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 45Input : 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++
// C++ program to print all the numbers// divisible by 3 and 5 for a given number#include <iostream>using namespace std;// Result function with Nvoid 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 codeint main(){ // input goes here int N = 100; // Calling function result(N); return 0;}// This code is contributed by Manish Shaw// (manishshaw1) |
Java
// Java program to print all the numbers// divisible by 3 and 5 for a given numberclass 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); }} |
Python3
# Python program to print all the numbers# divisible by 3 and 5 for a given number# Result function with Ndef 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 codeif __name__ == "__main__": # input goes here N = 100 # Calling function result(N) |
C#
// C# program to print all the numbers// divisible by 3 and 5 for a given numberusing 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// PHP program to print all the numbers// divisible by 3 and 5 for a given number// Result function with Nfunction 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 functionresult($N);// This code is contributed by ajit?> |
Javascript
<script>// Javascript program to// print all the numbers// divisible by 3 and 5// for a given number// Result function with Nfunction 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 herelet N = 100;// Calling functionresult(N);// This code is contributed by Bobby</script> |
Output
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++
// 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 |
Java
/*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 |
Python3
# python code to print numbers that# are divisible by 3 and 5n=50for i in range(0,n): # lcm of 3 and 5 is 15 if i%15==0: print(i,end=" ") |
C#
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 |
Javascript
<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> |
Output
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++
// 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
// Java code to print numbers that are divisible by 3 and 5class 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#
// C# code to print numbers that are divisible by 3 and 5using 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 |
Python3
#Python code to print numbers that are divisible by 3 and 5n = 50#lcm of 3 and 5 is 15#LCM(3, 5) = 15#start loop from 0 to n, by increment of +15 every timefor i in range(0,n,15): print(i,end=" ") #This code is contributed by Vinay Pinjala |
Javascript
//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 timelet n = 50;for (let i = 0; i <= n; i += 15) { console.log(i);}// This code is contributed by Edula Vinay Kumar Reddy |
Output
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.
- 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.
- Use a for loop to iterate over all the numbers less than N.
- For each number, check if it is divisible by both 3 and 5 using the modulo operator %.
- 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.
C++
#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
// Java code to print all numbers divisible by 3 and 5 from 0 to Npublic 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 + " "); } } }} |
Python3
N = 100for i in range(N): if i % 3 == 0 and i % 5 == 0: print(i, end=' ') |
Javascript
let N = 100;for(let i = 0; i<N; i++){ if(i%3 == 0 && i%5 == 0){ console.log(i+" "); }} |
C#
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 + " "); } } }} |
Output
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)
