# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 `

Output:

```19 28 37 46 55 64 73 82 91
```

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