# Multiply a given Integer with 3.5 using Python3

Hey folks,

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:

`# Python3 program for the above approach`

`# Function to multiple number`

`# with 3.5`

`def` `multiplyWith3Point5(x): `

` ` `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. `

` ` `x1Shift ` `=` `x << ` `1`

` ` `x2Shifts ` `=` `x << ` `2`

` ` `r ` `=` `(x ^ x1Shift) ^ x2Shifts`

` ` `c ` `=` `(x & x1Shift) | (x & x2Shifts) | (x1Shift & x2Shifts)`

` ` `while` `(c > ` `0` `):`

` ` `c <<` `=` `1`

` ` `t ` `=` `r`

` ` `r ^` `=` `c`

` ` `c &` `=` `t`

` ` `# Then divide by 2`

` ` `# r / 2`

` ` `r ` `=` `r >> ` `1`

` ` `return` `r`

` ` `# Driver Code`

`if` `__name__ ` `=` `=` `'__main__'` `:`

` ` `print` `(multiplyWith3Point5(` `5` `))`

Output

17