Prime Numbers

What are prime numbers?

• 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…..

Prime numbers

• In other words, the prime number is a positive integer greater than 1 that has exactly two factors, 1 and the number itself.
• There are many prime numbers, such as 2, 3, 5, 7, 11, 13, etc. 
• Keep in mind that 1 cannot be either prime or composite. 
• The remaining numbers, except for 1, are classified as prime and composite numbers. 

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Some interesting facts about Prime numbers:

• Except for 2, which is the smallest prime number and the only even prime number, all prime numbers are odd numbers.
• Every prime number can be represented in form of 6n + 1 or 6n – 1 except the prime numbers 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

• Fermat’s Little Theorem: If n is a prime number, then for every a, 1 <= a < n,

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.

Properties of prime numbers:

• Every number greater than 1 can be divided by at least one prime number.
• Every even positive integer greater than 2 can be expressed as the sum of two primes.
• Except 2, all other prime numbers are odd. In other words, we can say that 2 is the only even prime number.
• Two prime numbers are always coprime to each other.
• Each composite number can be factored into prime factors and individually all of these are unique in nature.

Prime numbers and co-prime numbers:

It is important to distinguish between prime numbers and co-prime numbers. Listed below are the differences between prime and co-prime numbers.

• A coprime number is always considered as a pair, whereas a prime number is considered as a single number.
• Co-prime numbers are numbers that have no common factor except 1. In contrast, prime numbers do not have such a condition.
• A co-prime number can be either prime or composite, but its greatest common factor (GCF) must always be 1. Unlike composite numbers, prime numbers have only two factors, 1 and the number itself.
• Example of co-prime: 13 and 15 are co-primes. The factors of 13 are 1 and 13 and the factors of 15 are 1, 3 and 5. We can see that they have only 1 as their common factor, therefore, they are coprime numbers.
• Example of prime: A few examples of prime numbers are 2, 3, 5, 7 and 11 etc.

How do we check whether a number is Prime or not?

Naive Approach: 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.

Below is the implementation:

C++14

// A school method based C++ program to
// check if a number is prime
#include <bits/stdc++.h>
using namespace std;

// function check whether a number
// is prime or not
bool 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 Code
int main()
{
    isPrime(11) ? cout << " true\n" : cout << " false\n";
    return 0;
}

Java

// A school method based Java program to
// check if a number is prime
import java.lang.*;
import java.util.*;

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 module
from 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 Code
if 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 prime
using 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 sqrt(n)
        for (int i = 2; i < Math.Sqrt(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

<?php
// A school method based PHP program to
// check if a number is prime

// function check whether a number
// is prime or not
function isPrime($n)
{
    // Corner case
    if ($n <= 1)
        return false;

    // Check from 2 to n-1
    for ($i = 2; $i < $n; $i++)
        if ($n % $i == 0)
            return false;

    return true;
}

// Driver Code
if(isPrime(11))
    echo("true");
else
    echo("false");

// This code is contributed by Ajit.
?>

Javascript

// A school method based Javascript program to
// check if a number is prime


// function check whether a number
// is prime or not
function isPrime(n)
{
    // Corner case
    if (n <= 1)
        return false;

    // Check from 2 to n-1
    for (let i = 2; i < n; i++)
        if (n % i == 0)
            return false;

    return true;
}

// Driver Code

    isPrime(11) ? console.log(" true" + "<br>") : console.log(" false" + "<br>");

// This code is contributed by Mayank Tyagi

Output

 true

Time Complexity: O(sqrt(n))
Auxiliary space: O(1)

Efficient approach: To check whether  the number is prime or not follow the below idea:

In the previous approach given if the size of the given number is too large then its square root will be also very large, so to deal with large size input we will deal with a few numbers such as 1, 2, 3, and the numbers which are divisible by 2 and 3 in separate cases and for remaining numbers, we will iterate our loop from 5 to sqrt(n) and check for each iteration whether that  (iteration) or (that iteration + 2) divides n or not. If we find any number that divides, we return false.

Below is the implementation for the above idea:

C++

// A school method based C++ program to
// check if a number is prime
#include <bits/stdc++.h>
using namespace std;

// function check whether a number
// is prime or not
bool isPrime(int n)
{
    // Check if n=1 or n=0
    if (n <= 1)
        return false;
    // Check if n=2 or n=3
    if (n == 2 || n == 3)
        return true;
    // Check whether n is divisible by 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return false;
    // Check from 5 to square root of n
    // Iterate i by (i+6)
    for (int i = 5; i <= sqrt(n); i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

// Driver Code
int main()
{
    isPrime(11) ? cout << "true\n" : cout << "false\n";
    return 0;
}
//  This code is contributed by Suruchi kumari

C

// A school method based C program to
// check if a number is prime
#include <math.h>
#include <stdio.h>
// function check whether a number
// is prime or not
int isPrime(int n)
{
    // Check if n=1 or n=0
    if (n <= 1)
        return 0;
    // Check if n=2 or n=3
    if (n == 2 || n == 3)
        return 1;
    // Check whether n is divisible by 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return 0;
    // Check from 5 to square root of n
    // Iterate i by (i+6)
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return 0;

    return 1;
}

// Driver Code
int main()
{
    if (isPrime(11) == 1)
        printf("true\n");
    else
        printf("false\n");
    return 0;
}
// This code is contributed by Suruchi Kumari

Java

// Java program to check whether a number
import java.lang.*;
import java.util.*;

class GFG {

    // Function check whether a number
    // is prime or not
    public static boolean isPrime(int n)
    {
        if (n <= 1)
            return false;

        // Check if n=2 or n=3
        if (n == 2 || n == 3)
            return true;

        // Check whether n is divisible by 2 or 3
        if (n % 2 == 0 || n % 3 == 0)
            return false;

        // Check from 5 to square root of n
        // Iterate i by (i+6)
        for (int i = 5; i <= Math.sqrt(n); i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;

        return true;
    }

    // Driver Code
    public static void main(String[] args)
    {
        if (isPrime(11)) {
            System.out.println("true");
        }
        else {
            System.out.println("false");
        }
    }
}

// This code is contributed by Sayan Chatterjee

Python3

import math

def is_prime(n: int) -> bool:
    # Check if n=1 or n=0
    if n <= 1:
        return False
    # Check if n=2 or n=3
    if n == 2 or n == 3:
        return True
    # Check whether n is divisible by 2 or 3
    if n % 2 == 0 or n % 3 == 0:
        return False
    # Check from 5 to square root of n
    # Iterate i by (i+6)
    for i in range(5, int(math.sqrt(n))+1, 6):
        if n % i == 0 or n % (i + 2) == 0:
            return False

    return True

print(is_prime(11))

C#

// C# program to check whether a number
using System;
class GFG {

    // Function check whether a number
    // is prime or not
    public static bool isPrime(int n)
    {
        if (n <= 1)
            return false;

        // Check if n=2 or n=3
        if (n == 2 || n == 3)
            return true;

        // Check whether n is divisible by 2 or 3
        if (n % 2 == 0 || n % 3 == 0)
            return false;

        // Check from 5 to square root of n
        // Iterate i by (i+6)
        for (int i = 5; i <= Math.Sqrt(n); i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;

        return true;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        if (isPrime(11)) {
            Console.WriteLine("true");
        }
        else {
            Console.WriteLine("false");
        }
    }
}

// This code is contributed by Abhijeet
// Kumar(abhijeet_19403)

Javascript

// A school method based JS program to
// check if a number is prime


// function check whether a number
// is prime or not
function isPrime(n)
{
    // Check if n=1 or n=0
    if (n <= 1)
        return false;
    // Check if n=2 or n=3
    if (n == 2 || n == 3)
        return true;
    // Check whether n is divisible by 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return false;
    // Check from 5 to square root of n
    // Iterate i by (i+6)
    for (var i = 5; i <= Math.sqrt(n); i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

// Driver Code
isPrime(11) ? console.log("true") : console.log("false");


//  This code is contributed by phasing17

Output

true

Time complexity: O(sqrt(n))
Auxiliary space: O(1)

Approach 3: To check the number is prime or not using recursion follow the below idea:

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.

Below is the implementation for the below idea:

C++

// C++ program to check whether a number
// is prime or not using recursion
#include <iostream>
using namespace std;

// function check whether a number
// is prime or not
bool 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 Code
int 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 recursion
import java.io.*;

class GFG {

    static int i = 2;

    // Function check whether a number
    // is prime or not
    public 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 not


def 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 Code
if (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 recursion
using 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

<script>
      // JavaScript program to check whether a number
      // is prime or not using recursion

      // function check whether a number
      // is prime or not
      var i = 2;

      function isPrime(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

      isPrime(35) ? document.write(" true\n") : document.write(" false\n");

      // This code is contributed by rdtank.
    </script>

Output

 false

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

Approach 4: To check the number is prime or not using Fermat’s little theorem with out using loop

Python3

# function defination
def isprime(n):
    # 2 and 1 will not work for fermat's little theorem
    if n == 2 and n == 1:
        print('true')
    else:
        # formula for cheacking prime or not
        p = (2**n-1) % n
        if p == 1:
            print('true')
        else:
            print('false')


# function call
isprime(4)
isprime(7)
isprime(2)

Java

import java.io.*;

class GFG {
    public static boolean isPrime(int n)
    {
          // 0,1 and 2 will not work for fermat's little theorem
        // Corner cases
        if (n == 0 || n == 1) {
            return false;
        }
        if ( n == 2) {
            return true;
        }
        // Checking Prime
        else
        {
            int p = (int)(Math.pow(2, n-1))%n;
            if (p==1)
                return true;
            else
                return false;
        }
    }
    //Driver Code
    public static void main(String[] args)
    {
        if (isPrime(35)) {
            System.out.println("Prime");
        }
        else {
            System.out.println("Not Prime");
        }
    }
}

//contributed by raj898rki

Output

false
true
true

Time complexity: O(1)
Auxiliary space: O(1)

Efficient solutions

• Primality Test | Set 1 (Introduction and School Method)
• Primality Test | Set 2 (Fermat Method)
• Primality Test | Set 3 (Miller–Rabin)
• Primality Test | Set 4 (Solovay-Strassen)
• Lucas Primality Test

Algorithms to find all prime numbers smaller than the N. 

• Sieve of Eratosthenes
• Sieve of Eratosthenes in 0(n) time complexity
• Segmented Sieve
• Sieve of Sundaram
• Bitwise Sieve
• Recent Articles on Sieve!

More problems related to Prime number 

• Find two distinct prime numbers with a given product
• Print all prime numbers less than or equal to N
• Recursive program for prime number
• Find two prime numbers with a given sum
• Find the highest occurring digit in prime numbers in a range
• Prime Factorization using Sieve O(log n) for multiple queries
• Program to print all prime factors of a given number
• Least prime factor of numbers till n
• Prime factors of LCM of array elements – GeeksforGeeks
• Program for Goldbach’s Conjecture
• Prime numbers and Fibonacci
• Composite Number
• Recent Articles on Prime Numbers!