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Multiply a number by 15 without using * and / operators
• Last Updated : 01 Apr, 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 ``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

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## Javascript

 ``
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));``    ``}``}`

## 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`

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