First N terms whose sum of digits is a multiple of 10

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++

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#include <bits/stdc++.h>
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 requried 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;
}

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Java

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// 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 requried 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

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Python3

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# 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 requried 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

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C#

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// 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 requried 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

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Output:

19 28 37 46 55 64 73 82 91 109


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