# 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``; ` `} ` ` `  `// 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; ` `} `

## 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``; ` `    ``} ` ` `  `    ``// 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); ` `    ``} ` `} `

## 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` ` `  `# 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) `

## 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``; ` `    ``} ` ` `  `    ``// 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 `

## 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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