Program to print all the numbers divisible by 3 and 5 for a given number

Given the integer N, the task is to print all the numbers less than N, which are divisible by 3 and 5.
Examples : 

Input : 50
Output : 0 15 30 45 

Input : 100
Output : 0 15 30 45 60 75 90 

Approach: For example, let's take N = 20 as a limit, then the program should print all numbers less than 20 which are divisible by both 3 and 5. For this divide each number from 0 to N by both 3 and 5 and check their remainder. If remainder is 0 in both cases then simply print that number.

Below is the implementation : 

// C++ program to print all the numbers
// divisible by 3 and 5 for a given number
#include <iostream>
using namespace std;

// Result function with N
void result(int N)
{     
    // iterate from 0 to N
    for (int num = 0; num < N; num++)
    {     
        // Short-circuit operator is used 
        if (num % 3 == 0 && num % 5 == 0)
            cout << num << " ";
    }
}

// Driver code
int main()
{     
    // input goes here
    int N = 100;
    
    // Calling function
    result(N);
    return 0;
}

// This code is contributed by Manish Shaw
// (manishshaw1)

// Java program to print all the numbers
// divisible by 3 and 5 for a given number

class GFG{
    
    // Result function with N
    static void result(int N)
    {     
        // iterate from 0 to N
        for (int num = 0; num < N; num++)
        {     
            // Short-circuit operator is used 
            if (num % 3 == 0 && num % 5 == 0)
                System.out.print(num + " ");
        }
    }
     
    // Driver code
    public static void main(String []args)
    {
        // input goes here
        int N = 100;
         
        // Calling function
        result(N);
    }
}

# Python program to print all the numbers
# divisible by 3 and 5 for a given number

# Result function with N
def result(N):
    
    # iterate from 0 to N
    for num in range(N):
        
            # Short-circuit operator is used 
            if num % 3 == 0 and num % 5 == 0:
                print(str(num) + " ", end = "")
                
            else:
                pass

# Driver code
if __name__ == "__main__":
    
    # input goes here
    N = 100
    
    # Calling function
    result(N)

// C# program to print all the numbers
// divisible by 3 and 5 for a given number
using System;
public class GFG{
    
    // Result function with N
    static void result(int N)
    {     
        // iterate from 0 to N
        for (int num = 0; num < N; num++)
        {     
            // Short-circuit operator is used 
            if (num % 3 == 0 && num % 5 == 0)
                Console.Write(num + " ");
        }
    }
    
    // Driver code
    static public void Main (){
        // input goes here
        int N = 100;
        // Calling function
        result(N);
    }
//This code is contributed by ajit.    
}

<?php
// PHP program to print all the numbers
// divisible by 3 and 5 for a given number

// Result function with N
function result($N)
{ 
    // iterate from 0 to N
    for ($num = 0; $num < $N; $num++)
    { 
        // Short-circuit operator is used 
        if ($num % 3 == 0 && $num % 5 == 0)
            echo $num, " ";
    }
}

// Driver code
    
// input goes here
$N = 100;

// Calling function
result($N);

// This code is contributed by ajit
?>

<script>

// Javascript  program to 
// print all the numbers
// divisible by 3 and 5 
// for a given number

// Result function with N
function result(N)
{
    // iterate from 0 to N
    for (let num = 0; num < N; num++)
    {
        // Short-circuit operator is used
        if (num % 3 == 0 && num % 5 == 0)
            document.write( num+ " ");
    }
}

// Driver code
    
// input goes here
let N = 100;

// Calling function
result(N);

// This code is contributed by Bobby

</script>

0 15 30 45 60 75 90 

Time Complexity: O(N)
Auxiliary Space: O(1)

Method: This can also be done by checking if the number is divisible by 15, since the LCM of 3 and 5 is 15 and any number divisible by 15 is divisible by 3 and 5 and vice versa also. 

// C++ code to print numbers that
// are divisible by 3 and 5
#include <iostream>
using namespace std;

int main()
{
  int n = 50; 
  for(int i = 0; i < n; i++)
  {

    //lcm of 3 and 5 is 15 
    if(i % 15 == 0){
      cout << i << " ";
    }
  }
  return 0;
}

// This code is contributed by laxmigangarajula03

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    public static void main (String[] args) {
      
       int n = 50;
        for (int i = 0; i < 50; i++) {

            // lcm of 3 and 5 is 15
            if (i % 15 == 0) {
                System.out.println(i+" ");
            }
        }
      
    }
}

// This code is contributed by laxmigangarajula03

# python code to print numbers that
# are divisible by 3 and 5 
n=50 
for i in range(0,n):
  # lcm of 3 and 5 is 15 
  if i%15==0:
    print(i,end=" ")

using System;

public class GFG {

    static public void Main()
    {

        int n = 50;
        for (int i = 0; i < 50; i++) {

            // lcm of 3 and 5 is 15
            if (i % 15 == 0) {
                Console.Write(i+" ");
            }
        }
    }
}
  
  // This code is contributed by laxmigangarajula03

<script>
let n = 50; 
  for(let i = 0; i < 50; i++)
  {

    //lcm of 3 and 5 is 15 
    if(i % 15 == 0){
      document.write(i+" ");
    }
  }
</script>

0 15 30 45 

Time Complexity: O(n)
Auxiliary Space: O(1)

Method 3 : we noticed that the LCM of 3 & 5 is 15, so we do not need to iterate the whole loop from 0 to n but we need to iterate from 0 and every time increase i by 15 so this way we can decrease time complexity as we do not iterate from 0 to n by step of +1 but we iterate from 15 to n by step of +15

// C++ code to print numbers that are divisible by 3 and 5
#include <iostream>
using namespace std;

int main()
{
    int n = 50;
    // lcm of 3 and 5 is 15
    // LCM(3, 5) = 15

    // start loop from 0 to n, by increment of +15 every time
    for (int i = 0; i < n; i += 15) {
        cout << i << " ";
    }
    return 0;
}

// Java code to print numbers that are divisible by 3 and 5
class GFG {
  public static void main(String[] args) {
    int n = 50;

    // lcm of 3 and 5 is 15
    // LCM(3, 5) = 15

    // start loop from 0 to n, by increment of +15 every time
    for (int i = 0; i < n; i += 15) {
      System.out.print(i + " ");
    }
  }
}

// This code is contributed by ajaymakavana.

// C# code to print numbers that are divisible by 3 and 5
using System;

public class GFG {
    public static void Main(string[] args) {
        int n = 50;

        // lcm of 3 and 5 is 15
        // LCM(3, 5) = 15

        // start loop from 0 to n, by increment of +15 every time
        for (int i = 0; i < n; i += 15) {
            Console.Write(i + " ");
        }
    }
}

// This code is contributed by ajaymakvana

#Python code to print numbers that are divisible by 3 and 5
n = 50
#lcm of 3 and 5 is 15
#LCM(3, 5) = 15

#start loop from 0 to n, by increment of +15 every time
for i in range(0,n,15):
  print(i,end=" ")
  
#This code is contributed by Vinay Pinjala

//Javascript code to print numbers that are divisible by 3 and 5
//lcm of 3 and 5 is 15
//LCM(3, 5) = 15

//start loop from 0 to n, by increment of +15 every time
let n = 50;
for (let i = 0; i <= n; i += 15) {
    console.log(i);
}
// This code is contributed by Edula Vinay Kumar Reddy

0 15 30 45 

Time Complexity: O(n/15) ~= O(n) (it's far better than above both method as we need to iterate i for only n/15 times)

Auxiliary Space: O(1) (constant extra space required)

Method 4: "for loop" approach in Python to print all numbers less than a given number that is divisible by both 3 and 5.

1. Take the input for the value of N from the user using the input() function and convert it to an integer using the int() function.
2. Use a for loop to iterate over all the numbers less than N.
3. For each number, check if it is divisible by both 3 and 5 using the modulo operator %.
4. If the remainder is 0, print the number using the print() function and the end parameter to ensure that the numbers are printed on the same line with a space in between them.

#include <iostream>
using namespace std;

int main() {
    int N = 100;

    for(int i = 0; i<N; i++){
        if(i%3 == 0 && i%5 == 0){
            cout << i << " ";
        }
    }
    return 0;
}

// Java code to print all numbers divisible by 3 and 5 from 0 to N

public class Main {
    public static void main(String[] args) {
        int N = 100;

        for(int i = 0; i<N; i++){
            if(i%3 == 0 && i%5 == 0){
                System.out.print(i + " ");
            }
        }
    }
}

N = 100

for i in range(N):
    if i % 3 == 0 and i % 5 == 0:
        print(i, end=' ')

let N = 100;

for(let i = 0; i<N; i++){
    if(i%3 == 0 && i%5 == 0){
        console.log(i+" ");
    }
}

using System;

class Gfg {
  public static void Main (string[] args) {
    int N = 100;

    for(int i = 0; i<N; i++){
        if(i%3 == 0 && i%5 == 0){
            Console.Write(i + " ");
        }
    }
  }
}

0 15 30 45 60 75 90 

The time complexity is O(N) where N is the given input number. 

The auxiliary space is O(1)