Program to print multiplication table of a number

Last Updated : 13 Feb, 2025

Given a number n, we need to print its table.

Examples :

Input: 5
Output:
5 * 1 = 5
5 * 2 = 10
5 * 3 = 15
5 * 4 = 20
5 * 5 = 25
5 * 6 = 30
5 * 7 = 35
5 * 8 = 40
5 * 9 = 45
5 * 10 = 50

Input: 2
Output:
2 * 1 = 2
2 * 2 = 4
2 * 3 = 6
2 * 4 = 8
2 * 5 = 10
2 * 6 = 12
2 * 7 = 14
2 * 8 = 16
2 * 9 = 18
2 * 10 = 20

Iterative Approach

The iterative approach for printing a multiplication table involves using a loop to calculate and print the product of a given number and the numbers in range from 1 to 10. In this method, you begin with the number whose table you want to print and use a loop to multiply it with increasing values.

C++

// CPP program to print table of a number #include <iostream> using namespace std;  void printTable(int n) {    for (int i = 1; i <= 10; ++i)          cout << n << " * " << i << " = "                << n * i << endl; }  int main() {     int n = 5;       printTable(n);     return 0; }

#include <stdio.h>  void printTable(int n) {      for (int i = 1; i <= 10; ++i)          printf("%d * %d = %d\n", n, i, n * i); }  int main() {     int n = 5;       printTable(n);     return 0; }

Java

// Java program to print table of a number  import java.io.*;  class GfG {      public static void printTable(int n)  {                        for (int i = 1; i <= 10; ++i)              System.out.println(n + " * " + i +                                " = " + n * i);     }        public static void main(String arg[]){            int n = 5;  		printTable(n);     } }

Python

# Python Program to print table of a number  def printTable(n):      for i in range (1, 11):                   # multiples from 1 to 10         print ("%d * %d = %d" % (n, i, n * i))   if __name__ == "__main__":   n = 5   printTable(n)

// C# program to print table of a number  using System;  class GfG {     public static void printTable(int n) {                  for (int i = 1; i <= 10; ++i)              Console.Write(n + " * " + i +                               " = " + n *                                 i + "\n");     }        public static void Main()   {        int n = 5;        printTable(n);      } }

JavaScript

// Javascript program to print // table of a number  function printTable(n) {  for (let i = 1; i <= 10; ++i)     console.log( n + " * " +i +             " = " + n *                 i); }  // Driver Code let n = 5;  printTable(n);

Output :

5 * 1 = 5
5 * 2 = 10
5 * 3 = 15
5 * 4 = 20
5 * 5 = 25
5 * 6 = 30
5 * 7 = 35
5 * 8 = 40
5 * 9 = 45
5 * 10 = 50

Time Complexity – O(1)
Space Complexity – O(1)

Illustration
Step by step execution of loop for the multiplication table of n = 5.

We have n = 5, and the loop will iterate from i = 1 to i = 10.

First Iteration (i = 1):

The loop multiplies n = 5 by i = 1.
Result: 5 * 1 = 5.
Output: 5 * 1 = 5.
Second Iteration (i = 2):

The loop multiplies n = 5 by i = 2.
Result: 5 * 2 = 10.
Output: 5 * 2 = 10.
Third Iteration (i = 3):

The loop multiplies n = 5 by i = 3.
Result: 5 * 3 = 15.
Output: 5 * 3 = 15.

….
….

Tenth Iteration (i = 10):

The loop multiplies n = 5 by i = 10.
Result: 5 * 10 = 50.
Output: 5 * 10 = 50.

Recursive Approach

In this method, we pass i as an additional parameter with initial value as 1. We print n * i and then recursively call for i+1. We stop the recursion when i becomes 11 as we need to print only 10 multiples of given number and i.

C++

#include <iostream> using namespace std;  // printTable() prints table of number and takes // 1 required value that is number of whose teble // to be printed and an optional input i whose d // efault value is 1 void printTable(int n, int i = 1) {     if (i == 11)         return;     cout << n << " * " << i << " = " << n * i << endl;     i++;     printTable(n, i); }  int main() {     int n = 5;     printTable(n); }

#include <stdio.h>  // printTable() prints table of number and takes // 1 required value that is number of whose table // to be printed and an optional input i whose default value is 1 void printTable(int n, int i) {     if (i == 11)         return;     printf("%d * %d = %d\n", n, i, n * i);     i++;     printTable(n, i); }  int main() {     int n = 5;     printTable(n, 1);     return 0; }

Java

import java.util.*;   class GfG {       // printTable() prints table of number and takes     // 1 required value that is number of whose teble to be     // printed and an optional input i whose default value is 1     static void printTable(int n, Integer... val)  {           int i = 1;         if (val.length != 0)             i = val[0];         if (i == 11) // base case             return;         System.out.println(n + " * " + i + " = " + n * i);         i++;          printTable(n, i);     }       public static void main(String[] args) {         int n = 5;         printTable(n);     } }

Python

# printTable() prints table of number and takes # 1 required value that is number of whose teble to be printed # and an optional input i whose default value is 1 def printTable(n, i=1):      if (i == 11):  # base case         return     print(n, "*", i, "=", n * i)     i += 1       printTable(n, i)  if __name__ == "__main__":   n = 5   printTable(n)

using System; using System.Collections.Generic;  class GfG {    // print_table() prints table of number and takes   // 1 required value that is number of whose teble to be   // printed and an optional input i whose default value is 1   static void printTable(int n, int i = 1) {     if (i == 11) // base case       return;     Console.WriteLine(n + " * " + i + " = " + n * i);     i++;     printTable(n, i);   }      public static void Main(string[] args) {     int n = 5;     printTable(n);   } }

JavaScript

// printTable() prints table of number and takes //1 required value that is number of whose teble to be printed //and an optional input i whose default value is 1  function printTable(n, i = 1) {     if (i == 11)// base case         return;     console.log(n + " * " + i + " = " + n * i);     i++;     printTable(n,i); }  // Driver Code let n = 5; printTable(n);

Output

5 * 1 = 5 5 * 2 = 10 5 * 3 = 15 5 * 4 = 20 5 * 5 = 25 5 * 6 = 30 5 * 7 = 35 5 * 8 = 40 5 * 9 = 45 5 * 10 = 50

Time Complexity – O(1)
Space Complexity – O(1)

Recursive Program to print multiplication table of a number

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Program to print multiplication table of a number

Iterative Approach

Illustration

Recursive Approach

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