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Multiplying a variable with a constant without using multiplication operator

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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#

// 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.


 



Last Updated : 05 Nov, 2021
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