Hackerrank Day 5 loops 30 days of code solution YASH PAL, 31 July 202413 October 2025 In this HackerRank Day 5 loops 30 days of code problem solution we need to develop a program that can read an integer input and then print the multiplication table on the output screen.ObjectiveIn this challenge, we will use loops to do some math. TaskGiven an integer, n, print its first 10 multiples. Each multiple n x i (where 2 <= i <= 10 ) should be printed on a new line in the form: n x i = result.Examplen=3The printout should look like this:3 x 1 = 33 x 2 = 63 x 3 = 93 x 4 = 123 x 5 = 153 x 6 = 183 x 7 = 213 x 8 = 243 x 9 = 273 x 10 = 30 Input FormatA single integer, n.Constraints2 <= n <= 20Output FormatPrint 10 lines of output; each line i (where 1<=i<=10) contains the result of nxi in the form:n x i = result.Problem solution in Python 2.#!/bin/python import sys N = int(raw_input().strip()) for i in range(10): print "{} x {} = {}".format(N, i+1, N * (i + 1))Problem solution in Python 3.#!/bin/python3 import math import os import random import re import sys if __name__ == '__main__': n = int(input()) for i in range(1,11): s = n * i print(n,"x",i,"=",s)Problem solution in javaimport java.io.*; import java.util.*; public class Solution { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); for(int i=1;i<11;i++){ int ans = n * i; System.out.println(n + " x " + i + " = " + ans); } } }Problem solution in C++#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <string> #include <bitset> #include <cstdio> #include <limits> #include <vector> #include <climits> #include <cstring> #include <cstdlib> #include <fstream> #include <numeric> #include <sstream> #include <iostream> #include <algorithm> #include <unordered_map> using namespace std; int main(){ int N; cin >> N; for (int i=1;i<=10;i++){ cout<<N<<" x "<<i<<" = "<<N*i<<endl; } return 0; }Problem solution in c.#include <math.h> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <assert.h> #include <limits.h> #include <stdbool.h> int main(){ int N,i; scanf("%d",&N); if(N>=2 && N<= 20){ for(i=1;i<=10;i++){ printf("%d x %d = %d n",N,i,(N*i)); } } return 0; }Problem solution in Javascript.process.stdin.resume(); process.stdin.setEncoding('ascii'); var input_stdin = ""; var input_stdin_array = ""; var input_currentline = 0; process.stdin.on('data', function (data) { input_stdin += data; }); process.stdin.on('end', function () { input_stdin_array = input_stdin.split("n"); main(); }); function readLine() { return input_stdin_array[input_currentline++]; } function printdis(N) { var i, temp; for (i = 1; i <= 10; i++) { temp = N * i; console.log(N + " x " + i + " = " + temp); } } /////////////// ignore above this line //////////////////// function main() { var N = parseInt(readLine()); printdis(N); }Also, readHackerRank Day 6 Let’s Review 30 days of code problem solution 30 days of code coding problems solutions HackerRank