# Find the Product of first N Prime Numbers

Given a positive integer N, calculate the product of the first N prime numbers.

Examples:

```Input : N = 3
Output : 30
Explanation : First 3 prime numbers are 2, 3, 5.

Input : N = 5
Output : 2310 ```

Approach:

• Create a sieve which will help us to identify if the number is prime or not in O(1) time.
• Run a loop starting from 1 until and unless we find n prime numbers.
• Multiply all the prime numbers and neglect those which are not prime.
• Then, display the product of 1st N prime numbers.

Time Complexity – O( Nlog(logN) )

Below is the implementation of above approach:

## C++

 `// C++ implementation of above solution``#include "cstring"``#include ``using` `namespace` `std;``#define MAX 10000` `// 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[MAX + 1];``void` `SieveOfEratosthenes()``{``    ``memset``(prime, ``true``, ``sizeof``(prime));` `    ``prime[1] = ``false``;` `    ``for` `(``int` `p = 2; p * p <= MAX; p++) {` `        ``// If prime[p] is not changed, then it is a prime``        ``if` `(prime[p] == ``true``) {` `            ``// Set all multiples of p to non-prime``            ``for` `(``int` `i = p * 2; i <= MAX; i += p)``                ``prime[i] = ``false``;``        ``}``    ``}``}` `// find the product of 1st N prime numbers``int` `solve(``int` `n)``{``    ``// count of prime numbers``    ``int` `count = 0, num = 1;` `    ``// product of prime numbers``    ``long` `long` `int` `prod = 1;` `    ``while` `(count < n) {` `        ``// if the number is prime add it``        ``if` `(prime[num]) {``            ``prod *= num;` `            ``// increase the count``            ``count++;``        ``}` `        ``// get to next number``        ``num++;``    ``}``    ``return` `prod;``}` `// Driver code``int` `main()``{``    ``// create the sieve``    ``SieveOfEratosthenes();` `    ``int` `n = 5;` `    ``// find the value of 1st n prime numbers``    ``cout << solve(n);` `    ``return` `0;``}`

## Java

 `// Java implementation of above solution` `class` `GFG``{``    ``static` `int` `MAX=``10000``;` `    ``// 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.``    ``static` `boolean``[] prime=``new` `boolean``[MAX + ``1``];``    ` `static` `void` `SieveOfEratosthenes()``{` `    ``prime[``1``] = ``true``;` `    ``for` `(``int` `p = ``2``; p * p <= MAX; p++) {` `        ``// If prime[p] is not changed, then it is a prime``        ``if` `(prime[p] == ``false``) {` `            ``// Set all multiples of p to non-prime``            ``for` `(``int` `i = p * ``2``; i <= MAX; i += p)``                ``prime[i] = ``true``;``        ``}``    ``}``}` `// find the product of 1st N prime numbers``static` `int` `solve(``int` `n)``{``    ``// count of prime numbers``    ``int` `count = ``0``, num = ``1``;` `    ``// product of prime numbers``    ``int` `prod = ``1``;` `    ``while` `(count < n) {` `        ``// if the number is prime add it``        ``if` `(!prime[num]) {``            ``prod *= num;` `            ``// increase the count``            ``count++;``        ``}` `        ``// get to next number``        ``num++;``    ``}``    ``return` `prod;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``// create the sieve``    ``SieveOfEratosthenes();` `    ``int` `n = ``5``;` `    ``// find the value of 1st n prime numbers``    ``System.out.println(solve(n));` `}``}``// This code is contributed by mits`

## C#

 `// C# implementation of above solution` `class` `GFG``{``    ``static` `int` `MAX=10000;` `    ``// 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.``    ``static` `bool``[] prime=``new` `bool``[MAX + 1];``    ` `static` `void` `SieveOfEratosthenes()``{` `    ``prime[1] = ``true``;` `    ``for` `(``int` `p = 2; p * p <= MAX; p++) {` `        ``// If prime[p] is not changed, then it is a prime``        ``if` `(prime[p] == ``false``) {` `            ``// Set all multiples of p to non-prime``            ``for` `(``int` `i = p * 2; i <= MAX; i += p)``                ``prime[i] = ``true``;``        ``}``    ``}``}` `// find the product of 1st N prime numbers``static` `int` `solve(``int` `n)``{``    ``// count of prime numbers``    ``int` `count = 0, num = 1;` `    ``// product of prime numbers``    ``int` `prod = 1;` `    ``while` `(count < n) {` `        ``// if the number is prime add it``        ``if` `(!prime[num]) {``            ``prod *= num;` `            ``// increase the count``            ``count++;``        ``}` `        ``// get to next number``        ``num++;``    ``}``    ``return` `prod;``}` `// Driver code``public` `static` `void` `Main()``{``    ``// create the sieve``    ``SieveOfEratosthenes();` `    ``int` `n = 5;` `    ``// find the value of 1st n prime numbers``    ``System.Console.WriteLine(solve(n));` `}``}``// This code is contributed by mits`

## Python

 `'''``python3 implementation of above solution'''``import` `math as mt` `MAX``=``10000` `'''``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``(``MAX``+``1``)]` `def` `SieveOfErastosthenes():``    ` `    ``prime[``1``]``=``False``    ` `    ``for` `p ``in` `range``(``2``,mt.ceil(mt.sqrt(``MAX``))):``        ``#if prime[p] is not changes, then it is a prime``        ` `        ``if` `prime[p]:``            ``#set all multiples of p to non-prime``            ``for` `i ``in` `range``(``2``*``p,``MAX``+``1``,p):``                ``prime[i]``=``False``                ` `#find the product of 1st N prime numbers` `def` `solve(n):``    ``#count of prime numbers``    ``count,num``=``0``,``1``    ` `    ``#product of prime numbers``    ` `    ``prod``=``1``    ``while` `count

## PHP

 ``

## Javascript

 ``

Output:
`2310`

Auxiliary Space: O(MAX)

NOTE: For larger values of N, the product may be give integer overflow errors.
Also for multiple queries, prefix array technique can be used which will give output of each query in O(1) after making the prefix array first which will take O(N) time.

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