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First N terms whose sum of digits is a multiple of 10
• Difficulty Level : Medium
• Last Updated : 11 Jun, 2021

Given an integer N, the task is to print the first N terms whose sum of digits is a multiple of 10. First few terms of the series are 19, 28, 37, 46, 55, …
Examples:

Input: N = 5
Output: 19 28 37 46 55
Input: N = 10
Output: 19 28 37 46 55 64 73 82 91 109

Approach: It can be observed that to get the Nth term of the required series, find the sum of the digits of N. If the sum is already a multiple of 10 then append digit 0 in the end of N else append the minimum possible digit in the end such that the new sum of digits is a multiple of 10

For example, to get the 19th term, since the sum of digits is already a multiple of 10 then append 0 and 190 is the 19th term of the series.
For N = 5, the minimum digit that can be appended to make the sum of digits as a multiple of 10 is 5 and 55 is the 5th term of the series.

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `const` `int` `TEN = 10;` `// Function to return the``// sum of digits of n``int` `digitSum(``int` `n)``{``    ``int` `sum = 0;``    ``while` `(n > 0) {` `        ``// Add last digit to the sum``        ``sum += n % TEN;` `        ``// Remove last digit``        ``n /= TEN;``    ``}` `    ``return` `sum;``}` `// Function to return the nth term``// of the required series``int` `getNthTerm(``int` `n)``{``    ``int` `sum = digitSum(n);` `    ``// If sum of digit is already``    ``// a multiple of 10 then append 0``    ``if` `(sum % TEN == 0)``        ``return` `(n * TEN);` `    ``// To store the minimum digit``    ``// that must be appended``    ``int` `extra = TEN - (sum % TEN);` `    ``// Return n after appending``    ``// the required digit``    ``return` `((n * TEN) + extra);``}` `// Function to print the first n terms``// of the required series``void` `firstNTerms(``int` `n)``{``    ``for` `(``int` `i = 1; i <= n; i++)``        ``cout << getNthTerm(i) << ``" "``;``}` `// Driver code``int` `main()``{``    ``int` `n = 10;` `    ``firstNTerms(n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG``{``    ``final` `static` `int` `TEN = ``10``;``    ` `    ``// Function to return the``    ``// sum of digits of n``    ``static` `int` `digitSum(``int` `n)``    ``{``        ``int` `sum = ``0``;``        ``while` `(n > ``0``)``        ``{``    ` `            ``// Add last digit to the sum``            ``sum += n % TEN;``    ` `            ``// Remove last digit``            ``n /= TEN;``        ``}``        ``return` `sum;``    ``}``    ` `    ``// Function to return the nth term``    ``// of the required series``    ``static` `int` `getNthTerm(``int` `n)``    ``{``        ``int` `sum = digitSum(n);``    ` `        ``// If sum of digit is already``        ``// a multiple of 10 then append 0``        ``if` `(sum % TEN == ``0``)``            ``return` `(n * TEN);``    ` `        ``// To store the minimum digit``        ``// that must be appended``        ``int` `extra = TEN - (sum % TEN);``    ` `        ``// Return n after appending``        ``// the required digit``        ``return` `((n * TEN) + extra);``    ``}``    ` `    ``// Function to print the first n terms``    ``// of the required series``    ``static` `void` `firstNTerms(``int` `n)``    ``{``        ``for` `(``int` `i = ``1``; i <= n; i++)``            ``System.out.print(getNthTerm(i) + ``" "``);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `n = ``10``;``    ` `        ``firstNTerms(n);``    ``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 code for above implementation``TEN ``=` `10` `# Function to return the``# sum of digits of n``def` `digitSum(n):``    ``sum` `=` `0``    ``while` `(n > ``0``):` `        ``# Add last digit to the sum``        ``sum` `+``=` `n ``%` `TEN` `        ``# Remove last digit``        ``n ``/``/``=` `TEN` `    ``return` `sum` `# Function to return the nth term``# of the required series``def` `getNthTerm(n):``    ``sum` `=` `digitSum(n)` `    ``# If sum of digit is already``    ``# a multiple of 10 then append 0``    ``if` `(``sum` `%` `TEN ``=``=` `0``):``        ``return` `(n ``*` `TEN)` `    ``# To store the minimum digit``    ``# that must be appended``    ``extra ``=` `TEN ``-` `(``sum` `%` `TEN)` `    ``# Return n after appending``    ``# the required digit``    ``return` `((n ``*` `TEN) ``+` `extra)` `# Function to print the first n terms``# of the required series``def` `firstNTerms(n):``    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ``print``(getNthTerm(i), end ``=` `" "``)` `# Driver code``n ``=` `10` `firstNTerms(n)` `# This code is contributed by Mohit Kumar`

## C#

 `// C# Program to Find the Unique elements``// in linked lists``using` `System;``    ` `class` `GFG``{``    ``readonly` `static` `int` `TEN = 10;``    ` `    ``// Function to return the``    ``// sum of digits of n``    ``static` `int` `digitSum(``int` `n)``    ``{``        ``int` `sum = 0;``        ``while` `(n > 0)``        ``{``    ` `            ``// Add last digit to the sum``            ``sum += n % TEN;``    ` `            ``// Remove last digit``            ``n /= TEN;``        ``}``        ``return` `sum;``    ``}``    ` `    ``// Function to return the nth term``    ``// of the required series``    ``static` `int` `getNthTerm(``int` `n)``    ``{``        ``int` `sum = digitSum(n);``    ` `        ``// If sum of digit is already``        ``// a multiple of 10 then append 0``        ``if` `(sum % TEN == 0)``            ``return` `(n * TEN);``    ` `        ``// To store the minimum digit``        ``// that must be appended``        ``int` `extra = TEN - (sum % TEN);``    ` `        ``// Return n after appending``        ``// the required digit``        ``return` `((n * TEN) + extra);``    ``}``    ` `    ``// Function to print the first n terms``    ``// of the required series``    ``static` `void` `firstNTerms(``int` `n)``    ``{``        ``for` `(``int` `i = 1; i <= n; i++)``            ``Console.Write(getNthTerm(i) + ``" "``);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main (String[] args)``    ``{``        ``int` `n = 10;``    ` `        ``firstNTerms(n);``    ``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`19 28 37 46 55 64 73 82 91 109`

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