Skip to content
Related Articles

Related Articles

Count of numbers satisfying m + sum(m) + sum(sum(m)) = N

View Discussion
Improve Article
Save Article
  • Difficulty Level : Hard
  • Last Updated : 07 Jun, 2022

Given an integer N, find out the count of numbers(m) that satisfy the condition m + sum(m) + sum (sum(m)) = N, where sum(m) denotes the sum of digits in m. Given N <= 10e9. 
Examples: 
 

Input: 9
Output: 1
Explanation: Only 1 positive integer satisfies 
             the condition that is 3, 
            3 + sum(3) + sum(sum(3)) = 3 + 3 + 3 
                                     = 9 

Input: 9939
Output: 4
Explanation: m can be 9898, 9907, 9910 and 9913. 
             9898 + sum(9898) + sum(sum(9898))  
              = 9898 + 34 + 7 = 9939.
             9907 + sum(9907) + sum(sum(9907)) 
              = 9907 + 25 + 7 = 9939.
             9910 + sum(9910) + sum(sum(9910)) 
              = 9910 + 19 + 10 = 9939.
             9913 + sum(9913) + sum(sum(9913))  
              = 9913 + 22 + 4 = 9939.

 

Approach: The first thing to note is that we are given the constraint N<=10e9. This means sum(x) can at maximum be 81 for any number, This is because the largest number below 10e9 is 999999999 whose digits add up to 81. The maximum case for sum(sum(x)) will be 16 (number being 79<=81), so at max it 81 + 16 which is 97 needs to be checked. We just need to iterate from N – 97 to N and check which integers satisfy the equation because no integer smaller than N-97 can satisfy the equation and neither can any integer greater than N can.
Below is the implementation of the above approach. 
 

C++




// CPP program to count numbers
// satisfying equation.
#include <bits/stdc++.h>
using namespace std;
 
// function that returns sum of
// digits in a number
int sum(int n)
{
    int rem = 0;
 
    // initially sum of digits is 0
    int sum_of_digits = 0;
 
    // loop runs till all digits
    // have been extracted
    while (n > 0) {
     
        // last digit from backside
        rem = n % 10;
 
        // sums up the digits
        sum_of_digits += rem;
 
        // the number is reduced to the
        // number removing the last digit
        n = n / 10;
    }
     
    // returns the sum of digits in a number
    return sum_of_digits;
}
 
// function to calculate the count of
// such occurrences
int count(int n)
{
    // counter to calculate the occurrences
    int c = 0;
 
    // loop to traverse from n-97 to n
    for (int i = n - 97; i <= n; i++) {
 
        // calls the function to calculate
        // the sum of digits of i
        int a = sum(i);
 
        // calls the function to calculate
        // the sum of digits of a
        int b = sum(a);
 
        // if the summation is equal to n
        // then increase counter by 1
        if ((i + a + b) == n) {
            c += 1;
        }
    }
     
    // returns the count
    return c;
}
 
// driver program to test the above function
int main()
{
    int n = 9939;
     
    // calls the function to get the answer
    cout << count(n) << endl;
 
    return 0;
}

Java



// Java program to count numbers
// satisfying equation.
import java.io.*;
 
class GFG {
     
    // function that returns sum of
    // digits in a number
    static int sum(int n)
    {
        int rem = 0;
     
        // initially sum of digits is 0
        int sum_of_digits = 0;
     
        // loop runs till all digits
        // have been extracted
        while (n > 0)
        {
         
            // last digit from backside
            rem = n % 10;
     
            // sums up the digits
            sum_of_digits += rem;
     
            // the number is reduced to the
            // number removing the last digit
            n = n / 10;
        }
         
        // returns the sum of digits in a number
        return sum_of_digits;
    }
     
    // function to calculate the count of
    // such occurrences
    static int count(int n)
    {
        // counter to calculate the occurrences
        int c = 0;
     
        // loop to traverse from n-97 to n
        for (int i = n - 97; i <= n; i++)
        {
     
            // calls the function to calculate
            // the sum of digits of i
            int a = sum(i);
     
            // calls the function to calculate
            // the sum of digits of a
            int b = sum(a);
     
            // if the summation is equal to n
            // then increase counter by 1
            if ((i + a + b) == n)
            {
                c += 1;
            }
        }
         
        // returns the count
        return c;
    }
 
    // driver program to test the above function
    public static void main (String[] args)
    {
        int n = 9939;
        // calls the function to get the answer
        System.out.println ( count(n) );
     
    }
}
 
// This article is contributed by vt_m
Python3



# Python program
# to count numbers
# satisfying equation.
 
# function that returns sum of
# digits in a number
def sum(n):
 
    rem = 0
  
    #initially sum of digits is 0
    sum_of_digits = 0
  
    # loop runs till all digits
    # have been extracted
    while (n > 0):
      
        # last digit from backside
        rem = n % 10
  
        # sums up the digits
        sum_of_digits += rem
  
        # the number is reduced to the
        # number removing the last digit
        n = n // 10
 
    # returns the sum
    # of digits in a number
    return sum_of_digits
  
# function to calculate
# the count of
# such occurrences
def count(n):
 
    # counter to calculate the occurrences
    c = 0
  
    # loop to traverse from n - 97 to n
    for i in range(n - 97,n+1):
  
        # calls the function to calculate
        # the sum of digits of i
        a = sum(i)
  
        # calls the function to calculate
        # the sum of digits of a
        b = sum(a)
  
        # if the summation is equal to n
        # then increase counter by 1
        if ((i + a + b) == n):
            c += 1
 
    # returns the count
    return c
  
# driver program to test
# the above function
 
n = 9939
      
# calls the function
# to get the answer
print(count(n))
 
# This code is contributed
# by Anant Agarwal.
C#



// C# program to count numbers
// satisfying equation.
using System;
 
class GFG {
     
    // function that returns sum 
    // of digits in a number
    static int sum(int n)
    {
        int rem = 0;
     
        // initially sum of
        // digits is 0
        int sum_of_digits = 0;
     
        // loop runs till all digits
        // have been extracted
        while (n > 0)
        {
            // last digit from
            // backside
            rem = n % 10;
     
            // sums up the digits
            sum_of_digits += rem;
     
            // the number is reduced
            // to the number removing
            // the last digit
            n = n / 10;
        }
         
        // returns the sum of
        // digits in a number
        return sum_of_digits;
    }
     
    // function to calculate the
    // count of such occurrences
    static int count(int n)
    {
         
        // counter to calculate
        // the occurrences
        int c = 0;
     
        // loop to traverse from n-97 to n
        for (int i = n - 97; i <= n; i++)
        {
     
            // calls the function to calculate
            // the sum of digits of i
            int a = sum(i);
     
            // calls the function to calculate
            // the sum of digits of a
            int b = sum(a);
     
            // if the summation is equal to n
            // then increase counter by 1
            if ((i + a + b) == n)
            {
                c += 1;
            }
        }
         
        // returns the count
        return c;
    }
 
    // Driver Code
    public static void Main ()
    {
        int n = 9939;
         
        // calling the function
        Console.Write ( count(n) );
     
    }
}
 
// This code is contributed by Nitin Mittal.
PHP



<?php
// PHP program to count numbers
// satisfying equation.
 
// function that returns sum of
// digits in a number
function sum($n)
{
    $rem = 0;
 
    // initially sum of
    // digits is 0
    $sum_of_digits = 0;
 
    // loop runs till all digits
    // have been extracted
    while ($n > 0)
    {
     
        // last digit from backside
        $rem = $n % 10;
 
        // sums up the digits
        $sum_of_digits += $rem;
 
        // the number is reduced to the
        // number removing the last digit
        $n = $n / 10;
    }
     
    // returns the sum of
    // digits in a number
    return $sum_of_digits;
}
 
// function to calculate the
// count of such occurrences
function countt($n)
{
     
    // counter to calculate
    // the occurrences
    $c = 0;
 
    // loop to traverse
    // from n-97 to n
    for ($i = $n - 97; $i <= $n; $i++)
    {
 
        // calls the function to calculate
        // the sum of digits of i
        $a = sum($i);
 
        // calls the function to calculate
        // the sum of digits of a
        $b = sum($a);
 
        // if the summation is equal to n
        // then increase counter by 1
        if (($i + $a + $b) == $n)
        {
            $c += 1;
        }
    }
     
    // returns the count
    return $c;
}
 
    // Driver Code
    $n = 9939;
     
    // calls the function
    // to get the answer
    echo countt($n) ;
 
//This code is contributed by nitin mittal.
?>
Javascript



<script>
// javascript program to count numbers
// satisfying equation.
 
    // function that returns sum of
    // digits in a number
    function sum(n)
    {
        var rem = 0;
 
        // initially sum of digits is 0
        var sum_of_digits = 0;
 
        // loop runs till all digits
        // have been extracted
        while (n > 0)
        {
 
            // last digit from backside
            rem = n % 10;
 
            // sums up the digits
            sum_of_digits += rem;
 
            // the number is reduced to the
            // number removing the last digit
            n = parseInt(n / 10);
        }
 
        // returns the sum of digits in a number
        return sum_of_digits;
    }
 
    // function to calculate the count of
    // such occurrences
    function count(n)
    {
     
        // counter to calculate the occurrences
        var c = 0;
 
        // loop to traverse from n-97 to n
        for (i = n - 97; i <= n; i++)
        {
 
            // calls the function to calculate
            // the sum of digits of i
            var a = sum(i);
 
            // calls the function to calculate
            // the sum of digits of a
            var b = sum(a);
 
            // if the summation is equal to n
            // then increase counter by 1
            if ((i + a + b) == n) {
                c += 1;
            }
        }
 
        // returns the count
        return c;
    }
 
    // driver program to test the above function   
        var n = 9939;
         
        // calls the function to get the answer
        document.write(count(n));
 
// This code is contributed by Rajput-Ji
</script>
Output: 
 
4
Time Complexity: O(log10N+97), as we are using two loops to traverse log10N and 97 times.
Auxiliary Space: O(1), as we are not using any extra space.
This article is contributed by Raja Vikramaditya. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

My Personal Notes
arrow_drop_up
Next

Program for Sum of the digits of a given number
Recommended Articles
Page :
Split N as the sum of K numbers satisfying the given conditions
21, Aug 20
Sort a permutation of first N Natural Numbers by swapping pairs satisfying given conditions
09, Jun 21
Count valid pairs in the array satisfying given conditions
11, Dec 18
Count of pairs satisfying the given condition
22, Jan 20
Queries to count distinct Binary Strings of all lengths from N to M satisfying given properties
30, Jul 20
Count possible permutations of given array satisfying the given conditions
06, Oct 20
Count of Ordered Pairs (X, Y) satisfying the Equation 1/X + 1/Y = 1/N
14, Aug 20
Count number of coordinates from an array satisfying the given conditions
30, Apr 21
Maximum possible sum of squares of stack elements satisfying the given properties
09, Oct 20
Generate N integers satisfying the given conditions
06, Sep 19
Number of K's such that the given array can be divided into two sets satisfying the given conditions
29, Oct 19
Find the maximum sum (a+b) for a given input integer N satisfying the given condition
14, Apr 20
Smallest positive integer X satisfying the given equation
05, Aug 20
Smallest value of X satisfying the condition X % A[i] = B[i] for two given arrays
15, Mar 21
Length of longest strictly increasing subset with each pair of adjacent elements satisfying the condition 2 * A[i] ≥ A[i + 1]
22, Mar 21
Lexicographically smallest permutation of length 2N that can be obtained from an N-length array satisfying given conditions
17, Jun 21
Find range of values for S in given Array with values satisfying [ arr[i] = floor((i*S)/K) ]
01, Sep 21
Sum of minimum value of x and y satisfying the equation ax + by = c
20, Jun 20
Check if elements of an array can be arranged satisfying the given condition
11, Dec 18
Maximum length of same indexed subarrays from two given arrays satisfying the given condition
09, Dec 20
Maximize occupied cells in given Matrix satisfying the conditions
11, Jan 22
Find x and y satisfying ax + by = n
16, Dec 17
Count of numbers of length N having prime numbers at odd indices and odd numbers at even indices
25, Aug 21
Count of binary strings of length N having equal count of 0's and 1's and count of 1's ≥ count of 0's in each prefix substring
25, Jul 20
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
GeeksforGeeks
Vote for difficulty
Current difficulty :
Hard





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!