Multiples of 3 and 5 without using % operator

02 Dec, 2021
Write a short program that prints each number from 1 to n on a new line.  
For each multiple of 3, print “Multiple of 3” instead of the number.For each multiple of 5, print “Multiple of 5” instead of the number.For numbers which are multiples of both 3 and 5, print “Multiple of 3. Multiple of 5.” instead of the number.
For each multiple of 3, print “Multiple of 3” instead of the number.
For each multiple of 5, print “Multiple of 5” instead of the number.
For numbers which are multiples of both 3 and 5, print “Multiple of 3. Multiple of 5.” instead of the number.
Examples:  
Input  : 15 
Output : 1
         2
         Multiple of 3.
         4
         Multiple of 5.
         Multiple of 3.
         7
         8
         Multiple of 3.
         Multiple of 5.
         11
         Multiple of 3.
         13
         14
         Multiple of 3. Multiple of 5.
 
The idea is iterate from 1 to n and keep track of multiples of 3 and 5 by adding 3 and 5 to current multiple. If current number matches with a multiple, we update our output accordingly. 
C++
Java
C#
PHP
Javascript
Python3
// C++ program to print multiples// of 3 and 5 without using % operator.#include <iostream>using namespace std; void findMultiples(int n){    int a = 3; // To keep track of multiples of 3    int b = 5; // To keep track of multiples of 5    for (int i = 1; i <= n; i++)    {        string s = "";         // Found multiple of 3        if (i == a)        {            a = a + 3; // Update next multiple of 3            s = s + "Multiple of 3. ";        }         // Found multiple of 5        if (i == b)        {            b = b + 5; // Update next multiple of 5            s = s + "Multiple of 5.";        }         if (s == "")            cout << (i) << endl;        else        cout << (s) << endl;    }} // Driver Codeint main(){    findMultiples(20);     return 0;} // This code is contributed// by Sach_Code
// Java program to print multiples of 3 and// 5 without using % operator.import java.io.*; class GFG{    static void findMultiples(int n)    {        int a = 3;  // To keep track of multiples of 3        int b = 5;  // To keep track of multiples of 5        for (int i=1; i<=n; i++)        {            String s = "";             // Found multiple of 3            if (i==a)            {                a = a + 3;  // Update next multiple of 3                s = s + "Multiple of 3. ";            }             // Found multiple of 5            if (i==b)            {                b = b+5;  // Update next multiple of 5                s = s + "Multiple of 5.";            }             if (s == "")                System.out.println(i);            else  System.out.println(s);        }    }     public static void main (String[] args)    {        findMultiples(20);    }}
// C# program to print multiples of 3 and// 5 without using % operator.using System; public class GFG {         static void findMultiples(int n)    {                 // To keep track of multiples of 3        int a = 3;                 // To keep track of multiples of 5        int b = 5;        for (int i = 1; i <= n; i++)        {            String s = "";             // Found multiple of 3            if (i == a)            {                                 // Update next multiple of 3                a = a + 3;                s = s + "Multiple of 3. ";            }             // Found multiple of 5            if (i == b)            {                                 // Update next multiple of 5                b = b + 5;                s = s + "Multiple of 5.";            }             if (s == "")                Console.WriteLine(i);            else                Console.WriteLine(s);        }    }     // Driver code    public static void Main ()    {        findMultiples(20);    }} // This code is contributed by Sam007.
<?php// PHP program to print multiples// of 3 and 5 without using % operator. function findMultiples($n){    $a = 3; // To keep track of multiples of 3    $b = 5; // To keep track of multiples of 5    for ($i = 1; $i <= $n; $i++)    {        $s = "";         // Found multiple of 3        if ($i == $a)        {            $a = $a + 3; // Update next multiple of 3            $s = $s . "Multiple of 3. ";        }         // Found multiple of 5        if ($i == $b)        {            $b = $b + 5; // Update next multiple of 5            $s = $s . "Multiple of 5.";        }         if ($s == "")            echo ($i). "\n";        else            echo ($s). "\n";    }} // Driver CodefindMultiples(20); // This code is contributed// by Akanksha Rai(Abby_akku)
<script> // JavaScript program for the above approach     function findMultiples(n)    {        let a = 3;  // To keep track of multiples of 3        let b = 5;  // To keep track of multiples of 5        for (let i=1; i<=n; i++)        {            let s = "";             // Found multiple of 3            if (i==a)            {                a = a + 3;  // Update next multiple of 3                s = s + "Multiple of 3. ";            }             // Found multiple of 5            if (i==b)            {                b = b+5;  // Update next multiple of 5                s = s + "Multiple of 5.";            }             if (s == "") {                document.write(i);                document.write("<br />");             }            else {                document.write(s);                document.write("<br />");            }        }    } // Driver Code    findMultiples(20);       // This code is contributed by susmitakundugoaldanga.</script>
# Python 3 program to print multiples# of 3 and 5 without using % operator.  def findMultiples(n):    a = 3 # To keep track of multiples of 3    b = 5 # To keep track of multiples of 5    for i in range(1,n+1):        s = ""         # Found multiple of 3        if (i == a):            a = a + 3 # Update next multiple of 3            s = s + "Multiple of 3. "                 # Found multiple of 5        if (i == b):            b = b + 5 # Update next multiple of 5            s = s + "Multiple of 5."        if (s == ""):            print(i)        else:            print(s)  # Driver Codeif __name__ == '__main__':    findMultiples(20)
Output:  
1
2
Multiple of 3. 
4
Multiple of 5.
Multiple of 3. 
7
8
Multiple of 3. 
Multiple of 5.
11
Multiple of 3. 
13
14
Multiple of 3. Multiple of 5.
16
17
Multiple of 3. 
19
Multiple of 5.
Time Complexity: O(N)Auxiliary Space: O(1) This article is contributed by Nimish Jain. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Sam007
Sach_Code
Akanksha_Rai
susmitakundugoaldanga
pankajsharmagfg
amartyaghoshgfg
Algorithms
Java
Java
Algorithms
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