# Find a sequence of N prime numbers whose sum is a composite number

Given an integer N and the task is to find a sequence of N prime numbers whose sum is a composite number.

Examples:

Input: N = 5
Output: 2 3 5 7 11
2 + 3 + 5 + 7 + 11 = 28 which is composite.

Input: N = 6
Output: 3 5 7 11 13 17

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

Approach: The sum of two prime numbers is always even which is composite as they are odd numbers except 2. There are two cases now,

1. When N is even then we can print any N prime numbers except 2 and their sum will always be even i.e. odd numbers when added even number of times will give an even sum.
2. When N is odd then we can print 2 and any other N – 1 primes to make sure that the sum is even. Since, N – 1 is even so the sum will be even for any primes except 2 then we add 2 as the Nth number to make sure that the sum remains even.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define MAXN 100000 ` ` `  `// To store prime numbers ` `vector<``int``> v; ` ` `  `// Function to find and store ` `// all the primes <= n ` `void` `SieveOfEratosthenes(``int` `n) ` `{ ` `    ``// Create a boolean array "prime[0..n]" and initialize ` `    ``// all entries it as true. A value in prime[i] will ` `    ``// finally be false if i is Not a prime, else true. ` `    ``bool` `prime[n + 1]; ` `    ``memset``(prime, ``true``, ``sizeof``(prime)); ` ` `  `    ``for` `(``int` `p = 2; p * p <= n; p++) { ` ` `  `        ``// If prime[p] is not changed, then it is a prime ` `        ``if` `(prime[p] == ``true``) { ` ` `  `            ``// Update all multiples of p greater than or ` `            ``// equal to the square of it ` `            ``// numbers which are multiple of p and are ` `            ``// less than p^2 are already been marked. ` `            ``for` `(``int` `i = p * p; i <= n; i += p) ` `                ``prime[i] = ``false``; ` `        ``} ` `    ``} ` ` `  `    ``// Store all the prime numbers ` `    ``for` `(``int` `p = 2; p <= n; p++) ` `        ``if` `(prime[p]) ` `            ``v.push_back(p); ` `} ` ` `  `// Function to print the required sequence ` `void` `GetSequence(``int` `n) ` `{ ` ` `  `    ``// If n is even then we do not include 2 ` `    ``// and start the sequence from the 2nd ` `    ``// smallest prime else we include 2 ` `    ``int` `i = (n % 2 == 0) ? 1 : 0; ` ` `  `    ``int` `cnt = 0; ` `    ``// Print the sequence ` `    ``while` `(cnt < n) { ` `        ``cout << v[i] << ``" "``; ` `        ``i++; ` `        ``cnt++; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 6; ` `    ``SieveOfEratosthenes(MAXN); ` ` `  `    ``GetSequence(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `     `  `static` `int` `MAXN = ``100000``; ` ` `  `// To store prime numbers ` `static` `Vector v = ``new` `Vector(); ` ` `  `// Function to find and store ` `// all the primes <= n ` `static` `void` `SieveOfEratosthenes(``int` `n) ` `{ ` `    ``// Create a boolean array "prime[0..n]" and initialize ` `    ``// all entries it as true. A value in prime[i] will ` `    ``// finally be false if i is Not a prime, else true. ` `    ``boolean``[] prime = ``new` `boolean``[n + ``1``]; ` `    ``Arrays.fill(prime,``true``); ` ` `  `    ``for` `(``int` `p = ``2``; p * p <= n; p++) ` `    ``{ ` ` `  `        ``// If prime[p] is not changed, then it is a prime ` `        ``if` `(prime[p] == ``true``)  ` `        ``{ ` ` `  `            ``// Update all multiples of p greater than or ` `            ``// equal to the square of it ` `            ``// numbers which are multiple of p and are ` `            ``// less than p^2 are already been marked. ` `            ``for` `(``int` `i = p * p; i <= n; i += p) ` `                ``prime[i] = ``false``; ` `        ``} ` `    ``} ` ` `  `    ``// Store all the prime numbers ` `    ``for` `(``int` `p = ``2``; p <= n; p++) ` `        ``if` `(prime[p]) ` `            ``v.add(p); ` `} ` ` `  `// Function to print the required sequence ` `static` `void` `GetSequence(``int` `n) ` `{ ` ` `  `    ``// If n is even then we do not include 2 ` `    ``// and start the sequence from the 2nd ` `    ``// smallest prime else we include 2 ` `    ``int` `i = (n % ``2` `== ``0``) ? ``1` `: ``0``; ` ` `  `    ``int` `cnt = ``0``; ` `     `  `    ``// Print the sequence ` `    ``while` `(cnt < n) ` `    ``{ ` `        ``System.out.print(v.get(i) + ``" "``); ` `        ``i++; ` `        ``cnt++; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``6``; ` `    ``SieveOfEratosthenes(MAXN); ` ` `  `    ``GetSequence(n); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

## python

 `# Python3 implementation of the approach ` ` `  `MAXN``=``100000` ` `  `# To store prime numbers ` `v``=``[] ` ` `  `# Function to find and store ` `# all the primes <= n ` `def` `SieveOfEratosthenes(n): ` ` `  `    ``# Create a boolean array "prime[0..n]" and initialize ` `    ``# all entries it as true. A value in prime[i] will ` `    ``# finally be false if i is Not a prime, else true. ` `    ``prime``=``[``True` `for` `i ``in` `range``(n ``+` `1``)] ` ` `  `    ``for` `p ``in` `range``(``2``,n``+``1``): ` ` `  `        ``# If prime[p] is not changed, then it is a prime ` `        ``if` `(prime[p] ``=``=` `True``): ` ` `  `            ``# Update all multiples of p greater than or ` `            ``# equal to the square of it ` `            ``# numbers which are multiple of p and are ` `            ``# less than p^2 are already been marked. ` `            ``for` `i ``in` `range``(``2` `*` `p, n ``+` `1``, p): ` `                ``prime[i] ``=` `False` ` `  `    ``# Store all the prime numbers ` `    ``for` `p ``in` `range``(``2``, n ``+` `1``): ` `        ``if` `(prime[p]): ` `            ``v.append(p) ` ` `  `# Function to print the required sequence ` `def` `GetSequence(n): ` ` `  `    ``# If n is even then we do not include 2 ` `    ``# and start the sequence from the 2nd ` `    ``# smallest prime else we include 2 ` `    ``if` `n ``%` `2` `=``=` `0``: ` `        ``i ``=` `1` `    ``else``: ` `        ``i ``=` `0` ` `  `    ``cnt ``=` `0` `    ``# Print the sequence ` `    ``while` `(cnt < n): ` `        ``print``(v[i],end``=``" "``) ` `        ``i ``+``=` `1` `        ``cnt ``+``=` `1` ` `  ` `  `# Driver code ` `n ``=` `6` `SieveOfEratosthenes(MAXN) ` ` `  `GetSequence(n) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{  ` `     `  `static` `int` `MAXN = 100000;  ` ` `  `// To store prime numbers  ` `static` `List<``int``> v = ``new` `List<``int``>();  ` ` `  `// Function to find and store  ` `// all the primes <= n  ` `static` `void` `SieveOfEratosthenes(``int` `n)  ` `{  ` `    ``// Create a boolean array "prime[0..n]" and initialize  ` `    ``// all entries it as true. A value in prime[i] will  ` `    ``// finally be false if i is Not a prime, else true.  ` `    ``Boolean[] prime = ``new` `Boolean[n + 1];  ` `    ``for``(``int` `i = 0; i < n + 1; i++) ` `        ``prime[i] = ``true``; ` ` `  `    ``for` `(``int` `p = 2; p * p <= n; p++)  ` `    ``{  ` ` `  `        ``// If prime[p] is not changed, then it is a prime  ` `        ``if` `(prime[p] == ``true``)  ` `        ``{  ` ` `  `            ``// Update all multiples of p greater than or  ` `            ``// equal to the square of it  ` `            ``// numbers which are multiple of p and are  ` `            ``// less than p^2 are already been marked.  ` `            ``for` `(``int` `i = p * p; i <= n; i += p)  ` `                ``prime[i] = ``false``;  ` `        ``}  ` `    ``}  ` ` `  `    ``// Store all the prime numbers  ` `    ``for` `(``int` `p = 2; p <= n; p++)  ` `        ``if` `(prime[p])  ` `            ``v.Add(p);  ` `}  ` ` `  `// Function to print the required sequence  ` `static` `void` `GetSequence(``int` `n)  ` `{  ` ` `  `    ``// If n is even then we do not include 2  ` `    ``// and start the sequence from the 2nd  ` `    ``// smallest prime else we include 2  ` `    ``int` `i = (n % 2 == 0) ? 1 : 0;  ` ` `  `    ``int` `cnt = 0;  ` `     `  `    ``// Print the sequence  ` `    ``while` `(cnt < n)  ` `    ``{  ` `        ``Console.Write(v[i] + ``" "``);  ` `        ``i++;  ` `        ``cnt++;  ` `    ``}  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main(String[] args)  ` `{  ` `    ``int` `n = 6;  ` `    ``SieveOfEratosthenes(MAXN);  ` ` `  `    ``GetSequence(n);  ` `}  ` `}  ` ` `  `/* This code is contributed by PrinciRaj1992 */`

Output:

```3 5 7 11 13 17
```

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

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.

Article Tags :
Practice Tags :

1

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