Multiplying a variable with a constant without using multiplication operator

As we know that every number can be represented as sum(or difference) of powers of 2, therefore what we can do is represent the constant as a sum of powers of 2.
For this purpose we can use the bitwise left shift operator. When a number is bitwise left shifted it is multiplied by 2 for every bit shift.
For example, suppose we want to multiply a variable say "a" by 10 then what we can do is

a = a << 3 + a << 1;

The expression a << 3 multiplies a by 8 ans expression a<<1 multiplies it by 2.
So basically what we have here is a = a*8 + a*2 = a*10
Similarly for multiplying with 7 what we can do is

a = a<<3 - a;
or
a = a<<2 + a<<1 + a;

Both these statements multiply a by 7.

C++

#include<iostream>
using namespace std;

// Returns n * 7
int multiplyBySeven(int n)
{
    // OR (n << 2) + (n << 1) + n
    return (n << 3) - n;
}

// Returns n * 12
int multiplyByTwelve(int n)
{
    return (n << 3) + (n << 2);
}

int main()
{
    cout << multiplyBySeven(5) << endl;
    cout << multiplyByTwelve(5) << endl;
    return 0;
}

Java

class GFG {
    
    // Returns n * 7
    static int multiplyBySeven(int n)
    {
        
        // OR (n << 2) + (n << 1) + n
        return (n << 3) - n;
    }

    // Returns n * 12
    static int multiplyByTwelve(int n)
    {
        return (n << 3) + (n << 2);
    }
    
    // Driver code
    public static void main(String[] args)
    {
        System.out.println(multiplyBySeven(5));
        System.out.println(multiplyByTwelve(5));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 program to Multiplying a
# variable with a constant

# Returns n * 7
def multiplyBySeven(n):
    
    # OR (n << 2) + (n << 1) + n
    return (n << 3) - n

# Returns n * 12
def multiplyByTwelve(n):
    return (n << 3) + (n << 2)
    
# Driver code
print(multiplyBySeven(5))
print(multiplyByTwelve(5))

# This code is contributed by Anant Agarwal.

// C# program to Multiplying a
// variable with a constant
using System;

class GFG
{
    // Returns n * 7
    static int multiplyBySeven(int n)
    {
        // OR (n << 2) + (n << 1) + n
        return (n << 3) - n;
    }
     
    // Returns n * 12
    static int multiplyByTwelve(int n)
    {
        return (n << 3) + (n << 2);
    }
    
    // Driver code
    public static void Main()
    {
        Console.WriteLine(multiplyBySeven(5));
        Console.WriteLine(multiplyByTwelve(5));
    }
}

// This code is contributed by Anant Agarwal.

PHP

<?php
// PHP program of multiply operator 
// Returns n * 7

function multiplyBySeven($n)
{
    return ($n << 3) - $n;
}

// Returns n * 12
function multiplyByTwelve($n)
{
    return ($n << 3) + ($n << 2);
}

// Driver Code
echo multiplyBySeven(5), "\n";
echo multiplyByTwelve(5), "\n";

// This code is contributed by Ajit
?>

JavaScript

<script>
// Python3 program to Multiplying a 
// variable with a constant

// Returns n * 7 
function multiplyBySeven(n) 
{ 
    // OR (n << 2) + (n << 1) + n 
    return (n << 3) - n; 
} 

// Returns n * 12 
function multiplyByTwelve(n) 
{ 
    return (n << 3) + (n << 2); 
} 

// Driver code 
document.write(multiplyBySeven(5) + "<br>"); 
document.write(multiplyByTwelve(5) + "<br>"); 
 
// This code is contributed by Surbhi Tyagi.

</script>

Output :

35
60

Time Complexity: O(1)

Auxiliary Space: O(1)

We just need to find the combination of powers of 2. Also, this comes really handy when we have a very large dataset and each one of them requires multiplication with the same constant as bitwise operators are faster as compared to mathematical operators.

Multiplying a variable with a constant without using multiplication operator

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