Introduction to Primality Test and School Method
Last Updated :
13 Feb, 2025
Given a positive integer, check if the number is prime or not. A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself. Examples of the first few prime numbers are {2, 3, 5, ...}
Examples :
Input: n = 11
Output: true
Input: n = 15
Output: false
Input: n = 1
Output: false
School Method
A simple solution is to iterate through all numbers from 2 to n-1 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 <bits/stdc++.h>
using namespace std;
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 Program to test above function
int main()
{
isPrime(11) ? cout << " true\n" : cout << " false\n";
isPrime(15) ? cout << " true\n" : cout << " false\n";
return 0;
}
// A school method based JAVA program
// to check if a number is prime
class GFG {
static boolean 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 Program
public static void main(String args[])
{
if (isPrime(11))
System.out.println(" true");
else
System.out.println(" false");
if (isPrime(15))
System.out.println(" true");
else
System.out.println(" false");
}
}
# A school method based Python3
# program to check if a number
# is prime
def isPrime(n):
# Corner case
if n <= 1:
return False
# Check from 2 to n-1
for i in range(2, n):
if n % i == 0:
return False
return True
# Driver Program to test above function
print("true") if isPrime(11) else print("false")
print("true") if isPrime(14) else print("false")
// A optimized school method based C#
// program to check if a number is prime
using System;
namespace prime {
public class GFG {
public static bool isprime(int n)
{
// Corner cases
if (n <= 1)
return false;
for (int i = 2; i < n; i++)
if (n % i == 0)
return false;
return true;
}
// Driver program
public static void Main()
{
if (isprime(11))
Console.WriteLine("true");
else
Console.WriteLine("false");
if (isprime(15))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
}
<script>
// A school method based Javascript program to check if a
// number is prime
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 Program to test above function
isPrime(11)? document.write(" true" + "<br>"): document.write(" false" + "<br>");
isPrime(15)? document.write(" true" + "<br>"): document.write(" false" + "<br>");
</script>
<?php
// A school method based PHP
// program to check if a number
// is prime
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
$tet = isPrime(11) ? " true\n" :
" false\n";
echo $tet;
$tet = isPrime(15) ? " true\n" :
" false\n";
echo $tet;
?>
Time complexity: O(n)
Auxiliary Space: O(1)
Optimized School Method
We can do the following optimizations: Instead of checking till n, we can check till √n because a larger factor of n must be a multiple of a smaller factor that has been already checked. The implementation of this method is as follows:
C++
// Optimised school method based C++ program to check if a
// number is prime
#include <bits/stdc++.h>
using namespace std;
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 Program to test above function
int main()
{
isPrime(11) ? cout << " true\n" : cout << " false\n";
isPrime(15) ? cout << " true\n" : cout << " false\n";
return 0;
}
// This code is contributed by Vikash Sangai
// Optimised school method based JAVA program
// to check if a number is prime
class GFG {
static boolean isPrime(int n)
{
// Corner case
if (n <= 1)
return false;
// Check from 2 to square root of n
for (int i = 2; i * i <= n; i++)
if (n % i == 0)
return false;
return true;
}
// Driver Program
public static void main(String args[])
{
if (isPrime(11))
System.out.println(" true");
else
System.out.println(" false");
if (isPrime(15))
System.out.println(" true");
else
System.out.println(" false");
}
}
// This code is contributed by Vikash Sangai
# Optimised school method based PYTHON program
# to check if a number is prime
# import the math module
import math
# function to check whether the number is prime or not
def isPrime(n):
# Corner case
if (n <= 1):
return False
# Check from 2 to square root of n
for i in range(2, int(math.sqrt(n)) + 1):
if (n % i == 0):
return False
return True
# Driver Program to test above function
print("true") if isPrime(11) else print("false")
print("true") if isPrime(15) else print("false")
# This code is contributed by bhoomikavemula
// Optimised school method based C#
// program to check if a number is prime
using System;
namespace prime {
public class GFG {
public static bool isprime(int n)
{
// Corner cases
if (n <= 1)
return false;
for (int i = 2; i * i <= n; i++)
if (n % i == 0)
return false;
return true;
}
// Driver program
public static void Main()
{
if (isprime(11))
Console.WriteLine("true");
else
Console.WriteLine("false");
if (isprime(15))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
}
// This code is contributed by Vikash Sangai
<script>
// JavaScript code for the above approach
function isPrime(n)
{
// Corner case
if (n <= 1) return false;
// Check from 2 to square root of n
for (let i = 2; i*i <= n; i++)
if (n % i == 0)
return false;
return true;
}
// Driver Code
if(isPrime(11))
document.write(" true" + "<br/>");
else
document.write(" false" + "<br/>");
if(isPrime(15))
document.write(" true" + "<br/>");
else
document.write(" false" + "<br/>");
// This code is contributed by sanjoy_62.
</script>
Time Complexity: O(√n)
Auxiliary Space: O(1)
Another approach
It is based on the fact that all primes greater than 3 are of the form 6k ± 1, where k is any integer greater than 0. This is because all integers can be expressed as (6k + i), where i = −1, 0, 1, 2, 3, or 4. And note that 2 divides (6k + 0), (6k + 2), and (6k + 4) and 3 divides (6k + 3). So, a more efficient method is to test whether n is divisible by 2 or 3, then to check through all numbers of the form 6k ± 1 <= √n. This is 3 times faster than testing all numbers up to √n.
C++
// C++ program to check the given number
// is prime or not
#include <bits/stdc++.h>
using namespace std;
// Function to check if the given number
// is prime or not.
bool isPrime(int n)
{
if (n == 2 || n == 3)
return true;
if (n <= 1 || n % 2 == 0 || n % 3 == 0)
return false;
// To check through all numbers of the form 6k ± 1
for (int i = 5; i * i <= n; 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";
isPrime(15) ? cout << " true\n" : cout << " false\n";
return 0;
}
// JAVA program to check the given number
// is prime or not
class GFG {
static boolean isPrime(int n)
{
// since 2 and 3 are prime
if (n == 2 || n == 3)
return true;
// if n<=1 or divided by 2 or 3 then it can not be prime
if (n <= 1 || n % 2 == 0 || n % 3 == 0)
return false;
// To check through all numbers of the form 6k ± 1
for (int i = 5; i * i <= n; i += 6)
{
if (n % i == 0 || n % (i + 2) == 0)
return false;
}
return true;
}
// Driver Program
public static void main(String args[])
{
if (isPrime(11))
System.out.println(" true");
else
System.out.println(" false");
if (isPrime(15))
System.out.println(" true");
else
System.out.println(" false");
}
}
// This code is contributed by Ujjwal Kumar Bhardwaj
# Python program to check the given number
# is prime or not
# Function to check if the given number
# is prime or not.
import math
def isPrime(n):
if n == 2 or n == 3:
return True
elif n <= 1 or n % 2 == 0 or n % 3 == 0:
return False
# To check through all numbers of the form 6k ± 1
# until i <= square root of n, with step value 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
# # Driver code
print(isPrime(11))
print(isPrime(20))
# # This code is contributed by Harsh Master
// C# program to check the given number
// is prime or not
using System;
class GFG {
static bool isPrime(int n)
{
// since 2 and 3 are prime
if (n == 2 || n == 3)
return true;
// if n<=1 or divided by 2 or 3 then it can not be prime
if (n <= 1 || n % 2 == 0 || n % 3 == 0)
return false;
// To check through all numbers of the form 6k ± 1
for (int i = 5; i * i <= n; i += 6)
{
if (n % i == 0 || n % (i + 2) == 0)
return false;
}
return true;
}
// Driver Program
public static void Main(String[] args)
{
if (isPrime(11))
Console.WriteLine(" true");
else
Console.WriteLine(" false");
if (isPrime(15))
Console.WriteLine(" true");
else
Console.WriteLine(" false");
}
}
// This code is contributed by Aman Kumar
<script>
// JavaScript program to check the given number
// is prime or not
// Function to check if the given number
// is prime or not.
function isPrime(n)
{
if (n == 2 || n == 3)
return true;
if (n <= 1 || n % 2 == 0 || n % 3 == 0)
return false;
// To check through all numbers of the form 6k ± 1
for (let i = 5; i * i <= n; i += 6) {
if (n % i == 0 || n % (i + 2) == 0)
return false;
}
return true;
}
// Driver Code
isPrime(11) ? document.write(" true" + "<br/>") : document.write(" false" + "<br/>");
isPrime(15) ? document.write(" true" + "<br/>") : document.write(" false" + "<br/>");
</script>
Time Complexity: O(√n)
Auxiliary Space: O(1)
Primality Test (Introduction and School Method)
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