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# Find a range of composite numbers of given length

Given an integer n, we need to find a range of positive integers such that all the number in that range are composite and length of that range is n. You may print anyone range in the case of more than one answer. A composite number is a positive integer that has at least one divisor other than 1 and itself (Source : wiki

Examples :

```Input : 3
Output : [122, 124]
Explanation 122, 123, 124 are all composite numbers```

The solution is little tricky. Since there are many possible answers, we discuss a generalized solution here.

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
.... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```

Example for above algorithm

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```

## C++

 `// C++ program to find a range of``// composite numbers of given length``#include ``using` `namespace` `std;` `// method to find factorial``// of given number``int` `factorial (``int` `n)``{``    ``if` `(n == 0)``        ``return` `1;` `    ``return` `n * factorial(n-1);``}` `// to print range of length n``// having all composite integers``int` `printRange(``int` `n)``{``int` `a = factorial(n + 2) + 2;``int` `b = a + n - 1;``cout << ``"["` `<< a << ``", "` `<< b << ``"]"``;``return` `0;``}` `// Driver method``int` `main()``{``    ``int` `n = 3 ;``    ``printRange(n);``    ``return` `0;``}` `// This code is contributed by Anshika Goyal`

## Java

 `// Java program to find a range of composite``// numbers of given length` `class` `Test``{``    ``// method to find factorial of given number``    ``static` `int` `factorial(``int` `n)``    ``{``        ``if` `(n == ``0``)``          ``return` `1``;``         ` `        ``return` `n*factorial(n-``1``);``    ``}``    ` `    ``// to print range of length n``    ``//  having all composite integers``    ``static` `void` `printRange(``int` `n)``    ``{``       ``int`  `a = factorial(n + ``2``) + ``2``;``       ``int`  `b = a + n - ``1``;``       ``System.out.println(``"["` `+ a + ``", "` `+ b + ``"]"``);``    ``}``    ` `    ``// Driver method``    ``public` `static` `void` `main(String args[]) ``throws` `Exception``    ``{``        ``int` `n = ``3` `;``        ``printRange(n);``    ``}``}`

## Python3

 `# Python program to find a range of composite``# numbers of given length` `# function to calculate factorial``def` `factorial(n):``    ``a ``=` `1``    ``for` `i ``in` `range``(``2``, n ``+` `1``):``        ``a ``*``=` `i``    ``return` `a` `# to print range of length n``# having all composite integers``def` `printRange(n):``    ``a ``=` `factorial(n ``+` `2``) ``+` `2``    ``b ``=` `a ``+` `n ``-` `1``    ``print``(``"["``+``str``(a)``+``", "``+``str``(b)``+``"]"``)` `# driver code to test above functions``n ``=` `3``printRange(n)`

## C#

 `// C# program to find a range of``// composite numbers of given``// length``using` `System;` `class` `GFG {``    ` `    ``// method to find factorial``    ``// of given number``    ``static` `int` `factorial(``int` `n)``    ``{``        ``if` `(n == 0)``        ``return` `1;``        ` `        ``return` `n*factorial(n-1);``    ``}``    ` `    ``// to print range of length n``    ``// having all composite integers``    ``static` `void` `printRange(``int` `n)``    ``{``    ``int` `a = factorial(n + 2) + 2;``    ``int` `b = a + n - 1;``    ``Console.WriteLine(``"["` `+ a +``                   ``", "` `+ b + ``"]"``);``    ``}``    ` `    ``// Driver method``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 3 ;``        ``printRange(n);``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

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## Javascript

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Output :

`[122, 124]`

Analysis of above algorithm
Time Complexity : O(n)
Auxiliary Space : O(n)

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