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

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

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

19 28 37 46 55 64 73 82 91


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