25
    if (lastTwo % 25 == 0):
        return True

    return False

# Function to check divisibleBy75 or not
def divisibleBy75(number):
    
    # check if divisibleBy3 and divisibleBy25
    if (divisibleBy3(number) and 
        divisibleBy25(number)):
        return True

    return False

# Driver Code
if __name__ == '__main__':
    number = "754586672150"

    # divisible
    divisible = divisibleBy75(number)

    # if divisibleBy75
    if (divisible):
        print("Yes")
    else:
        print("No")

# This code is contributed by 
# Surendra_Gangwar

C#

// C# implementation to check the number
// is divisible by 75 or not
using System;

class GFG 
{

// check divisibleBy3
static bool divisibleBy3(string number)
{
    // to store sum of Digit
    int sumOfDigit = 0;

    // traversing through each digit
    for (int i = 0; i < number.Length; i++)
    
        // summing up Digit
        sumOfDigit += number[i] - '0';

    // check if sumOfDigit is divisibleBy3
    if (sumOfDigit % 3 == 0)
        return true;

    return false;
}

// check divisibleBy25
static bool divisibleBy25(string number)
{
    // if a single digit number
    if (number.Length < 2)
        return false;

    // length of the number
    int length = number.Length;

    // taking the last two digit
    int lastTwo = (number[length - 2] - '0') * 10 + 
                  (number[length - 1] - '0');

    // checking if the lastTwo digit
    // is divisibleBy25
    if (lastTwo % 25 == 0)
        return true;

    return false;
}

// Function to check divisibleBy75 or not
static bool divisibleBy75(string number)
{

    // check if divisibleBy3 and divisibleBy25
    if (divisibleBy3(number) && divisibleBy25(number))
        return true;

    return false;
}

// Driver Code
public static void Main () 
{
    string number = "754586672150";
    
    // divisible
    bool divisible = divisibleBy75(number);
    
    // if divisibleBy75
    if (divisible)
        Console.WriteLine( "Yes");
    else
        Console.WriteLine( "No");
}
}

// This code is contributed 
// by inder_verma..

PHP

<?php
// PHP implementation to check the 
// number is divisible by 75 or not

// check divisibleBy3
function divisibleBy3($number)
{
    // to store sum of Digit
    $sumOfDigit = 0;

    // traversing through each digit
    for ($i = 0; 
         $i < strlen($number); $i++)
        // summing up Digit
        $sumOfDigit += $number[$i] - '0';

    // check if sumOfDigit is 
    // divisibleBy3
    if ($sumOfDigit % 3 == 0)
        return true;

    return false;
}

// check divisibleBy25
function divisibleBy25($number)
{
    // if a single digit number
    if (strlen($number) < 2)
        return false;

    // length of the number
    $length = strlen($number);

    // taking the last two digit
    $lastTwo = ($number[$length - 2] - '0') * 10 + 
               ($number[$length - 1] - '0');

    // checking if the lastTwo digit
    // is divisibleBy25
    if ($lastTwo % 25 == 0)
        return true;

    return false;
}

// Function to check divisibleBy75 or not
function divisibleBy75($number)
{

    // check if divisibleBy3 and 
    // divisibleBy25
    if (divisibleBy3($number) && 
        divisibleBy25($number))
        return true;

    return false;
}

// Driver Code
$number = "754586672150";

// divisible
$divisible = divisibleBy75($number);

// if divisibleBy75
if ($divisible)
    echo "Yes";
else
    echo "No";
    
// This code is contributed by ANKITRAI1
?>

JavaScript

<script>

// JavaScript implementation to check the number
// is divisible by 75 or not

// check divisibleBy3
function divisibleBy3(number)
{
    // to store sum of Digit
    let sumOfDigit = 0;

    // traversing through each digit
    for (let i = 0; i < number.length; i++)
        // summing up Digit
        sumOfDigit += number[i] - '0';

    // check if sumOfDigit is divisibleBy3
    if (sumOfDigit % 3 == 0)
        return true;

    return false;
}

// check divisibleBy25
function divisibleBy25(number)
{
    // if a single digit number
    if (number.length < 2)
        return false;

    // length of the number
    let length = number.length;

    // taking the last two digit
    let lastTwo = (number[length - 2] - '0') * 10
                  + (number[length - 1] - '0');

    // checking if the lastTwo digit is divisibleBy25
    if (lastTwo % 25 == 0)
        return true;

    return false;
}

// Function to check divisibleBy75 or not
function divisibleBy75(number)
{

    // check if divisibleBy3 and divisibleBy25
    if (divisibleBy3(number) && divisibleBy25(number))
        return true;

    return false;
}

// Drivers code
    let number = "754586672150";

    // divisible
    let divisible = divisibleBy75(number);

    // if divisibleBy75
    if (divisible)
        document.write("Yes");
    else
        document.write("No");

</script>

Output

No

Time Complexity: O(length(number))
Auxiliary Space: O(1)

Approach 2: Using Sum of Digits Method

Here is steps to this approach:

1. In this implementation, we first check if the number has at least two digits, because any number less than 75 can't be divisible by 75.
2.  Then, we calculate the last two digits of the number and check if they are divisible by 25. If not, the number can't be divisible by 75.
3.  Next, we calculate the sum of all the digits of the number and check if it is divisible by 3. If not, the number can't be divisible by 75. Finally, if both conditions are satisfied, the number is divisible by 75.

Here is the code below:

C++

#include <bits/stdc++.h>
using namespace std;

bool isDivisibleBy75(string number) {
    int n = number.length();

    // if number is less than 75, it can't be divisible by 75
    if (n < 2) {
        return false;
    }

    // calculate the last two digits of the number
    int lastTwoDigits = (number[n-2]-'0')*10 + (number[n-1]-'0');

    // if last two digits are not divisible by 25, the number can't be divisible by 75
    if (lastTwoDigits % 25 != 0) {
        return false;
    }

    // calculate the sum of all the digits
    int sumOfDigits = 0;
    for (int i = 0; i < n; i++) {
        sumOfDigits += number[i]-'0';
    }

    // if sum of digits is not divisible by 3, the number can't be divisible by 75
    if (sumOfDigits % 3 != 0) {
        return false;
    }

    // if all conditions are satisfied, the number is divisible by 75
    return true;
}

int main() {
    string number = "754586672150";
    if (isDivisibleBy75(number)) {
        cout << "Yes\n";
    } else {
        cout << "No\n";
    }
    return 0;
}

Java

// Java program for the above approach
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    String number = "754586672150";
    if (isDivisibleBy75(number)) {
      System.out.println("Yes");
    } else {
      System.out.println("No");
    }
  }

  private static boolean isDivisibleBy75(String number) {
    int n = number.length();

    // if number is less than 75, it can't be divisible by 75
    if (n < 2) {
      return false;
    }

    // calculate the last two digits of the number
    int lastTwoDigits = (number.charAt(n - 2) - '0') * 10 + (number.charAt(n - 1) - '0');

    // if last two digits are not divisible by 25, the number can't be divisible by 75
    if (lastTwoDigits % 25 != 0) {
      return false;
    }

    // calculate the sum of all the digits
    int sumOfDigits = 0;
    for (int i = 0; i < n; i++) {
      sumOfDigits += number.charAt(i) - '0';
    }

    // if sum of digits is not divisible by 3, the number can't be divisible by 75
    if (sumOfDigits % 3 != 0) {
      return false;
    }

    // if all conditions are satisfied, the number is divisible by 75
    return true;
  }
}

// This code is contributed by rishabmalhdijo

Python3

def isDivisibleBy75(number):
    n = len(number)

    # if number is less than 75, it can't be divisible by 75
    if n < 2:
        return False

    # calculate the last two digits of the number
    lastTwoDigits = int(number[n-2:])

    # if last two digits are not divisible by 25, the number can't be divisible by 75
    if lastTwoDigits % 25 != 0:
        return False

    # calculate the sum of all the digits
    sumOfDigits = 0
    for digit in number:
        sumOfDigits += int(digit)

    # if sum of digits is not divisible by 3, the number can't be divisible by 75
    if sumOfDigits % 3 != 0:
        return False

    # if all conditions are satisfied, the number is divisible by 75
    return True

# Driver code
number = "754586672150"
if isDivisibleBy75(number):
    print("Yes")
else:
    print("No")

C#

using System;

public class MainClass
{

    private static bool IsDivisibleBy75(string number)
    {
        int n = number.Length;

        // if number is less than 75, it can't be divisible by 75
        if (n < 2)
        {
            return false;
        }

        // calculate the last two digits of the number
        int lastTwoDigits = (number[n - 2] - '0') * 10 + (number[n - 1] - '0');

        // if last two digits are not divisible by 25, the number can't be divisible by 75
        if (lastTwoDigits % 25 != 0)
        {
            return false;
        }

        // calculate the sum of all the digits
        int sumOfDigits = 0;
        for (int i = 0; i < n; i++)
        {
            sumOfDigits += number[i] - '0';
        }

        // if sum of digits is not divisible by 3, the number can't be divisible by 75
        if (sumOfDigits % 3 != 0)
        {
            return false;
        }

        // if all conditions are satisfied, the number is divisible by 75
        return true;
    }
   public static void Main(string[] args)
    {
        string number = "754586672150";
        if (IsDivisibleBy75(number))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}

JavaScript

function isDivisibleBy75(number) {
    let n = number.length;
    // if number is less than 75, it can't be divisible by 75
    if (n < 2) {
        return false;
    }
    // calculate the last two digits of the number
    let lastTwoDigits = (number[n-2]-'0')*10 + (number[n-1]-'0');
    // if last two digits are not divisible by 25, the number can't be divisible by 75
    if (lastTwoDigits % 25 != 0) {
        return false;
    }
    // calculate the sum of all the digits
    let sumOfDigits = 0;
    for (let i = 0; i < n; i++) {
        sumOfDigits += number[i]-'0';
    }
    // if sum of digits is not divisible by 3, the number can't be divisible by 75
    if (sumOfDigits % 3 != 0) {
        return false;
    }
    // if all conditions are satisfied, the number is divisible by 75
    return true;
}

let number = "754586672150";
if (isDivisibleBy75(number)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output

No

Time Complexity: O(N)  where n is the length of the input string.
Auxiliary Space: O(1)

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