# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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

 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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