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Prime Numbers

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What are Prime Numbers?

A prime number is defined as a natural number greater than 1 and is divisible by only 1 and itself. 

In other words, the prime number is a positive integer greater than 1 that has exactly two factors, 1 and the number itself. First few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 . . .

Note: 1 is not either prime or composite. The remaining numbers, except for 1, are classified as prime and composite numbers. 

Prime numbers

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 any natural number.
  • 2 and 3 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.

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.

  • Coprime numbers are always considered as a pair, whereas a prime number is 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 to check whether a number is Prime or not? 

Naive Approach: The naive approach is to

Iterate from 2 to  (n-1) and check if any number in this range divides n. If the number divides n, then it is not a prime number.

Time Complexity: O(N)
Auxiliary Space: O(1)

Naive approach (recursive): 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 of the above 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) if we consider the recursion stack. Otherwise, it is O(1).

Efficient Approach: An efficient solution is to:

Iterate through all numbers from 2 to ssquare root of n and for every number check if it divides n [because if a number is expressed as n = xy and any of the x or y is greater than the root of n, the other must be less than the root value]. 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
 
 
# import sqrt from math module
from math import sqrt
 
 
 
# Function check whether a number
# is prime or not
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 __name__ == '__main__':
    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

                    

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") : console.log(" false");
 
// This code is contributed by Mayank Tyagi

                    

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.
?>

                    

Output
true

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

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

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. Iterate from 5 to sqrt(n) and check for each iteration whether (that value) or (that value + 2) divides n or not and increment the value by 6 [because any prime can be expressed as 6n+1 or 6n-1]. 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"
 
if __name__ == '__main__':
    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)

Efficient solutions

Algorithms to find all prime numbers smaller than the N. 

More problems related to Prime number 



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