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 = 50print(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 = 50print (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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