Last Updated : 29 Mar, 2024
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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
- 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.
- 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
- 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!
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