Answers for "python prime number checker"

determine if number is prime python

# Time Efficient Primality Check in Python

def primeCheck(n):
    # 0, 1, even numbers greater than 2 are NOT PRIME
    if n==1 or n==0 or (n % 2 == 0 and n > 2):
        return "Not prime"
    else:
        # Not prime if divisable by another number less
        # or equal to the square root of itself.
        # n**(1/2) returns square root of n
        for i in range(3, int(n**(1/2))+1, 2):
            if n%i == 0:
                return "Not prime"
        return "Prime"

python prime check

def isPrime(n):
  if n<2:		#1, 0 and all negative numbers are not prime
    return False
  elif n==2:	#2 is prime but cannot be calculated with the formula below becuase of the range function
    return True
  else:
    for i in range(2, n):
      if (n % i) == 0:	#if you can precisely divide a number by another number, it is not prime
        return False
    return True			#if the progam dont return False and arrives here, it means it has checked all the numebrs smaller than n and nono of them divides n. So n is prime

prime checker in python

#make the function
#to do this all hte vairibles go in side the function

def CheckIfPrime ():
    a1 = input("which number do you want to check")
    a = int(a1)#you need the checking number as an int not an str
    b = 2 #the number to check againts
    c = ("yes")
    while b < a:#run the loop
        if a%b == 0:#check if the division has a remainder
            c = ("no")#set the answer
        b = b+1
    print(c)#print the output
CheckIfPrime ()#call the function

check for prime in python

def is_prime(n: int) -> bool:
    """Primality test using 6k+-1 optimization."""
    import math
    if n <= 3:
        return n > 1
    if n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    while i <= math.sqrt(n):
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6
    return True