# 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

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));` `    ``}` `}`

## 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

 ``

## Javascript

 ``

Output:

`105`

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++

 `// 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));` `    ``}` `}`

## 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

 ``

Output:

`105`

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

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