Composite numbers with digit sum 1

Given a range [L, R], the task is to find all the numbers from the range which are composite as well as the eventual sum of their digits is 1.

Examples:

Input: L = 10, R = 100
Output: 10 28 46 55 64 82 91 100
10 = 1 + 0 = 1
28 = 2 + 8 = 10 = 1 + 0 = 1

91 = 9 + 1 = 10 = 1 + 0 = 1
100 = 1 + 0 + 0 = 1

Input: L = 250, R = 350
Output: 253 262 280 289 298 316 325 334 343

Approach: For every number in the range check if the number is composite i.e. it has a divisor other than 1 and the number itself. If the current number is a composite number then keep on calculating the sum of its digits until the number is reduced to a single digit, if this digit is 1 then the chosen number is a valid number.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the above approach
#include <iostream>
using namespace std;
  
// Function that returns true if number n
// is a composite number
bool isComposite(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return false;
  
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return true;
  
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return true;
  
    return false;
}
  
// Functin that returns true if the eventual
// digit sum of number nm is 1
bool isDigitSumOne(int nm)
{
  
    // Loop till the sum is not single digit number
    while (nm > 9) {
  
        // Intitialize the sum as zero
        int sum_digit = 0;
  
        // Find the sum of digits
        while (nm > 0) {
            int digit = nm % 10;
            sum_digit = sum_digit + digit;
            nm = nm / 10;
        }
        nm = sum_digit;
    }
  
    // If sum is eventually 1
    if (nm == 1)
        return true;
    else
        return false;
}
  
// Function to print the required numbers
// from the given range
void printValidNums(int l, int r)
{
    for (int i = l; i <= r; i++) {
  
        // If i is one of the required numbers
        if (isComposite(i) && isDigitSumOne(i))
            cout << i << " ";
    }
}
  
// Driver code
int main(void)
{
    int l = 10, r = 100;
  
    printValidNums(l, r);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the above approach
public class GFG {
  
    // Function that returns true if number n
    // is a composite number
    static boolean isComposite(int n)
    {
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return false;
  
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return true;
  
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return true;
  
        return false;
    }
  
    // Functin that returns true if the eventual
    // digit sum of number nm is 1
    static boolean isDigitSumOne(int nm)
    {
  
        // Loop till the sum is not single 
        // digit number
        while (nm > 9) {
  
            // Intitialize the sum as zero
            int sum_digit = 0;
  
            // Find the sum of digits
            while (nm > 0) {
                int digit = nm % 10;
                sum_digit = sum_digit + digit;
                nm = nm / 10;
            }
            nm = sum_digit;
        }
  
        // If sum is eventually 1
        if (nm == 1)
            return true;
        else
            return false;
    }
  
    // Function to print the required numbers
    // from the given range
    static void printValidNums(int l, int r)
    {
        for (int i = l; i <= r; i++) {
  
            // If i is one of the required numbers
            if (isComposite(i) && isDigitSumOne(i))
                System.out.print(i + " ");
        }
    }
  
    // Driver code
    public static void main(String arg[])
    {
        int l = 10, r = 100;
        printValidNums(l, r);
    }
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
  
# Function that returns true if number n 
# is a composite number
def isComposite(n): 
    
    # Corner cases 
    if (n <= 1): 
        return False
    if (n <= 3): 
        return False
    
    # This is checked so that we can skip  
    # middle five numbers in below loop 
    if (n % 2 == 0 or n % 3 == 0): 
        return True
    i = 5
    while(i * i <= n): 
            
        if (n % i == 0 or n % (i + 2) == 0): 
            return True
        i = i + 6
            
    return False
  
# Functin that returns true if the eventual 
# digit sum of number nm is 1
def isDigitSumOne(nm) :
      
    # Loop till the sum is not single
    # digit number
    while(nm>9) :
          
        # Intitialize the sum as zero
        sum_digit = 0
          
        # Find the sum of digits
        while(nm != 0) :
            digit = nm % 10
            sum_digit = sum_digit + digit
            nm = nm // 10
        nm = sum_digit
      
    # If sum is eventually 1
    if(nm == 1):
        return True
    else:
        return False
          
# Function to print the required numbers 
# from the given range
def printValidNums(m, n ):
    for i in range(m, n + 1):
          
        # If i is one of the required numbers
        if(isComposite(i) and isDigitSumOne(i)) :
            print(i, end =" ")
  
# Driver code
l = 10
r = 100
printValidNums(m, n)

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the above approach
using System;
  
class GFG
{
      
    // Function that returns true if number n
    // is a composite number
    static bool isComposite(int n)
    {
          
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return false;
  
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return true;
  
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return true;
  
        return false;
    }
  
    // Functin that returns true if the 
    // eventual digit sum of number nm is 1
    static bool isDigitSumOne(int nm)
    {
  
        // Loop till the sum is not single 
        // digit number
        while (nm > 9)
        {
  
            // Intitialize the sum as zero
            int sum_digit = 0;
  
            // Find the sum of digits
            while (nm > 0)
            {
                int digit = nm % 10;
                sum_digit = sum_digit + digit;
                nm = nm / 10;
            }
            nm = sum_digit;
        }
  
        // If sum is eventually 1
        if (nm == 1)
            return true;
        else
            return false;
    }
  
    // Function to print the required numbers
    // from the given range
    static void printValidNums(int l, int r)
    {
        for (int i = l; i <= r; i++)
        {
  
            // If i is one of the required numbers
            if (isComposite(i) && isDigitSumOne(i))
                Console.Write(i + " ");
        }
    }
  
    // Driver code
    static public void Main ()
    {
        int l = 10, r = 100;
        printValidNums(l, r);
    }
}
  
// This code is contributed by jit_t

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the above approach 
  
// Function that returns true if number n 
// is a composite number 
function isComposite($n
    // Corner cases 
    if ($n <= 1) 
        return false; 
    if ($n <= 3) 
        return false; 
  
    // This is checked so that we can skip 
    // middle five numbers in below loop 
    if ($n % 2 == 0 || $n % 3 == 0) 
        return true; 
  
    for ($i = 5; $i * $i <= $n; $i = $i + 6) 
        if ($n % $i == 0 || $n % ($i + 2) == 0) 
            return true; 
  
    return false; 
  
// Functin that returns true if the eventual 
// digit sum of number nm is 1 
function isDigitSumOne($nm
  
    // Loop till the sum is not single 
    // digit number 
    while ($nm > 9) 
    
  
        // Intitialize the sum as zero 
        $sum_digit = 0; 
  
        // Find the sum of digits 
        while ($nm > 0) 
        
            $digit = $nm % 10; 
            $sum_digit = $sum_digit + $digit
            $nm = floor($nm / 10); 
        
        $nm = $sum_digit
    
  
    // If sum is eventually 1 
    if ($nm == 1) 
        return true; 
    else
        return false; 
  
// Function to print the required numbers 
// from the given range 
function printValidNums($l, $r
    for ($i = $l; $i <= $r; $i++)
    
  
        // If i is one of the required numbers 
        if (isComposite($i) && isDigitSumOne($i)) 
            echo $i, " "
    
  
// Driver code 
$l = 10; $r = 100; 
  
printValidNums($l, $r); 
  
// This code is contributed by Ryuga
?>

chevron_right


Output:

10 28 46 55 64 82 91 100

Optimizations : We can precompute composite numbers using Sieve Algorithms.



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : jit_t, AnkitRai01



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.