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

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  • 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 <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 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




<script>
 
const TEN = 10;
 
// Function to return the
// sum of digits of n
function digitSum(n)
{
    let sum = 0;
    while (n > 0) {
 
        // Add last digit to the sum
        sum += n % TEN;
 
        // Remove last digit
        n = Math.floor(n / TEN);
    }
 
    return sum;
}
 
// Function to return the nth term
// of the required series
function getNthTerm(n)
{
    let 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
    let 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
function firstNTerms(n)
{
    for (let i = 1; i <= n; i++)
        document.write(getNthTerm(i) + " ");
}
 
// Driver code
 
    let n = 10;
 
    firstNTerms(n);
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output: 

19 28 37 46 55 64 73 82 91 109

 


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