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# 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++

 `// C++ program to print all the numbers``// divisible by 3 and 5 for a given number``#include ``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

 `// 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);``    ``}``}`

## Python3

 `# 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#

 `// 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

 ``

## Javascript

 ``

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 ``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 5``n``=``50``for` `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

 ``

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 ``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 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#

 `// 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`

## Python3

 `#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

 `//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`

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.

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.

## C++

 `#include ``using` `namespace` `std;` `int` `main() {``    ``int` `N = 100;` `    ``for``(``int` `i = 0; i

## Java

 `// 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

## Python3

 `N ``=` `100` `for` `i ``in` `range``(N):``    ``if` `i ``%` `3` `=``=` `0` `and` `i ``%` `5` `=``=` `0``:``        ``print``(i, end``=``' '``)`

## Javascript

 `let N = 100;` `for``(let i = 0; i

## C#

 `using` `System;` `class` `Gfg {``  ``public` `static` `void` `Main (``string``[] args) {``    ``int` `N = 100;` `    ``for``(``int` `i = 0; 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)