Multiply a number by 15 without using * and / operators

Given a integer N, the task is to multiply the number with 15 without using multiplication * and division / operators.

Examples:

Input: N = 10
Output: 150

Input: N = 7
Output: 105

Method 1: We can multiply integer N by 15 using bitwise operators. First left shift the number by 4 bits which is equal to (16 * N) then subtract the original number N from the shifted number i.e. ((16 * N) – N) which is equal to 15 * N.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return (15 * N) without
// using '*' or '/' operator
long multiplyByFifteen(long n)
{
    // prod = 16 * n
    long prod = (n << 4);
  
    // ((16 * n) - n) = 15 * n
    prod = prod - n;
  
    return prod;
}
  
// Driver code
int main()
{
    long n = 7;
  
    cout << multiplyByFifteen(n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
class GFG {
  
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 16 * n
        long prod = (n << 4);
  
        // ((16 * n) - n) = 15 * n
        prod = prod - n;
  
        return prod;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        long n = 7;
        System.out.print(multiplyByFifteen(n));
    }
}

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Function to return (15 * N) without 
# using '*' or '/' operator
def multiplyByFifteen(n):
      
    # prod = 16 * n
    prod = (n << 4)
      
    # ((16 * n) - n) = 15 * n
    prod = prod - n
      
    return prod
      
# Driver code
n = 7
print(multiplyByFifteen(n))

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
class GFG {
  
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 16 * n
        long prod = (n << 4);
  
        // ((16 * n) - n) = 15 * n
        prod = prod - n;
  
        return prod;
    }
  
    // Driver code
    public static void Main()
    {
        long n = 7;
        Console.Write(multiplyByFifteen(n));
    }
}

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach
  
// Function to return (15 * N) without
// using '*' or '/' operator
function multiplyByFifteen($n)
{
    // prod = 16 * n
    $prod = ($n << 4);
  
    // ((16 * n) - n) = 15 * n
    $prod = $prod - $n;
  
    return $prod;
}
  
// Driver code
  
    $n = 7;
  
    echo multiplyByFifteen($n);
  
// This code is contributed by anuj_67..
?>

chevron_right


Output:

105

Method 2: We can also calculate the multiplication (15 * N) as sum of (8 * N) + (4 * N) + (2 * N) + N which can be obtained by performing left shift operations as (8 * N) = (N << 3), (4 * N) = (n << 2) and (2 * N) = (n << 1).

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return (15 * N) without
// using '*' or '/' operator
long multiplyByFifteen(long n)
{
    // prod = 8 * n
    long prod = (n << 3);
  
    // Add (4 * n)
    prod += (n << 2);
  
    // Add (2 * n)
    prod += (n << 1);
  
    // Add n
    prod += n;
  
    // (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
    return prod;
}
  
// Driver code
int main()
{
    long n = 7;
  
    cout << multiplyByFifteen(n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
class GFG {
  
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 8 * n
        long prod = (n << 3);
  
        // Add (4 * n)
        prod += (n << 2);
  
        // Add (2 * n)
        prod += (n << 1);
  
        // Add n
        prod += n;
  
        // (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
        return prod;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        long n = 7;
        System.out.print(multiplyByFifteen(n));
    }
}

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Function to perform Multiplication
def multiplyByFifteen(n):
      
    # prod = 8 * n
    prod = (n << 3)
      
    # Add (4 * n)
    prod += (n << 2)
      
    # Add (2 * n)
    prod += (n << 1)
      
    # Add n
    prod += n
      
    # (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
    return prod
      
# Driver code
n = 7
print(multiplyByFifteen(n))

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
class GFG {
  
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 8 * n
        long prod = (n << 3);
  
        // Add (4 * n)
        prod += (n << 2);
  
        // Add (2 * n)
        prod += (n << 1);
  
        // Add n
        prod += n;
  
        // (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
        return prod;
    }
  
    // Driver code
    public static void Main()
    {
        long n = 7;
        Console.Write(multiplyByFifteen(n));
    }
}

chevron_right


Output:

105


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : vt_m