A number, $x$, is said to be a multiple of another number, $y$, if the remainder of $x/y$ is $0$. Similarly, a number, $n$, will be a multiple of 3 and 5 if the remainder of $n/3$ and $n/5$ is equal to $0$.

In order to find all the numbers from $0$ to $N$ that are multiples of 3 and 5, each number needs to be considered separately, and the remainders of both $n/3$ and $n/5$ should be compared to $0$.

> Since the range $0$ to $N$ is traversed only once, the [time complexity](https://www.educative.io/answers/what-is-big-o-notation) of this algorithm is $O(n)$.

# Implementation

The following code implements this algorithm in C++, Java, and Python:

#include <iostream>
using namespace std;

void findMultiples(int n){
  for(int i = 0; i <= n; i++)
    if(i % 3 == 0 && i % 5 == 0)
      cout << i << endl;
}

int main() {
  findMultiples(120);

  return 0;
}

class Multiples{
  static void findMultiples(int n){
    for(int i = 0; i <= n; i++)
      if(i % 3 == 0 && i % 5 == 0)
        System.out.println(i);
  }

  public static void main(String[] args){
    findMultiples(120);
  }
}

Java

def findMultiples(x):
  for x in range(0, x + 1):
    if x % 3 == 0 and x % 5 == 0:
        print(x);

findMultiples(120);

Python

How to find all multiples of 3 and 5 up to number N

To find multiples of 3 and 5 up to N, check if each number from 0 to N is divisible by both 3 and 5.