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Program to print numbers having remainder 3 when divided by 11

Last Updated : 20 Feb, 2024
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Write a program to print all numbers from 1 to 1000 which leave a remainder of 3 when divided by 11.

Output format:

3, 14, 25, 36, 47…. 993

Approach: To solve the problem, follow the below idea:

The problem can be solved by simply using for loop and iterate from 3 up to the 1000 and increment by 11 as we are dividing by 11.

Step-by-step approach:

  • Iterate from 3 to limit, in each iteration increment by 11.
  • Print the number for each iteration.

Below is the implementation of the above approach:

C++




#include <iostream>
using namespace std;
 
void printNumbers(int limit)
{
    // Input the limit from the user
    cout << "Numbers leaving a remainder of 3 when divided "
            "by 11 up to "
         << limit << ":" << endl;
 
    // Iterate through numbers and print those meeting the
    // condition
    for (int number = 3; number < limit; number += 11) {
        cout << number << " ";
    }
 
    cout << endl;
}
 
int main()
{
    int limit = 1000;
 
    printNumbers(limit);
 
    return 0;
}


Java




import java.util.Scanner;
 
public class Main {
    // Function to print numbers leaving a remainder of 3 when divided by 11
    static void printNumbers(int limit) {
        // Input the limit from the user
        System.out.println("Numbers leaving a remainder of 3 when divided by 11 up to " + limit + ":");
 
        // Iterate through numbers and print those meeting the condition
        for (int number = 3; number < limit; number += 11) {
            System.out.print(number + " ");
        }
 
        System.out.println();
    }
 
    public static void main(String[] args) {
        // Create a Scanner object for user input
        Scanner scanner = new Scanner(System.in);
 
        // Set the limit to 1000 (as in the original C++ code)
        int limit = 1000;
 
        // Call the function to print numbers
        printNumbers(limit);
 
        // Close the Scanner
        scanner.close();
    }
}


Python3




def print_numbers(limit):
    # Input the limit from the user
    print(f"Numbers leaving a remainder of 3 when divided by 11 up to {limit}:")
     
    # Iterate through numbers and print those meeting the condition
    for number in range(3, limit, 11):
        print(number, end=" ")
 
    print()
 
if __name__ == "__main__":
    limit = 1000
 
    print_numbers(limit)


C#




using System;
 
class Program
{
    // Function to print numbers leaving a remainder of 3 when divided by 11 up to a given limit
    static void PrintNumbers(int limit)
    {
        // Input the limit from the user
        Console.WriteLine($"Numbers leaving a remainder of 3 when divided by 11 up to {limit}:");
 
        // Iterate through numbers and print those meeting the condition
        for (int number = 3; number < limit; number += 11)
        {
            Console.Write(number + " ");
        }
 
        Console.WriteLine();
    }
 
    static void Main()
    {
        int limit = 1000;
 
        // Call the function to print numbers
        PrintNumbers(limit);
 
        // Ensure the console window stays open
        Console.ReadLine();
    }
}


Javascript




// Function to print numbers leaving a remainder of 3 when divided by 11
function printNumbers(limit) {
    // Input the limit from the user
    console.log(`Numbers leaving a remainder of 3 when divided by 11 up to ${limit}:`);
 
    // Iterate through numbers and print those meeting the condition
    for (let number = 3; number < limit; number += 11) {
        process.stdout.write(`${number} `);
    }
 
    console.log(); // Move to the next line after printing all numbers
}
 
// Set the limit to 1000 (as in the original Java code)
const limit = 1000;
 
// Call the function to print numbers
printNumbers(limit);


Output

Numbers leaving a remainder of 3 when divided by 11 up to 1000:
3 14 25 36 47 58 69 80 91 102 113 124 135 146 157 168 179 190 201 212 223 234 245 256 267 278 289 300 311 322 333 344 355 366 377 388 39...







Time complexity: O(1)
Auxiliary Space: O(1)



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