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++

// 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;
}

Java

// 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));
    }
}

Python3

# 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))

C#

// 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));
    }
}

PHP

<?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..
?>

Javascript

<script>

// JavaScript implementation of the approach

// Function to return (15 * N) without
// using '*' or '/' operator
function multiplyByFifteen(n)
{
    // prod = 16 * n
    let prod = (n << 4);

    // ((16 * n) - n) = 15 * n
    prod = prod - n;

    return prod;
}

// Driver code

    let n = 7;

    document.write(multiplyByFifteen(n));

// This code is contributed by Surbhi Tyagi.

</script>

Output:

105

Time Complexity: O(1)
Auxiliary Space: O(1)

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++

// 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;
}

Java

// 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));
    }
}

Python3

# 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))

C#

// 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));
    }
}

Javascript

<script>

// Javascript implementation of the approach

// Function to return (15 * N) without
// using '*' or '/' operator
function multiplyByFifteen(n)
{

    // prod = 8 * n
    var 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
var n = 7;

document.write(multiplyByFifteen(n));

// This code is contributed by shikhasingrajput

</script>

Output:

105

Time Complexity: O(1)
Auxiliary Space: O(1)