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) |
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)