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# Prime Numbers

• Difficulty Level : Easy

A prime number is a natural number greater than 1, which is only divisible by 1 and itself. First few prime numbers are : 2 3 5 7 11 13 17 19 23 ….. Some interesting fact about Prime numbers

• Two is the only even Prime number.
• Every prime number can be represented in form of 6n+1 or 6n-1 except the prime number 2 and 3, where n is a natural number.
• Two and Three are only two consecutive natural numbers that are prime.
• Goldbach Conjecture: Every even integer greater than 2 can be expressed as the sum of two primes.
• Wilson Theorem: Wilson’s theorem states that a natural number p > 1 is a prime number if and only if
(p - 1) ! ≡  -1   mod p
OR  (p - 1) ! ≡  (p-1) mod p
an-1 ≡ 1 (mod n)
OR
an-1 % n = 1
• Prime Number Theorem: The probability that a given, randomly chosen number n is prime is inversely proportional to its number of digits, or to the logarithm of n.
• Lemoine’s Conjecture: Any odd integer greater than 5 can be expressed as a sum of an odd prime (all primes other than 2 are odd) and an even semiprime. A semiprime number is a product of two prime numbers. This is called Lemoine’s conjecture.

How we check whether a number is Prime or not?

• Naive solution
A naive solution is to iterate through all numbers from 2 to sqrt(n) and for every number check if it divides n. If we find any number that divides, we return false.

## C++

 // A school method based C++ program to// check if a number is prime#include using namespace std; // function check whether a number// is prime or notbool isPrime(int n){    // Corner case    if (n <= 1)        return false;     // Check from 2 to square root of n    for (int i = 2; i <= sqrt(n); i++)        if (n % i == 0)            return false;     return true;} // Driver Codeint main(){    isPrime(11) ? cout << " true\n" : cout << " false\n";    return 0;}

## Java

 // A school method based Java program to// check if a number is primeimport java.util.*;import java.lang.*; class GFG {     // Check for number prime or not    static boolean isPrime(int n)    {         // Check if number is less than        // equal to 1        if (n <= 1)            return false;         // Check if number is 2        else if (n == 2)            return true;         // Check if n is a multiple of 2        else if (n % 2 == 0)            return false;         // If not, then just check the odds        for (int i = 3; i <= Math.sqrt(n); i += 2)        {            if (n % i == 0)                return false;        }        return true;    }     // Driver code    public static void main(String[] args)    {        if (isPrime(19))            System.out.println("true");         else            System.out.println("false");    }} // This code is contributed by Ronak Bhensdadia

## Python3

 # A school method based Python3 program# to check if a number is prime # function check whether a number# is prime or not #import sqrt from math modulefrom math import sqrt def isPrime(n):     # Corner case    if (n <= 1):        return False     # Check from 2 to sqrt(n)    for i in range(2, int(sqrt(n))+1):        if (n % i == 0):            return False     return True  # Driver Codeif isPrime(11):    print("true")else:    print("false") # This code is contributed by Sachin Bisht

## C#

 // A school method based C# program to// check if a number is primeusing System; class GFG {    // function check whether a    // number is prime or not    static bool isPrime(int n)    {        // Corner case        if (n <= 1)            return false;         // Check from 2 to n-1        for (int i = 2; i < n; i++)            if (n % i == 0)                return false;         return true;    }     // Driver Code    static void Main()    {        if (isPrime(11))            Console.Write(" true");         else            Console.Write(" false");    }} // This code is contributed by Sam007

## PHP

 

## Javascript

 
Output

 true

Time Complexity: O( Using Recursion :

Recursion can also be used to check if a number between 2 to n – 1 divides n. If we find any number that divides, we return false.

## C++

 // C++ program to check whether a number// is prime or not using recursion#include using namespace std; // function check whether a number// is prime or notbool isPrime(int n){    static int i = 2;     // corner cases    if (n == 0 || n == 1) {        return false;    }     // Checking Prime    if (n == i)        return true;     // base cases    if (n % i == 0) {        return false;    }    i++;    return isPrime(n);} // Driver Codeint main(){     isPrime(35) ? cout << " true\n" : cout << " false\n";    return 0;} // This code is contributed by yashbeersingh42

## Java

 // Java program to check whether a number// is prime or not using recursionclass GFG{     static int i = 2;  // Function check whether a number// is prime or notpublic static boolean isPrime(int n){         // Corner cases    if (n == 0 || n == 1)    {        return false;    }       // Checking Prime    if (n == i)        return true;             // Base cases    if (n % i == 0)    {        return false;    }    i++;    return isPrime(n);} // Driver Code  public static void main(String[] args){    if (isPrime(35))    {        System.out.println("true");    }    else    {        System.out.println("false");    }}} // This code is contributed by divyeshrabadiya07

## Python3

 # Python3 program to check whether a number# is prime or not using recursion # Function check whether a number# is prime or notdef isPrime(n, i):         # Corner cases    if (n == 0 or n == 1):        return False         # Checking Prime     if (n == i):        return True         # Base cases    if (n % i == 0):        return False         i += 1         return isPrime(n, i) # Driver Codeif (isPrime(35, 2)):  print("true")else:  print("false")   #  This code is contributed by bunnyram19

## C#

 // C# program to check whether a number// is prime or not using recursionusing System;class GFG {         static int i = 2;         // function check whether a number    // is prime or not    static bool isPrime(int n)    {              // corner cases        if (n == 0 || n == 1) {            return false;        }              // Checking Prime        if (n == i)            return true;              // base cases        if (n % i == 0) {            return false;        }        i++;        return isPrime(n);    }     static void Main() {    if(isPrime(35))    {        Console.WriteLine("true");    }    else{        Console.WriteLine("false");    }  }} // This code is contributed by divyesh072019

## Javascript

 
Output
 false

Time Complexity :O(N), Space Complexity :O(N)

• Efficient solutions

Algorithms to find all prime numbers smaller than the N.

More problems related to Prime number

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