Efficient way to multiply with 7

Difficulty Level : Easy
Last Updated : 01 Apr, 2021

We can multiply a number by 7 using bitwise operator. First left shift the number by 3 bits (you will get 8n) then subtract the original numberfrom the shifted number and return the difference (8n – n).
Program:

CPP

# include<bits/stdc++.h>
 
using namespace std;
 //c++ implementation 
long multiplyBySeven(long n)
{  
    /* Note the inner bracket here. This is needed 
       because precedence of '-' operator is higher 
       than '<<' */
    return ((n<<3) - n);
}
   
/* Driver program to test above function */
int main()
{
    long n = 4;
     
    cout<<multiplyBySeven(n);
   
    return 0;
}

C

# include<stdio.h>
 
int multiplyBySeven(unsigned int n)
{ 
    /* Note the inner bracket here. This is needed
       because precedence of '-' operator is higher
       than '<<' */
    return ((n<<3) - n);
}
 
/* Driver program to test above function */
int main()
{
    unsigned int n = 4;
    printf("%u", multiplyBySeven(n));
 
    getchar();
    return 0;
}

Java

// Java program to multiply any
// positive number to 7
 
class GFG {
     
    static int multiplyBySeven(int n)
    {
        /* Note the inner bracket here.
        This is needed because precedence
        of '-' operator is higher
        than '<<' */
        return ((n << 3) - n);
    }
     
    // Driver code
    public static void main (String arg[])
    {
        int n = 4;
         
        System.out.println(multiplyBySeven(n));
    }
}
 
// This code is contributed by Anant Agarwal.

Python

# Python program to multiply any
# positive number to 7
 
# Function to mutiply any number with 7
def multiplyBySeven(n):
 
    # Note the inner bracket here.
    # This is needed because
    # precedence of '-' operator is
    # higher than '<<'
    return ((n << 3) - n)
 
# Driver code
n = 4
print(multiplyBySeven(n))
 
# This code is contributed by Danish Raza

C#

// C# program to multiply any
// positive number to 7
using System;
 
class GFG
{
    static int multiplyBySeven(int n)
    {
        /* Note the inner bracket here. This is needed
        because precedence of '-' operator is higher
        than '<<' */
        return ((n << 3) - n);
    }
     
    // Driver code
    public static void Main ()
    {
        int n = 4;
        Console.Write(multiplyBySeven(n));
    }
}
 
// This code is contributed by Sam007

PHP

<?php
 
 
function multiplyBySeven($n)
{
 
    // Note the inner bracket here.
    // This is needed because
    // precedence of '-' operator 
    // is higherthan '<<'
    return (($n<<3) - $n);
}
 
    // Driver Code
    $n = 4;
    echo multiplyBySeven($n);
 
// This code is contributed by vt_m.
?>

Javascript

<script>
   
function multiplyBySeven(n)
{ 
 
    /* Note the inner bracket here. This is needed
      because precedence of '-' operator is higher
      than '<<' */
 
    return ((n << 3) - n);
}
   
// Driver program to test above function
 
     n = 4;
    document.write(multiplyBySeven(n));
     
  // This code is contributed by simranarora5sos
</script>

Output:

Time Complexity: O(1)
Space Complexity: O(1)
Note: Works only for positive integers.
Same concept can be used for fast multiplication by 9 or other numbers.

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