# Multiply a number by 15 without using * and / operators

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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));     } }

## Python

 # 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



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 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));     } }

## Python

 # 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));     } }

Output:

105

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