Efficient way to multiply with 7

Given a number n, find the result after multiplying it by 7.

Example:

Input: n=8
Output: 56

Naive Approach: The basic way to solve this is to use a multiplication operator, add the number 7 times or use recursive function.

#include <bits/stdc++.h>

using namespace std;
// c++ implementation
long multiplyBySeven(long n) { return (n * 7); }

/* Driver program to test above function */
int main()
{
    long n = 4;

    cout << multiplyBySeven(n);

    return 0;
}

// Java implementation
import java.io.*;

class GFG {
    
    static long multiplyBySeven(long n) { return (n * 7); }
    
    /* Driver program to test above function */
    public static void main (String[] args) {
        
        long n = 4;
        
        System.out.println(multiplyBySeven(n));
        
    }
}

// This code is contributed by Aman Kumar

// C# implementation
using System;

public class GFG{
    
    static long multiplyBySeven(long n) { return (n * 7); }
    
    /* Driver program to test above function */
    static public void Main (){
        long n = 4;

        Console.Write(multiplyBySeven(n));
    }
}

// This code is contributed by Pushpesh Raj.

def multiplyBySeven(n: int) -> int:
    return n * 7

n = 4
print(multiplyBySeven(n))

<?php
function multiplyBySeven($n) {
    return $n * 7;
}
$n = 4;
echo multiplyBySeven($n);
?>

<script>
function multiplyBySeven(n) {
    return n * 7;
}

let n = 4;
console.log(multiplyBySeven(n));
</script>

28

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

Efficient Approach: We can multiply a number by 7 using a bitwise operator. 

• First left shift the number by 3 bits (you will get 8n) 
• Then subtract the original number from the shifted number 
• Return the difference (8n - n). 

Below is the implementation of the above approach:

# 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;
}

# 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; 
}

// 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 program to multiply any 
# positive number to 7 

# Function to multiply 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# 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


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.
?>

<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>

28

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

Method 3: 

1. We know that 7 can be represented as 2^3 - 1, which means that 7 can be expressed as (2^3) - 1.
2. So, we can write the multiplication of a number by 7 as:
7*n = (2^3 - 1)*n = (2^3)*n - n = (n << 3) - n
3. Here, << operator is used for left shift operation by 3 bits, which is equivalent to multiplying by 2^3.
We can further simplify the above expression as:
7*n = (n << 3) - n = (n << 2 + n << 1 + n)
Here, << operator is used for left shift operation by 1 bit, which is equivalent to multiplying by 2.

#include <iostream>

using namespace std;

int multiplyBy7(int num) {
    return (num << 2) + (num << 1) + num;
}

int main() {
    int n = 4;
    cout << multiplyBy7(n) << endl; // Output: 28
    return 0;
}

public class GFG {
    public static int multiplyBy7(int num) {
        return (num << 2) + (num << 1) + num;
    }

    public static void main(String[] args) {
        int n = 4;
        System.out.println(multiplyBy7(n)); // Output: 28
    }
}

def multiply_by_7(num):
    return (num << 2) + (num << 1) + num
#Driver code
n = 4
print(multiply_by_7(n))

using System;

namespace MultiplyBy7 {
    class Program {
        static int MultiplyBy7(int num) {
            return (num << 2) + (num << 1) + num;
        }

        static void Main(string[] args) {
            int n = 4;
            Console.WriteLine(MultiplyBy7(n)); // Output: 28
        }
    }
}

function multiplyBy7(num) {
  return (num << 2) + (num << 1) + num;
}

let n = 4;
console.log(multiplyBy7(n)); // Output: 28

28

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

Note: Works only for positive integers. 
Same concept can be used for fast multiplication by 9 or other numbers.