# Code for binary multiplication by 7 then divide by 2

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 x3.5 by adding 2x, x and x/2. To calculate 2x, left shift x by 1 and to calculate x/2, right shift x by 2. The function should not use any of the arithmetic operators (+, ++, –, -, … etc). Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits. We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result. The idea is to use subtractor logic. Write a program to subtract one from a given number. The idea is to use bitwise operators. Like addition, Write a function Add() that returns sum of two integers. The function should not use any of the arithmetic operators (+, ++, –, -, … etc). Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits. We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result. The idea is to use subtractor logic. The use of operators like ‘+’, ‘-‘, ‘’, ‘/’, ‘++’, ‘–‘ …etc. are not allowed.

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:

`// C# program for above approach`

`using` `System;`

`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)`

`{`

` ` `Console.WriteLine(multiplyWith3Point5(5));`

`}`

`}`

Output

17