Skip to content
Related Articles

Related Articles

Improve Article

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

  • Difficulty Level : Medium
  • Last Updated : 27 Apr, 2021
Geek Week

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 
 

 

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 a number of times will give an even sum.
  2. When N is odd then we can print 2 and any other N – 1 prime 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 <bits/stdc++.h>
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<Integer> v = new Vector<Integer>();
 
// 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 */

Javascript




<script>
 
// Javascript implementation
// of the approach
 
var MAXN = 100000;
 
// To store prime numbers
var v = [];
 
// Function to find and store
// all the primes <= n
function 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.
    var prime = Array(n + 1).fill(true);
    var p;
    for (p = 2; p * p <= n; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true) {
             var i;
            // 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 = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
 
    // Store all the prime numbers
    for (p = 2; p <= n; p++)
        if (prime[p])
            v.push(p);
}
 
// Function to print
// the required sequence
function 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
    var i = (n % 2 == 0) ? 1 : 0;
 
    var cnt = 0;
    // Print the sequence
    while (cnt < n) {
        document.write(v[i] + " ");
        i++;
        cnt++;
    }
}
 
// Driver code
    n = 6;
    SieveOfEratosthenes(MAXN);
    GetSequence(n);
 
</script>
Output: 
3 5 7 11 13 17

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :