Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

Find the first N integers such that the sum of their digits is equal to 10

  • Last Updated : 08 Apr, 2021

Given an integer N, the task is to print first N integers whose sum of digits is 10.
Examples: 
 

Input: N = 4 
Output: 19 28 37 46
Input: N = 6 
Output: 19 28 37 46 55 64 
 

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

 

Approach: Initialise num = 19 to get the first number of the series, now add 9 to the previous number and check whether the sum of the digits of the new number is 10. If yes then this is the next number of the series, this is because the difference between any two consecutive numbers of the required series is at least 9.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the
// sum of digits of n
int sum(int n)
{
    int sum = 0;
    while (n) {
 
        // Add the last digit to the sum
        sum = sum + n % 10;
 
        // Remove last digit
        n = n / 10;
    }
 
    // Return the sum of digits
    return sum;
}
 
// Function to print the first n numbers
// whose sum of digits is 10
void firstN(int n)
{
 
    // First number of the series is 19
    int num = 19, cnt = 1;
    while (cnt != n) {
 
        // If the sum of digits of the
        // current number is equal to 10
        if (sum(num) == 10) {
 
            // Print the number
            cout << num << " ";
            cnt++;
        }
 
        // Add 9 to the previous number
        num += 9;
    }
}
 
// Driver code
int main()
{
    int n = 10;
    firstN(n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the
// sum of digits of n
static int sum(int n)
{
    int sum = 0;
    while (n > 0)
    {
 
        // Add the last digit to the sum
        sum = sum + n % 10;
 
        // Remove last digit
        n = n / 10;
    }
 
    // Return the sum of digits
    return sum;
}
 
// Function to print the first n numbers
// whose sum of digits is 10
static void firstN(int n)
{
 
    // First number of the series is 19
    int num = 19, cnt = 1;
    while (cnt != n)
    {
 
        // If the sum of digits of the
        // current number is equal to 10
        if (sum(num) == 10)
        {
 
            // Print the number
            System.out.print(num + " ");
            cnt++;
        }
 
        // Add 9 to the previous number
        num += 9;
    }
}
 
// Driver code
public static void main(String[] args)
{
    int n = 10;
    firstN(n);
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 implementation of the approach
 
# Function to return the
# sum of digits of n
def sum(n) :
 
    sum = 0;
    while (n) :
 
        # Add the last digit to the sum
        sum = sum + n % 10;
 
        # Remove last digit
        n = n // 10;
 
    # Return the sum of digits
    return sum;
 
# Function to print the first n numbers
# whose sum of digits is 10
def firstN(n) :
 
    # First number of the series is 19
    num = 19; cnt = 1;
     
    while (cnt != n) :
 
        # If the sum of digits of the
        # current number is equal to 10
        if (sum(num) == 10) :
 
            # Print the number
            print(num,end= " ");
            cnt += 1;
 
        # Add 9 to the previous number
        num += 9;
 
# Driver code
if __name__ == "__main__" :
 
    n = 10;
    firstN(n);
     
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the
// sum of digits of n
static int sum(int n)
{
    int sum = 0;
    while (n > 0)
    {
 
        // Add the last digit to the sum
        sum = sum + n % 10;
 
        // Remove last digit
        n = n / 10;
    }
 
    // Return the sum of digits
    return sum;
}
 
// Function to print the first n numbers
// whose sum of digits is 10
static void firstN(int n)
{
 
    // First number of the series is 19
    int num = 19, cnt = 1;
    while (cnt != n)
    {
 
        // If the sum of digits of the
        // current number is equal to 10
        if (sum(num) == 10)
        {
 
            // Print the number
            Console.Write(num + " ");
            cnt++;
        }
 
        // Add 9 to the previous number
        num += 9;
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 10;
    firstN(n);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript implementation of the approach
 
    // Function to return the
    // sum of digits of n
    function sum(n) {
        var sum = 0;
        while (n > 0) {
 
            // Add the last digit to the sum
            sum = sum + n % 10;
 
            // Remove last digit
            n = parseInt(n / 10);
        }
 
        // Return the sum of digits
        return sum;
    }
 
    // Function to prvar the first n numbers
    // whose sum of digits is 10
    function firstN(n) {
 
        // First number of the series is 19
        var num = 19, cnt = 1;
        while (cnt != n) {
 
            // If the sum of digits of the
            // current number is equal to 10
            if (sum(num) == 10) {
 
                // Print the number
                document.write(num + " ");
                cnt++;
            }
 
            // Add 9 to the previous number
            num += 9;
        }
    }
 
    // Driver code
     
        var n = 10;
        firstN(n);
 
// This code contributed by Rajput-Ji
 
</script>
Output: 
19 28 37 46 55 64 73 82 91

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :