Multiply a number by 15 without using * and / operators
Last Updated :
30 Nov, 2022
Given a integer N, the task is to multiply the number with 15 without using multiplication * and division / operators.
Examples:
Input: N = 10
Output: 150
Input: N = 7
Output: 105
Method 1: We can multiply integer N by 15 using bitwise operators. First left shift the number by 4 bits which is equal to (16 * N) then subtract the original number N from the shifted number i.e. ((16 * N) – N) which is equal to 15 * N.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
long multiplyByFifteen( long n)
{
long prod = (n << 4);
prod = prod - n;
return prod;
}
int main()
{
long n = 7;
cout << multiplyByFifteen(n);
return 0;
}
|
Java
class GFG {
static long multiplyByFifteen( long n)
{
long prod = (n << 4 );
prod = prod - n;
return prod;
}
public static void main(String[] args)
{
long n = 7 ;
System.out.print(multiplyByFifteen(n));
}
}
|
Python3
def multiplyByFifteen(n):
prod = (n << 4 )
prod = prod - n
return prod
n = 7
print (multiplyByFifteen(n))
|
C#
using System;
class GFG {
static long multiplyByFifteen( long n)
{
long prod = (n << 4);
prod = prod - n;
return prod;
}
public static void Main()
{
long n = 7;
Console.Write(multiplyByFifteen(n));
}
}
|
PHP
<?php
function multiplyByFifteen( $n )
{
$prod = ( $n << 4);
$prod = $prod - $n ;
return $prod ;
}
$n = 7;
echo multiplyByFifteen( $n );
?>
|
Javascript
<script>
function multiplyByFifteen(n)
{
let prod = (n << 4);
prod = prod - n;
return prod;
}
let n = 7;
document.write(multiplyByFifteen(n));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Method 2: We can also calculate the multiplication (15 * N) as sum of (8 * N) + (4 * N) + (2 * N) + N which can be obtained by performing left shift operations as (8 * N) = (N << 3), (4 * N) = (n << 2) and (2 * N) = (n << 1).
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
long multiplyByFifteen( long n)
{
long prod = (n << 3);
prod += (n << 2);
prod += (n << 1);
prod += n;
return prod;
}
int main()
{
long n = 7;
cout << multiplyByFifteen(n);
return 0;
}
|
Java
class GFG {
static long multiplyByFifteen( long n)
{
long prod = (n << 3 );
prod += (n << 2 );
prod += (n << 1 );
prod += n;
return prod;
}
public static void main(String[] args)
{
long n = 7 ;
System.out.print(multiplyByFifteen(n));
}
}
|
Python3
def multiplyByFifteen(n):
prod = (n << 3 )
prod + = (n << 2 )
prod + = (n << 1 )
prod + = n
return prod
n = 7
print (multiplyByFifteen(n))
|
C#
using System;
class GFG {
static long multiplyByFifteen( long n)
{
long prod = (n << 3);
prod += (n << 2);
prod += (n << 1);
prod += n;
return prod;
}
public static void Main()
{
long n = 7;
Console.Write(multiplyByFifteen(n));
}
}
|
Javascript
<script>
function multiplyByFifteen(n)
{
var prod = (n << 3);
prod += (n << 2);
prod += (n << 1);
prod += n;
return prod;
}
var n = 7;
document.write(multiplyByFifteen(n));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...