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++ 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 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)
# 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# 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 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++ 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 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 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# 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 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.
?>
<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)