Multiply a given Integer with 3.5

Difficulty Level : Easy
Last Updated : 16 Oct, 2020

Given an integer x, write a function that multiplies x with 3.5 and returns the integer result. You are not allowed to use %, /, *.

Examples :
Input: 2
Output: 7

Input: 5
Output: 17 (Ignore the digits after decimal point)

Solution:
1. We can get x*3.5 by adding 2*x, x and x/2. To calculate 2*x, left shift x by 1 and to calculate x/2, right shift x by 2.
Below is the implementation of the above approach:

C++

// C++ program to multiply 
// a number with 3.5 
#include <bits/stdc++.h> 
  
int multiplyWith3Point5(int x) 
{ 
    return (x<<1) + x + (x>>1); 
}  
  
/* Driver program to test above functions*/
int main() 
{ 
    int x = 4;  
    printf("%d", multiplyWith3Point5(x)); 
    getchar(); 
    return 0; 
} 

Java

// Java Program to multiply 
// a number with 3.5 
  
class GFG { 
          
    static int multiplyWith3Point5(int x) 
    { 
        return (x<<1) + x + (x>>1); 
    }  
      
    /* Driver program to test above functions*/
    public static void main(String[] args) 
    { 
        int x = 2;  
        System.out.println(multiplyWith3Point5(x)); 
    } 
} 
  
// This code is contributed by prerna saini. 

Python3

# Python 3 program to multiply 
# a number with 3.5 
  
def multiplyWith3Point5(x): 
  
    return (x<<1) + x + (x>>1) 
   
  
# Driver program to  
# test above functions 
x = 4 
print(multiplyWith3Point5(x)) 
  
# This code is contributed by 
# Smitha Dinesh Semwal 

C#

// C# Program to multiply 
// a number with 3.5 
using System; 
  
class GFG 
{ 
          
    static int multiplyWith3Point5(int x) 
    { 
        return (x<<1) + x + (x>>1); 
    }  
      
    /* Driver program to test above functions*/
    public static void Main() 
    { 
        int x = 2;  
        Console.Write(multiplyWith3Point5(x)); 
    } 
      
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// PHP program to multiply 
// a number with 3.5 
  
  
function multiplyWith3Point5( $x) 
{ 
    return ($x << 1) + $x + ($x >> 1); 
}  
  
// Driver Code 
$x = 4;  
echo multiplyWith3Point5($x); 
      
// This code is contributed by vt_m. 
?> 

Output

2. Another way of doing this could be (8*x – x)/2 (See below code). Thanks to Ajaym for suggesting this.

C

#include <stdio.h> 
int multiplyWith3Point5(int x) 
{ 
  return ((x<<3) - x)>>1; 
}     

Another Approach:

Another way of doing this could be by doing a binary multiplication by 7 then divide by 2 using only <<, ^, &, and >>.

But here we have to mention that only positive numbers can be passed to this method.

Below is the implementation of the above approach:

Java

// Java program for above approach 
import java.io.*; 
  
class GFG  
{ 
     
    // Function to multiple number 
    // with 3.5 
    static int multiplyWith3Point5(int x) 
    { 
        int r = 0; 
  
        // The 3.5 is 7/2, so multiply  
        // by 7 (x * 7) then 
        // divide the result by 2  
        // (result/2) x * 7 -> 7 is 
        // 0111 so by doing mutiply  
        // by 7 it means we do 2 
        // shifting for the number  
        // but since we doing 
        // multiply we need to take  
        // care of carry one. 
        int x1Shift = x << 1; 
        int x2Shifts = x << 2; 
  
        r = (x ^ x1Shift) ^ x2Shifts; 
        int c = (x & x1Shift) | (x & x2Shifts) 
                | (x1Shift & x2Shifts); 
        while (c > 0) { 
            c <<= 1; 
            int t = r; 
            r ^= c; 
            c &= t; 
        } 
  
        // Then divide by 2  
        // r / 2 
        r = r >> 1; 
        return r; 
    } 
    
    // Driver Code 
    public static void main(String[] args) 
    { 
        System.out.println(multiplyWith3Point5(5)); 
    } 
}

Output

Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem

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