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Multiply a number with 10 without using multiplication operator

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Given a number, the task is to multiply it with 10 without using multiplication operator?
Examples: 
 

Input : n = 50
Output: 500
// multiplication of 50 with 10 is = 500

Input : n = 16
Output: 160
// multiplication of 16 with 10 is = 160

 

A simple solution for this problem is to run a loop and add n with itself 10 times. Here we need to perform 10 operations. 

C++




// C++ program to multiply a number with 10
// without using multiplication operator
#include<bits/stdc++.h>
using namespace std;
  
// Function to find multiplication of n with
// 10 without using multiplication operator
int multiplyTen(int n)
{
    int sum=0;
    // Running a loop and add n with itself 10 times
    for(int i=0;i<10;i++)
    {
        sum=sum+n;
    }
    return sum;
}
  
// Driver program to run the case
int main()
{
    int n = 50;
    cout << multiplyTen(n);
    return 0;
}


Java




// Java program to multiply a number with 10
// without using multiplication operator
import java.util.*;
  
public class GFG {
      // Function to find multiplication of n with
    // 10 without using multiplication operator
      public static int multiplyTen(int n) {
        int sum = 0;
        // Running a loop and add n with itself 10 times
        for (int i = 0; i < 10; i++) {
              sum = sum + n;
        }
        return sum;
      }
      
      // Driver program to run the case
      public static void main(String[] args) {
        int n = 50;
        System.out.println(multiplyTen(n));
      }
}
// This code is contributed by Prasad Kandekar(prasad264)


Python3




# python program to multiply a number with 10
# without using multiplication operator
  
# Function to find multiplication of n with
# 10 without using multiplication operator
def multiplyTen(n):
    sum = 0
    # Running a loop and add n with itself 10 times
    for i in range(10):
        sum += n
    return sum
  
# Driver code
n = 50
print(multiplyTen(n))
# This code is contributed by Prasad Kandekar(prasad264)


C#




// C# program to multiply a number with 10
// without using multiplication operator
using System;
public class GFG {
    
      // Function to find multiplication of n with
    // 10 without using multiplication operator
      static int MultiplyTen(int n) {
        int sum = 0;
        
        // Running a loop and add n with itself 10 times
        for (int i = 0; i < 10; i++) {
              sum += n;
        }
        return sum;
      }
    
    // Driver program to run the case
      static void Main(string[] args) {
        int n = 50;
        Console.WriteLine(MultiplyTen(n));
      }
}
// This code is contributed by Prasad Kandekar(prasad264)


Javascript




// Javascript program to multiply a number with 10
// without using multiplication operator
  
// Function to find multiplication of n with
// 10 without using multiplication operator
function multiplyTen(n) {
    let sum = 0;
    // Running a loop and add n with itself 10 times
    for (let i = 0; i < 10; i++) {
        sum += n;
    }
    return sum;
}
  
// Driver code
let n = 50;
console.log(multiplyTen(n));
// This code is contributed by Prasad Kandekar(prasad264)


Output

500

Time Complexity: O(1)

Auxiliary Space: O(1)

A better solution is to use bit manipulation. We have to multiply n with 10 i.e; n*10, we can write this as n*(2+8) = n*2 + n*8 and since we are not allowed to use multiplication operator we can do this using left shift bitwise operator. So n*10 = n<<1 + n<<3.
 

C++




// C++ program to multiply a number with 10 using
// bitwise operators
#include<bits/stdc++.h>
using namespace std;
  
// Function to find multiplication of n with
// 10 without using multiplication operator
int multiplyTen(int n)
{
    return (n<<1) + (n<<3);
}
  
// Driver program to run the case
int main()
{
    int n = 50;
    cout << multiplyTen(n);
    return 0;
}


Java




// Java Code to Multiply a number with 10
// without using multiplication operator
import java.util.*;
  
class GFG {
      
    // Function to find multiplication of n 
    // with 10 without using multiplication
    // operator
    public static int multiplyTen(int n)
    {
        return (n << 1) + (n << 3);
    }
      
    /* Driver program to test above function */
    public static void main(String[] args) 
    {
        int n = 50;
        System.out.println(multiplyTen(n));
         
    }
}
    
// This code is contributed by Arnav Kr. Mandal.


Python 3




# Python 3 program to multiply a 
# number with 10 using bitwise
# operators
  
# Function to find multiplication
# of n with 10 without using
# multiplication operator
def multiplyTen(n):
  
    return (n << 1) + (n << 3)
  
# Driver program to run the case
n = 50
print (multiplyTen(n))
  
# This code is contributed by 
# Smitha


C#




// C# Code to Multiply a number with 10
// without using multiplication operator
using System;
  
class GFG {
      
    // Function to find multiplication of n 
    // with 10 without using multiplication
    // operator
    public static int multiplyTen(int n)
    {
        return (n << 1) + (n << 3);
    }
      
    // Driver Code
    public static void Main() 
    {
        int n = 50;
        Console.Write(multiplyTen(n));
          
    }
}
      
// This code is contributed by Nitin Mittal.


PHP




<?php
// PHP program to multiply a 
// number with 10 using
// bitwise operators
  
// Function to find multiplication 
// of n with 10 without using 
// multiplication operator
function multiplyTen($n)
{
    return ($n << 1) + ($n << 3);
}
  
    // Driver Code
    $n = 50;
    echo multiplyTen($n);
  
// This code is contributed by nitin mittal. 
?>


Javascript




<script>
// JavaScript program to multiply a number with 10 using 
// bitwise operators 
  
// Function to find multiplication of n with 
// 10 without using multiplication operator 
function multiplyTen(n) 
    return (n<<1) + (n<<3); 
  
// Driver program to run the case 
  
    let n = 50; 
    document.write(multiplyTen(n)); 
   
// This code is contributed by Surbhi Tyagi.
  
</script>


Output: 

500

Time Complexity: O(1)

Auxiliary Space: O(1)

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Last Updated : 11 Sep, 2023
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