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Composite numbers with digit sum 1
  • Last Updated : 26 Jul, 2019

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++




// 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;
}
  
// Function 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;
}


Java




// 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;
    }
  
    // Function 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);
    }
}


Python3




# 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
  
# Function 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)


C#




// 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;
    }
  
    // Function 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


PHP




<?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; 
  
// Function 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
?>


Output:

10 28 46 55 64 82 91 100

Optimizations : We can precompute composite numbers using Sieve Algorithms.

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