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Multiply a number by 15 without using * and / operators

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  • Difficulty Level : Basic
  • Last Updated : 15 Dec, 2021

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++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return (15 * N) without
// using '*' or '/' operator
long multiplyByFifteen(long n)
{
    // prod = 16 * n
    long prod = (n << 4);
 
    // ((16 * n) - n) = 15 * n
    prod = prod - n;
 
    return prod;
}
 
// Driver code
int main()
{
    long n = 7;
 
    cout << multiplyByFifteen(n);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG {
 
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 16 * n
        long prod = (n << 4);
 
        // ((16 * n) - n) = 15 * n
        prod = prod - n;
 
        return prod;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        long n = 7;
        System.out.print(multiplyByFifteen(n));
    }
}

Python3




# Python3 implementation of the approach
 
# Function to return (15 * N) without
# using '*' or '/' operator
def multiplyByFifteen(n):
     
    # prod = 16 * n
    prod = (n << 4)
     
    # ((16 * n) - n) = 15 * n
    prod = prod - n
     
    return prod
     
# Driver code
n = 7
print(multiplyByFifteen(n))

C#




// C# implementation of the approach
using System;
class GFG {
 
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 16 * n
        long prod = (n << 4);
 
        // ((16 * n) - n) = 15 * n
        prod = prod - n;
 
        return prod;
    }
 
    // Driver code
    public static void Main()
    {
        long n = 7;
        Console.Write(multiplyByFifteen(n));
    }
}

PHP




<?php
// PHP implementation of the approach
 
// Function to return (15 * N) without
// using '*' or '/' operator
function multiplyByFifteen($n)
{
    // prod = 16 * n
    $prod = ($n << 4);
 
    // ((16 * n) - n) = 15 * n
    $prod = $prod - $n;
 
    return $prod;
}
 
// Driver code
 
    $n = 7;
 
    echo multiplyByFifteen($n);
 
// This code is contributed by anuj_67..
?>

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to return (15 * N) without
// using '*' or '/' operator
function multiplyByFifteen(n)
{
    // prod = 16 * n
    let prod = (n << 4);
 
    // ((16 * n) - n) = 15 * n
    prod = prod - n;
 
    return prod;
}
 
// Driver code
 
    let n = 7;
 
    document.write(multiplyByFifteen(n));
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output: 

105

 

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++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return (15 * N) without
// using '*' or '/' operator
long multiplyByFifteen(long n)
{
    // prod = 8 * n
    long prod = (n << 3);
 
    // Add (4 * n)
    prod += (n << 2);
 
    // Add (2 * n)
    prod += (n << 1);
 
    // Add n
    prod += n;
 
    // (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
    return prod;
}
 
// Driver code
int main()
{
    long n = 7;
 
    cout << multiplyByFifteen(n);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG {
 
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 8 * n
        long prod = (n << 3);
 
        // Add (4 * n)
        prod += (n << 2);
 
        // Add (2 * n)
        prod += (n << 1);
 
        // Add n
        prod += n;
 
        // (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
        return prod;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        long n = 7;
        System.out.print(multiplyByFifteen(n));
    }
}

Python3




# Python3 implementation of the approach
 
# Function to perform Multiplication
def multiplyByFifteen(n):
     
    # prod = 8 * n
    prod = (n << 3)
     
    # Add (4 * n)
    prod += (n << 2)
     
    # Add (2 * n)
    prod += (n << 1)
     
    # Add n
    prod += n
     
    # (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
    return prod
     
# Driver code
n = 7
print(multiplyByFifteen(n))

C#




// C# implementation of the approach
using System;
class GFG {
 
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 8 * n
        long prod = (n << 3);
 
        // Add (4 * n)
        prod += (n << 2);
 
        // Add (2 * n)
        prod += (n << 1);
 
        // Add n
        prod += n;
 
        // (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
        return prod;
    }
 
    // Driver code
    public static void Main()
    {
        long n = 7;
        Console.Write(multiplyByFifteen(n));
    }
}

Javascript




<script>
 
// Javascript implementation of the approach  
 
// Function to return (15 * N) without
// using '*' or '/' operator
function multiplyByFifteen(n)
{
     
    // prod = 8 * n
    var prod = (n << 3);
 
    // Add (4 * n)
    prod += (n << 2);
 
    // Add (2 * n)
    prod += (n << 1);
 
    // Add n
    prod += n;
 
    // (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
    return prod;
}
 
// Driver code
var n = 7;
 
document.write(multiplyByFifteen(n));
 
// This code is contributed by shikhasingrajput
 
</script>

Output: 

105

 

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


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